The number of functions f, from the set A = { x ∈ N : x 2 − 1 0 x + 9 ≤ 0 } to the set B = { n 2 : n ∈ N } such that f(x) ≤ (x – 3)2 + 1, for every x ∈ A, is…………

x 2 − 1 0 x + 9 ≤ 0 ( x − 1 ) ( x − 9 ) ≤ 0 A = {1, 2, 3, ……, 9} for set B, f ( x ) ≤ ( x − 3 ) 2 + 1 total number of such function = 2 × 1 × 1 × 1 × 2 × 3 × 4 × 5 × 6 = 2 × 6! = 1440

2 3 − 1 3 1 × 7 + 4 3 − 3 + 2 3 − 1 3 2 × 1 1 + 6 3 − 5 3 + 4 3 + 2 3 − 1 3 3 × 1 5 + . . . . . . . + 3 0 3 − 2 9 3 + 2 8 3 − 2 7 3 + . . . . + 2 3 − 1 3 1 5 × 6 3 i s e q u a l t o . . . . . . . . . . . . .

t n = 2 ( 2 3 + 4 3 + 6 3 + . . . . + ( 2 n ) 3 ) − ( 1 3 − 2 3 + 3 3 + . . . . . ( 2 n ) 3 ) n ( 4 n + 3 ) 2 × 2 3 × ( n × ( n + 1 ) 2 ) 2 − ( 2 n × ( 2 n + 1 ) 2 ) 2 n ( 4 n + 3 ) tn = n n o w ∑ n = 1 1 5 T n = 1 5 . 1 6 2

Find the values of p and q so that f(x)={x2+3x+p,x≤1qx+2,x>1 is differentiable at x=1 .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t f ( x ) = { x 2 + 3 x + p , x ≤ 1 q x + 2 , x > 1 a t x = 1 . L . H . L . f ' ( c ) = l i m x → 1 − f ( x ) − f ( c ) x − c f ' ( 1 ) = l i m x → 1 − f ( x ) − f ( 1 ) x − 1 = l i m x → 1 − ( x 2 + 3 x + p ) − ( 1 + 3 + p ) x − 1 = l i m h → 0 [ ( 1 − h ) 2 + 3 ( 1 − h ) + p ] − ( 1 + 3 + p ) 1 − h − 1 = l i m h → 0 [ 1 + h 2 − 2 h + 3 − 3 h + p ] − ( 4 + p ) − h = l i m h → 0 [ h 2 − 5 h + 4 + p ] − [ 4 + p ] − h = l i m h → 0 h 2 − 5 h + 4 + p − 4 − p − h = l i m h → 0 h 2 − 5 h − h = l i m h → 0 h [ h − 5 ] − h = 5 R . H . L . f ' ( 1 ) = l i m x → 1 + f ( x ) − f ( 1 ) x − 1 = l i m x → 1 + ( q x + 2 ) − ( 1 + 3 + p ) x − 1 = l i m h → 0 [ q ( 1 + h ) + 2 ] − [ 4 + p ] 1 + h − 1 = l i m h → 0 q + q h + 2 − 4 − p h = l i m h → 0 q + q h − 2 − p h For existing the limit q − 2 − p = 0 ⇒ q − p = 0 … ( 1 ) ⇒ l i m h → 0 q h − 0 h = q I f L . H . L . f ' ( 1 ) = R . H . L . f ' ( 1 ) t h e n q = 5 . N o w p

If xmyn=(x+y)m+n , prove that (i) dydx=yx (ii) d2ydx2=0

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: ( i ) G i v e n t h a t x m . y n = ( x + y ) m + n T a k i n g l o g o n b o t h s i d e s , w e g e t , ⇒ l o g x m . y n = l o g ( x + y ) m + n [ ? l o g x y = l o g x + l o g y ] ⇒ l o g x m + l o g y n = ( m + n ) l o g ( x + y ) ⇒ m l o g x + n l o g y = ( m + n ) l o g ( x + y ) D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x ⇒ m . d d x l o g x + n . d d x l o g y = ( m + n ) d d x l o g ( x + y ) ⇒ m . 1 x + n . 1 y . d y d x = ( m + n ) 1 x + y ( 1 + d y d x ) ⇒ m x + n y . d y d x = m + n x + y . ( 1 + d y d x ) ⇒ m x + n y . d y d x = m + n x + y + m + n x + y . d y d x ⇒ n y . d y d x − m + n x + y . d y d x = m + n x + y − m x ⇒ ( n y − m + n x + y ) . d y d x = m + n x + y − m x ⇒ ( n x + n y − m y − n y y ( x + y ) ) . d y d x = ( m x + n x − m x − m y x ( x + y ) ) ⇒ ( n x − m y y ( x + y ) ) . d y d x = ( n x − m y x ( x + y ) ) ⇒ d y d x = n x − m y x ( x + y ) × y ( x + y ) n x − m y = y x ⇒ d y d x = y x H e n c e , p r o v e d . ( i i ) G i v e n t h a t d y d x = y x D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x ⇒ d d x ( d y d x ) = d d x ( y x ) ⇒ d 2 y d x 2 = x . d y d x − y . 1 x 2 ⇒ d 2 y d x 2 = x . y x − y x 2 [ ? d y d x = y x ] ⇒ d 2 y d x 2 = y − y x 2 = 0 x 2 = 0 H e n c e , d 2 y d x 2 = 0 H e n c e , p r o v e d .

If x=sint and y=sin pt , prove that (1−x2)d2ydx2−xdydx+ p2y=0

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: Given that x=sint and y=sinptDifferentiating both sides w.r.t. t⇒ dxdt=cost and dydt=cospt.p=p.cospt⇒ dydx=dydtdxdt=p.cosptcost⇒ dydx=p.cosptcostAgain differentiating both sides w.r.t. x⇒ ddx(dydx)=p.ddx(cosptcost)⇒ d2ydx2=p.[cost.ddx(cospt)−cospt.ddx(cost)cos2t]⇒ d2ydx2=p.[cost.(−sinpt).pdtdx−cospt.(−sint).dtdxcos2t]⇒ d2ydx2=p.[−pcost.sinpt+cospt.sintcos2t]dtdx⇒ d2ydx2=p.[−pcost.sinpt+cospt.sintcos2t]1cost⇒ d2ydx2=p.(−pcost.sinpt+cospt.sintcos3t)Now we have to prove that (1−x2)d2ydx2−xdydx+ p2y=0L.H.S.=(1−x2)[p.(−pcost.sinpt+cospt.sintcos3t)]−x(p.cosptcost)+ p2y⇒ =(1−sin2t)[p.(−pcost.sinpt+cospt.sintcos3t)]−p.sintcosptcost+ p2.sinpt⇒ =cos2t[−p2cost.sinpt+pcospt.sintcos3t]−p.sintcosptcost+ p2.sinpt⇒ =−p2cost.sinpt+pcospt.sintcost−p.sintcosptcost+ p2.sinpt⇒ =−p2cost.sinpt+pcospt.sint−p.sintcospt+p2.sinptcostcost⇒ =0cost=0=R.H.S. Hence, proved.

Let the solution curve of the differential equation xdydx−y=y2+16x2, y(1)=3 be y=y(x). Then y(2) is equal to:

dydx=yx+y2+16x2x Put y = V⋅x differentiable worst = x dydx=V+x⋅dVdx V + xdVdx=V+V2+16 xdVdx=V2+16 apply variable separable method ∫dVV2+16=∫dxx+lnC ⇒ln|V+V2+16|=ln Cx yx+y2+16x2x=Cx Given y(1) = 3 C = 8 Now at x = 2 y2+y2+16.42=8.2 y2 + 64 = (32 – y)2 y = 15

Let a→=αi^a+3j^−k^, b→=3i^−βj^+4k^ and c→=i^+2j^−2k^ where α,β∈R, be three vectors. If the projection of a→ on c→ is 103 and b→×c→=−6i^+10j^+7k^, then the value of + is equal to:

a→⋅c^=α+6+21+4+4 103=8+α3⇒α=2 b→×c→=i^ (2β−8)+j^ (10)+k^ (6+β) =−6i^+10j^+7k^ So = 1

Maths NCERT Exemplar Solutions Class 12th Chapter Five: Overview, Questions, Preparation

Q: A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is t a n − 1 3 4 . Water is poured in it a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is………..

V = 1 3 π r 2 h V = π 8 h 3 t a n 2 α d v d t = π h 2 t a n 2 α . d h d t CSA = π r l d ( C S A ) d t = 1 5 1 6 π 2 h . d h d t 1 5 8 × 2 3 = 5 m 2 / h r s

Updated on Jul 28, 2025 12:17 IST

By Vishal Baghel, Executive Content Operations

Table of contents

Continuity and Differentiability Questions and Answers
JEE Mains Solutions 2022,29th june , Maths, first shift
JEE Mains 2022
JEE Mains 2022

Continuity and Differentiability Questions and Answers

Q1. Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix} a t x = 1 . \\ L . H . L . f' (c) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (c)}{x - c} \\ f' (1) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 3 x + p) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[{(1 - h)}^{2} + 3 (1 - h) + p] - (1 + 3 + p)}{1 - h - 1} \\ = \underset{h \to 0}{l i m} \frac{[1 + h^{2} - 2 h + 3 - 3 h + p] - (4 + p)}{- h} \\ = \underset{h \to 0}{l i m} \frac{[h^{2} - 5 h + 4 + p] - [4 + p]}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h + 4 + p - 4 - p}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h}{- h} = \underset{h \to 0}{l i m} \frac{h [h - 5]}{- h} = 5 \\ R . H . L . f' (1) = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{+}}{l i m} \frac{(q x + 2) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[q (1 + h) + 2] - [4 + p]}{1 + h - 1} \\ = \underset{h \to 0}{l i m} \frac{q + q h + 2 - 4 - p}{h} = \underset{h \to 0}{l i m} \frac{q + q h - 2 - p}{h} \\ F o r e x i s t i n g t h e limit \\ q - 2 - p = 0 \Rightarrow q - p = 0 \dots (1) \\ \underset{h \to 0}{\Rightarrow l i m} \frac{q h - 0}{h} = q \\ I f L . H . L . f' (1) = R . H . L . f' (1) t h e n q = 5 . \\ N o w p u t t i n g t h e v a l u e o f q i n e q n . (1) \\ 5 - p = 2 \Rightarrow p = 3 \\ H e n c e, t h e v a l u e o f p i s 3 a n d q i s 5 . \end{array}$

Q2. If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$ .

Sol:

$\begin{array}{l} (i) G i v e n t h a t x^{m} . y^{n} = {(x + y)}^{m + n} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t, \\ \Rightarrow l o g x^{m} . y^{n} = l o g {(x + y)}^{m + n} [∵ l o g x y = l o g x + l o g y] \\ \Rightarrow l o g x^{m} + l o g y^{n} = (m + n) l o g (x + y) \\ \Rightarrow m l o g x + n l o g y = (m + n) l o g (x + y) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow m . \frac{d}{d x} l o g x + n . \frac{d}{d x} l o g y = (m + n) \frac{d}{d x} l o g (x + y) \\ \Rightarrow m . \frac{1}{x} + n . \frac{1}{y} . \frac{d y}{d x} = (m + n) \frac{1}{x + y} (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} . (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} + \frac{m + n}{x + y} . \frac{d y}{d x} \\ \Rightarrow \frac{n}{y} . \frac{d y}{d x} - \frac{m + n}{x + y} . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n}{y} - \frac{m + n}{x + y}) . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n x + n y - m y - n y}{y (x + y)}) . \frac{d y}{d x} = (\frac{m x + n x - m x - m y}{x (x + y)}) \\ \Rightarrow (\frac{n x - m y}{y (x + y)}) . \frac{d y}{d x} = (\frac{n x - m y}{x (x + y)}) \\ \Rightarrow \frac{d y}{d x} = \frac{n x - m y}{x (x + y)} \times \frac{y (x + y)}{n x - m y} = \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} H e n c e, p r o v e d . \end{array}$

$\begin{array}{l} (i i) G i v e n t h a t \frac{d y}{d x} = \frac{y}{x} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{y}{x}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{d y}{d x} - y . 1}{x^{2}} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{y}{x} - y}{x^{2}} [∵ \frac{d y}{d x} = \frac{y}{x}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{y - y}{x^{2}} = \frac{0}{x^{2}} = 0 \\ H e n c e, \frac{d^{2} y}{d x^{2}} = 0 H e n c e, p r o v e d . \end{array}$

Q3. If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

Sol:

$\begin{array}{l} G i v e n t h a t x = s i n t a n d y = s i n p t \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t \\ \Rightarrow \frac{d x}{d t} = c o s t a n d \frac{d y}{d t} = c o s p t . p = p . c o s p t \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{p . c o s p t}{c o s t} \\ \Rightarrow \frac{d y}{d x} = \frac{p . c o s p t}{c o s t} \\ A g a i n d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = p . \frac{d}{d x} (\frac{c o s p t}{c o s t}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . \frac{d}{d x} (c o s p t) - c o s p t . \frac{d}{d x} (c o s t)}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . (- s i n p t) . p \frac{d t}{d x} - c o s p t . (- s i n t) . \frac{d t}{d x}}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{d t}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{1}{c o s t} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t}) \\ N o w w e h a v e t o p r o v e t h a t \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \\ L . H . S . = (1 - x^{2}) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - x (\frac{p . c o s p t}{c o s t}) + p^{2} y \\ \Rightarrow = (1 - s i n^{2} t) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = c o s^{2} t [\frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s^{3} t}] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s t} - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t - p . s i n t c o s p t + p^{2} . s i n p t c o s t}{c o s t} \\ \Rightarrow = \frac{0}{c o s t} = 0 = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

Q4.

Sol:

Commonly asked questions

Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t x^{m} . y^{n} = {(x + y)}^{m + n} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t, \\ \Rightarrow l o g x^{m} . y^{n} = l o g {(x + y)}^{m + n} [? l o g x y = l o g x + l o g y] \\ \Rightarrow l o g x^{m} + l o g y^{n} = (m + n) l o g (x + y) \\ \Rightarrow m l o g x + n l o g y = (m + n) l o g (x + y) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow m . \frac{d}{d x} l o g x + n . \frac{d}{d x} l o g y = (m + n) \frac{d}{d x} l o g (x + y) \\ \Rightarrow m . \frac{1}{x} + n . \frac{1}{y} . \frac{d y}{d x} = (m + n) \frac{1}{x + y} (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} . (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} + \frac{m + n}{x + y} . \frac{d y}{d x} \\ \Rightarrow \frac{n}{y} . \frac{d y}{d x} - \frac{m + n}{x + y} . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n}{y} - \frac{m + n}{x + y}) . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n x + n y - m y - n y}{y (x + y)}) . \frac{d y}{d x} = (\frac{m x + n x - m x - m y}{x (x + y)}) \\ \Rightarrow (\frac{n x - m y}{y (x + y)}) . \frac{d y}{d x} = (\frac{n x - m y}{x (x + y)}) \\ \Rightarrow \frac{d y}{d x} = \frac{n x - m y}{x (x + y)} \times \frac{y (x + y)}{n x - m y} = \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} H e n c e, p r o v e d . \end{array}$

$\begin{array}{l} (i i) G i v e n t h a t \frac{d y}{d x} = \frac{y}{x} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{y}{x}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{d y}{d x} - y . 1}{x^{2}} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{y}{x} - y}{x^{2}} [? \frac{d y}{d x} = \frac{y}{x}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{y - y}{x^{2}} = \frac{0}{x^{2}} = 0 \\ H e n c e, \frac{d^{2} y}{d x^{2}} = 0 H e n c e, p r o v e d . \end{array}$

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

Examine the continuity of the function
$f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t y = f (x) w i l l b e c o n t i n u o u s a t x = a i f \\ \underset{x \to a^{-}}{l i m} f (x) = \underset{x \to a}{l i m} f (x) = \underset{x \to a^{+}}{l i m} f (x) \\ G i v e n f (x) = x^{3} + 2 x^{2} - 1 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1}{l i m} f (x) = {(1)}^{3} + 2 {(1)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{+}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = 2 . \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

$f (x) = {\begin{matrix} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{matrix}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 2^{+}}{l i m} f (x) = 3 x + 5 \\ = \underset{h \to 0}{l i m} 3 (2 + h) + 5 = 11 \\ \underset{x \to 2}{l i m} f (x) = 3 x + 5 = 3 (2) + 5 = 11 \\ \underset{x \to 2^{-}}{l i m} f (x) = x^{2} = \underset{h \to 0}{l i m} {(2 - h)}^{2} \\ = \underset{h \to 0}{l i m} {(2)}^{2} + h^{2} - 4 h = {(2)}^{2} = 4 \\ Since \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \neq \underset{x \to 2^{-}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 2 . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} \frac{1 - c o s 2 x}{x^{2}}, & i f x \neq 0 \\ 5, & i f x = 0 \end{matrix}$ at $x = 2$ , $x = 0$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 - h)}{{(0 - h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s (- 2 h)}{h^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 [\underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 + h)}{{(0 + h)}^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 \\ \underset{x \to 0}{l i m} f (x) = 5 \\ A s \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 . \end{array}$

Kindly consider the following
$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

Kindly consider the following
$f (x) = {\begin{matrix} \frac{| x - 4 |}{2 (x - 4)}, & x \neq 4 \\ 0, & x = 4 \end{matrix}$ at $x = 2$ , $x = 4$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 4^{-}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} [\begin{array}{l} f o r x < 4, | x - 4 | = - (x - 4) \\ f o r x > 4, | x - 4 | = (x - 4) \end{array}] \\ = \underset{h \to 0}{l i m} \frac{- [4 - h - 4]}{2 [4 - h - 4]} = \underset{h \to 0}{l i m} \frac{h}{- 2 h} = - \frac{1}{2} \\ \underset{x \to 4^{+}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} = \underset{h \to 0}{l i m} \frac{[4 + h - 4]}{2 [4 + h - 4]} = \underset{h \to 0}{l i m} \frac{h}{2 h} = \frac{1}{2} \\ \underset{x \to 4}{l i m} f (x) = 0 \\ ∴ \underset{x \to 4^{-}}{l i m} f (x) \neq \underset{x \to 4^{+}}{l i m} f (x) \neq \underset{x \to 4}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 4 . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} | x | c o s \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = | x | c o s \frac{1}{x} \\ = \underset{h \to 0}{l i m} | 0 - h | c o s \frac{1}{(0 - h)} = \underset{h \to 0}{l i m} h . c o s \frac{1}{h} = 0 [? c o s \frac{1}{x} o s c i l l a t e b e t w e e n - 1 a n d 1] \\ \underset{x \to 0^{+}}{l i m} f (x) = | x | c o s \frac{1}{x} = \underset{h \to 0}{l i m} | 0 + h | c o s \frac{1}{(0 + h)} = \underset{h \to 0}{l i m} h . c o s \frac{1}{h} = 0 \\ \underset{x \to 0}{l i m} f (x) = 0 \\ ∴ \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (x) = \underset{x \to 0}{l i m} f (x) = 0 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 0 . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} | x - a | s i n \frac{1}{x - a}, & x \neq a \\ 0, & x = a \end{matrix}$ at $x = 0$ , $x = a$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to a^{-}}{l i m} f (x) = | x - a | s i n \frac{1}{x - a} \\ = \underset{h \to 0}{l i m} | a - h - a | s i n \frac{1}{(a - h - a)} = \underset{h \to 0}{l i m} h . s i n \frac{1}{- h} \\ = \underset{h \to 0}{l i m} - h . s i n \frac{1}{h} [? s i n (- θ) = - s i n θ] \\ = 0 \times [a n u m b e r o s c i l l a t e b e t w e e n - 1 a n d 1] = 0 \\ \underset{x \to a^{+}}{l i m} f (x) = | x - a | s i n \frac{1}{x - a} = \underset{h \to 0}{l i m} | a + h - a | . s i n \frac{1}{(a + h - a)} = \underset{h \to 0}{l i m} h . s i n \frac{1}{h} \\ = 0 \times [a n u m b e r o s c i l l a t e b e t w e e n - 1 a n d 1] = 0 \\ \underset{x \to a}{l i m} f (x) = 0 \\ A s \underset{x \to a^{-}}{l i m} f (x) = \underset{x \to a^{+}}{l i m} f (x) = \underset{x \to a}{l i m} f (x) = 0 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = a . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{e^{1 / x}}{1 + e^{1 / x}} \\ = \underset{h \to 0}{l i m} \frac{e^{\frac{1}{0 - h}}}{1 + e^{\frac{1}{0 - h}}} = \underset{h \to 0}{l i m} \frac{e^{- 1 / h}}{1 + e^{- 1 / h}} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{1 / h} (1 - e^{- 1 / h})} = \underset{h \to 0}{l i m} \frac{1}{e^{1 / h} - 1} = \underset{h \to 0}{l i m} \frac{1}{e^{1 / 0} - 1} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{\infty} - 1} = \underset{h \to 0}{l i m} \frac{1}{0 - 1} = - 1 [? e^{\infty} = 0] \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{e^{1 / x}}{1 + e^{1 / x}} = \underset{h \to 0}{l i m} \frac{e^{\frac{1}{0 + h}}}{1 + e^{\frac{1}{0 + h}}} = \underset{h \to 0}{l i m} \frac{e^{1 / h}}{1 + e^{1 / h}} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / h} (1 + e^{1 / h})} = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / h} + 1} = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / 0} + 1} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{- \infty} + 1} = \underset{h \to 0}{l i m} \frac{1}{0 + 1} = 1 [? e^{- \infty} = 0] \\ \underset{x \to 0}{l i m} f (x) = 0 \\ A s \underset{x \to 0^{-}}{l i m} f (x) \neq \underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} \frac{x^{2}}{2}, i f & 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, i f & 1 < x \leq 2 \end{matrix}$ at $x = 0$ , $x = 1$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 1^{-}}{l i m} f (x) = \frac{x^{2}}{2} = \underset{h \to 0}{l i m} \frac{{(1 - h)}^{2}}{2} = \frac{1}{2} \\ \underset{x \to 1}{l i m} f (x) = \frac{x^{2}}{2} = \frac{{(1)}^{2}}{2} = \frac{1}{2} \\ \underset{x \to 1^{+}}{l i m} f (x) = 2 x^{2} - 3 x + \frac{3}{2} = 2 {(1)}^{2} - 3 (1) + \frac{3}{2} = 2 - 3 + \frac{3}{2} = \frac{1}{2} \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) = \frac{1}{2} \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

Kindly consider the following
$f (x) = | x | + | x - 1 |$ at $x = 1$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 1^{-}}{l i m} f (x) = | x | + | x - 1 | = \underset{h \to 0}{l i m} | 1 - h | + | 1 - h - 1 | \\ = | 1 - 0 | + | 1 - 0 - 1 | = 1 + 0 = 1 \\ \underset{x \to 1}{l i m} f (x) = | x | + | x - 1 | = | 1 | + | 1 - 1 | = 1 + 0 = 1 \\ \underset{x \to 1^{+}}{l i m} f (x) = | x | + | x - 1 | = \underset{h \to 0}{l i m} | 1 + h | + | 1 + h - 1 | \\ = | 1 + 0 | + | 1 + 0 - 1 | = 1 + 0 = 1 \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

Find the value of $k$ in each of the Exercises 11 to 14 so that the function $f$ is continuous at the indicated point:

$f (x) = {\begin{matrix} 3 x - 8, & i f x \leq 5 \\ 2 k, & i f x > 5 \end{matrix}$ at $x = 5$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 5}{l i m} f (x) = 3 x - 8 = \underset{h \to 0}{l i m} 3 (5 - h) - 8 = 15 - 8 = 7 \\ \underset{x \to 5^{+}}{l i m} f (x) = 2 k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 5 \\ ∴ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 5^{+}}{l i m} f (x) \\ ∴ 7 = 2 k \Rightarrow k = \frac{7}{2} \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) \\ H e n c e, t h e v a l u e o f k i s \frac{7}{2} . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} \frac{2^{x + 2} - 16}{4^{x} - 16}, & i f x \neq 2 \\ k, & i f x = 2 \end{matrix}$ at $x = 2$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{2^{x + 2} - 16}{4^{x} - 16} = \frac{2^{2} . 2^{x} - 16}{{(2^{x})}^{2} - {(4)}^{2}} = \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)} \\ = \frac{4}{(2^{x} + 4)} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} \frac{4}{(2^{2 - h} + 4)} = \frac{4}{2^{2} + 4} = \frac{4}{4 + 4} = \frac{4}{8} = \frac{1}{2} \\ \underset{x \to 2}{l i m} f (x) = k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 2 \\ ∴ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \\ ∴ k = \frac{1}{2} \\ H e n c e, t h e v a l u e o f k i s \frac{1}{2} . \end{array}$

Kindly consider the following

$f (x) = {\begin{matrix} \frac{1 - c o s k x}{x s i n x}, & i f x \neq 0 \\ \frac{1}{2}, & i f x = 0 \end{matrix}$ at $x = 0$

Kindly go through the solution

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{1 - c o s k x}{x s i n x} = \underset{h \to 0}{l i m} \frac{1 - c o s k (0 - h)}{(0 - h) s i n (0 - h)} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s (- k h)}{(- h) s i n (- h)} = \underset{h \to 0}{l i m} \frac{1 - c o s k h}{h s i n h} [\begin{array}{l} ? s i n (- θ) = - s i n θ \\ c o s (- θ) = c o s θ \end{array}] \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} \frac{k h}{2}}{h s i n h} = \underset{\begin{array}{l} h \to 0 \\ k h \to 0 \end{array}}{l i m} \frac{2 s i n \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} \times \frac{s i n \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} . \frac{1}{h . \frac{s i n h}{h} . h} \\ = 2.1 . \frac{k h}{2} . 1 . \frac{k h}{2} . \frac{1}{h^{2}} . 1 [\begin{array}{l} \underset{h \to 0}{l i m} \frac{s i n h}{h} = 1 a n d \\ \underset{k h \to 0}{l i m} \frac{s i n k h}{k h} = 1 \end{array}] \\ = \frac{k^{2}}{2} \\ \underset{x \to 0}{l i m} f (x) = \frac{1}{2} \\ ∴ \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0}{l i m} f (x) \\ ∴ \frac{k^{2}}{2} = \frac{1}{2} \Rightarrow k^{2} = 1 \Rightarrow k = \pm 1 \\ H e n c e, t h e v a l u e o f k i s \pm 1 . \end{array}$

Prove that the function $f$ defined by $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix}$ remains discontinuous at $x = 0$ , regardless of the choice of $k$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{x}{| x | + 2 x^{2}} = \underset{h \to 0}{l i m} \frac{0 - h}{| 0 - h | + 2 {(0 - h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{- h}{h + 2 h^{2}} = \underset{h \to 0}{l i m} \frac{- h}{h (1 + 2 h)} \\ = \underset{h \to 0}{l i m} \frac{- 1}{1 + 2 h} = \frac{- 1}{1 + 2 (0)} = - 1 \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{x}{| x | + 2 x^{2}} = \underset{h \to 0}{l i m} \frac{0 + h}{| 0 + h | + 2 {(0 + h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{h}{h + 2 h^{2}} = \underset{h \to 0}{l i m} \frac{h}{h (1 + 2 h)} = \frac{1}{1 + 0} = 1 \\ A s \underset{x \to 0^{-}}{l i m} f (x) \neq \underset{x \to 0^{+}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 r e g a r d l e s s t h e c h o i c e o f k . \end{array}$

Find the values of $a$ and $b$ such that the function $f$ defined by

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 4^{-}}{l i m} f (x) = \frac{x - 4}{| x - 4 |} + a = \underset{h \to 0}{l i m} \frac{4 - h - 4}{| 4 - h - 4 |} + a \\ = \underset{h \to 0}{l i m} \frac{- h}{h} + a = - 1 + a \\ \underset{x \to 4}{l i m} f (x) = a + b \\ \underset{x \to 4^{+}}{l i m} f (x) = \frac{x - 4}{| x - 4 |} + b = \underset{h \to 0}{l i m} \frac{4 + h - 4}{| 4 + h - 4 |} + b \\ = \underset{h \to 0}{l i m} \frac{h}{h} + b = 1 + b \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 4 . \\ ∴ \underset{x \to 4^{-}}{l i m} f (x) = \underset{x \to 4}{l i m} f (x) = \underset{x \to 4^{+}}{l i m} f (x) \\ - 1 + a = a + b = 1 + b \\ ∴ - 1 + a = a + b \Rightarrow b = - 1 \\ 1 + b = a + b \Rightarrow a = 1 \\ H e n c e, t h e v a l u e o f a = 1 a n d b = - 1 . \end{array}$

Given the function $f (x) = \frac{1}{x + 2}$ . Find the points of discontinuity of the composite function $y = f (f (x))$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{1}{x + 2} \\ f [f (x)] = \frac{1}{f (x) + 2} = \frac{1}{\frac{1}{x + 2} + 2} = \frac{1}{\frac{1 + 2 x + 4}{x + 2}} = \frac{x + 2}{2 x + 5} \\ ∴ f [f (x)] = \frac{x + 2}{2 x + 5} \\ T h i s f u n c t i o n w i l l n o t b e d e f i n e d a n d c o n t i n u o u s w h e r e 2 x + 5 = 0 \\ \Rightarrow x = \frac{- 5}{2} . \\ H e n c e, x = \frac{- 5}{2} i s t h e point o f d i s c o n t i n u i t y . \end{array}$

Find all points of discontinuity of the function $f (t) = \frac{1}{t^{2} + t - 2}$ where $t = \frac{1}{x - 1}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = \frac{1}{t^{2} + t - 2} \\ \Rightarrow f (t) = \frac{1}{{(\frac{1}{x - 1})}^{2} + \frac{1}{x - 1} - 2} [P u t t i n g t = \frac{1}{x - 1}] \\ = \frac{1}{\frac{1 + x - 1 - 2 {(x - 1)}^{2}}{{(x - 1)}^{2}}} = \frac{{(x - 1)}^{2}}{x - 2 x^{2} - 2 + 4 x} \\ = \frac{{(x - 1)}^{2}}{- 2 x^{2} + 5 x - 2} = \frac{{(x - 1)}^{2}}{- (2 x^{2} - 5 x + 2)} \\ = \frac{{(x - 1)}^{2}}{- [2 x^{2} - 4 x - x + 2]} = \frac{{(x - 1)}^{2}}{- [2 x (x - 2) - 1 (x - 2)]} \\ = \frac{{(x - 1)}^{2}}{- (x - 2) (2 x - 1)} = \frac{{(x - 1)}^{2}}{(2 - x) (2 x - 1)} \\ S o, i f f (t) i s d i s c o n t i n u o u s, t h e n 2 - x = 0 ∴ x = 2 \\ a n d 2 x - 1 = 0 ∴ x = \frac{1}{2} \\ H e n c e, t h e r e q u i r e d o f d i s c o n t i n u i t y a r e 2 a n d \frac{1}{2} . \end{array}$

Show that the function $f (x) = | s i n x + c o s x |$ is continuous at $x = π$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | s i n x + c o s x | a t x = π \\ P u t g (x) = s i n x + c o s x a n d h (x) = | x | \\ ∴ h [g (x)] = h (s i n x + c o s x) = | s i n x + c o s x | \\ N o w, g (x) = s i n x + c o s x i s a c o n t i n u o u s f u n c t i o n since s i n x a n d c o s x a r e t w o c o n t i n u o u s \\ f u n c t i o n s a t x = π . \\ W e k n o w t h a t e v e r y modulus f u n c t i o n i s a c o n t i n u o u s f u n c t i o n e v e r y w h e r e . \\ H e n c e, f (x) = | s i n x + c o s x | i s a c o n t i n u o u s f u n c t i o n a t x = π . \end{array}$

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x [x], & 0 \leq x < 2 \\ (x - 1) x, & 2 \leq x < 3 \end{matrix}$ at $x = 2$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t a f u n c t i o n f i s d i f f e r e n t i a b l e a t a point' a' i n i t s d o m a i n i f \\ L f' (c) = R f' (c) \\ w h e r e L f' (c) = \underset{h \to 0}{l i m} \frac{f (a - h) - f (a)}{- h} a n d R f' (c) = \underset{h \to 0}{l i m} \frac{f (a + h) - f (a)}{h} \\ H e r e, f (x) = {\begin{cases} x [x], i f 0 \leq x \leq 2 \\ (x - 1) x, i f 2 \leq x \leq 2 \end{cases} a t x = 2 . \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (2 - h) - f (2)}{- h} = \underset{h \to 0}{l i m} \frac{(2 - h) [2 - h] - (2 - 1) 2}{- h} \\ = \underset{h \to 0}{l i m} \frac{(2 - h) . 1 - 2}{- h} [? [2 - h] = 1] \\ = \underset{h \to 0}{l i m} \frac{2 - h - 2}{- h} = 1 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (2 + h) - f (2)}{h} = \underset{h \to 0}{l i m} \frac{(2 + h - 1) (2 + h) - (2 - 1) . 2}{h} \\ = \underset{h \to 0}{l i m} \frac{(1 + h) (2 + h) - 2}{h} = \underset{h \to 0}{l i m} \frac{2 + h + 2 h + h^{2} - 2}{h} \\ = \underset{h \to 0}{l i m} \frac{3 h + h^{2}}{h} \underset{h \to 0}{= l i m} \frac{h (3 + h)}{h} = 3 \\ L f' (c) \neq R f' (c) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 2 . \end{array}$

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x^{2} s i n x \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$ at $x = 0$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{cases} x^{2} s i n \frac{1}{x}, i f x \neq 0 \\ 0, i f x = 0 \end{cases} a t x = 0 . \\ F o r d i f f e r e n t i a b i l i t y w e k n o w t h a t \\ L f' (c) = R f' (c) \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (0 - h) - f (0)}{- h} = \underset{h \to 0}{l i m} \frac{{(0 - h)}^{2} s i n \frac{1}{(0 - h)} - 0}{- h} = \frac{h^{2} s i n (\frac{1}{- h})}{- h} \\ = \underset{h \to 0}{l i m} h . s i n (\frac{1}{h}) = 0 \times [- 1 \leq s i n (\frac{1}{h}) \leq 1] = 0 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (0 + h) - f (0)}{h} = \underset{h \to 0}{l i m} \frac{{(0 + h)}^{2} s i n \frac{1}{(0 + h)} - 0}{h} = \underset{h \to 0}{l i m} \frac{h^{2} s i n (\frac{1}{h})}{h} \\ = \underset{h \to 0}{l i m} h . s i n (\frac{1}{h}) = 0 \times [- 1 \leq s i n (\frac{1}{h}) \leq 1] = 0 \\ S o, L f' (c) = R f' (c) = 0 \\ H e n c e, f (x) i s d i f f e r e n t i a b l e a t x = 0 . \end{array}$

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} 1 + x, & i f x \leq 2 \\ 5 - x, & i f x > 2 \end{matrix}$ at $x = 2$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) i s d i f f e r e n t i a b l e a t x = 2 i f \\ L f' (2) = R f' (2) \\ ∴ L f' (c) = \underset{h \to 0}{l i m} \frac{f (2 - h) - f (2)}{- h} = \underset{h \to 0}{l i m} \frac{(1 + 2 - h) - (1 + 2)}{- h} \\ = \underset{h \to 0}{l i m} \frac{3 - h - 3}{- h} = \frac{- h}{- h} = 1 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (2 + h) - f (2)}{h} = \underset{h \to 0}{l i m} \frac{[5 - (2 + h)] - (1 + 2)}{h} \\ = \underset{h \to 0}{l i m} \frac{3 - h - 3}{h} = \frac{- h}{h} = - 1 \\ S o, L f' (c) \neq R f' (c) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 2 . \end{array}$

Show that $f (x) = ∣ x - 5 ∣$ is continuous but not differentiable at $x = 5$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = | x - 5 | \\ \Rightarrow f (x) = {\begin{cases} - (x - 5) i f x - 5 < 0 o r x < 5 \\ x - 5 i f x - 5 > 0 o r x > 5 \end{cases} \\ F o r c o n t i n u i t y a t x = 5 \\ L . H . L . \underset{h \to 5^{-}}{l i m} f (x) = - (x - 5) \\ = \underset{h \to 0}{l i m} - (5 - h - 5) = \underset{h \to 0}{l i m} h = 0 \\ R . H . L . \underset{h \to 5^{+}}{l i m} f (x) = x - 5 \\ = \underset{h \to 0}{l i m} (5 + h - 5) = \underset{h \to 0}{l i m} h = 0 \\ L . H . L . = R . H . L . \\ S o, f (x) i s c o n t i n u o u s a t x = 5 . \\ N o w, f o r d i f f e r e n t i a b i l i t y \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (5 - h) - f (5)}{- h} = \underset{h \to 0}{l i m} \frac{- (5 - h - 5) - (5 - 5)}{- h} \\ = \underset{h \to 0}{l i m} \frac{h}{- h} = \frac{h}{- h} = 1 \\ R f' (5) = \underset{h \to 0}{l i m} \frac{f (5 + h) - f (5)}{h} = \underset{h \to 0}{l i m} \frac{(5 + h - 5) - (5 - 5)}{h} \\ = \underset{h \to 0}{l i m} \frac{h - 0}{h} = \frac{h}{h} = 1 \\ S o, L f' (5) \neq R f' (5) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 5 . \end{array}$

A function $f : R \to R$ satisfies the equation $f (x + y) = f (x) f (y)$ for all $x, y \in R$ , where $f (x) \neq 0$ . Suppose that the function is differentiable at $x = 0$ and $f^{'} (0) = 2$ .
Prove that $f^{'} (x) = 2 f (x)$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f : R \to R s a t i s f i e s t h e e q u a t i o n f (x + y) = f (x) . f (y) \forall x, y \in R, f (x) \neq 0 . \\ L e t u s t a k e a n y point x = 0 a t w h i c h t h e f u n c t i o n f (x) i s d i f f e r e n t i a b l e . \\ ∴ f' (0) = \underset{h \to 0}{l i m} \frac{f (0 + h) - f (0)}{h} \\ 2 = \underset{h \to 0}{l i m} \frac{f (0) . f (h) - f (0)}{h} [? f (0) = f (h)] \dots (1) \\ 2 = \underset{h \to 0}{l i m} \frac{f (0) . [f (h) - 1]}{h} \\ N o w, f' (x) = \underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h} \\ = \underset{h \to 0}{l i m} \frac{f (x) . f (h) - f (x)}{h} [? f (x + y) = f (x) . f (y)] \\ = \underset{h \to 0}{l i m} \frac{f (x) . [f (h) - 1]}{h} = 2 f (x) [F r o m e q (1)] \\ H e n c e, f' (x) = 2 f (x) \end{array}$

Differentiate each of the following w.r.t. $x$ (Exercises 25 to 43):

$2^{c o s^{2}} x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = 2^{c o s^{2} x} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t \\ l o g y = l o g 2^{c o s^{2} x} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = l o g 2 . \frac{d}{d x} c o s^{2} x \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = l o g 2 . [2 c o s x . \frac{d}{d x} c o s x] \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = l o g 2 . [2 c o s x (- s i n x)] \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = l o g 2 (- s i n 2 x) \\ \Rightarrow \frac{d y}{d x} = - y . l o g 2 s i n 2 x \\ H e n c e, \frac{d y}{d x} = - 2^{c o s^{2} x} (l o g 2 s i n 2 x) \end{array}$

Kindly consider the following

$\frac{8^{x}}{x^{8}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{8^{x}}{x^{8}} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t, l o g y = l o g \frac{8^{x}}{x^{8}} \\ \Rightarrow l o g y = l o g 8^{x} - l o g x^{8} \Rightarrow l o g y = x l o g 8 - 8 l o g x \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = l o g 8.1 - \frac{8}{x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = y [l o g 8 - \frac{8}{x}] \\ H e n c e, \frac{d y}{d x} = \frac{8^{x}}{x^{8}} [l o g 8 - \frac{8}{x}] \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$l o g [l o g (l o g x^{5})]$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = l o g [l o g (l o g x^{5})] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d y}{d x} = \frac{d}{d x} l o g [l o g (l o g x^{5})] \\ \Rightarrow = \frac{1}{log (l o g x^{5})} \times \frac{d}{d x} log (l o g x^{5}) \\ = \frac{1}{log (l o g x^{5})} \times \frac{1}{l o g (x^{5})} \times \frac{d}{d x} l o g x^{5} \\ = \frac{1}{log (l o g x^{5})} \times \frac{1}{l o g (x^{5})} \times \frac{1}{x^{5}} . \frac{d}{d x} x^{5} \\ = \frac{1}{log (l o g x^{5})} \times \frac{1}{l o g (x^{5})} \times \frac{1}{x^{5}} . 5 x^{4} \\ = \frac{5}{x log (x^{5}) . l o g l o g (x^{5})} \\ H e n c e, \frac{d y}{d x} = \frac{5}{x log (x^{5}) . l o g l o g (x^{5})} . \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$s i n^{2} (a x^{2} + b x + c)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = s i n^{n} (a x^{2} + b x + c) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d y}{d x} = \frac{d}{d x} s i n^{n} (a x^{2} + b x + c) \\ \Rightarrow = n . s i n^{n - 1} (a x^{2} + b x + c) . \frac{d}{d x} s i n (a x^{2} + b x + c) \\ = n . s i n^{n - 1} (a x^{2} + b x + c) . c o s (a x^{2} + b x + c) . \frac{d}{d x} (a x^{2} + b x + c) \\ = n . s i n^{n - 1} (a x^{2} + b x + c) . c o s (a x^{2} + b x + c) . (2 a x + b) \\ H e n c e, \frac{d y}{d x} = n (2 a x + b) . s i n^{n - 1} (a x^{2} + b x + c) . c o s (a x^{2} + b x + c) . \end{array}$

Kindly consider the following

$c o s (t a n√x+1)$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2})$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2}) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d y}{d x} = \frac{d}{d x} s i n (x^{2}) + 2 s i n x . \frac{d}{d x} (s i n x) + 2 s i n (x^{2}) \frac{d}{d x} s i n (x^{2}) \\ = c o s x^{2} . \frac{d}{d x} (x^{2}) + 2 s i n x . \frac{d}{d x} (s i n x) + 2 s i n (x^{2}) \frac{d}{d x} s i n (x^{2}) \\ = c o s x^{2} . 2 x + 2 s i n x . c o s x + 2 s i n x^{2} . c o s x^{2} . \frac{d}{d x} (x^{2}) \\ = 2 x . c o s x^{2} + s i n 2 x + 2 s i n x^{2} . c o s x^{2} . 2 x \\ = 2 x . c o s x^{2} + s i n 2 x + 2 x s i n 2 x^{2} \\ H e n c e, \frac{d y}{d x} = 2 x . c o s x^{2} + s i n 2 x + 2 x s i n 2 x^{2} \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

${(s i n x)}^{c o s x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(s i n x)}^{c o s x} \\ T a k i n g l o g o n b o t h s i d e s, \\ l o g y = l o g {(s i n x)}^{c o s x} \\ l o g y = c o s x . l o g (s i n x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{1}{y} . \frac{d y}{d x} = \frac{d}{d x} c o s x . l o g (s i n x) \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = c o s x . \frac{d}{d x} l o g (s i n x) + l o g (s i n x) . \frac{d}{d x} c o s x \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = c o s x . \frac{1}{s i n x} . \frac{d}{d x} (s i n x) + l o g (s i n x) . (- s i n x) \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = c o t x . c o s x - s i n x . l o g (s i n x) \\ \frac{d y}{d x} = y [c o t x . c o s x - s i n x . l o g (s i n x)] \\ H e n c e, \frac{d y}{d x} = {(s i n x)}^{c o s x} [\frac{c o s^{2} x}{s i n x} - s i n x . l o g (s i n x)] \end{array}$

Kindly consider the following

$s i n^{m} x c o s^{n} x$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = s i n^{m} x c o s^{n} x \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d y}{d x} = \frac{d}{d x} (s i n^{m} x c o s^{n} x) \\ = s i n^{m} x . \frac{d}{d x} (c o s^{n} x) + c o s^{n} x . \frac{d}{d x} (s i n^{m} x) \\ = s i n^{m} x . n . c o s^{n - 1} x \frac{d}{d x} (c o s x) + c o s^{n} x . m . s i n^{m - 1} x \frac{d}{d x} (s i n x) \\ = n . s i n^{m} x . c o s^{n - 1} x . (- s i n x) + m . c o s^{n} x . s i n^{m - 1} x . c o s x \\ = - n . s i n^{m + 1} x . c o s^{n - 1} x + m . c o s^{n + 1} x . s i n^{m - 1} x \\ = s i n^{m} x . c o s^{n} x [- n \frac{s i n x}{c o s x} + m . \frac{c o s x}{s i n x}] \\ H e n c e, \frac{d y}{d x} = s i n^{m} x . c o s^{n} x [- n t a n x + m . c o t x] \end{array}$

Kindly consider the following

${(x + 1)}^{2} {(x + 2)}^{3} {(x + 3)}^{4}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(x + 1)}^{2} {(x + 2)}^{3} {(x + 3)}^{4} \\ T a k i n g l o g o n b o t h s i d e s, \\ l o g y = l o g [{(x + 1)}^{2} . {(x + 2)}^{3} . {(x + 3)}^{4}] \\ \Rightarrow l o g y = l o g {(x + 1)}^{2} + l o g {(x + 2)}^{3} + l o g {(x + 3)}^{4} [? l o g x y = l o g x + l o g y] \\ \Rightarrow l o g y = 2 l o g (x + 1) + 3 l o g (x + 2) + 4 l o g (x + 3) [? l o g x^{y} = y l o g x] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{1}{y} . \frac{d y}{d x} = 2 . \frac{d}{d x} l o g (x + 1) + 3 . \frac{d}{d x} l o g (x + 2) + 4 . \frac{d}{d x} l o g (x + 3) \\ \frac{1}{y} . \frac{d y}{d x} = 2 . \frac{1}{x + 1} + 3 . \frac{1}{x + 2} + 4 . \frac{1}{x + 3} \\ \frac{d y}{d x} = y [\frac{2}{x + 1} + \frac{3}{x + 2} + \frac{4}{x + 3}] \\ = {(x + 1)}^{2} {(x + 2)}^{3} {(x + 3)}^{4} [\frac{2}{x + 1} + \frac{3}{x + 2} + \frac{4}{x + 3}] \\ = {(x + 1)}^{2} {(x + 2)}^{3} {(x + 3)}^{4} [\frac{2 (x + 2) (x + 3) + 3 (x + 1) (x + 3) + 4 (x + 1) (x + 2)}{(x + 1) (x + 2) (x + 3)}] \\ = (x + 1) {(x + 2)}^{2} {(x + 3)}^{3} [2 x^{2} + 10 x + 12 + 3 x^{2} + 12 x + 9 + 4 x^{2} + 12 x + 8] \\ = (x + 1) {(x + 2)}^{2} {(x + 3)}^{3} [9 x^{2} + 34 x + 29] \\ H e n c e, \frac{d y}{d x} = (x + 1) {(x + 2)}^{2} {(x + 3)}^{3} [9 x^{2} + 34 x + 29] \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$t a n^{- 1} (s e c x + t a n x), - \frac{π}{2} < x < \frac{π}{2}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = t a n^{- 1} (s e c x + t a n x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d y}{d x} = \frac{d}{d x} [t a n^{- 1} (s e c x + t a n x)] \\ = \frac{1}{1 + {(s e c x + t a n x)}^{2}} . \frac{d}{d x} (s e c x + t a n x) \\ = \frac{1}{1 + s e c^{2} x + t a n^{2} x + 2 s e c x t a n x} . (s e c x t a n x + s e c^{2} x) \\ = \frac{1}{(1 + t a n^{2} x) + s e c^{2} x + 2 s e c x t a n x} . s e c x (t a n x + s e c x) \\ = \frac{1}{s e c^{2} x + s e c^{2} x + 2 s e c x t a n x} . s e c x (t a n x + s e c x) \\ = \frac{1}{2 s e c^{2} x + 2 s e c x t a n x} . s e c x (t a n x + s e c x) \\ = \frac{1}{2 s e c x (s e c x + t a n x)} . s e c x (t a n x + s e c x) = \frac{1}{2} \\ H e n c e, \frac{d y}{d x} = \frac{1}{2} \end{array}$

Kindly consider the following

$t a n^{- 1} (\frac{a c o s x - b s i n x}{b c o s x + a s i n x}), - \frac{π}{2} < x ⟨ \frac{π}{2} a n d \frac{a}{b} t a n x ⟩ - 1$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = t a n^{- 1} (\frac{a c o s x - b s i n x}{b c o s x + a s i n x}) \\ \Rightarrow y = t a n^{- 1} [\frac{\frac{a c o s x}{b c o s x} - \frac{b s i n x}{b c o s x}}{\frac{b c o s x}{b c o s x} + \frac{a s i n x}{b c o s x}}] \\ \Rightarrow y = t a n^{- 1} [\frac{\frac{a}{b} - t a n x}{1 + \frac{a}{b} t a n x}] \\ \Rightarrow y = t a n^{- 1} \frac{a}{b} - t a n^{- 1} (t a n x) [? t a n^{- 1} (\frac{x - y}{1 + x y}) = t a n^{- 1} x - t a n^{- 1} y] \\ \Rightarrow y = t a n^{- 1} \frac{a}{b} - x \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d y}{d x} = \frac{d}{d x} [t a n^{- 1} \frac{a}{b}] - \frac{d}{d x} (x) = 0 - 1 = - 1 \\ H e n c e, \frac{d y}{d x} = - 1 \end{array}$

Kindly consider the following

$s e c^{- 1} (\frac{1}{4 x^{3} - 3 x}), 0 < x <1\sqrt2$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$t a n^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 a x^{2}}), -1\sqrt3$
$\frac{}{} < x / a <1 √3$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = t a n^{- 1} [\frac{3 a^{2} x - x^{3}}{a^{3} - 3 a x^{2}}] \\ P u t x = a t a n θ ∴ θ = t a n^{- 1} \frac{x}{a} \\ \Rightarrow y = t a n^{- 1} [\frac{3 a^{2} a t a n θ - a^{3} t a n^{3} θ}{a^{3} - 3 a . a^{2} t a n^{2} θ}] \\ \Rightarrow y = t a n^{- 1} [\frac{3 a^{3} t a n θ - a^{3} t a n^{3} θ}{a^{3} - 3 a^{3} t a n^{2} θ}] \\ \Rightarrow y = t a n^{- 1} [\frac{3 t a n θ - t a n^{3} θ}{1 - 3 t a n^{2} θ}] \\ \Rightarrow y = t a n^{- 1} (t a n 3 θ) [? t a n 3 θ = \frac{3 t a n θ - t a n^{3} θ}{1 - 3 t a n^{2} θ}] \\ \Rightarrow y = 3 θ \Rightarrow y = 3 t a n^{- 1} \frac{x}{a} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d y}{d x} = 3 . \frac{d}{d x} (t a n^{- 1} \frac{x}{a}) = 3 . \frac{1}{1 + \frac{x^{2}}{a^{2}}} . \frac{d}{d x} . (\frac{x}{a}) = 3 . \frac{a^{2}}{a^{2} + x^{2}} . \frac{1}{a} = \frac{3 a}{a^{2} + x^{2}} \\ H e n c e, \frac{d y}{d x} = \frac{3 a}{a^{2} + x^{2}} \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Find $\frac{d y}{d x}$ of each of the functions expressed in parametric form in Exercises from 44 to 48.

$x = t + \frac{1}{t}, y = t - \frac{1}{t}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = t + \frac{1}{t}, y = t - \frac{1}{t} \\ D i f f e r e n t i a t i n g b o t h t h e g i v e n p a r a m e t r i c f u n c t i o n s w . r . t . t \\ \frac{d x}{d t} = 1 - \frac{1}{t^{2}}, \frac{d y}{d t} = 1 + \frac{1}{t^{2}} \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{1 + \frac{1}{t^{2}}}{1 - \frac{1}{t^{2}}} = \frac{t^{2} + 1}{t^{2} - 1} \\ H e n c e, \frac{d y}{d x} = \frac{t^{2} + 1}{t^{2} - 1} \end{array}$

Kindly consider the following

$x = e^{θ} (θ + \frac{1}{θ}), y = e^{- θ} (θ - \frac{1}{θ})$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = e^{θ} (θ + \frac{1}{θ}), y = e^{- θ} (θ - \frac{1}{θ}) \\ D i f f e r e n t i a t i n g b o t h t h e p a r a m e t r i c f u n c t i o n s w . r . t . θ \\ \frac{d x}{d θ} = e^{θ} (1 - \frac{1}{θ^{2}}) + (θ + \frac{1}{θ}) . e^{θ} \\ \frac{d x}{d θ} = e^{θ} (1 - \frac{1}{θ^{2}} + θ + \frac{1}{θ}) \Rightarrow e^{θ} (\frac{θ^{2} - 1 + θ^{3} + θ}{θ^{2}}) \\ = e^{θ} \frac{(θ^{3} + θ^{2} + θ - 1)}{θ^{2}} \\ y = e^{- θ} (θ - \frac{1}{θ}) \\ \frac{d y}{d θ} = e^{- θ} (1 + \frac{1}{θ^{2}}) + (θ - \frac{1}{θ}) . (- e^{- θ}) \\ \frac{d y}{d θ} = e^{- θ} (1 + \frac{1}{θ^{2}} - θ + \frac{1}{θ}) \Rightarrow e^{- θ} (\frac{θ^{2} + 1 - θ^{3} + θ}{θ^{2}}) \\ = e^{- θ} \frac{(- θ^{3} + θ^{2} + θ + 1)}{θ^{2}} \\ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{e^{- θ} \frac{(- θ^{3} + θ^{2} + θ + 1)}{θ^{2}}}{e^{θ} \frac{(θ^{3} + θ^{2} + θ - 1)}{θ^{2}}} \\ = e^{- 2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1}) \\ H e n c e, \frac{d y}{d x} = e^{- 2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1}) \end{array}$

Kindly consider the following

$x = 3 c o s θ - 2 c o s^{3} θ, y = 3 s i n θ - 2 s i n^{3} θ$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = 3 c o s θ - 2 c o s^{3} θ, y = 3 s i n θ - 2 s i n^{3} θ \\ D i f f e r e n t i a t i n g b o t h t h e g i v e n p a r a m e t r i c f u n c t i o n s w . r . t . θ \\ \frac{d x}{d θ} = - 3 s i n θ - 6 c o s^{2} θ . \frac{d}{d θ} (c o s θ) \\ = - 3 s i n θ - 6 c o s^{2} θ . (- s i n θ) \\ = - 3 s i n θ + 6 c o s^{2} θ . s i n θ \\ \frac{d y}{d θ} = 3 c o s θ - 6 s i n^{2} θ . \frac{d}{d θ} (s i n θ) \\ = 3 c o s θ - 6 s i n^{2} θ . c o s θ \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 c o s θ - 6 s i n^{2} θ . c o s θ}{- 3 s i n θ + 6 c o s^{2} θ . s i n θ} \\ = \frac{c o s θ (3 - 6 s i n^{2} θ)}{s i n θ (- 3 + 6 c o s^{2} θ)} = \frac{c o s θ [3 - 6 (1 - c o s^{2} θ)]}{s i n θ [- 3 + 6 c o s^{2} θ]} \\ = c o t θ (\frac{3 - 6 + 6 c o s^{2} θ}{- 3 + 6 c o s^{2} θ}) = c o t θ (\frac{- 3 + 6 c o s^{2} θ}{- 3 + 6 c o s^{2} θ}) = c o t θ \\ H e n c e, \frac{d y}{d x} = c o t θ \end{array}$

Kindly consider the following

$s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} N o w, t a k i n g t a n y = \frac{2 t}{1 - t^{2}} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t, w e g e t \\ \frac{d}{d t} (t a n y) = \frac{d}{d t} (\frac{2 t}{1 - t^{2}}) \\ \Rightarrow s e c^{2} y . \frac{d y}{d t} = \frac{(1 - t^{2}) . \frac{d}{d t} (2 t) - 2 t . \frac{d}{d t} (1 - t^{2})}{{(1 - t^{2})}^{2}} \\ \Rightarrow s e c^{2} y . \frac{d y}{d t} = \frac{2 . (1 - t^{2}) - 2 t . (- 2 t)}{{(1 - t^{2})}^{2}} \\ \Rightarrow s e c^{2} y . \frac{d y}{d t} = \frac{2 - 2 t^{2} + 4 t^{2}}{{(1 - t^{2})}^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{2 + 2 t^{2}}{{(1 - t^{2})}^{2}} \times \frac{1}{s e c^{2} y} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{1}{1 + t a n^{2} y} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{1}{1 + {(\frac{2 t}{1 - t^{2}})}^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{1}{\frac{{(1 - t^{2})}^{2} + 4 t^{2}}{{(1 - t^{2})}^{2}}} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{{(1 - t^{2})}^{2}}{1 + t^{4} - 2 t^{2} + 4 t^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{{(1 - t^{2})}^{2}}{1 + t^{4} + 2 t^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{{(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{2}{1 + t^{2}} \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{2}{1 + t^{2}}}{\frac{2}{1 + t^{2}}} = 1 \\ H e n c e, \frac{d y}{d x} = 1 \end{array}$

Kindly consider the following

$x = \frac{1 + l o g t}{t^{2}}, y = \frac{3 + 2 l o g t}{t}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = \frac{1 + l o g t}{t^{2}}, y = \frac{3 + 2 l o g t}{t} \\ D i f f e r e n t i a t i n g b o t h t h e p a r a m e t r i c f u n c t i o n s w . r . t . t, w e g e t \\ \Rightarrow \frac{d x}{d t} = \frac{t^{2} . \frac{d}{d t} (1 + l o g t) - (1 + l o g t) . \frac{d}{d t} (t^{2})}{t^{4}} \\ \Rightarrow = \frac{t^{2} . (\frac{1}{t}) - (1 + l o g t) . 2 t}{t^{4}} \\ \Rightarrow = \frac{t - (1 + l o g t) . 2 t}{t^{4}} \\ \Rightarrow = \frac{t [1 - 2 - 2 l o g t]}{t^{4}} = \frac{[- 1 - 2 l o g t]}{t^{3}} \\ \Rightarrow \frac{d x}{d t} = \frac{- (1 + 2 l o g t)}{t^{3}} \\ N o w, y = \frac{3 + 2 l o g t}{t} \\ \Rightarrow \frac{d y}{d t} = \frac{t . \frac{d}{d t} (3 + 2 l o g t) - (3 + 2 l o g t) . \frac{d}{d t} (t)}{t^{2}} \\ \Rightarrow = \frac{t . (\frac{2}{t}) - (3 + 2 l o g t) . 1}{t^{2}} \\ \Rightarrow = \frac{2 - 3 - 2 l o g t}{t^{2}} \\ \Rightarrow = \frac{- (1 + 2 l o g t)}{t^{2}} \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{- (1 + 2 l o g t)}{t^{2}}}{\frac{- (1 + 2 l o g t)}{t^{3}}} = \frac{t^{3}}{t^{2}} = t \\ H e n c e, \frac{d y}{d x} = t \end{array}$

If $x = e^{c o s 2 t}$ and $y = e^{s i n 2 t}$ , prove that $\frac{d y}{d x} = \frac{- y l o g x}{x l o g y}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = e^{c o s 2 t}, y = e^{s i n 2 t} \\ \Rightarrow c o s 2 t = l o g x a n d s i n 2 t = l o g y \\ D i f f e r e n t i a t i n g b o t h t h e p a r a m e t r i c f u n c t i o n s w . r . t . t, w e g e t \\ \Rightarrow \frac{d x}{d t} = e^{c o s 2 t} . \frac{d}{d t} (c o s 2 t) \\ \Rightarrow = e^{c o s 2 t} (- s i n 2 t) . \frac{d}{d t} (2 t) \\ \Rightarrow = e^{c o s 2 t} (- s i n 2 t) . 2 = - 2 e^{c o s 2 t} s i n 2 t \\ N o w, y = e^{s i n 2 t} \\ \Rightarrow \frac{d y}{d t} = e^{s i n 2 t} . \frac{d}{d t} (s i n 2 t) \\ \Rightarrow = e^{s i n 2 t} (c o s 2 t) . \frac{d}{d t} (2 t) \\ \Rightarrow = e^{s i n 2 t} (c o s 2 t) . 2 \\ \Rightarrow = 2 e^{s i n 2 t} c o s 2 t \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 e^{s i n 2 t} c o s 2 t}{- 2 e^{c o s 2 t} s i n 2 t} = \frac{e^{s i n 2 t} c o s 2 t}{- e^{c o s 2 t} s i n 2 t} \\ = \frac{y c o s 2 t}{- x s i n 2 t} = \frac{y l o g x}{- x l o g y} [\begin{array}{l} ? c o s 2 t = l o g x \\ s i n 2 t = l o g y \end{array}] \\ H e n c e, \frac{d y}{d x} = \frac{y l o g x}{- x l o g y} \end{array}$

If $x = a s i n 2 t (1 + c o s 2 t)$ and $y = b c o s 2 t (1 - c o s 2 t)$ , show that ${(\frac{d y}{d x})}_{a t t = \frac{π}{4}} = \frac{b}{a}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = a s i n 2 t (1 + c o s 2 t), y = b c o s 2 t (1 - c o s 2 t) \\ D i f f e r e n t i a t i n g b o t h t h e p a r a m e t r i c f u n c t i o n s w . r . t . t, w e g e t \\ \Rightarrow \frac{d x}{d t} = a [s i n 2 t . \frac{d}{d t} (1 + c o s 2 t) + (1 + c o s 2 t) . \frac{d}{d t} (s i n 2 t)] \\ \Rightarrow = a [s i n 2 t . (- s i n 2 t) . 2 + (1 + c o s 2 t) (c o s 2 t) . 2] \\ \Rightarrow = a [- 2 s i n^{2} 2 t + 2 c o s 2 t + 2 c o s^{2} 2 t] \\ \Rightarrow = a [2 (c o s^{2} 2 t - s i n^{2} 2 t) + 2 c o s 2 t] \\ \Rightarrow = a [2 c o s 4 t + 2 c o s 2 t] [? c o s 2 x = c o s^{2} x - s i n^{2} x] \\ \Rightarrow = 2 a [c o s 4 t + c o s 2 t] \\ N o w, y = b c o s 2 t (1 - c o s 2 t) \\ \Rightarrow \frac{d y}{d t} = b [c o s 2 t . \frac{d}{d t} (1 - c o s 2 t) + (1 - c o s 2 t) . \frac{d}{d t} (c o s 2 t)] \\ \Rightarrow = b [c o s 2 t . (s i n 2 t) . 2 + (1 - c o s 2 t) (- s i n 2 t) . 2] \\ \Rightarrow = b [2 s i n 2 t c o s 2 t - 2 s i n 2 t + 2 s i n 2 t c o s 2 t] \\ \Rightarrow = b [s i n 4 t - 2 s i n 2 t + s i n 4 t] [? s i n 2 x = 2 s i n x c o s x] \\ \Rightarrow = b [2 s i n 4 t - 2 s i n 2 t] = 2 b (s i n 4 t - s i n 2 t) \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 b [s i n 4 t - s i n 2 t]}{2 a [c o s 4 t + c o s 2 t]} = \frac{b}{a} [\frac{s i n 4 t - s i n 2 t}{c o s 4 t + c o s 2 t}] \\ P u t t = \frac{π}{4} \\ ∴ {(\frac{d y}{d x})}_{a t t = \frac{π}{4}} = \frac{b}{a} [\frac{s i n 4 (\frac{π}{4}) - s i n 2 (\frac{π}{4})}{c o s 4 (\frac{π}{4}) + c o s 2 (\frac{π}{4})}] = \frac{b}{a} [\frac{s i n π - s i n \frac{π}{2}}{c o s π + c o s \frac{π}{2}}] \\ &thi \end{array}$

If $x = 3 s i n t - s i n 3 t, y = 3 c o s t - c o s 3 t$ , find $\frac{d y}{d x}$ at $t = \frac{π}{3}$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Differentiate $\frac{x}{s i n x}$ w.r.t. $s i n x$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{x}{s i n x} a n d z = s i n x \\ D i f f e r e n t i a t i n g b o t h t h e p a r a m e t r i c f u n c t i o n s w . r . t . x \\ \frac{d y}{d x} = \frac{s i n x . \frac{d}{d x} (x) - x . \frac{d}{d x} (s i n x)}{{(s i n x)}^{2}} \\ = \frac{s i n x . 1 - x . c o s x}{(s i n^{2} x)} = \frac{s i n x - x . c o s x}{s i n^{2} x} \\ \frac{d z}{d x} = c o s x \\ ∴ \frac{d y}{d z} = \frac{\frac{d y}{d x}}{\frac{d z}{d x}} = \frac{\frac{s i n x - x . c o s x}{s i n^{2} x}}{c o s x} = \frac{s i n x - x c o s x}{s i n^{2} x c o s x} \\ = \frac{s i n x}{s i n^{2} x c o s x} - \frac{x c o s x}{s i n^{2} x c o s x} \\ = \frac{t a n x}{s i n^{2} x} - \frac{x}{s i n^{2} x} = \frac{t a n x - x}{s i n^{2} x} \\ H e n c e, \frac{d y}{d z} = \frac{t a n x - x}{s i n^{2} x} \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Find $\frac{d y}{d x}$ when $x$ and $y$ are connected by the relation given in each of the Exercises 54 to 57.

$s i n (x y) + \frac{x}{y} = x^{2} - y$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t s i n x y + \frac{x}{y} = x^{2} - y \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} s i n (x y) + \frac{d}{d x} (\frac{x}{y}) = \frac{d}{d x} (x^{2}) - \frac{d}{d x} (y) \\ \Rightarrow c o s x y . \frac{d}{d x} (x y) + \frac{y . \frac{d}{d x} (x) - x . \frac{d y}{d x}}{y^{2}} = 2 x - \frac{d y}{d x} \\ \Rightarrow c o s x y [x . \frac{d y}{d x} + y . 1] + \frac{y . 1}{y^{2}} - \frac{x}{y^{2}} \frac{d y}{d x} = 2 x - \frac{d y}{d x} \\ \Rightarrow x c o s x y . \frac{d y}{d x} + y c o s x y + \frac{1}{y} - \frac{x}{y^{2}} \frac{d y}{d x} = 2 x - \frac{d y}{d x} \\ \Rightarrow x c o s x y . \frac{d y}{d x} - \frac{x}{y^{2}} \frac{d y}{d x} + \frac{d y}{d x} = 2 x - y c o s x y - \frac{1}{y} \\ \Rightarrow [x c o s x y - \frac{x}{y^{2}} + 1] \frac{d y}{d x} = 2 x - y c o s x y - \frac{1}{y} \\ \Rightarrow [\frac{x y^{2} c o s x y - x + y^{2}}{y^{2}}] \frac{d y}{d x} = \frac{2 x y - y^{2} c o s x y - 1}{y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 x y - y^{2} c o s x y - 1}{y} \times \frac{y^{2}}{x y^{2} c o s x y - x + y^{2}} \\ = \frac{2 x y^{2} - y^{3} c o s (x y) - y}{x y^{2} c o s (x y) - x + y^{2}} \\ H e n c e, \frac{d y}{d x} = \frac{2 x y^{2} - y^{3} c o s (x y) - y}{x y^{2} c o s (x y) - x + y^{2}} \end{array}$

Kindly consider the following

$s e c (x + y) = x y$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t s e c (x + y) = x y \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} s e c (x + y) = \frac{d}{d x} (x y) \\ \Rightarrow s e c (x + y) . t a n (x + y) . \frac{d}{d x} (x + y) = x . \frac{d y}{d x} + y . 1 \\ \Rightarrow s e c (x + y) . t a n (x + y) . (1 + \frac{d y}{d x}) = x . \frac{d y}{d x} + y \\ \Rightarrow s e c (x + y) . t a n (x + y) + s e c (x + y) . t a n (x + y) . \frac{d y}{d x} = x . \frac{d y}{d x} + y \\ \Rightarrow s e c (x + y) . t a n (x + y) . \frac{d y}{d x} - x . \frac{d y}{d x} = y - s e c (x + y) . t a n (x + y) \\ \Rightarrow [s e c (x + y) . t a n (x + y) - x] \frac{d y}{d x} = y - s e c (x + y) . t a n (x + y) \\ \Rightarrow \frac{d y}{d x} = \frac{y - s e c (x + y) . t a n (x + y)}{s e c (x + y) . t a n (x + y) - x} \\ H e n c e, \frac{d y}{d x} = \frac{y - s e c (x + y) . t a n (x + y)}{s e c (x + y) . t a n (x + y) - x} \end{array}$

Kindly consider the following

$t a n^{- 1} (x^{2} + y^{2}) = a$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t a n^{- 1} (x^{2} + y^{2}) = a \\ \Rightarrow x^{2} + y^{2} = t a n a \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (t a n a) \\ \Rightarrow 2 x + 2 y . \frac{d y}{d x} = 0 \Rightarrow 2 y . \frac{d y}{d x} = - 2 x \\ \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{2 y} = \frac{- x}{y} \\ H e n c e, \frac{d y}{d x} = \frac{- x}{y} \end{array}$

Kindly consider the following

${(x^{2} + y^{2})}^{2} = x y$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t {(x^{2} + y^{2})}^{2} = x y \\ \Rightarrow x^{4} + y^{4} + 2 x^{2} y^{2} = x y \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} (x^{4}) + \frac{d}{d x} (y^{4}) + 2 . \frac{d}{d x} (x^{2} y^{2}) = \frac{d}{d x} (x y) \\ \Rightarrow 4 x^{3} + 4 y^{3} . \frac{d y}{d x} + 2 [x^{2} . 2 y . \frac{d y}{d x} + y^{2} . 2 x] = x . \frac{d y}{d x} + y . 1 \\ \Rightarrow 4 x^{3} + 4 y^{3} . \frac{d y}{d x} + 4 x^{2} y . \frac{d y}{d x} + 4 x y^{2} = x . \frac{d y}{d x} + y \\ \Rightarrow 4 y^{3} . \frac{d y}{d x} + 4 x^{2} y . \frac{d y}{d x} - x . \frac{d y}{d x} = y - 4 x^{3} - 4 x y^{2} \\ \Rightarrow [4 y^{3} + 4 x^{2} y - x] \frac{d y}{d x} = y - 4 x^{3} - 4 x y^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{y - 4 x^{3} - 4 x y^{2}}{4 y^{3} + 4 x^{2} y - x} \\ H e n c e, \frac{d y}{d x} = \frac{y - 4 x^{3} - 4 x y^{2}}{4 y^{3} + 4 x^{2} y - x} \end{array}$

Kindly consider the following

If $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , then show that $\frac{d y}{d x} \cdot \frac{d x}{d y} = 1$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} (a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c) = \frac{d}{d x} (0) \\ \Rightarrow a . 2 x + 2 h (x . \frac{d y}{d x} + y . 1) + b . 2 y . \frac{d y}{d x} + 2 g . 1 + 2 f . \frac{d y}{d x} + 0 = 0 \\ \Rightarrow 2 a x + 2 h x . \frac{d y}{d x} + 2 h y + 2 b y . \frac{d y}{d x} + 2 g + 2 f . \frac{d y}{d x} = 0 \\ \Rightarrow 2 h x . \frac{d y}{d x} + 2 b y . \frac{d y}{d x} + 2 f . \frac{d y}{d x} = - 2 a x - 2 h y - 2 g \\ \Rightarrow (2 h x + 2 b y + 2 f) . \frac{d y}{d x} = - 2 (a x + h y + g) \\ \Rightarrow \frac{d y}{d x} = \frac{- 2 (a x + h y + g)}{2 (h x + b y + f)} = \frac{- (a x + h y + g)}{(h x + b y + f)} \\ \Rightarrow \frac{d y}{d x} = \frac{- (a x + h y + g)}{(h x + b y + f)} \\ N o w, d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \frac{d}{d y} (a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c) = \frac{d}{d y} (0) \\ \Rightarrow 2 a x . \frac{d x}{d y} + 2 h (y . \frac{d x}{d y} + x . 1) + 2 b y + 2 g . \frac{d x}{d y} + 2 f . 1 + 0 = 0 \\ \Rightarrow 2 a x . \frac{d x}{d y} + 2 h y . \frac{d x}{d y} + 2 h x + 2 b y + 2 g . \frac{d x}{d y} + 2 f = 0 \\ \Rightarrow 2 a x \frac{d x}{d y} + 2 h y . \frac{d x}{d y} + 2 g . \frac{d x}{d y} = - 2 h x - 2 b y - 2 f \\ \Rightarrow (2 a x + 2 h y + 2 g) . \frac{d x}{d y} = - 2 (h x + b y + f) \\ \Rightarrow \frac{d x}{d y} = \frac{- 2 (h x + b y + f)}{2 (a x + h y + g)} = \frac{- (h x + b y + f)}{(a x + h y + g)} \\ \Rightarrow \frac{d x}{d y} = \frac{- (h x + b y + f)}{(a x + h y + g)} \\ ∴ \frac{d y}{d x} . \frac{d x}{d y} = [\frac{- (a x + h y + g)}{(h x + b y + f)}] . [\frac{- (h x + b y + f)}{(a x + h y + g)}] = 1 \\ H e n c e, \frac{d y}{d x} . \frac{d x}{d y} = 1 . H e n c e, p r o v e d . \end{array}$

Kindly consider the following

If $x = e^{\frac{x}{y}}$ , prove that $\frac{d y}{d x} = \frac{x - y}{x l o g x}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = e^{\frac{x}{y}} \\ T a k i n g l o g o n b o t h t h e s i d e s, \\ l o g x = l o g e^{\frac{x}{y}} \\ \Rightarrow l o g x = \frac{x}{y} l o g e \Rightarrow l o g x = \frac{x}{y} [? l o g e = 1] \dots (1) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} l o g x = \frac{d}{d x} (\frac{x}{y}) \\ \Rightarrow \frac{1}{x} = \frac{y . 1 - x . \frac{d y}{d x}}{y^{2}} \Rightarrow y^{2} = x (y - x . \frac{d y}{d x}) \\ \Rightarrow y^{2} = x y - x^{2} . \frac{d y}{d x} \Rightarrow - x^{2} . \frac{d y}{d x} = y^{2} - x y \\ \Rightarrow \frac{d y}{d x} = \frac{y^{2} - x y}{- x^{2}} = \frac{- y^{2} + x y}{x^{2}} = \frac{y (x - y)}{x^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} . \frac{(x - y)}{x} = \frac{1}{l o g x} . (\frac{x - y}{x}) [? l o g x = \frac{x}{y} f r o m e q n (1)] \\ H e n c e, \frac{d y}{d x} = \frac{x - y}{x l o g x} . H e n c e, p r o v e d . \end{array}$

Kindly consider the following

If $x^{y} = e^{y - x}$ , prove that $\frac{d y}{d x} = \frac{{(1 + l o g y)}^{2}}{l o g y}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y^{x} = e^{y - x} \\ T a k i n g l o g o n b o t h t h e s i d e s, \\ l o g y^{x} = l o g e^{y - x} \\ \Rightarrow x l o g y = (y - x) l o g e \Rightarrow x l o g y = y - x [? l o g e = 1] \\ \Rightarrow x l o g y + x = y \\ \Rightarrow x (l o g y + 1) = y \\ \Rightarrow x = \frac{y}{l o g y + 1} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \frac{d x}{d y} = \frac{d}{d y} (\frac{y}{l o g y + 1}) \\ \Rightarrow = \frac{(l o g y + 1) . 1 - y . \frac{d}{d y} (l o g y + 1)}{{(l o g y + 1)}^{2}} \\ \Rightarrow = \frac{(l o g y + 1) - y . \frac{1}{y}}{{(l o g y + 1)}^{2}} = \frac{l o g y + 1 - 1}{{(l o g y + 1)}^{2}} = \frac{l o g y}{{(l o g y + 1)}^{2}} \\ W e k n o w t h a t \\ H e n c e, \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{\frac{l o g y}{{(l o g y + 1)}^{2}}} = \frac{{(l o g y + 1)}^{2}}{l o g y} \\ H e n c e, p r o v e d . \end{array}$

Kindly consider the following

If $y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}}$ , show that $\frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}} \\ \Rightarrow y = {(c o s x)}^{y} [? y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}}] \\ T a k i n g l o g o n b o t h t h e s i d e s, l o g y = y . l o g (c o s x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{1}{y} . \frac{d y}{d x} = y . \frac{d}{d x} l o g (c o s x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = y . \frac{1}{c o s x} . \frac{d}{d x} (c o s x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = y . \frac{1}{c o s x} . (- s i n x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} - l o g (c o s x) . \frac{d y}{d x} = - y t a n x \\ \Rightarrow [\frac{1}{y} - l o g (c o s x)] \frac{d y}{d x} = - y t a n x \\ \Rightarrow \frac{d y}{d x} = \frac{- y t a n x}{\frac{1}{y} - l o g (c o s x)} = \frac{y^{2} t a n x}{y l o g c o s x - 1} \\ H e n c e, \frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1} \\ H e n c e, p r o v e d . \end{array}$

Kindly consider the following

If $x s i n (a + y) + s i n a c o s (a + y) = 0$ , prove that

$\frac{d y}{d x} = \frac{2 s i n^{2} (a + y)}{s i n a}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

If $y = t a n^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = t a n^{- 1} x \\ \Rightarrow x = t a n y \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \Rightarrow \frac{d x}{d y} = s e c^{2} y \Rightarrow \frac{d y}{d x} = \frac{1}{s e c^{2} y} = co s^{2} y \\ A g a i n d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (co s^{2} y) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 2 c o s y (- s i n y) . \frac{d y}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 s i n y c o s y . co s^{2} y \\ ∴ \frac{d^{2} y}{d x^{2}} = - 2 s i n y co s^{3} y \end{array}$

Verify Rolle's theorem for each of the functions in Exercises 65 to 69.

$f (x) = x {(x - 1)}^{2}$ in $[0, 1]$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x {(x - 1)}^{2} i n [0, 1] \\ (i) f (x) = x {(x - 1)}^{2}, b e i n g a n algebraic p o l y n o m i a l, i s c o n t i n u o u s i n [0, 1] . \\ (i i) f^{'} (x) = x . 2 (x - 1) + {(x - 1)}^{2} . 1 \\ = 2 x^{2} - 2 x + x^{2} + 1 - 2 x \\ = 3 x^{2} - 4 x + 1 w h i c h e x i s t s i n (0, 1) \\ (i i i) f (x) = x {(x - 1)}^{2} \\ = 0 {(0 - 1)}^{2} = 0; f (1) = 1 {(1 - 1)}^{2} = 0 \\ \Rightarrow f (0) = f (1) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (0, 1) s u c h t h a t f^{'} (c) = 0 \\ ∴ f^{'} (c) = 3 c^{2} - 4 c + 1 = 0 \Rightarrow 3 c^{2} - 3 c - c + 1 = 0 \\ \Rightarrow 3 c (c - 1) - 1 (c - 1) = 0 \Rightarrow (c - 1) (3 c - 1) = 0 \\ c - 1 = 0 \Rightarrow c = 1 \\ 3 c - 1 = 0 \Rightarrow 3 c = 1 ∴ c = \frac{1}{3} \in (0, 1) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

Kindly consider the following

$f (x) = s i n^{4} x + c o s^{4} x$ in $[0, \frac{π}{2}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = s i n^{4} x + c o s^{4} x i n [0, \frac{π}{2}] \\ (i) f (x) = s i n^{4} x + c o s^{4} x, b e i n g sine a n d cosine f u n c t i o n s, f (x) i s c o n t i n u o u s i n [0, \frac{π}{2}] . \\ (i i) f^{'} (x) = 4 s i n^{3} x . c o s x + 4 c o s^{3} x (- s i n x) \\ = 4 s i n^{3} x . c o s x - 4 s i n x c o s^{3} x \\ = 4 s i n x . c o s x (s i n^{2} x - c o s^{2} x) \\ = - 4 s i n x . c o s x (c o s^{2} x - s i n^{2} x) \\ = - 2.2 s i n x . c o s x c o s 2 x [\begin{array}{l} ? c o s 2 x = c o s^{2} x - s i n^{2} x \\ s i n 2 x = 2 s i n x c o s x \end{array}] \\ = - 2 s i n 2 x . c o s 2 x = - s i n 4 x w h i c h e x i s t s i n (0, \frac{π}{2}) \\ S o f (x) i s d i f f e r e n t i a b l e i n (0, \frac{π}{2}) . \\ (i i i) f (0) = s i n^{4} (0) + c o s^{4} (0) = 1 \\ f (\frac{π}{2}) = s i n^{4} (\frac{π}{2}) + c o s^{4} (\frac{π}{2}) = 1 \\ ∴ f (0) = f (\frac{π}{2}) = 1 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (0, \frac{π}{2}) s u c h t h a t f^{'} (c) = 0 \\ ∴ f^{'} (c) = 0 \Rightarrow - s i n 4 c = 0 \Rightarrow s i n 4 c = 0 \\ \Rightarrow s i n 4 c = s i n 0 \Rightarrow 4 c = n π \\ ∴ c = \frac{n π}{4}, n \in I \\ F o r n = 1, c = \frac{π}{4} \in (0, \frac{π}{2}) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

Kindly consider the following

$f (x) = l o g (x^{2} + 2) - l o g 3$ in $[- 1, 1]$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = l o g (x^{2} + 2) - l o g 3 i n [- 1, 1] \\ (i) f (x) = l o g (x^{2} + 2) - l o g 3, b e i n g a logarithm f u n c t i o n s, f (x) i s c o n t i n u o u s i n [- 1, 1] . \\ (i i) f^{'} (x) = \frac{1}{x^{2} + 2} . 2 x - 0 = \frac{2 x}{x^{2} + 2} w h i c h e x i s t s i n (- 1, 1) \\ S o f (x) i s d i f f e r e n t i a b l e i n (- 1, 1) . \\ (i i i) f (- 1) = l o g (1 + 2) - l o g 3 \Rightarrow l o g 3 - l o g 3 = 0 \\ f (1) = l o g (1 + 2) - l o g 3 \Rightarrow l o g 3 - l o g 3 = 0 \\ ∴ f (- 1) = f (1) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (- 1, 1) s u c h t h a t f^{'} (c) = 0 \\ ∴ \frac{2 c}{c^{2} + 2} = 0 \Rightarrow 2 c = 0 \Rightarrow c = 0 \in (- 1, 1) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

Kindly consider the following

$f (x) = x (x + 3) e^{- x / 2}$ in $[- 3, 0]$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x (x + 3) e^{- x / 2} i n [- 3, 0] \\ (i) Algebraic f u n c t i o n s a n d exponential f u n c t i o n s a r e c o n t i n u o u s i n t h e i r d o m a i n s . \\ ∴ f (x) i s c o n t i n u o u s i n [- 3, 0] . \\ (i i) f^{'} (x) = x (x + 3) . \frac{d}{d x} e^{- x / 2} + x . e^{- x / 2} . \frac{d}{d x} (x + 3) + (x + 3) . e^{- x / 2} \frac{d}{d x} . x \\ = x (x + 3) . e^{- x / 2} . (- \frac{1}{2}) + x . e^{- x / 2} . 1 + (x + 3) . e^{- x / 2} . 1 \\ = e^{- x / 2} [\frac{- x (x + 3)}{2} + x + x + 3] \\ = e^{- x / 2} [\frac{- x (x + 3)}{2} + 2 x + 3] = e^{- x / 2} [\frac{- x^{2} - 3 x + 4 x + 6}{2}] \\ = e^{- x / 2} [\frac{- x^{2} + x + 6}{2}] w h i c h e x i s t s i n (- 3, 0) \\ S o f (x) i s d i f f e r e n t i a b l e i n (- 3, 0) . \\ (i i i) f (- 3) = (- 3) (- 3 + 3) e^{- 3 / 2} = 0 \\ f (0) = (0) (0 + 3) e^{- 0 / 2} = 0 \\ ∴ f (- 3) = f (0) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (- 3, 0) s u c h t h a t f^{'} (c) = 0 \\ ∴ e^{- c / 2} [\frac{- c^{2} + c + 6}{2}] = 0 \Rightarrow - \frac{e^{- c / 2}}{2} [(c - 3) (c + 2)] = 0 \Rightarrow - e^{- c / 2} \neq 0 \\ ∴ (c - 3) (c + 2) = 0 \Rightarrow c = 3, - 2 \\ c = 3, - 2 \in (- 3, 0) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Discuss the applicability of Rolleâ€™s theorem on the function given by $f (x) = {\begin{matrix} x^{2} + 1, & 0 \leq x \leq 1 \\ 3 - x, & 1 \leq x \leq 2 \end{matrix}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) f (x) b e i n g a n algebraic p o l y n o m i a l i s c o n t i n u o u s e v e r y w h e r e . \\ (i i) f (x) m u s t b e d i f f e r e n t i a b l e a t x = 1 \\ L . H . L . = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 1) - (1 + 1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{x^{2} + 1 - 2}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{x^{2} - 1}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x - 1) (x + 1)}{x - 1} = \underset{x \to 1^{-}}{l i m} (x + 1) = (1 + 1) = 2 \\ R . H . L . = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(3 - x) - (1 + 1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(3 - x) - 2}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{1 - x}{x - 1} = - 1 \\ ∴ L . H . L . \neq R . H . L . \\ S o, f (x) i s n o t d i f f e r e n t i a b l e a t x = 1 . \\ H e n c e, R o l l e' s T h e o r e m i s n o t a p p l i c a b l e i n [0, 2] . \end{array}$

Find the points on the curve $y = c o s x - 1$ in $[0, 2 π]$ , where the tangent is parallel to the x-axis.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = c o s x - 1 o n [0, 2 π] \\ F i r s t l y, w e h a v e t o f i n d a point c o n t h e g i v e n c u r v e y = c o s x - 1 o n [0, 2 π] s u c h t h a t t h e tangent at \\ c = 0 \in [0, 2 π] i s p a r a l l e l t o x - a x i s i . e ., f^{'} (c) = 0 w h e r e f^{'} (c) i s t h e s l o p e o f t h e tangent . \\ S o, w e h a v e t o v e r i f y t h e R o l l e' s T h e o r e m . \\ (i) y = c o s x - 1, i s t h e c o m b i n a t i o n o f cosine a n d constant f u n c t i o n s . S o, i t i s c o n t i n u o u s o n [0, 2 π] . \\ (i i) \frac{d y}{d x} = - s i n x w h i c h e x i s t s i n (0, 2 π) \\ S o i t i s d i f f e r e n t i a b l e o n (0, 2 π) . \\ (i i i) L e t f (x) = c o s x - 1 \\ f (0) = c o s 0 - 1 \Rightarrow 1 - 1 = 0 \\ f (2 π) = c o s 2 π - 1 \Rightarrow 1 - 1 = 0 \\ ∴ f (0) = f (2 π) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e l i e s a point c \in (0, 2 π) s u c h t h a t f^{'} (c) = 0 \\ ∴ - s i n c = 0 \Rightarrow s i n c = 0 \\ ∴ c = n π, n \in I \\ \Rightarrow c = π \in (0, 2 π) \\ H e n c e, c = π i s t h e p o i n t o n t h e c u r v e i n (0, 2 π) a t w h i c h t h e tangent i s p a r a l l e l t o x - a x i s . \end{array}$

Using Rolleâ€™s theorem, find the point on the curve $y = x (x - 4)$ , $x \in [0, 4]$ , where the tangent is parallel to the x-axis.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = x (x - 4), x \in [0, 4] \\ L e t f (x) = x (x - 4), x \in [0, 4] \\ (i) f (x) b e i n g a n algebraic p o l y n o m i a l i s c o n t i n u o u s f u n c t i o n e v e r y w h e r e . \\ S o, f (x) = x (x - 4) i s c o n t i n u o u s x i n [0, 4] . \\ (i i) f^{'} (x) = 2 x - 4 w h i c h e x i s t s i n (0, 4) \\ S o, f (x) i s d i f f e r e n t i a b l e . \\ (i i i) f (0) = 0 (0 - 4) = 0 \\ f (4) = 0 (4 - 4) = 0 \\ ∴ f (0) = f (4) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point c \in (0, 4) s u c h t h a t f^{'} (c) = 0 \\ ∴ 2 c - 4 = 0 \Rightarrow c = 2 \in (0, 4) \\ H e n c e, c = 2 i s t h e point i n (0, 4) o n t h e g i v e n c u r v e a t w h i c h t h e tangent i s p a r a l l e l t o t h e x - a x i s . \end{array}$

Verify the Mean Value Theorem for each of the functions given in Exercises 73 to 76.

$f (x) = \frac{1}{4 x - 1}$ in $[1, 4]$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

$f (x) = x^{3} - 2 x^{2} - x + 3$ in $[0, 1]$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{3} - 2 x^{2} - x + 3 i n [0, 1] \\ (i) f (x) i s a n f u n c t i o n, s o i t i s c o n t i n u o u s i n [0, 1] . \\ (i i) f^{'} (x) = 3 x^{2} - 4 x - 1 w h i c h e x i s t s i n (0, 1) \\ S o, f (x) i s d i f f e r e n t i a b l e . \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e \in (0, 1) s u c h t h a t \\ f^{'} (c) = \frac{f (b) - f (a)}{b - a} \\ \Rightarrow 3 c^{2} - 4 c - 1 = \frac{[{(1)}^{3} - 2 {(1)}^{2} - (1) + 3] - [0 - 0 - 0 - 3]}{1 - 0} \\ \Rightarrow 3 c^{2} - 4 c - 1 = \frac{(1 - 2 - 1 + 3) - (3)}{1} \\ \Rightarrow 3 c^{2} - 4 c - 1 = 1 - 3 \Rightarrow 3 c^{2} - 4 c - 1 = - 2 \\ \Rightarrow 3 c^{2} - 4 c + 1 = 0 \Rightarrow 3 c^{2} - 3 c - c + 1 = 0 \\ \Rightarrow 3 c (c - 1) - 1 (c - 1) = 0 \Rightarrow (c - 1) (3 c - 1) = 0 \\ \Rightarrow c - 1 = 0 ∴ c = 1 \\ 3 c - 1 = 0 ∴ c = \frac{1}{3} \in (0, 1) \\ H e n c e, M e a n V a l u e T h e o r e m i s v e r i f i e d . \end{array}$

Kindly consider the following

$f (x) = s i n x - s i n 2 x$ in $[0, π]$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Find a point on the curve $y = {(x - 3)}^{2}$ , where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

Using the Mean Value Theorem, prove that there is a point on the curve $y = 2 x^{2} - 5 x + 3$ between the points $A (1, 0)$ and $B (2, 1)$ , where the tangent is parallel to the chord $A B$ . Also, find that point.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = 2 x^{2} - 5 x + 3 \\ L e t f (x) = 2 x^{2} - 5 x + 3 \\ (i) B e i n g a n p o l y n o m i a l, f (x) i s c o n t i n u o u s i n [1, 2] . \\ (i i) f^{'} (x) = 4 x - 5 w h i c h e x i s t s i n (1, 2) . \\ H e n c e, b y m e a n v a l u e t h e o r e m, t h e r e m u s t e x i s t a \in (1, 2) \\ A (1, 0) a n d B (2, 1) \\ ∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} \\ \Rightarrow 4 c - 5 = \frac{(8 - 10 + 3) - (2 - 5 + 3)}{2 - 1} \\ \Rightarrow 4 c - 5 = \frac{1 - 0}{1} = 1 \Rightarrow 4 c = 5 + 1 = 6 \\ ∴ c = \frac{6}{4} = \frac{3}{2} \\ ∴ y = 2 {(\frac{3}{2})}^{2} - 5 (\frac{3}{2}) + 3 \\ \Rightarrow = 2 \times \frac{9}{4} - \frac{15}{2} + 3 = \frac{9}{2} - \frac{15}{2} + 3 \\ \Rightarrow = \frac{9 - 15 + 6}{2} = 0 \\ H e n c e, (\frac{3}{2}, 0) i s t h e \\ A (1, 0) a n d B (2, 1) . \end{array}$

Choose the correct answer from given four options in each of the Exercises from 83 to 96.

If $f (x) = 2 x$ and $g (x) = \frac{x^{2}}{2} + 1$ , then which of the following can be a discontinuous function:

(A) $f (x) + g (x)$

(B) $f (x) - g (x)$

(C) $f (x) \cdot g (x)$

(D) $\frac{g (x)}{f (x)}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t t h e algebraic polynomials are continuous functions everywhere. \\ ∴ f (x) + g (x) i s c o n t i n u o u s [\begin{array}{l} ? S u m, d i f f e r e n c e a n d p r o d u c t o f t w o c o n t i n u o u s \\ f u n c t i o n s i s a l s o c o n t i n u o u s \end{array}] \\ f (x) - g (x) i s c o n t i n u o u s \\ f (x) . g (x) i s c o n t i n u o u s \\ \frac{g (x)}{f (x)} i s o n l y c o n t i n u o u s i f g (x) \neq 0 \\ ∴ \frac{f (x)}{g (x)} = \frac{2 x}{\frac{x^{2}}{2} + 1} = \frac{4 x}{x^{2} + 2} \\ H e r e, \frac{g (x)}{f (x)} = \frac{\frac{x^{2}}{2} + 1}{2 x} = \frac{x^{2} + 2}{4 x} w h i c h i s d i s c o n t i n u o u s a t x = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) \end{array}$

The function $f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$ is

(A) Discontinuous at only one point

(B) Discontinuous at exactly two points

(C) Discontinuous at exactly three points

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{4 - x^{2}}{4 x - x^{3}} \\ F o r d i s c o n t i n u o u s f u n c t i o n \\ \Rightarrow 4 x - x^{3} = 0 \\ \Rightarrow 4 (x - x^{2}) = 0 \\ \Rightarrow x (2 - x) (2 + x) = 0 \\ \Rightarrow x = 0, x = 2, x = - 2 \\ H e n c e, t h e g i v e n f u c t i o n i s d i s c o n t i n u o u s e x a c t l y a t t h r e e points . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) \end{array}$

The set of points where the function $f (x) =\sqrt2 - s i n x$ is differentiable is

(A) R

(B) R $- {\frac{1}{2}}$

(C) $(0, \infty)$

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | 2 x - 1 | s i n x \\ C l e a r l y, f (x) i s n o t d i f f e r e n t i a b l e a t x = \frac{1}{2} \\ R . H . L . = f^{'} (\frac{1}{2}) = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h} \\ = \underset{h \to 0}{l i m} \frac{| 2 (\frac{1}{2} + h) - 1 | s i n (\frac{1}{2} + h) - 0}{h} \\ = \underset{h \to 0}{l i m} \frac{| 2 h | s i n (\frac{1 + 2 h}{2})}{h} = 2 s i n (\frac{1}{2}) \\ A l s o, L . H . L . = f^{'} (\frac{1}{2}) = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} - h) - f (\frac{1}{2})}{- h} \\ = \underset{h \to 0}{l i m} \frac{| 2 (\frac{1}{2} - h) - 1 | [- s i n (\frac{1}{2} - h)] - 0}{- h} \\ = \underset{h \to 0}{l i m} \frac{| - 2 h | [- s i n (\frac{1}{2} - h)]}{- h} = - 2 s i n (\frac{1}{2}) \\ ∴ L . H . L . \neq R . H . L . \\ S o, t h e g i v e n f u n c t i o n f (x) i s n o t d i f f e r e n t i a b l e a t x = \frac{1}{2} . \\ ∴ f (x) i s d i f f e r e n t i a b l e i n R - {\frac{1}{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

The function $f (x) = c o t x$ is discontinuous on the set

(A) ${x = n π : n \in}$

(B) ${x = 2 n π : n \in}$

(C) ${x = (2 n + 1) \frac{π}{2} : n \in}$

(D) ${x = \frac{n π}{2} : n \in}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = c o t x \\ \Rightarrow f (x) = \frac{c o s x}{s i n x} \\ W e k n o w t h a t s i n x = 0 i f f (x) i s d i s c o n t i n u o u s . \\ ∴ I f s i n x = 0 \\ ∴ x = n π, n \in n π . \\ S o, t h e g i v e n f u n c t i o n f (x) i s d i s c o n t i n u o u s o n t h e s e t {x = n π; n \in Z} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) \end{array}$

The function $f (x) = e^{| x |}$ is

(A) Continuous everywhere but not differentiable at $x = 0$

(B) Continuous and differentiable everywhere

(C) Not continuous at $x = 0$

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = e^{| x |} \\ W e k n o w t h a t modulus f u n c t i o n i s c o n t i n u o u s b u t n o t d i f f e r e n t i a b l e i n i t s d o m a i n . \\ L e t g (x) = | x | a n d t (x) = e^{| x |} \\ ∴ f (x) = g o t (x) = g [t (x)] = e^{| x |} \\ Since g (x) a n d t (x) b o t h a r e c o n t i n u o u s a t x = 0 b u t f (x) i s n o t d i f f e r e n t i a b l e a t x = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) \end{array}$

If $f (x) = \frac{x^{2} - 1}{x} s i n x$ , where $x \neq 0$ , then the value of the function $f$ at $x = 0$ , so that the function is continuous at $x = 0$ , is

(A) 0

(B) -1

(C) 1

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{2} s i n \frac{1}{x} w h e r e x \neq 0 . \\ S o, t h e V a l u e o f t h e f u n c t i o n f a t x = 0, s o t h a t f (x) i s c o n t i n u o u s i s 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) \end{array}$

If $f (x) = {\begin{matrix} m x + 1, & x \leq \frac{π}{2} \\ s i n x + n, & x > \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , then

(A) $m = 1, n = 0$

(B) $m = \frac{n π}{2} + 1$

(C) $n = \frac{m π}{2}$

(D) $m = n = \frac{π}{2}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} m x + 1, & x \leq \frac{π}{2} \\ s i n x + n, & x > \frac{π}{2} \end{matrix} i s c o n t i n u o u s a t x = \frac{π}{2} \\ R . H . L . = \underset{x \to {\frac{π}{2}}^{+}}{l i m} (s i n x + n) = \underset{h \to 0}{l i m} [s i n (\frac{π}{2} + h) + n] \\ = \underset{h \to 0}{l i m} c o s h + n = 1 + n \\ A l s o, L . H . L . = \underset{x \to {\frac{π}{2}}^{-}}{l i m} (m x + 1) = \underset{h \to 0}{l i m} [m (\frac{π}{2} - h) + 1] \\ = \frac{m π}{2} + 1 \\ W h e n f (x) i s c o n t i n u o u s a t x = \frac{π}{2} . \\ ∴ L . H . L . = R . H . L . \\ \frac{m π}{2} + 1 = 1 + n \Rightarrow n = \frac{m π}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) \end{array}$

Let $f (x) = ∣ s i n x ∣$ . Then

(A) $f$ is everywhere differentiable

(B) $f$ is everywhere continuous but not differentiable at $x = n π, n \in$ .

(C) $f$ is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in$ .

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | s i n x | \\ L e t g (x) = | s i n x | a n d t (x) = | x | \\ ∴ f (x) = t o g (x) = t [g (x)] = t (s i n x) = | s i n x | \\ w h e r e g (x) a n d t (x) b o t h a r e c o n t i n u o u s . \\ ∴ f (x) = g o t (x) i s c o n t i n u o u s b u t t (x) i s n o t d i f f e r e n t i a b l e a t x = 0 . \\ S o, f (x) i s n o t c o n t i n u o u s a t s i n x = 0 \Rightarrow x = n π, n \in Z . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

If $y = l o g_{} (\frac{1 - x^{2}}{1 + x^{2}})$ , then $\frac{d y}{d x}$ is equal to

(A) $\frac{4 x^{3}}{1 - x^{4}}$

(B) $\frac{- 4 x}{1 - x^{4}}$

(C) $\frac{1}{4 - x^{4}}$

(D) $\frac{- 4 x^{3}}{1 - x^{4}}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = l o g (\frac{1 - x^{2}}{1 + x^{2}}) \\ \Rightarrow y = l o g (1 - x^{2}) - l o g (1 + x^{2}) [? l o g \frac{x}{y} = l o g x - l o g y] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d y}{d x} = \frac{1}{1 - x^{2}} . \frac{d}{d x} (1 - x^{2}) - \frac{1}{1 + x^{2}} . \frac{d}{d x} (1 + x^{2}) \\ \Rightarrow = \frac{- 2 x}{1 - x^{2}} - \frac{2 x}{1 + x^{2}} = \frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{(1 - x^{2}) (1 + x^{2})} = \frac{- 4 x}{(1 - x^{4})} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

If $y =\sqrtsinx+y$ , then $\frac{d y}{d x}$ is equal to

(A) $\frac{x}{2 y - 1}$

(B) $\frac{x}{1 - 2 y}$

(C) $\frac{x}{1 - 2 y}$

(D) $\frac{x}{2 y - 1}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

If $x = t^{2}, y = t^{3}$ , then $\frac{d^{2} y}{d x^{2}}$ is

(A) $\frac{3}{2}$

(B) $\frac{3}{4 t}$

(C) $\frac{3}{2 t}$

(D) $\frac{3}{4}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = t^{2} a n d y = t^{3} \\ D i f f e r e n t i a t i n g b o t h f u n c t i o n s w . r . t . t \\ \Rightarrow \frac{d x}{d t} = 2 t a n d \frac{d y}{d t} = 3 t^{2} \\ ∴ \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2} \\ D i f f e r e n t i a t i n g a g a i n w . r . t . x \\ \frac{d}{d x} (\frac{d y}{d x}) = \frac{3}{2} . \frac{d t}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{3}{2} . \frac{1}{2 t} = \frac{3}{4 t} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

The value of $c$ in Rolle's theorem for the function $f (x) = x^{3} - 3 x$ in the interval $[0, \sqrt3]$ is

(A) 1

(B) -1

(C) $\frac{3}{2}$

(D) $\frac{1}{3}$

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

For the function $f (x) = x + \frac{1}{x}$ , $x \in [1, 3]$ , the value of $c$ for the Mean Value Theorem is

(A) 1

(B) 3

(C) 2

(D) None of these

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is __________.

This is a Fill in the Blanks Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} | x | = | x + 1 | i s t h e f u n c t i o n w h i c h i s c o n t i n u o u s e v e r y w h e r e b u t f a i l s t o b e d i f f e r e n t i a b l e \\ a t x = 0 a n d x = 1 . \end{array}$

Derivative of $x^{2}$ w.r.t. $x^{3}$ is __________.

This is a Fill in the Blanks Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = x^{2} a n d t = x^{3} \\ D i f f e r e n t i a t i n g b o t h f u n c t i o n s w . r . t . x \\ \Rightarrow \frac{d y}{d x} = 2 x a n d \frac{d t}{d x} = 3 x^{2} \\ ∴ \frac{d y}{d t} = \frac{d y / d x}{d t / d x} = \frac{2 x}{3 x^{2}} = \frac{2}{3 x} \\ S o, t h e d e r i v a t i v e o f x^{2} w . r . t . x^{3} i s \frac{2}{3 x} \end{array}$

If $f (x) = ∣ c o s x ∣$ , then $f^{'} (\frac{π}{4}) =$ __________.

This is a Fill in the Blanks Type questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Fill in the Blanks Type questions as classified in NCERT Exemplar

Sol:

State True or False for the statements in each of the Exercises 102 to 106:

Rolleâ€™s theorem is applicable for the function $f (x) = ∣ x - 1 ∣$ in $[0, 2]$ .

This is a True or False Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} F a l s e . G i v e n t h a t f (x) = | x - 1 | i n [0, 2] \\ W e k n o w t h a t t h e f u n c t i o n i s n o t d i f f e r e n t i a b l e . S o . i t i s f a l s e . \end{array}$

If $f$ is continuous on its domain $D$ , then $? f ?$ is also continuous on $D$ .

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $T r u e . W e k n o w t h a t t h e f u n c t i o n i s c o n t i n u o u s f u n c t i o n o n i t s d o m a i n .$

The composition of two continuous functions is a continuous function.

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $T r u e . W e k n o w t h a t t h e s u m a n d d i f f e r e n c e o f t w o o r m o r e f u n c t i o n s i s a l w a y s c o n t i n u o u s .$

Trigonometric and inverse-trigonometric functions are differentiable in their respective domains.

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $T r u e .$

If $f \cdot g$ is continuous at $x = a$ , then $f$ and $g$ are separately continuous at x=a.

This is a True or False Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} F a l s e . \\ L e t u s t a k e a n e x a m p l e : f (x) = s i n x a n d g (x) = c o t x \\ ∴ f (x) . g (x) = s i n x . c o t x = s i n x . \frac{c o s x}{s i n x} = c o s x w h i c h i s c o n t i n u o u s a t x = 0 b u t c o t x i s n o t \\ c o n t i n u o u s a t x = 0 . \end{array}$

JEE Mains Solutions 2022,29th june , Maths, first shift

Commonly asked questions

The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to:

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$= \frac{10 + 4 \times^{10} ? C_{2}}{1 0^{4}} = \frac{10 + \frac{4 \times 10.9}{2}}{1 0^{4}} = \frac{19}{1000}$

Let the solution curve of the differential equation

$x \frac{d y}{d x} - y = \sqrt[]{y^{2} + 16 x^{2}}, y (1) = 3 b e y = y (x) .$ Then y(2) is equal to:

$\frac{d y}{d x} = \frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x}$

Put y = $V \cdot x$

differentiable worst = x

$\frac{d y}{d x} = V + x \cdot \frac{d V}{d x}$

V + $x \frac{d V}{d x} = V + \sqrt[]{V^{2} + 16}$

$x \frac{d V}{d x} = \sqrt[]{V^{2} + 16}$

apply variable separable method

$\int \frac{d V}{\sqrt[]{V^{2} + 16}} = \int \frac{d x}{x} + l n C$

$\Rightarrow l n | V + \sqrt[]{V^{2} + 16} | = l n C x$

$\frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x} = C x$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2} + \frac{\sqrt[]{y^{2} + 16.4}}{2} = 8.2$

y2 + 64 = (32 – y)2

y = 15

If the mirror image of the point (2, 4,7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to:

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x - 2}{3} = \frac{y - 4}{- 1} = \frac{z - 7}{4} = \frac{- 2 (6 - 4 + 28 - 2)}{3^{2} + {(- 1)}^{2} + 4^{2}}$

Let $\frac{x - 2}{3} = \frac{y - 4}{- 1} = \frac{z - 7}{4} = \frac{- 28}{13} = λ$

$x = 3 λ + 2 y = - λ + 4 z = 4 λ + 7$

Now according to the question

(a, b, c) = $(3 λ + 2, - λ + 4, 4 λ + 7)$

Now $2 a + b + 2 c = 6 λ + 4 - λ + 4 + 8 λ + 14$

= $13 λ + 22$

= 28 + 22 [Use $λ = \frac{- 28}{13}$ ]

= 6

Let f : R R be a function defined by:

$f (x) = [\begin{matrix} m a x (t^{3} - 3 t) & ; & t \leq x; x \leq 2 & x^{2} + 2 x - 6 \\ ; & 2 < x < 3 & [x - 3] + 9 & ; \\ 3 \leq x \leq 5 & 2 x + 1 & ; & x > 5 \end{matrix}$

where [t] is the greatest integer less than or equal to t. Let me be the number of points where f is not differentiable and $I = \int_{- 2}^{2} f (x) d x .$ Then the ordered pair (m, I) is equal to:

Draw g(t) = t³ – 3t

g’(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³– 3t) = ${\begin{matrix} t^{3} - 3 t & ; - 2 < t < - 1 \\ 2 & ; - 1 < t < 2 \end{matrix}$

$I = \int_{- 2}^{2} f (x) d x$

= $\int_{- 2}^{- 1} (t^{3} - 3 t) d t + \int_{- 1}^{2} 2 d t$

I = $\frac{27}{4}$

again rewrite the f(x)

$f (x) = {\begin{matrix} \begin{matrix} x^{3} - 3 x & 2 \end{matrix} & ; & \begin{matrix} x \leq - 1 & - 1 < x \leq 2 \end{matrix} \\ \begin{matrix} x^{2} + 2 x - 6 & 9 \end{matrix} & ; & \begin{matrix} 2 < x < 3 & 3 \leq x < 4 \end{matrix} \\ \begin{matrix} 10 & 11 & 2 x + 1 \end{matrix} & ; & \begin{matrix} 4 \leq x < 5 & x = 5 & x > 5 \end{matrix} \end{matrix}}$

$f' (x) = {\begin{matrix} 3 x^{2} - 3 & ; & x < - 1 & 0 & ; & - 1 < x < 2 \\ 2 x + 2 & ; & 2 < x < 3 & 0 & ; & 3 < x < 4 \\ 0 & ; & 4 < x < 5 & 2 & ; & x > 5 \end{matrix}}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

Let $\vec{a} = α \hat{i} a + 3 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} - β \hat{j} + 4 \hat{k} a n d \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} w h e r e α, β \in R,$ be three vectors. If the projection of $\vec{a} o n \vec{c} i s \frac{10}{3} a n d \vec{b} \times \vec{c} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k},$ then the value of + is equal to:

$\vec{a} \cdot \hat{c} = \frac{α + 6 + 2}{\sqrt[]{1} + 4 + 4}$

$\frac{10}{3} = \frac{8 + α}{3} \Rightarrow α = 2$

$\vec{b} \times \vec{c} = \hat{i} (2 β - 8) + \hat{j} (10) + \hat{k} (6 + β)$

$= - 6 \hat{i} + 10 \hat{j} + 7 \hat{k}$

So = 1

The area enclosed by y2 = 8x and y = $\sqrt[]{2} x$ that lies outside the triangle formed by y = $\sqrt[]{2} x,$ x = 1, $y = 2 \sqrt[]{2},$ is equal to:

Area of $Δ A B C$

= $\frac{1}{2} A B \cdot B C$

$= \frac{1}{2} \cdot \sqrt[]{2} \cdot 1$

$= \frac{1}{\sqrt[]{2}}$

now required area

$= \int_{0}^{4} (2 \sqrt[]{2} - \sqrt[]{2} x) d x - \frac{1}{\sqrt[]{2}}$

$= \frac{32 \sqrt[]{2}}{3} = 8 \sqrt[]{2} - \frac{1}{\sqrt[]{2}}$

$= \frac{13 \sqrt[]{2}}{6}$

If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + $δ z = k,$ where $δ, k \in R$

has infinitely many solutions, then $δ + k$ is equal to:

$= | \begin{matrix} 2 & 1 & - 1 \\ 1 & - 3 & 2 \\ 1 & 4 & δ \end{matrix} | = 0 \Rightarrow δ = - 3$

and $Δ_{1} = | \begin{matrix} 7 & 1 & - 1 \\ 1 & - 3 & 2 \\ k & 4 & - 3 \end{matrix} | = 0 \Rightarrow k = 6$

Let a and b be the roots of the equation x² + (2i – 1) = 0. Then, the value of $| α^{8} + β^{8} |$ is equal to:

x² = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

now $| α^{8} + β^{8} | = 2 | α^{8} |$

$= 2 {| (α^{2}) |}^{4}$

$= 2 {| α^{2} |}^{4}$

$= 2 {| 1 - 2 i |}^{4}$

$= 2 \times 25$

Let $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ be such that $(p \land q) Δ ((p \lor q) \Rightarrow q)$ is a tautology. Then $Δ$ is equal to:

$(p \lor q) \Rightarrow q$

= $\sim (p \lor q) \lor q$

= $(\sim p \land \sim q) \lor q$

= $(\sim p \lor q) \land (\sim q \lor q)$

= $\sim p \lor q$

now $(p \land q) Δ (\sim p \lor q) i s t a u t o l o g ? y$

pq $p \land q$ $\sim p \lor q$ $(p \land q) Δ (\sim p \lor q)$

TTTTT

TFFFT

FTFTT

FFFTT

So, $Δ = \Rightarrow$

Let A = $[a_{i j}]$ be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix $A^{2} + A^{3} + . . . + A^{10}$ is equal to:

$A = [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

$A^{2} =$ $[\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

$= 3 [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

A² = 3A

A³ = 3A²

A³ = 3²A

A⁴ = 3³A

Aⁿ = 3^n-1A

now, A² + A³ +….+A¹⁰

= 3A + 3² A +….. + 3⁹A

= 3A (1 + 3 +….+ 3⁸

$= 3 A \cdot \frac{(3^{9} - 1)}{3 - 1}$

$= \frac{3^{10} - 3}{2} A$

Let a set A = A1 $\cup$ A2 $\cup$ …. $\cup$ Ak, where Ai $\cap$ Aj = $?$ for $i \neq j, 1 \leq i, j \leq k .$ Define the reaction R from A to A by R = ${(x, y) : y \in A_{i} i f a n d o n l y i f x \in A_{i}, 1 \leq i \leq k} .$ Then, R is:

Use aRb = a is related to b, belongs to A iff a belongs to A.

In simple terms, aRb is true if both a & b belongs to the same set.

For reflexive

aRa, a $\in$ A, so it is true.

For symmetric

Let aRb be true

Þ a & b belongs to the same set.

Þ b & a also belongs to the same set

Þ bRa will be true

For transitive

Let aRb and bRc be true.

aRb Þ a, b belongs to the same set

bRc Þ b, c belongs to the same set

Þ (a, c) belongs to the same set

Þ so aRc will be true.

So R is an equivalence relation.

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all $n \geq 0 .$ Then, $\sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to:

$a_{n + 2} = 2 \cdot a_{n + 1} - a_{n + 1}$

$a_{2} = 2 a_{1} - a_{0} + 1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\geq 2, a_{n} = \frac{n (n - 1)}{2}$

$\sum_{n = 2}^{} \frac{n (n - 1)}{2 \cdot 7^{n}} = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . . .$

Let S = $\frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . .$

$\frac{\frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + . . . .}{S - \frac{S}{7} = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + . . . .}$

$\frac{\frac{6}{7} S \cdot \frac{1}{7} = \frac{1}{7^{3}} + \frac{2}{7^{3}} + . . . .}{\frac{6}{7} S - \frac{6}{49} S = \frac{1}{7^{2}} + \frac{1}{7^{3}} + . . . .}$

$\frac{36}{49} S = \frac{\frac{1}{7^{2}}}{1 - \frac{1}{7}}$

$S = \frac{1}{7^{2}} \times \frac{7}{6} \times \frac{49}{36}$

$S = \frac{7}{216}$

The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle $\frac{π}{4}$ at the origin, is equal to:

Let A, A’ be (, 2) AB and A’B subtends $\frac{π}{4}$ angle at (0, 0) slope of OA = $\frac{2}{α}$

slope of OB = $\frac{3}{2}$

$t a n \frac{π}{4} = | \frac{\frac{2}{α} - \frac{3}{2}}{1 + \frac{2}{α} \cdot \frac{3}{2}} |$

$\Rightarrow 1 + \frac{3}{α} = \pm (\frac{4 - 3 α}{2 α})$

$\Rightarrow \frac{α + 3}{α} = \pm (\frac{4 - 3 α}{2 α}) \Rightarrow α = 10, - \frac{2}{5}$

now distance between A’A, (10, 2) & $(\frac{- 2}{5}, 2) i s \frac{52}{5}$

A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:

Let perimeter of $Δ$ is x and that of square is 22 – x

now area $= \frac{\sqrt[]{3}}{4} {(\frac{x}{3})}^{2} + {(\frac{22 - x}{4})}^{2}$

for maximum or minimum, $\frac{d A}{d x} = 0$

x $= \frac{22 \sqrt[]{3}}{\sqrt[]{3} + 4}$

now side of a $Δ = \frac{x}{3}$

$= \frac{22 \sqrt[]{3}}{3 (\sqrt[]{3} + 4)}$

$= 66$ $9$ $+$ $4$

The domain of the function $c o s^{-} (\frac{2 s i n^{- 1} (\frac{1}{4 x^{2} - 1})}{π}) i s :$

Case 1: 1 $\leq \frac{1}{4 x^{2} - 1} \leq 1$

4x2 – 1 $\geq 1 o r 4 x^{2} - 1 \leq - 1$

$x^{2} \geq \frac{2}{4} o r x^{2} \leq 0$

So x $\in (- \infty, - \frac{1}{\sqrt[]{2}}] \cup [\frac{1}{\sqrt[]{2}}, \infty) \cup {0}$

If the constant term in the expansion of ${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} i s 2^{k} . l,$ where $l$ is an odd integer, then the value of k is equal to:

${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10}$

General term $10! \frac{{(3 x^{3})}^{α} \cdot {(- 2^{x})}^{β} \cdot {(5 x^{- 5})}^{γ}}{α! β! γ!}$

$= 10! \frac{3^{α} \cdot {(- 2)}^{β} \cdot 5^{γ} \cdot x^{3 α + 2 β - 5 γ}}{α! β! γ!}$

Now for constant term 3 + 2 $- 5 γ = 0$ …………..(i)

$α + β + γ = 10$ …………(ii)

From equation (i) & (ii)

$3 α + 2 (10 - α - γ) = 5 γ$

$α + 20 = 7 γ$

= 1, $γ$ = 3, = 6

Constant term $10! \cdot \frac{3^{1} \cdot {(- 2)}^{6} \cdot 5^{3}}{3! 6!} = 2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7^{1}$

$\int_{0}^{5} c o s (π (x - [\frac{x}{2}])) d x,$ where [t] denotes greatest integer less than or equal to t, is equal to:

$I = \int_{0}^{5} c o s (π x - π [\frac{x}{2}]) d x$

$I = \int_{0}^{2} c o s (π x) d x + \int_{2}^{4} c o s (π x - π) d x + \int_{4}^{5} c o s (π x - 2 π) d x$

$I = {\frac{s i n π x}{π} |}_{0}^{2} + {\frac{s i n (π x - π)}{π} |}_{2}^{4} + {\frac{s i n (π x - 2 π)}{π} |}_{4}^{5} = 0$

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $\frac{π}{2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ $a^{2} > b^{2} .$ If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^{2}}$ is equal to:

$A P ⊥ B P$

M₁ M₂ = 1

$\frac{2 t}{t^{2} - 3} \times \frac{2 / 6}{3 - \frac{1}{t^{2}}} = - 1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $= \frac{2 b^{2}}{a}$

$4 = \frac{2 b^{2}}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b²= a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e} (1 - e^{2})$

e²+ 2e – 1 = 0

$e = \frac{- 2 \pm \sqrt[]{4 + 4}}{2}$

$e = - 1 + \sqrt[]{2}$

now $\frac{1}{e^{2}}$ $=$ $3$ $+$ $2$

Let the tangent to the circle C₁ : x² + y² = 2 at the point M(1, 1) intersect the circle $C_{2} : {(x - 3)}^{2}$ + ${(y - 2)}^{2} = 5,$ at two distinct points A and B. If the tangents to C₂at the points A and B intersect at N, then the area of the triangle ANB is equal to:

Tangent to C₁ at (-1, 1) is T = 0

x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping from (3, 2) to centre

OP = $| \frac{3 - 2 + 2}{\sqrt[]{2}} | = \frac{3}{\sqrt[]{2}}$

AP = $\sqrt[]{r^{2} - O P^{2}}$

$= \sqrt[]{5 - \frac{9}{2}} = \frac{1}{\sqrt[]{2}}$

$t a n θ = \frac{O P}{A P} = \frac{3 / \sqrt[]{2}}{1 / \sqrt[]{2}} = 3$

area of $Δ A B N = \frac{1}{2} A N^{2} \cdot s i n 2 θ$

AN = $\frac{\sqrt[]{5}}{3}$

$= \frac{1}{2} \cdot \frac{5}{9} \cdot (\frac{2 t a n θ}{1 + t a n^{2} θ})$

$= \frac{5}{2.9} \times \frac{2.3}{1 + 3^{2}} = \frac{1}{6}$

sin = $\frac{A P}{A N}$

Let the mean and the variance of 5 observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} b e \frac{24}{5} a n d \frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then (4a + x₅) is equal to:

$\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{7}{2}$

$x_{1} + x_{2} + x_{3} + x_{4} = 14$

and $\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = \frac{24}{5}$

x₅= 10

Variance $\frac{\sum_{i = 1}^{4} x_{i}^{2}}{4} - {(\frac{Σ x i}{4})}^{2} = a$

$\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}}{4} - \frac{49}{4} = a$

$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = 4 a + 49$

and $\frac{x_{1}^{2} + x_{2}^{2} + x_{5}^{2} + x_{4}^{2} + x_{5}^{2}}{5} - {(\frac{24}{5})}^{2} = \frac{194}{25}$

$\frac{4 a + 49 + x_{5}^{2}}{5} = \frac{576 + 194}{25}$

49 + 49 + $x_{5}^{2} = \frac{770}{5}$

49 $+ x_{5}^{2} + 49 = 154$

4a + 149 = 154

4a = 5

now 4a + x₅= 15

Let S = ${z \in C : | z - 2 | \leq 1, z (1 + i) + \bar{z} (1 - i) \leq 2} . L e t | z - 4 i |$ attains minimum and maximum values, respectively, at $z_{1} \in S a n d z_{2} \in S .$ If $5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = α + β \sqrt[]{5},$ where and are integers, then the value of + is equal to……………

Let z = x + iy

$| z - 2 | \leq 1 \Rightarrow | x - 2 + i y | \leq 1$

(x – 2)² + y² $\leq 1$

$z (1 + i) + \bar{z} (1 - i) \leq 2$

$(x + i y) (1 + i) + (x - i y) (1 - i) \leq 2$

$x + i x + i y - y + x - i x - i y - y \leq 2$

x – y $\leq 1$

PD will be least as CP – r = DP, general pts on circle with centre (2, 0) is (2 + r cos, 0 + r sin )

here r = 1, (2 + cos , sin ) now slope of CP is $\frac{4 - 0}{0 - 2} = - 2$

tan = 2

so D point will be $(2 - \frac{1}{\sqrt[]{5}}, \frac{2}{\sqrt[]{5}}) A P$ will be the greatest. A(1, 0)

now $| z_{1}^{2} | + | z_{2}^{2} |$

$= {| 1 + 0 i |}^{2} + {| 2 - \frac{1}{\sqrt[]{5}} + \frac{2 i}{\sqrt[]{5}} |}^{2}$

$= 1 + {(2 - \frac{1}{\sqrt[]{5}})}^{2} + \frac{4}{5}$

$= 6 - \frac{4}{\sqrt[]{5}}$

now $5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = α + β \sqrt[]{5}$

$5 (6 - \frac{4}{\sqrt[]{5}}) = α + β \sqrt[]{5}$

= 30, = 4

+ = 26

Let y = y(x) be the solution of the differential equation

$\frac{d y}{d x} + \frac{\sqrt[]{2} y}{2 c o s^{4} x - c o s 2 x} = x e^{t a n^{- 1} (\sqrt[]{2} c o t 2 x)},$ $0 < x < \frac{π}{2} w i t h y (\frac{π}{4}) = \frac{π^{2}}{32} .$

2cos⁴ x – cos²x = $2 {(\frac{1 + c o s 2 x}{2})}^{2} - c o s 2 x$

$= \frac{1 + c o s^{2} 2 x}{2}$

$\sqrt[]{2} \int \frac{2 d x}{1 + c o s^{2} 2 x} = \sqrt[]{2} \int 2 \cdot \frac{s e c^{2} 2 x d x}{2 + t a n^{2} 2 x}$

$= \sqrt[]{2} \cdot \frac{1}{\sqrt[]{2}} \cdot t a n^{- 1} (\frac{t a n x}{\sqrt[]{2}})$

now IF = $e^{\int P d x}$

$e^{\sqrt[]{2} \int 2 \frac{s e c^{2} 2 x d x}{2 + t a n^{2} 2 x}} = e^{t a n^{- 1} (\frac{t a n 2 x}{\sqrt[]{2}})}$

Solution: $y \cdot e^{t a n^{- 1} (\frac{t a n 2 x}{\sqrt[]{2}})} = \int x . e^{t a n^{- 1} (\sqrt[]{2} \cdot c o t 2 x)} \cdot e^{t a n^{- 1} (\frac{t a n 2 x}{\sqrt[]{2}})} d x$

$y \cdot e^{t a n^{- 1} (\frac{t a n 2 x}{\sqrt[]{2}})} = \frac{x^{2}}{2} \cdot e^{π / 2} + C$

at $x = \frac{π}{4}, y = \frac{π^{2}}{32}, C = 0$

at $x = \frac{π}{3}, y = \frac{π^{2}}{18} e^{t a n^{- 1} α}$

$y \cdot e^{t a n^{- 1}} \frac{(t a n 2 \frac{π}{3})}{\sqrt[]{2}} = \frac{π^{2}}{18} \cdot e^{π / 2}$

$\frac{π^{2}}{18} \cdot e^{t a n^{- 1} α} \cdot e^{t a n^{- 1}} (- \frac{\sqrt[]{3}}{2}) = \frac{π^{2}}{18} \cdot e^{π / 2}$

tan^-1 a + tan^-1 $(- \sqrt[]{3} / 2) = \frac{π}{2}$

cot^-1a = tan^-1 $(- \frac{\sqrt[]{3}}{2})$

tan^-1 $\frac{1}{α} = t a n^{- 1} (- \frac{\sqrt[]{3}}{2})$

$\frac{1}{α} = - \sqrt[]{3} / 2$

2 = $\frac{2}{3}$

Let d be the distance between the foot of perpendiculars of the points P(1, 2, -1) and Q(2, -1, 3) on the plane -x + y + z = 1. Then d² is equal to…………..

Let $π \equiv - x + y + z - 1 = 0$

1 2 > 0 as both pt.lies on same side

now $P_{1} = | \frac{- 1 + 2 - 1 - 1}{\sqrt[]{3}} | = \frac{1}{\sqrt[]{3}}$

P2 = $| \frac{- 2 + (- 1) + 3 - 1}{\sqrt[]{3}} | = \frac{1}{\sqrt[]{3}}$

as P₁= P₂so distance between foot of perpendicular will be same as distance between the points

d = $\sqrt[]{{(1 - 2)}^{2} + {(2 + 1)}^{2} + {(- 1 - 3)}^{2}}$

= $\sqrt[]{1 + 9 + 16}$ $=$

The number of elements in the set

$S = {θ \in [- 4 π, 4 π] : 3 c o s^{2} 2 θ + 6 c o s 2 θ - 10 c o s^{2} θ + 5 = 0}$ is…………….

3cos²2θ + 6cos2θ - $10 \frac{(1 + c o s 2 θ)}{2} + 5 = 0$

$c o s 2 θ (3 c o s 2 θ + 1) = 0$

Þ cos2θ = 0, $\frac{- 1}{3}$

Draw y = cos2θ, y = 0 and y = $\frac{- 1}{3},$ find the pt. of intersection.

The number of solutions of the equation 2θ - cos²θ + $\sqrt[]{2}$ = 0 in R is equal to………..

Draw y = cos²x and y = 2x + $2$

$50 t a n (3 t a n^{- 1} (\frac{1}{2}) + 2 c o s^{- 1} (\frac{1}{\sqrt[]{5}})) + 4 \sqrt[]{2} t a n (\frac{1}{2} t a n (2 \sqrt[]{2}))$ is equal to……………..

$c o s^{- 1} \frac{1}{\sqrt[]{5}} = c o t^{- 1} \frac{1}{2}$

$t a n (2 (t a n^{- 1} \frac{1}{2} + c o t^{- 1} \frac{1}{2})) + t a n^{- 1} (\frac{1}{2}) = t a n (π + t a n^{- 1} \frac{1}{2}) = \frac{1}{2}$

$t a n 2 α = 2 \sqrt[]{2} \Rightarrow \frac{2 t a n α}{1 - t a n^{2} α} = 2 \sqrt[]{2}$

$\sqrt[]{2} t a n^{2} α + t a n α - \sqrt[]{2} = 0$

$t a n α = \frac{1}{\sqrt[]{2}}, (- \sqrt[]{2}) \to r e j e c t e d$

Let $c, k \in R . I f f (x) = (c + 1) x^{2} + (1 - c^{2}) x + 2 k$ and $f (x + y) = f (x) + f (y) - x y,$ for all x, y $\in$ R, then the value of $| 2 (f (1) + f (2) + f (3) + . . . . + f (20)) |$ is equal to………….

f (x) = (c + 1)x² + (1 – c²)x + 2k

$f' (x) = ((c + 1)) 2 x + (1 - c^{2}) . . . . . . . . . . . . . (i)$

$f' (x) = \underset{y \to 0}{l i m} I \frac{f (x + y) - f (x)}{x + y - x}$

$= \underset{y \to 0}{l i m} \frac{f (y) - x y}{y}$

$f' (x) = \underset{y \to 0}{l i m} \frac{f (y)}{y} - x$

$x + f' (x) = \underset{y \to 0}{l i m} \frac{f (y) - f (0)}{y - 0} a s f (c) = 0$

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0,$ be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (2 \sqrt[]{2} + \sqrt[]{14}) .$ If the eccentricity H is $\frac{\sqrt[]{11}}{2},$ then the value of a² + b² is equal to………….

Given 2a + 2b = $4 (2 \sqrt[]{2} + \sqrt[]{14})$

$= 4 \sqrt[]{2} (2 + \sqrt[]{7})$

$b^{2} = a^{2} (\frac{11}{4} - 1)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2} a$

Let $P_{1} : \vec{r} . (2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let P₂be another plane which passes through the points $(2, - 3, 2), (2, - 2, - 3) a n d (1, - 4, 2)$ . If the direction ratios of the line of intersection of P₁ and P₂ be 16, , then the value of + is equal to……………

2x + y – 3z = 4

$π_{2} = | \begin{matrix} x - 2 & y + 3 & z - 2 \\ 0 & 1 & - 5 \\ - 1 & - 1 & 0 \end{matrix} | = 0$

$π_{2} : - 5 x + 5 y + z + 23 = 0$

now line lying in both the planes have DR.

$\frac{a}{1 + 15} = \frac{b}{15 - 2} = \frac{c}{10 + 5}$

$\frac{a}{16} = \frac{b}{13} = \frac{c}{15}$

So direction ratio’s a : b : c = 16 : 13 : 15

$x + f' (x) = f' (0)$

$f' (0) = 1 - c^{2} f r o m e q u a t i o n (i)$

$x + f' (x) = 1 - c^{2}$

$\frac{x^{2}}{2} + f (x) = (1 - c^{2}) x + d$

f(0) = 0

$f (x) = \frac{- x^{2}}{α} + (1 - c^{2}) x$

c = 3/2

$f " (x) = - 1 = 2 (c + 1)$

$f (x) = \frac{- x^{2}}{2} + (\frac{- 5}{4}) x$

now $(f (1) + f (2) + . . . . . . . f (20))$

$= \frac{- 1}{2} (1^{2} + 2^{2} + . . . . . . + 2 0^{2}) - \frac{5}{4} (1 + 2 + . . . . + 20)$

$= \frac{20 \cdot 21}{2} \cdot \frac{(- 82 - 15)}{12}$

now $2 | f (1) + f (2) + . . . . + f (20) |$

$= \frac{20 \cdot 21 \cdot 97}{12} = 3395$

Let b₁ b₂ b₃ b₄be a 4-element permutation with $b_{i} \in {1, 2, 3, . . . ., 100} f o r 1 \leq i \leq 4 a n d b_{i} \neq b_{j}$ for $i \neq j,$ such that either b_1, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁ b₂b₃ b₄ is equal to……………

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

JEE Mains 2022

Commonly asked questions

Consider a matrix A = $[\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}]$ $\begin{matrix} ^{} & ^{} & ^{} \end{matrix}$ , where , , $γ$ are three distinct natural numbers.

If $\frac{d e t (a d j (a d j (a d j (a d j A))))}{{(α - β)}^{16} {(β - γ)}^{16} {(γ - α)}^{16}} = 2^{32} \times 3^{16},$ $\frac{}{^{}^{}^{}}^{}^{}$ then the number of such 3 – tuples (, , $γ$ ) is…………

Kindly consider the following figure

The number of functions f, from the set A = ${x \in N : x^{2} - 10 x + 9 \leq 0}$ to the set B = ${n^{2} : n \in N}$ such that f(x) $\leq$ (x – 3)² + 1, for every x $\in$ A, is…………

$x^{2} - 10 x + 9 \leq 0$

$(x - 1) (x - 9) \leq 0$

A = {1, 2, 3, ……, 9}

for set B, $f (x) \leq {(x - 3)}^{2} + 1$

total number of such function = 2 × 1 × 1 × 1 × 2 × 3 × 4 × 5 × 6 = 2 × 6! = 1440

Let for the 9^th term in the binomial expansion of ${(3 + 6 x)}^{n}$ , in the increasing powers of 6x, to be the greatest for x $= \frac{3}{2}$ , the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k + n₀ is equal to:

Kindly consider the following figure

$\frac{2^{3} - 1^{3}}{1 \times 7} + \frac{4^{3} - 3 + 2^{3} - 1^{3}}{2 \times 11} + \frac{6^{3} - 5^{3} + 4^{3} + 2^{3} - 1^{3}}{3 \times 15} + . . . . . . . +$

$\frac{3 0^{3} - 2 9^{3} + 2 8^{3} - 2 7^{3} + . . . . + 2^{3} - 1^{3}}{15 \times 63} i s e q u a l t o . . . . . . . . . . . . .$

$t n = \frac{2 (2^{3} + 4^{3} + 6^{3} + . . . . + {(2 n)}^{3}) - (1^{3} - 2^{3} + 3^{3} + . . . . . {(2 n)}^{3})}{n (4 n + 3)}$

^{$\frac{2 \times 2^{3} \times {(\frac{n \times (n + 1)}{2})}^{2} - {(\frac{2 n \times (2 n + 1)}{2})}^{2}}{n (4 n + 3)}$}

^{tn = n}

^{$n o w \sum_{n = 1}^{15} T_{n} = \frac{15.16}{2}$}

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $t a n^{- 1} \frac{3}{4} .$ Water is poured in it a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is………..

$V = \frac{1}{3} π r^{2} h$

$V = \frac{π}{8} h^{3} t a n^{2} α$

$\frac{d v}{d t} = π h^{2} t a n^{2} α . \frac{d h}{d t}$

CSA = $π r l$

$\frac{d (C S A)}{d t} = \frac{15}{16} π 2 h . \frac{d h}{d t}$

$\frac{15}{8} \times \frac{2}{3} = 5 m^{2} / h r s$

For the curve C : $(x^{2} + y^{2} - 3) + {(x^{2} - y^{2} - 1)}^{5} = 0,$ the value of $3 y' - y^{3} y ",$ at the point (a, a), a > 0, on C, is equal to………..

Find a, α

$(α^{2} + α^{2} - 3) + {(α^{2} - α^{2} - 1)}^{5} = 0$

$α = \sqrt[]{2}$

zx differentiate the curve

$2 x + 2 y . y^{1} + 5 {(x^{2} - y^{2} - 1)}^{4} (2 x - 2 y) = 0$ ……. (i)

differentiate equation (i)

$(α, α) \Rightarrow y'' = \frac{- 23}{4 \sqrt[]{2}}$

Let f(x) = $m i n {[x - 1], [x - 2], . . . ., [x - 10]}$ where [t] denotes the greatest integer $\leq$ t. Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} {(f (x))}^{2} d x + \int_{0}^{10} | f (x) | d x$ is equal to………..

Use [x + n] = n + [x], where $n \in z$ f (x) = [x] – 10 will be minimum for $x \in [0, 10]$ break the limits as G.I.F. is discontinuous at integral points.

$\int_{0}^{10} f (x) d x = - 10 - 9 - . . . . - 1$

$= \frac{- 10.11}{2}$

$\int_{0}^{10} | f (x) | d x = 10 + 9 + . . . . + 1$

Let f be a differentiable function satisfying f(x) = $\frac{2}{\sqrt[]{3}} \int_{0}^{\sqrt[]{3}} f (\frac{λ^{2} x}{3}) d λ,$ x > 0 and f(1) = $\sqrt[]{3}$ . If y = f(x) passes through the point (a, 6), then a is equal to………….

$f (x) = \frac{2}{\sqrt[]{3}} \int_{0}^{\sqrt[]{3}} f (\frac{λ^{2} x}{3}) d λ$

Substitute $\frac{λ^{2} x}{3} = t$ $\frac{^{}}{}$

$f (x) = \frac{1}{\sqrt[]{x}} \int_{0}^{x} \frac{f (t)}{\sqrt[]{t}} d t$

differentiate using leibneitz rule

$\sqrt[]{x} . f' (x) + f (x) . \frac{1}{2 \sqrt[]{x}} = \frac{f (x)}{\sqrt[]{x}}$

f (1) = $\sqrt[]{3} \Rightarrow f (x) = \sqrt[]{3 x}$

A common tangent T to the curves $C_{1} : \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ and C₂ = $\frac{x^{2}}{42} - \frac{y^{2}}{143} = 1$ does not pass through the fourth quadrant. If T touches C₁ at (x₁, y₁) and C₂ at (x₂, y₂), then $| 2 x_{1} + x_{2} |$ is equal to………

Common tangents

$T_{1} : y = m x \pm \sqrt[]{4 m^{2} + 9}$

$T_{2} : y = m x \pm \sqrt[]{42 m^{2} - 13}$

So, 4m² + 9 = 42 m² – 13 Þ m = $\pm 2$

$\Rightarrow c = \pm 5$

So tangent $y = \pm 2 x \pm 5$

y = 2x + 5 does not pass through 4^th quadrant compare this tangent with T = 0 to get pt. of intersection

$\frac{x x_{2}}{42} - \frac{y y_{2}}{143} = 1 \Rightarrow x_{2} = \frac{- 84}{5}$

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4 \vec{c}, \vec{b} \times \vec{c} = 9 \vec{a}$ and $\vec{c} \times \vec{a} = α \vec{b}, α > 0 .$ If $| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{1}{36},$ then a is equal to…….

Data contradiction.

Maths NCERT Exemplar Solutions Class 12th Chapter Five Exam

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