If the straight-line x cosα + y sinα = p touches the curve, x 2 a 2 + y 2 b 2 = 1 , then prove that a² cos²α + b² sin²α =p².

This is a Long Answer Type Question as classified in NCERT Exemplar Sol: W e k n o w t h a t y = m x + c w i l l t o u c h t h e e l l i p s e x 2 a 2 + y 2 b 2 = 1 i f c 2 = a 2 m 2 + b 2 H e r e e q u a t i o n o f s t r a i g h t l i n e i s x c o s α + y s i n α = p a n d t h a t o f e l l i p s e i s x 2 a 2 + y 2 b 2 = 1 x c o s α + y s i n α = p ⇒ y s i n α = − x c o s α + p ⇒ y = − x c o s α s i n α + p s i n α ⇒ y = − x c o t α + p s i n α C o m p a i n g w i t h y = m x + c , w e g e t m = − c o t α a n d c = p s i n α S o , a c c o r d i n g t o t h e c o n d i t i o n , w e g e t c 2 = a 2 m 2 + b 2 p 2 s i n 2 α = a 2 ( − c o t α ) 2 + b 2 ⇒ p 2 s i n 2 α = a 2 c o s 2 α s i n 2 α + b 2 ⇒ p 2 = a 2 c o s 2 α + b 2 s i n 2 α H e n c e , a 2 c o s 2 α + b 2 s i n 2 α = p 2 H e n c e p r o v e d .

Let f(x) = x−1x+1,x∈R−{0,−1,1}. If fn+1(x)=f(fn(x)) for all n∈N, then f6 (6) + f7 (7) is equal a to

f (x)=x−1x+1⇒f (f (x))=x−1x+1−1x−1x+1+1=−1x f3 (x)=−x+1x−1⇒f4 (x)=x−1x+1+1x−1x+1−1=x So, f6 (6)+f7 (7)=f2 (6)+f3 (7) = −16−7+17−1=−96=−32

The remainder when (2021)2023 is divided by 7 is

2021≡−2 mod (7) ⇒ (2021)2023≡ (−2)2023mod (7) …… (i) Now, (−2)3≡−1mod (7) ⇒ (−2)2023≡ (−2)mod (7)≡5mod (7) ……. (ii) (i) & (ii) ⇒ (2021)2023≡5mod (7) ∴ Remainder = 5

limx→12sin(cos−1x)−x1−tan(cos−1x) is equal to

limx→12sin (cos−1x)−x1−tan (cos−1x) let cos−1x=π4+θ = limθ→∞2sinθ−2tanθ (1−tanθ)=− 1

A line ‘I’ passing through origin is perpendicular to the lines l1:r→=(3+t)t^+(−1+2t)j^+(4+2t)k^ l2:r→=(3+2s)i^+(3+2s)j^+(2+s)k^ If the co-ordinates of the point in the first octant on 'l2' at a distance of 17 from the point of intersection of 'l' and 'l1' and (a,b,c), then 18(a + b + c) is equal to……..

l1:x−31=y+12=z−42=t l2:x−32=y−32=z−21=s |i^j^k^12221|= (−2i^+3j^−2k^) l:r→=0→+λ (−2i^+3j^−2k^) l&l1→ Point of intersection l&l1 is P (2, 3, 2) Point Q on l2 is (3 + 2s, 3 + 2s, 2 + s).

If limx→0ax−(e4x−1)ax(e4x−1) exists and is equal to b, then the value of a – 2b is……..

limx→0ax− (e4x−1)ax (e4x−1)=b, use of L’ Hospital rule implies limx→0a−4e4xa (e4x−1)+ax (4e4x) =a−40⇒a=4 ⇒limn? →04 (−4.e4x)4.4e4x+16e4x+16x.4e4x =−1616+16=−12=b a – 2b = 4 – (1) = 5

A line is a common tangent circle and the parabola y2 = 4x. If the two point of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to………

A tangent to y2 = 4x is x – ty + t2 = 0 3+t21+t2=3 (3 + t2)2 = 9 (1 + t2) 9+t4+6t2=9+9t2 Point of contact (3, 23)= (a, b) x−3y+3=0 3x+y−33=0]4x−6=0 x=32, y=32+3 & (32, 32+2)= (c, d), 2 (a+c)=2 (3+32)=9

If the curve, y = y(x) represented by the solution of the differential equation (2xy2−y)dx+xdy=0, passes through the intersection of the lines, 2x – 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to………….

2xy2−y)dx+xdy=0 ⇒dydx+2y2−yx=0 Put 1y=z Then −1y2dydx=dzdx dzdx+1xz=2 Point (2, 1) c = 2 – 4 = 2 y = xx2−2 |y (1)|=1

The value of ∫−22|3x2−3x−6|dx is..........

I=3∫−22|x2−x−2|dx =3 (∫−2−1 (x2−x−2)dx−∫−12 (x2−x−2)dx) =3 [ (76+23)− {−103−76}] =3 (116+276)=382=19

The integral ∫ e 3 l o g e 2 x + 5 e 2 l o g e 2 x e 4 l o g e x + 5 e 3 l o g e x − 7 e 2 l o g e x d x , x > 0 i s e q u a l t o : (where c is constant of integration)

4 l o g e | x 2 + 5 x − 7 | + c

1 4 l o g e | x 2 + 5 x − 7 | + c

l o g e | x 2 + 5 x − 7 | + c

The following system of linear equations 2x + 3y + 2z = 9 3x + 2y + 2z = 9 x – y + 4z = 8

has a solution (α, β, γ) satisfying α + β 2 + γ 3  = 12

The contrapositive of the statement “If you will work, you will earn money” is

If you will not earn money, you will not work

If you will earn money, you will work

To earn money, you need to work

You will earn money, if you will not work

If In = ∫ π 4 π 2 c o t n x d x , t h e n :

1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in A.P.

1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6

I 2 + I 4 , I 3 + I 5 , I 4 + I 6 are in A.P.

I 2 + I 4 , ( I 3 + I 5 ) 2 , I 4 + I 6 are in G.P.

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

Q: Let A be a 3 × 3 invertible matrix. If |adj(24A)|=|adj(3adj(2A))|, then |A|2 is equal to

|adj (24A)|=|adj (3adj (2A))| ⇒|24A|2=|3adj (2A)|2 246|A|2=36. (23)4|A|4 |A|2=24636.212=218.3636.212=26

Q: The ordered pair (a, b), for which the system of linear equations 3x – 2y + z = b 5x – 8y + 9z = 3 2x + y + az = -1 has no solution, is

System of equation can be written as (3−215−8921a) (xyz)= (b3−1) (3−2115−2427633a) (xyz)= (b9−3) R3−2R1, R2−5R1 for no solution 3a + 9 = 0 but 32−9b2≠0 ⇒a=−3 ⇒b≠13

Updated on Aug 13, 2025 12:34 IST

By alok kumar singh, Executive Content Operations

Table of contents

Applications of Derivatives Questions and Answers
26th June 2022 (First Shift)
JEE MAINS 25th Feb 2021

Applications of Derivatives Questions and Answers

1. If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{π}{3}$ .

Sol:

2. Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^{5} - 5 x ⁴ + 5 x ³ - 1$ Also, find the corresponding local maximum and local minimum values.

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{5} - 5 x^{4} + 5 x^{3} - 1 \\ f^{'} (x) = 5 x^{4} - 20 x^{3} + 15 x^{2} \\ F o r l o c a l maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 5 x^{4} - 20 x^{3} + 15 x^{2} = 0 \Rightarrow 5 x^{2} (x^{2} - 4 x + 3) = 0 \\ \Rightarrow 5 x^{2} (x^{2} - 3 x - x + 3) = 0 \Rightarrow x^{2} (x - 3) (x - 1) = 0 \\ ∴ x = 0, x = 1 a n d x = 3 \\ N o w f^{''} (x) = 20 x^{3} - 60 x^{2} + 30 x \\ \Rightarrow f^{''} {(x)}_{a t x = 0} = 20 {(0)}^{3} - 60 {(0)}^{2} + 30 (0) = 0 w h i c h i s n e i t h e r M a x i m a n o r M i n i m a . \\ ∴ f (x) h a s t h e point o f inflextion a t x = 0 \\ \Rightarrow f^{''} {(x)}_{a t x = 1} = 20 {(1)}^{3} - 60 {(1)}^{2} + 30 (1) \\ = 20 - 60 + 30 = - 10 < 0 M a x i m a \\ \Rightarrow f^{''} {(x)}_{a t x = 3} = 20 {(3)}^{3} - 60 {(3)}^{2} + 30 (3) \\ = 540 - 540 + 90 = 90 > 0 M i n i m a \\ T h e m a x i m u m v a l u e o f t h e f u n c t i o n a t x = 1 \\ f (x) = {(1)}^{5} - 5 {(1)}^{4} + 5 {(1)}^{3} - 1 \\ = 1 - 5 + 5 - 1 = 0 \\ T h e minimum v a l u e o f t h e f u n c t i o n a t x = 3 \\ f (x) = {(3)}^{5} - 5 {(3)}^{4} + 5 {(3)}^{3} - 1 \\ = 243 - 405 + 135 - 1 = 378 - 406 = - 28 \\ H e n c e, t h e f u n c t i o n h a s i t s maxima a t x = 1 a n d t h e maximum v a l u e = 0 a n d i t h a s minimum \\ v a l u e a t x = 3 a n d i t s minimum v a l u e = - 28 \\ x = 0 i s t h e point o f inflextion. \end{array}$

3.A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription, and it is believed that for every increase of Re 1/-, one subscriber will discontinue the service. Find what increase will bring maximum profit.

Sol:

$\begin{array}{l} L e t u s c o n s i d e r t h a t t h e c o m p a n y i n c r e a s e s t h e a n n u a l s u b s c r i p t i o n b y R s . x \\ S o, x i s t h e n u m b e r o f s u b s c r i b e r s w h o d i s c o n t i n u e t h e s e r v i c e s . \\ ∴ T o t a l r e v e n u e, R (x) = (500 - x) (300 + x) \\ = 150000 + 500 x - 300 x - x^{2} \\ = - x^{2} + 200 x + 150000 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t R^{'} (x) = - 2 x + 200 \\ F o r l o c a l maxima a n d l o c a l minima, R^{'} (x) = 0 \\ - 2 x + 200 = 0 \Rightarrow x = 100 \\ R^{''} (x) = - 2 < 0 M a x i m a \\ S o, R (x) i s maximum a t x = 100 \\ H e n c e, i n o r d e r t o g e t maximum profit, the company should increase its a n n u a l s u b s c r i p t i o n \\ b y R s . 100 . \end{array}$

4. If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then prove that a² cos²α + b² sin²α =p².

Sol:

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

Commonly asked questions

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{π}{3}$ .

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

Sol:

Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^{5} - 5 x ? + 5 x ³ - 1$ Also, find the corresponding local maximum and local minimum values.

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

Sol:

A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription, and it is believed that for every increase of Re 1/-, one subscriber will discontinue the service. Find what increase will bring maximum profit.

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

Sol:

If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then prove that a² cos²α + b² sin²α =p².

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

Sol:

An open box with a square base is to be made of a given quantity of cardboard of area c². Show that the maximum volume of the box is $\frac{c ³}{6 \sqrt 3}$ cubic units.

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

Sol:

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides. Also, find the maximum volume.

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

Sol:

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

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

Sol:

$\begin{array}{l} N o w p u t t i n g t h e v a l u e o f K i n e q . (i) w e g e t \\ 6 x^{2} + 4 π r^{2} = x^{2} (π + 6) \\ \Rightarrow 6 x^{2} + 4 π r^{2} = π x^{2} + 6 x^{2} \Rightarrow 4 π r^{2} = π x^{2} \Rightarrow 4 r^{2} = x^{2} \\ ∴ 2 r = x \\ ∴ x : 2 r = 1 : 1 \\ N o w d i f f e r e n t i a t i n g e q . (i i) w . r . t ., x, w e h a v e \\ \frac{d^{2} V}{d x^{2}} = 6 x - \frac{3}{} \frac{d}{d x} [x {(K - 6 x^{2})}^{1 / 2}] \\ = 6 x - \frac{3}{} [x . \frac{1}{2} \times (- 12 x) + {(K - 6 x^{2})}^{1 / 2} . 1] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2}}{} +] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2} + K - 6 x^{2}}{}] = 6 x + \frac{3}{} [\frac{12 x^{2} - K}{}] \\ = 6 + \frac{3}{} [\frac{\frac{12 K}{π + 6} - K}{}] [P u t x =] \\ = 6 + \frac{3}{} [\frac{12 K - π K - 6 K}{}] = 6 + \frac{3}{} [\frac{6 K - π K}{}] \\ = 6 + \frac{3}{π} [(6 K - π K)] > 0 s o i t i s minima . \\ H e n c e, t h e r e q u i r e d r a t i o i s 1 : 1 w h e n t h e c o m b i n e d v o l u m e i s minimum . \end{array}$

Minimum

AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

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

Sol:

A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm². Find the least cost of the box

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

Sol:

The sum of the surface areas of a rectangular parallelepiped with sides x, 2x, and and a sphere is given to be constant. Prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.

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

Sol:

$\begin{array}{l} L e t' r' b e t h e r a d i u s o f t h e s p h e r e . \\ ∴ S u r f a c e a r e a o f t h e s p h e r e = 4 π r^{2} \\ V o l u m e o f t h e s p h e r e = \frac{4}{3} π r^{3} \\ T h e s i d e s o f t h e p a r a l l e l o p i p e d a r e x, 2 x a n d \frac{x}{3} \\ ∴ I t s s u r f a c e a r e a = 2 [x \times 2 x + 2 x \times \frac{x}{3} + x \times \frac{x}{3}] \\ = 2 [2 x^{2} + \frac{2 x^{2}}{3} + \frac{x^{2}}{3}] = 2 [2 x^{2} + x^{2}] \\ = 2 [3 x^{2}] = 6 x^{2} \\ V o l u m e o f t h e p a r a l l e l o p i p e d = x \times 2 x \times \frac{x}{3} = \frac{2}{3} x^{3} \\ A s p e r t h e c o n d i t i o n s o f t h e q u e s t i o n, \\ s u r f a c e a r e a o f t h e p a r a l l e l o p i p e d + s u r f a c e a r e a o f t h e s p h e r e = constant \\ \Rightarrow 6 x^{2} + 4 π r^{2} = K (constant) \Rightarrow 4 π r^{2} = K - 6 x^{2} \\ ∴ r^{2} = \frac{K - 6 x^{2}}{4 π} \dots (i) \\ N o w l e t V = V o l u m e o f t h e p a r a l l e l o p i p e d + V o l u m e o f t h e s p h e r e \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π r^{3} \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π {[\frac{K - 6 x^{2}}{4 π}]}^{3 / 2} [F r o m e q . (i)] \\ \Rightarrow &thins \end{array}$

$\begin{array}{l} S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 π x^{2} = 9 (K - 6 x^{2}) \Rightarrow 4 π x^{2} = 9 K - 54 x^{2} \\ \Rightarrow 4 π x^{2} + 54 x^{2} = 9 K \\ \Rightarrow K = \frac{4 π x^{2} + 54 x^{2}}{9} (i i) \\ \Rightarrow 2 x^{2} (2 π + 27) = 9 K \\ ∴ x^{2} = \frac{9 K}{2 (2 π + 27)} = 3 \\ N o w f r o m e q . (i) w e h a v e \\ r^{2} = \frac{K - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{\frac{4 π x^{2} + 54 x^{2}}{9} - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{4 π x^{2} + 54 x^{2} - 54 x^{2}}{9 \times 4 π} = \frac{4 π x^{2}}{9 \times 4 π} = \frac{x^{2}}{9} \Rightarrow r = \frac{x}{3} ∴ x = 3 r \\ N o w w e h a v e \frac{d V}{d x} = 2 x^{2} - \frac{3 x}{} {(K - 6 x^{2})}^{1 / 2} \end{array}$

$\begin{array}{l} = 12 - \frac{3}{} \frac{\frac{4 K π + 54 K - 108 K}{4 π + 54}}{} \\ = 12 - \frac{3}{} [\frac{\frac{4 K π - 54 K}{4 π + 54}}{}] \\ = 12 - \frac{3}{} [\frac{4 K π - 54 K}{.}] \\ = 12 - \frac{6}{} [\frac{2 π - 27}{.}] \\ = 12 + \frac{6 K}{} [\frac{27 - 2 π}{.}] > 0 [? 27 - 2 π > 0] \\ ∴ \frac{d^{2} V}{d x^{2}} > 0 S o, i t i s minima. \\ H e n c e, t h e s u m o f v o l u m e i s min \end{array}$

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

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

Sol.

$\begin{array}{l} B a l l o f s a l t i s s p h e r i c a l \\ ∴ V o l u m e o f b a l l, V = \frac{4}{3} π r^{3}, w h e r e r = r a d i u s o f t h e b a l l . \\ A s p e r Q u e s t i o n, \frac{d V}{d t} \propto S, w h e r e S = s u r f a c e a r e a o f t h e b a l l \\ \Rightarrow \frac{d}{d t} (\frac{4}{3} π r^{3}) \propto 4 π r^{2} [? S = 4 π r^{2}] \\ \Rightarrow \frac{4}{3} π . 3 r^{2} . \frac{d}{d t} \propto 4 π r^{2} \\ \Rightarrow 4 π r^{2} . \frac{d r}{d t} = Κ . 4 π r^{2} [K = constant o f p r o p o r t i o n a l i t y] \\ \Rightarrow \frac{d r}{d t} = Κ . \frac{4 π r^{2}}{4 π r^{2}} \\ \Rightarrow \frac{d r}{d t} = Κ . 1 = K \\ H e n c e, t h e r a d i u s o f t h e b a l l i s decreasing a t constant r a t e . \end{array}$

If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

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

Sol:

$\begin{array}{l} W e k n o w t h a t \\ A r e a o f c i r c l e, A = π r^{2}, w h e r e r = r a d i u s o f t h e c i r c l e a n d p e r i m e t e r = 2 π r \\ A s p e r Q u e s t i o n, \frac{d A}{d t} = Κ, w h e r e K = constant \\ \Rightarrow \frac{d}{d t} (π r^{2}) = K \\ \Rightarrow π . 2 r . \frac{d r}{d t} = K \\ ∴ \frac{d r}{d t} = \frac{Κ}{4 π r^{2}} \dots (1) \\ N o w P e r i m e t e r c = 2 π r \\ D i f f e r e n t i a t i n g b o t h s i d e s ? w . r . t . t, w e g e t \\ \Rightarrow \frac{d c}{d t} = \frac{d}{d t} (2 π r) \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{d r}{d t} \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{K}{2 π r} = \frac{K}{r} [F r o m e q n (1)] \\ \Rightarrow \frac{d c}{d t} \propto \frac{1}{r} \\ H e n c e, t h e p e r i m e t e r o f t h e c i r c l e varies inversely a s the r a d i u s o f t h e c i r c l e . \end{array}$

Find an angle θ, 0 < θ < $\frac{π}{2}$ , which increases twice as fast as its sine.

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

Sol:

$\begin{array}{l} A s p e r t h e g i v e n c o n d i t i o n, \\ \frac{d θ}{d t} = 2 \frac{d}{d t} (s i n θ) \\ \Rightarrow \frac{d θ}{d t} = 2 c o s θ \frac{d θ}{d t} \Rightarrow 1 = 2 c o s θ \\ ∴ c o s θ = \frac{1}{2} \Rightarrow c o s θ = c o s \frac{π}{3} \Rightarrow θ = \frac{π}{3} \\ H e n c e, t h e r e q u i r e d a n g l e i s \frac{π}{3} . \end{array}$

Find the approximate value of (1.999) ?.

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

Sol:

$\begin{array}{l} {(1.999)}^{5} = {(2 - 0.001)}^{5} \\ L e t x = 2 a n d Δ x = - 0.001 \\ L e t y = x^{5} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., x, w e g e t \\ \frac{d y}{d x} = 5 x^{4} = 5 {(2)}^{4} = 80 \\ N o w Δ y = (\frac{d y}{d x}) . Δ x = 80 . (- 0.001) = - 0.080 \\ ∴ {(1.999)}^{5} = y + Δ y \\ = x^{5} - 0.080 = {(2)}^{5} - 0.080 = 32 - 0.080 = 31.92 \\ H e n c e, a p p r o x i m a t e v a l u e o f {(1.999)}^{5} i s 31.92 . \end{array}$

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

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

Sol:

$\begin{array}{l} I n t e r n a l r a d i u s r = 3 c m \\ E x t e r n a l r a d i u s R = r + Δ r = 3.0005 c m \\ ∴ Δ r = 3.0005 - 3 = 0.0005 c m \\ L e t y = r^{3} \Rightarrow y + Δ y = {(r + Δ r)}^{3} = R^{3} = {(3.0005)}^{3} \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., r, w e g e t \\ \frac{d y}{d x} = 3 r^{2} \\ ∴ Δ y = (\frac{d y}{d r}) . Δ r = 3 r^{2} \times 0.0005 \\ = 3 \times {(3)}^{2} \times 0.0005 = 27 \times 0.0005 = 0.0135 \\ ∴ {(3.0005)}^{3} = y + Δ y [F r o m e q n (i)] \\ = {(3)}^{3} + 0.0135 = 27 + 0.0135 = 27.0135 \\ V o l u m e o f t h e s h e l l = \frac{4}{3} π [R^{3} - r^{3}] \\ = \frac{4}{3} π [27.0135 - 27] = \frac{4}{3} π \times 0.0135 \\ = 4 π \times 0.0045 = 4 \times 3.14 \times 0.0045 = 0.018 π c m^{3} \\ H e n c e, a p p r o x i m a t e V o l u m e o f t h e m e t a l i n t h e s h e l l i s 0.018 π c m^{3} . \end{array}$

A swimming pool is to be drained for cleaning. If L represents the number of liters of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 – t)², how fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

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

Sol.

$\begin{array}{l} G i v e n t h a t L = 200 {(10 - t)}^{2} \\ w h e r e L r e p r e s e n t s t h e n u m b e r o f l i t r e s o f w a t e r i n t h e p o o l . \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., t, w e g e t \\ \frac{d L}{d t} = 200 \times 2 (10 - t) (- 1) = - 400 (10 - t) \\ B u t t h e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = - \frac{d L}{d t} = 400 (10 - t) \dots (i) \\ R a t e a t w h i c h t h e w a t e r i s r u n n i n g a f t e r 5 seconds \\ = 400 (10 - 5) = 2000 L / s (f i n a l r a t e) \\ F o r i n i t i a l r a t e p u t t = 0 \\ = 400 (10 - 0) = 4000 L / s \\ T h e a v e r a g e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = \frac{I n i t i a l r a t e + F i n a l r a t e}{2} = \frac{4000 + 2000}{2} = \frac{6000}{2} = 3000 L / s \\ H e n c e, t h e r e q u i r e d r a t e = 3000 L / s . \end{array}$

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

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

Sol.

$\begin{array}{l} L e t x b e t h e l e n g t h o f t h e c u b e \\ ∴ V o l u m e o f t h e c u b e V = x^{3} \dots (i) \\ G i v e n t h a t \frac{d V}{d t} = K \\ D i f f e r e n t i a t i n g e q (i) w . r . t . t, w e g e t \\ \frac{d V}{d t} = 3 x^{2} . \frac{d x}{d t} = K (constant) \\ ∴ \frac{d x}{d t} = \frac{K}{3 x^{2}} \\ N o w s u r f a c e ? ? a r e a o f t h e c u b e, S = 6 x^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t, w e g e t \\ \frac{d S}{d t} = 6.2 . x . \frac{d x}{d t} = 12 x . \frac{K}{3 x^{2}} \\ \Rightarrow \frac{d S}{d t} = \frac{4 K}{x} \Rightarrow \frac{d S}{d t} \propto \frac{1}{x} (4 K = constant) \\ H e n c e, s u r f a c e ? ? a r e a o f t h e c u b e v a r i e s i n v e r s e l y a s t h e l e n g t h o f t h e s i d e . \end{array}$

x and y are the sides of two squares such that y = x – $x^{2}$ . Find the rate of change of the area of the second square with respect to the area of the first square.

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

Sol:

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

Find the condition that the curves 2x = $y^{2}$ and 2xy = k intersect orthogonally.

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

Sol.

$\begin{array}{l} T h e t w o c i r c l e s intersect o r t h o g o n a l l y i f t h e a n g l e b e t w e e n t h e tangents d r a w n t o t h e t w o c i r c l e s \\ a t t h e point o f t h e i r intersection i s 9 0^{0} . \\ E q u a t i o n o f t h e t w o c i r c l e s a r e g i v e n a s \\ 2 x = y^{2} \dots (i) \\ a n d 2 x y = k \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) a n d e q (i i) w . r . t . x, w e g e t \\ 2.1 = 2 y . \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y} \Rightarrow m_{1} = \frac{1}{y} (m_{1} = s l o p e o f t h e tangent) \\ \Rightarrow 2 x y = k \\ \Rightarrow 2 [x . \frac{d y}{d x} + y . 1] = 0 \\ ∴ \frac{d y}{d x} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x} (m_{2} = s l o p e o f t h e o t h e r tangent) \\ I f t h e t w o tangents a r e p e r p e n d i c u l a r t o e a c h o t h e r, \\ t h e n m_{1} \times m_{2} = - 1 \\ \Rightarrow \frac{1}{y} \times - \frac{y}{x} = - 1 \Rightarrow \frac{1}{x} = 1 \Rightarrow x = 1 \\ N o w s o l v i n g 2 x = y^{2} [F r o m (i)] \\ a n d &thinsp \end{array}$

Prove that the curves xy = 4 and x² + y² = 8 touch each other.

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

Sol:

$\begin{array}{l} G i v e n c i r c l e s a r e 2 x y = 4 \dots (i) \\ a n d x^{2} + y^{2} = 8 \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) w . r . t . x, w e g e t \\ x . \frac{d y}{d x} + y . 1 = 0 \Rightarrow \frac{d y}{d x} = - \frac{y}{x} \Rightarrow m_{1} = \frac{1}{y} (m_{1} = s l o p e o f t h e tangent to the curve) \\ D i f f e r e n t i a t i n g e q (i i) w . r . t . x, w e g e t \\ ∴ 2 x + 2 y . \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \Rightarrow m_{2} = - \frac{x}{y} (m_{2} = s l o p e o f t h e o t h e r tangent to the curve) \\ T o f i n d t h e point o f c o n t a c t o f t h e t w o c i r c l e s \\ t h e n m_{1} = m_{2} \\ \Rightarrow - \frac{y}{x} = - \frac{x}{y} \Rightarrow x^{2} = y^{2} \\ P u t t i n g t h e v a l u e o f y^{2} i n e q . (i i) \\ x^{2} + x^{2} = 8 \Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \\ ∴ x = \pm 2 \\ ? x^{2} = y^{2} \Rightarrow y = \pm 2 \\ ∴ T h e point o f c o n t a c t o f t h e t w o c i r c l e s a r e (2, 2) a n d (- 2, 2) . \end{array}$

Find the coordinates of the point on the curve √x + √y = 4 at which the tangent is equally inclined to the axes

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

Sol:

Prove that the curves y² = 4x and x² + y² – 6x + 1 = 0 touch each other at the point (1, 2).

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e e q u a t i o n o f t h e t w o c u r v e s a r e y^{2} = 4 x \dots (i) \\ a n d x^{2} + y^{2} - 6 x + 1 = 0 \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) w . r . t . x, w e g e t \\ 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} \\ S l o p e o f t h e tangent at (1, 2), m_{1} = \frac{2}{2} = 1 \\ D i f f e r e n t i a t i n g e q (i i) w . r . t . x, w e g e t \\ \Rightarrow 2 x + 2 y . \frac{d y}{d x} - 6 = 0 \Rightarrow 2 y . \frac{d y}{d x} = 6 - 2 x \Rightarrow \frac{d y}{d x} = \frac{6 - 2 x}{2 y} \\ ∴ S l o p e o f t h e tangent at t h e s a m e point (1, 2), m_{2} = \frac{6 - 2 \times 1}{2 \times 2} = \frac{4}{4} = 1 \\ W e s e e t h a t m_{1} = m_{2} = 1 at t h e point (1, 2) . \\ H e n c e, t h e g i v e n c i r c l e s t o u c h e a c h o t h e r a t t h e s a m e point (1, 2) . \end{array}$

Find the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4.

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

Sol:

$\begin{array}{l} W e h a v e e q u a t i o n o f t h e c u r v e 3 x^{2} - y^{2} = 8 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ 6 x - 2 y . \frac{d y}{d x} = 0 \Rightarrow - 2 y \frac{d y}{d x} = - 6 x \Rightarrow \frac{d y}{d x} = \frac{3 x}{y} \\ S l o p e o f t h e tangent t o t h e g i v e n c u r v e = \frac{3 x}{y} \\ ∴ S l o p e o f t h e normal t o t h e c u r v e = - \frac{1}{\frac{3 x}{y}} = - \frac{y}{3 x} . \\ N o w d i f f e r e n t i a t i n g b o t h s i d e s t h e g i v e n l i n e x + 3 y = 4 \\ \Rightarrow 1 + 3 . \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{1}{3} \\ Since t h e normal t o t h e c u r v e i s p a r a l l e l t o t h e g i v e n l i n e x + 3 y = 4 . \\ ∴ - \frac{y}{3 x} = - \frac{1}{3} \Rightarrow y = x \\ P u t t i n g t h e v a l u e s o f y i n 3 x^{2} - y^{2} = 8, w e g e t \\ 3 x^{2} - x^{2} = 8 \Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2 \\ ∴ y = \pm 2 \\ ∴ T h e points o n t h e c u r v e a r e (2, 2) a n d (- 2, - 2) . \\ N o w e q u a t i o n o f t h e normal t o t h e c u r v e a t (2, 2) i s \\ y - 2 = - \frac{1}{3} (x - 2) \\ \Rightarrow 3 y - 6 = - x + 2 \Rightarrow x + 3 y = 8 \\ a t (- 2, - 2) y + 2 = - \frac{1}{3} (x + 2) \\ \Rightarrow 3 y + 6 = - x - 2 \Rightarrow x + 3 y = - 8 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e x + 3 y = 8 a n d x + 3 y = - 8 o r x + 3 y = \pm 8 . \end{array}$

At what points on the curve x² + y² – 2x – 4y + 1 = 0 are the tangents parallel to the y-axis?

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

Sol:

$\begin{array}{l} G i v e n t h a t e q u a t i o n o f t h e c u r v e i s x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ 2 x + 2 y . \frac{d y}{d x} - 2 - 4 . \frac{d y}{d x} = 0 \Rightarrow (2 y - 4) \frac{d y}{d x} = 2 - 2 x \\ \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y - 4} \dots (i i) \\ Since t h e tangent t o t h e c u r v e i s p a r a l l e l t o t h e y - a x i s . \\ ∴ S l o p e \frac{d y}{d x} = t a n \frac{π}{2} = \infty = \frac{1}{0} \\ S o, f r o m e q n . (i i) w e g e t \\ \frac{2 - 2 x}{2 y - 4} = \frac{1}{0} \Rightarrow 2 y - 4 = 0 \Rightarrow y = 2 \\ P u t t i n g t h e v a l u e o f y i n e q n . (i), w e g e t \\ \Rightarrow x^{2} + {(2)}^{2} - 2 x - 8 + 1 = 0 \\ \Rightarrow x^{2} - 2 x + 4 - 8 + 1 = 0 \Rightarrow x^{2} - 2 x - 3 = 0 \\ \Rightarrow x^{2} - 3 x - x - 3 = 0 \\ \Rightarrow x (x - 3) + 1 (x - 3) = 0 \Rightarrow (x - 3) (x + 1) = 0 \\ \Rightarrow \end{array}$

Show that the line x/a + y/b = 1 touches the curve y = b(-x/a) at the point where the curve intersects the y-axis.

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

Sol:

$\begin{array}{l} G i v e n t h a t e q u a t i o n o f t h e c u r v e i s y = b . e^{- x / a} \\ a n d t h e e q u a t i o n o f t h e l i n e i s \frac{x}{a} + \frac{y}{b} = 1 \\ L e t t h e c o o r d i n a t e s o f t h e point w h e r e t h e c u r v e intersects t h e y - a x i s b e (0, y_{1}) \\ N o w, d i f f e r e n t i a t i n g y = b . e^{- x / a} b o t h s i d e s w . r . t . x, w e g e t \\ \frac{d y}{d x} = b . e^{- x / a} (- \frac{1}{a}) = - \frac{b}{a} . e^{- x / a} \\ So, t h e s l o p e o f t h e tangent, m_{1} = - \frac{b}{a} . e^{- x / a} \\ D i f f e r e n t i a t i n g \frac{x}{a} + \frac{y}{b} = 1 b o t h s i d e s w . r . t . x, w e g e t \\ ∴ \frac{1}{a} + \frac{1}{b} . \frac{d y}{d x} = 0 \\ So, t h e s l o p e o f t h e l i n e, m_{2} = - \frac{b}{a} \\ I f t h e l i n e t o u c h e s t h e c u r v e, t h e n m_{1} = m_{2} \\ \Rightarrow - \frac{b}{a} . e^{- x / a} = - \frac{b}{a} \Rightarrow e^{- x / a} = 1 \\ \Rightarrow \frac{- x}{a} l o g e = l o g 1 (T a k i n g l o g o n b o t h s i d e s) \\ \Rightarrow \frac{- x}{a} = 0 \Rightarrow x = 0 \\ P u t t i n g x = 0 i n e q u a t i o n y = b . e^{- x / a} \\ \Rightarrow y = b . e^{0} = b \\ H e n c e, t h e g i v e n e q u a t i o n o f t h e c u r v e intersects a t (0, b) i . e . o n y - a x i s . \end{array}$

Kindly consider the following

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

Sol.

$\begin{array}{l} G i v e n t h a t f (x) = 2 x + c o t^{- 1} x + l o g (- x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = 2 - \frac{1}{1 + x^{2}} + \frac{1}{- 1} \times \frac{d}{d x} (- x) \\ = 2 - \frac{1}{1 + x^{2}} + \frac{(\frac{1}{2} \times (2 x - 1))}{- 1} \\ = 2 - \frac{1}{1 + x^{2}} + \frac{x -}{(- x)} \\ = 2 - \frac{1}{1 + x^{2}} - \frac{- x}{(- x)} \\ = 2 - \frac{1}{1 + x^{2}} - \frac{1}{} \\ F o r increasing f u n c t i o n, f^{'} (x) \geq 0 \\ ∴ 2 - \frac{1}{1 + x^{2}} - \frac{1}{} \geq 0 \\ \Rightarrow \frac{2 (1 + x^{2}) - 1 +}{1 + x^{2}} \geq 0 \Rightarrow 2 + 2 x^{2} - 1 + \geq 0 \\ \Rightarrow 2 x^{2} + 1 + \geq 0 2 x^{2} + 1 \geq - \\ S q u a r i n g b o t h s i d e s, w e g e t 4 x^{4} + 1 + 4 x^{2} \geq 1 + x^{2} \\ \Rightarrow 4 x^{4} + 4 x^{2} - x^{2} \geq 0 \Rightarrow 4 x^{4} + 3 x^{2} \geq 0 \Rightarrow x^{2} (4 x^{2} + 3) \geq 0 \\ w h i c h i s t r u e f o r a n y v a l u e o f x \in R \\ H e n c e, t h e g i v e n f u n c t i o n i s a n increasing f u n c t i o n o v e r R . \end{array}$

Kindly consider the following

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = s i n x - c o s x - 2 a x + b, a \geq 1 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = c o s x + s i n x - 2 a \\ F o r decreasing f u n c t i o n, f^{'} (x) < 0 \\ ∴ c o s x + s i n x - 2 a < 0 \\ \Rightarrow 2 (\frac{}{2} c o s x + \frac{1}{2} s i n x) - 2 a < 0 \\ \Rightarrow \frac{}{2} c o s x + \frac{1}{2} s i n x - a < 0 \\ \Rightarrow (c o s \frac{π}{6} c o s x + s i n \frac{π}{6} s i n x) - a < 0 \\ \Rightarrow c o s (x - \frac{π}{6}) - a < 0 \\ Since c o s x \in [- 1, 1] a n d a \geq 1 \\ ∴ f^{'} (x) < 0 \\ H e n c e, t h e g i v e n f u n c t i o n i s decreasing f u n c t i o n i n R . \end{array}$

Show that f(x) = tan?¹(sinx + cosx) is an increasing function in (0, $\frac{π}{4}$ ).

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = t a n^{- 1} (s i n x + c o s x) i n (0, \frac{π}{4}) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = \frac{1}{1 + {(s i n x + c o s x)}^{2}} . \frac{d}{d x} (s i n x + c o s x) \\ \Rightarrow f^{'} (x) = \frac{1 \times (c o s x - s i n x)}{1 + {(s i n x + c o s x)}^{2}} \\ \Rightarrow f^{'} (x) = \frac{(c o s x - s i n x)}{1 + s i n^{2} x + c o s^{2} x + 2 s i n x c o s x} \\ \Rightarrow f^{'} (x) = \frac{(c o s x - s i n x)}{1 + 1 + 2 s i n x c o s x} \Rightarrow f^{'} (x) = \frac{c o s x - s i n x}{2 + 2 s i n x c o s x} \\ F o r a n increasing f u n c t i o n, f^{'} (x) \geq 0 \\ ∴ \frac{c o s x - s i n x}{2 + 2 s i n x c o s x} \geq 0 \\ \Rightarrow c o s x - s i n x \geq 0 [? (2 + s i n 2 x) \geq 0 i n (0, \frac{π}{4})] \\ \Rightarrow c o s \geq s i n x, w h i c h i s t r u e f o r (0, \frac{π}{4}) \\ H e n c e, t h e g i v e n f u n c t i o n f (x) i s a n increasing f u n c t i o n i n (0, \frac{π}{4}) . \end{array}$

At what point is the slope of the curve y = –x³ + 3x² + 9x – 27 maximum? Also, find the maximum slope.

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

Sol.

$\begin{array}{l} G i v e n t h a t y = - x^{3} + 3 x^{2} + 9 x - 27 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{d y}{d x} = - 3 x^{2} + 6 x + 9 \\ L e t s l o p e o f t h e c u r v e \frac{d y}{d x} = Z \\ ∴ z = - 3 x^{2} + 6 x + 9 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \frac{d z}{d x} = - 6 x + 6 \\ F o r l o c a l maxima a n d l o c a l minima, \frac{d z}{d x} = 0 \\ ∴ - 6 x + 6 = 0 \Rightarrow x = 1 \\ \Rightarrow \frac{d^{2} z}{d x^{2}} = - 6 < 0 M a x i m a \\ P u t x = 1 i n e q u a t i o n o f t h e c u r v e y = {(- 1)}^{3} + 3 {(1)}^{2} + 9 (1) - 27 \\ = - 1 + 3 + 9 - 27 = - 16 \\ M a x i m u m s l o p e = - 3 {(1)}^{2} + 6 (1) + 9 = 12 \\ H e n c e, (1, - 16) i s t h e point a t w h i c h t h e s l o p e o f t h e g i v e n c u r v e i s maximum a n d \\ M a x i m u m s l o p e = 12 \end{array}$

Prove that f(x) = sinx + cosx has a maximum value at x = , $\frac{π}{6}$ .

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

Sol:

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases when the side is 10 cm is:

(A) 10 cm²/s

(B) √3 cm²/s

(C) 10 cm²/s

(D) 10/3 cm²/s

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

The curve y = x^1/? has at (0, 0):

(A) A vertical tangent (parallel to y-axis)

(B) A horizontal tangent (parallel to x-axis)

(C) An oblique tangent

(D) No tangent

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} E q u a t i o n o f c u r v e i s y = x^{1 / 5} \\ D i f f e r e n t i a t i n g w . r . t . x, w e g e t \frac{d y}{d x} = \frac{1}{5} x^{- 4 / 5} \\ (a t x = 0) \frac{d y}{d x} = \frac{1}{5} {(0)}^{- 4 / 5} = \frac{1}{5} \times \frac{1}{0} = \infty \\ \frac{d y}{d x} = \infty \\ ∴ T h e tangent is parallel to y-axis. \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The equation of the normal to the curve *3x² – y² = 8, which is parallel to the line *x + 3y = 8, is:

(A) 3x – y = 8

(B) 3x + y + 8 = 0

(C) x + 3y ± 8 = 0

(D) x + 3y = 0

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c u r v e 3 x^{2} - y^{2} = 8 \\ D i f f e r e n t i a t i n g b o t h w . r . t . x, w e g e t \\ 6 x - 2 y . \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{3 x}{y} \\ \frac{3 x}{y} i s t h e s l o p e o f t h e tangent . \\ ∴ s l o p e o f t h e n o r m a l = \frac{- 1}{\frac{d y}{d x}} = \frac{- y}{3 x} \\ N o w x + 3 y = 8 i s p a r a l l e l t o t h e n o r m a l \\ D i f f e r e n t i a t i n g b o t h w . r . t . t, w e h a v e \\ 1 + 3 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{1}{3} \\ ∴ \frac{- y}{3 x} = - \frac{1}{3} \Rightarrow y = x \\ P u t t i n g y = x i n e q . (i) w e g e t \\ 3 x^{2} - x^{2} = 8 \Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \\ ∴ x = \pm 2 a n d y = \pm 2 \\ S o t h e points a r e (2, 2) a n d (- 2, - 2) . \\ E q u a t i o n o f n o r m a l t o t h e g i v e n c u r v e a t (2, 2) i s \\ y - 2 = - \frac{1}{3} (x - 2) \\ \Rightarrow 3 y - 6 = - x + 2 \Rightarrow x + 3 y - 8 = 0 \\ E q u a t i o n o f n o r m a l a t (- 2, - 2) i s \\ &thin \end{array}$

If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D) 6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e g i v e n c u r v e s a r e a y + x^{2} = 7 \dots (i) \\ a n d x^{3} = y \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ a . \frac{d y}{d x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{2 x}{a} \\ ∴ m_{1} = - \frac{2 x}{a} (m_{1} = \frac{d y}{d x}) \\ N o w, d i f f e r e n t i a t i n g e q . (i i) w . r . t . t, w e h a v e \\ 3 x^{2} = \frac{d y}{d x} \Rightarrow m_{2} = 3 x^{2} (m_{2} = \frac{d y}{d x}) \\ T h e t w o c u r v e s a r e s a i d t o b e o r t h o g o n a l i f t h e a n g l e b e t w e e n t h e tangents a t t h e point o f \\ intersection i s 9 0^{0} . \\ ∴ m_{1} \times m_{2} = - 1 \\ \Rightarrow - \frac{2 x}{a} \times 3 x^{2} = - 1 \Rightarrow - \frac{6 x^{3}}{a} = - 1 \Rightarrow 6 x^{3} = a \\ (1, 1) i s t h e point o f intersection o f t w o c u r v e s . \\ ∴ 6 {(1)}^{3} = a \Rightarrow a = 6 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D) 6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

If y = x? – 10 and x changes from 2 to 1.99, what is the change in *y

(A) 0.32

(B) 0.032

(C) 5.68

(D) 5.968

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t y = x^{4} - 10 \\ \frac{d y}{d x} = 4 x^{3} \\ Δ x = 2.00 - 1.99 = 0.01 \\ ∴ Δ y = \frac{d y}{d x} . Δ x = 4 x^{3} . Δ x \\ = 4 \times {(2)}^{3} \times 0.01 = 32 \times 0.01 = 0.32 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The equation of the tangent to the curve y(1 + x²) = 2 – x, where it crosses the x-axis, is:

(A) x + 5y = 2

(B) x – 5y = 2

(C) 5x – y = 2

(D) 5x + y = 2

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y (1 + x^{2}) = 2 - x \dots (i) \\ I f i t c u t s x - a x i s, t h e n y - c o o r d i n a t e i s 0 . \\ ∴ 0 (1 + x^{2}) = 2 - x \Rightarrow x = 2 \\ P u t x = 2 i n e q u a t i o n (i) \\ y (1 + 4) = 2 - 2 \Rightarrow y (5) = 0 \Rightarrow y = 0 \\ Point o f c o n t a c t = (2, 0) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ y \times 2 x + (1 + x^{2}) \frac{d y}{d x} = - 1 \\ \Rightarrow 2 x y + (1 + x^{2}) \frac{d y}{d x} = - 1 \Rightarrow (1 + x^{2}) \frac{d y}{d x} = - 1 - 2 x y \\ ∴ \frac{d y}{d x} = \frac{- 1 - 2 x y}{(1 + x^{2})} \Rightarrow {\frac{d y}{d x}}_{(2, 0)} = \frac{- 1}{(1 + 4)} = \frac{- 1}{5} \\ E q u a t i o n o f tangent i s y - 0 = \frac{- 1}{5} (x - 2) \\ \Rightarrow 5 y = - x + 2 \Rightarrow x + 5 y = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The points at which the tangents to the curve y = x³ – 12x + 18 are parallel to the x-axis are:

(A) (2, –2), (–2, –34)

(B) (2, 34), (–2, 0)

(C) (0, 34), (–2, 0)

(D) (2, 2), (–2, 34)

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t y = x^{3} - 12 x + 18 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e h a v e \\ \Rightarrow \frac{d y}{d x} = 3 x^{2} - 12 \\ Since t h e tangents a r e p a r a l l e l t o x - a x i s, t h e n \frac{d y}{d x} = 0 \\ ∴ 3 x^{2} - 12 = 0 \Rightarrow x = \pm 2 \\ ∴ y_{x = 2} = {(2)}^{3} - 12 (2) + 18 = 8 - 24 + 18 = 2 \\ y_{x = - 2} = {(- 2)}^{3} - 12 (- 2) + 18 = - 8 + 24 + 18 = 34 \\ ∴ Points a r e (2, 2) a n d (- 2, 34) \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The tangent to the curve y = e²? at the point (0,1) meets the x-axis at:

(A) (0, 1)

(B) (–1/2, 0)

(C) (2, 0)

(D) (0, 2)

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e c u r v e i s y = e^{2 x} \\ S l o p e o f t h e tangent \frac{d y}{d x} = 2 e^{2 x} \Rightarrow {\frac{d y}{d x}}_{(0, 1)} = 2 e^{0} = 2 \\ ∴ E q u a t i o n o f tangent t o t h e c u r v e a t (0, 1) i s \\ y - 1 = 2 (x - 0) \\ \Rightarrow y - 1 = 2 x \Rightarrow y - 2 x = 1 \\ Since t h e tangent m e e t x - a x i s, w h e r e y = 0 \\ ∴ 0 - 2 x = 1 \Rightarrow x = - \frac{1}{2} \\ So, the Point i s (- \frac{1}{2}, 0) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, –1) is:

(A) 22/7

(B) 6/7

(C) -6/7

(D) –6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n c u r v e i s x = t^{2} + 3 t - 8 a n d y = 2 t^{2} - 2 t - 5 \\ D i f f e r e n t i a t i n g b o t h e q u a t i o n s w . r . t, t \\ \frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} \\ N o w (2, - 1) l i e s o n t h e c u r v e \\ ∴ 2 = t^{2} + 3 t - 8 \Rightarrow t^{2} + 3 t - 10 = 0 \\ \Rightarrow t^{2} + 5 t - 2 t - 10 = 0 \Rightarrow t (t + 5) - 2 (t + 5) = 0 \\ \Rightarrow (t + 5) (t - 2) = 0 \\ ∴ t = 2, t = - 5 a n d - 1 = 2 t^{2} - 2 t - 5 \\ \Rightarrow 2 t^{2} - 2 t - 4 = 0 \\ \Rightarrow t^{2} - t - 2 = 0 \Rightarrow t^{2} - 2 t + t - 2 = 0 \\ \Rightarrow t (t - 2) + 1 (t - 2) = 0 \Rightarrow (t + 1) (t - 2) = 0 \\ \Rightarrow t = - 1 a n d t = 2 \\ S o t = 2 i s c o m m o n v a l u e \\ ∴ S l o p e = {\frac{d y}{d x}}_{x = 2} = \frac{4 \times 2 - 2}{2 \times 2 + 3} = \frac{6}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of:

(A) π/4

(B) π/3

(C) π/2

(D) π/6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

The interval on which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing is:

(A) [–1, ∞)

(B) [–2, –1]

(C) (–∞, –2]

(D) [–1, 1]

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n f u n c t i o n i s f (x) = 2 x^{3} + 9 x^{2} + 12 x - 1 \\ f^{'} (x) = 6 x^{2} + 18 x + 12 \\ F o r increasing a n d decreasing f^{'} (x) = 0 \\ ∴ 6 x^{2} + 18 x + 12 = 0 \\ \Rightarrow x^{2} + 3 x + 2 = 0 \Rightarrow x^{2} + 2 x + x + 2 = 0 \\ \Rightarrow x (x + 2) + 1 (x + 2) = 0 \Rightarrow (x + 2) (x + 1) = 0 \\ \Rightarrow x = - 2, x = - 1 \\ T h e p o s s i b l e intervals a r e (- \infty, - 2), (- 2, - 1), (- 1, \infty) \\ N o w f^{'} (x) = (x + 2) (x + 1) \\ \Rightarrow f^{'} {(x)}_{(- \infty, - 2)} = (-) (-) = (+) increasing \\ \Rightarrow f^{'} {(x)}_{(- 2, - 1)} = (+) (-) = (-) decreasing \\ \Rightarrow f^{'} {(x)}_{(- 1, - \infty)} = (+) (+) = (+) increasing \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let the function f: R → R be defined by f(x) = 2x + cosx, then f:

(A) Has a minimum at x = π

(B) Has a maximum at x = 0

(C) Is a decreasing function

(D) Is an increasing function

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t f (x) = 2 x + c o s x \\ f^{'} (x) = 2 - s i n x \\ Since f^{'} (x) > 0 \forall x \\ S o f (x) i s a n increasing f u n c t i o n . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The function y = x (x – 3) ² decreases for the values of x given by:

(A) 1 < x < 3

(B) x < 0

(C) x > 0

(D) 0 < x < 3/2

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} T h e g i v e n f u n c t i o n i s y = x {(x - 3)}^{2} \\ \frac{d y}{d x} = x . 2 (x - 3) + {(x - 3)}^{2} . 1 \\ \Rightarrow \frac{d y}{d x} = 2 x (x - 3) + {(x - 3)}^{2} \\ F o r increasing a n d decreasing \frac{d y}{d x} = 0 \\ ∴ 2 x (x - 3) + {(x - 3)}^{2} = 0 \\ \Rightarrow (x - 3) (2 x + x - 3) = 0 \Rightarrow (x - 3) (3 x - 3) = 0 \\ \Rightarrow 3 (x - 3) (x - 1) = 0 \\ ∴ x = 1, x = 3 \\ T h e p o s s i b l e intervals a r e (- \infty, 1), (1, 3), (3, \infty) \\ N o w \frac{d y}{d x} = (x - 3) (x - 1) \\ \Rightarrow F o r (- \infty, 1) = (-) (-) = (+) increasing \\ \Rightarrow F o r (1, 3) = (-) (+) = (-) decreasing \\ \Rightarrow F o r (3, \infty) = (+) (+) = (+) increasing \\ So the function decreas e s i n (1, 3) o r 1 < x < 3 \\ H e n c e, &thin \end{array}$

The function f(x) = 4 sin³x – 6 sin²x + 12 sinx + 100 is strictly decreasing in:

(A) Increasing in (π, 3π/2)

(B) Decreasing in ( π/2,π)

(C) Decreasing in (-π/2,π/2 )

(D) Decreasing in (0, π/2)

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n f u n c t i o n i s f (x) = 4 s i n^{3} x - 6 s i n^{2} x + 12 s i n x + 100 \\ f^{'} (x) = 12 s i n^{2} x . c o s x - 12 s i n x c o s x + 12 c o s x \\ = 12 c o s x [s i n^{2} x - s i n x + 1] \\ = 12 c o s x [s i n^{2} x + (1 - s i n x)] \\ ? 1 - s i n x \geq 0 a n d s i n^{2} x \geq 0 \\ ∴ s i n^{2} x + 1 - s i n x \geq 0 (w h e n c o s x > 0) \\ H e n c e, f^{'} (x) > 0 w h e n c o s x > 0 i . e ., x \in (- \frac{π}{2}, \frac{π}{2}) \\ S o, f (x) i s increasing w h e r e x \in (- \frac{π}{2}, \frac{π}{2}) f^{'} (x) < 0 \\ w h e n c o s x < 0 i . e ., x \in (\frac{π}{2}, \frac{3 π}{2}) \\ H e n c e, f (x) i s decreasing w h e n x \in (\frac{π}{2}, \frac{3 π}{2}) \\ A s (\frac{π}{2}, π) \in (\frac{π}{2}, \frac{3 π}{2}) \\ S o, f (x) i s decreasing i n (\frac{π}{2}, π) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Which of the following functions is decreasing in (0,) π/2 )?

(A) sin2x

(B) tanx

(C) cosx

(D) cos3x

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = c o s x \\ S o, f^{'} (x) = - s i n x \Rightarrow f^{'} (x) < 0 i n (0, \frac{π}{2}) \\ S o, f (x) = c o s x i s decreasing i n (0, \frac{π}{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The function f(x) = tanx – x:

(A) Always increases

(B) Always decreases

(C) Never increases

(D) Sometimes increases and sometimes decreases

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = t a n x - x \\ S o, f^{'} (x) = s e c^{2} x - 1 \Rightarrow f^{'} (x) > 0 \forall \in R \\ S o, f (x) i s a l w a y s increasing \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If x is real, the minimum value of x² – 8x + 17 is:

(A) –1

(B) 0

(C) 1

(D) 2

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = x^{2} - 8 x + 17 \\ S o, f^{'} (x) = 2 x - 8 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 2 x - 8 = 0 \Rightarrow x = 4 \\ S o, x = 4 i s t h e point o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ f^{''} (x) = 2 > 0 minima a t x = 4 \\ ∴ f {(x)}_{x = 4} = {(4)}^{2} - 8 (4) + 17 \\ = 16 - 32 + 17 = 33 - 32 = 1 \\ S o t h e minimum v a l u e o f t h e f u n c t i o n i s 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The smallest value of the polynomial x³ – 18x² + 96x in [0, 9] is:

(A) 126

(B) 0

(C) 135

(D) 160

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = x^{3} - 18 x^{2} + 96 x \\ S o, f^{'} (x) = 3 x^{2} - 36 x + 96 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 3 x^{2} - 36 x + 96 = 0 \\ \Rightarrow x^{2} - 12 x + 32 = 0 \\ \Rightarrow x^{2} - 8 x - 4 x + 32 = 0 \Rightarrow x (x - 8) - 4 (x - 8) = 0 \\ \Rightarrow (x - 8) (x - 4) = 0 \\ ∴ x = 8, 4 \in [0, 9] \\ S o, x = 4, 8 a r e t h e points o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ Now we will calculate the absolute maxima o r absolute minima at x=0,4,8,9 \\ ∴ f (x) = x^{3} - 18 x^{2} + 96 x \\ f {(x)}_{x = 0} = 0 - 0 + 0 = 0 \\ f {(x)}_{x = 4} = {(4)}^{3} - 18 {(4)}^{2} + 96 (4) \\ = 64 - 288 + 384 = 448 - 288 = 160 \\ f {(x)}_{x = 8} = {(8)}^{3} - 18 {(8)}^{2} + 96 (8) \\ = 512 - 1152 + 768 = 1280 - 1152 = 128 \\ f {(x)}_{x = 9} = {(9)}^{3} - 18 {(9)}^{2} + 96 (9) \\ = 729 - 1458 + 864 = 1593 - 1458 = 135 \\ S o t h e absolute minimum v a l u e o f &th \end{array}$

The function f(x) = 2x³ – 3x² – 12x + 4 has:

(A) Two points of local maximum

(B) Two points of local minimum

(C) One maximum and one minimum

(D) No maximum or minimum

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4 \\ S o, f^{'} (x) = 6 x^{2} - 6 x - 12 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 6 x^{2} - 6 x - 12 = 0 \\ \Rightarrow x^{2} - x - 2 = 0 \\ \Rightarrow x^{2} - 2 x + x - 2 = 0 \Rightarrow x (x - 2) + 1 (x - 2) = 0 \\ \Rightarrow (x + 1) (x - 2) = 0 \\ ∴ x = - 1, 2 \\ S o, x = - 1, 2 a r e t h e points o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ Now f^{''} (x) = 12 x - 6 \\ f^{''} {(x)}_{x = - 1} = 12 (- 1) - 6 = - 12 - 6 = - 18 < 0, maxima \\ f^{''} {(x)}_{x = 2} = 12 (2) - 6 = 24 - 6 = 18 > 0, minima \\ S o t h e function is maximum at x = - 1 and minimum a t x = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The maximum value of sinx –cosx is:

(A) 1/4
(B) 1/2
(C) √2
(D) 2√2

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

At $x = \frac{5 π}{6},$ the function f(x) = 2sin³x + 3cos³x is:

(A) Maximum

(B) Minimum

(C) Zero

(D) Neither maximum nor minimum

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = 2 s i n 3 x + 3 c o s 3 x \\ S o, f^{'} (x) = 2 c o s 3 x . 3 - 3 s i n 3 x . 3 = 6 c o s 3 x - 9 s i n 3 x \\ f^{''} (x) = - 6 s i n 3 x . 3 - 9 c o s 3 x . 3 \\ = - 18 s i n 3 x - 27 c o s 3 x \\ Now f^{''} (\frac{5 π}{6}) = - 18 s i n 3 (\frac{5 π}{6}) - 27 c o s 3 (\frac{5 π}{6}) \\ = - 18 s i n (\frac{5 π}{2}) - 27 c o s (\frac{5 π}{2}) \\ = - 18 s i n (2 π + \frac{π}{2}) - 27 c o s (2 π + \frac{π}{2}) \\ = - 18 s i n \frac{π}{2} - 27 c o s \frac{π}{2} = - 18.1 - 27.0 \\ = - 18 < 0 maxima \\ M aximum v a l u e o f f (x) a t x = \frac{5 π}{6} \\ f (\frac{5 π}{6}) = 2 s i n 3 (\frac{5 π}{6}) + 3 c o s 3 (\frac{5 π}{6}) = 2 s i n (\frac{5 π}{2}) + 3 c o s (\frac{5 π}{2}) \\ = 2 s i n (2 π + \frac{π}{2}) + 3 c o s (2 π + \frac{π}{2}) = 2 s i n \frac{π}{2} + 3 c o s \frac{π}{2} = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The maximum slope of the curve y = –x³ + 3x² + 9x – 27 is:

(A) 0

(B) 12

(C) 16

(D) 32

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = - x^{3} + 3 x^{2} + 9 x - 27 \\ S o, \frac{d y}{d x} = - 3 x^{2} + 6 x + 9 \\ ∴ S l o p e o f t h e g i v e n c u r v e, \\ m = - 3 x^{2} + 6 x + 9 \\ \frac{d m}{d x} = - 6 x + 6 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, \frac{d m}{d x} = 0 \\ ∴ - 6 x + 6 = 0 \Rightarrow x = 1 \\ Now \frac{d^{2} m}{d x^{2}} = - 6 < 0 maxima \\ ∴ M aximum v a l u e o f t h e s l o p e at x =1 is \\ m_{x = 1} = - 3 {(1)}^{2} + 6 (1) + 9 = - 3 + 6 + 9 = 12 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The function f(x) = xˣ has a stationary point at:

(A) x = e

(B) x = 1/e

(C) x = 1

(D) x =√e

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} W e h a v e f (x) = x^{x} \\ T a k i n g l o g o f b o t h s i d e s, w e h a v e \\ l o g f (x) = x l o g x \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{1}{f (x)} f^{'} (x) = x . \frac{1}{x} + l o g x . 1 \\ \Rightarrow f^{'} (x) = f (x) [1 + l o g x] = x^{x} [1 + l o g x] \\ T o f i n d s t a t i o n a r y point, f^{'} (x) = 0 \\ ∴ x^{x} [1 + l o g x] = 0 \\ x^{x} \neq 0 ∴ 1 + l o g x = 0 \\ \Rightarrow l o g x = - 1 \Rightarrow x = e^{- 1} \Rightarrow x = \frac{1}{e} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The maximum value of (1/x)^x is:

(A) e

(B) $e^{e}$

(C) $\frac{1}{e^{e}}$

(D) $e^{\frac{1}{e}}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = {(\frac{1}{x})}^{x} \\ T a k i n g l o g o f b o t h s i d e s, w e h a v e \\ l o g f (x) = x l o g \frac{1}{x} \\ l o g f (x) = x l o g x^{- 1} \Rightarrow l o g [f (x)] = - [x l o g x] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{1}{f (x)} f^{'} (x) = - [x . \frac{1}{x} + l o g x . 1] \\ \Rightarrow f^{'} (x) = - f (x) [1 + l o g x] = - {(\frac{1}{x})}^{x} [1 + l o g x] \\ F o r l o c a l maxima a n d l o c a l minima f^{'} (x) = 0 \\ - {(\frac{1}{x})}^{x} [1 + l o g x] = 0 \Rightarrow {(\frac{1}{x})}^{x} [1 + l o g x] = 0 \\ {(\frac{1}{x})}^{x} \neq 0 ∴ 1 + l o g x = 0 \\ \Rightarrow l o g x = - 1 \Rightarrow x = e^{- 1} \Rightarrow x = \frac{1}{e} \\ S o, x = \frac{1}{e} i s t h e s t a t i o n a r y point. \\ N o w, f^{'} (x) = - {(\frac{1}{x})}^{x} [1 + l o g x] \\ f^{''} (x) = - [{(\frac{1}{x})}^{x} (\frac{1}{x}) + (1 + l o g x) . \frac{d}{d x} {(x)}^{x}] \\ f^{''} (x) = - [{(e)}^{1 / e} (e) + (1 + l o g \frac{1}{e}) \frac{d}{d x} {(\frac{1}{e})}^{1 / e}] \\ x = \frac{1}{e} = - e^{\frac{1}{e} + 1} < 0 maxima \\ ∴ Maximum value of the function at x = \frac{1}{e} i s \\ f (\frac{1}{e}) = {(\frac{1}{1 / e})}^{1 / e} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The curves y = 4x² + 2x – 8 and y = x³ – x + 13 touch each other at the point ___.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e y = 4 x^{2} + 2 x - 8 \dots (i) \\ a n d y = x^{3} - x + 13 \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ \frac{d y}{d x} = 8 x + 2 \Rightarrow m_{1} = 8 x + 2 [m_{1} i s s l o p e o f c u r v e (i)] \\ D i f f e r e n t i a t i n g e q . (i i) w . r . t . x, w e g e t \\ \frac{d y}{d x} = 3 x^{2} - 1 \Rightarrow m_{2} = 3 x^{2} - 1 [m_{2} i s s l o p e o f c u r v e (i i)] \\ I f t h e t w o c u r v e s t o u c h e a c h o t h e r, t h e n m_{1} = m_{2} \\ ∴ 8 x + 2 = 3 x^{2} - 1 \\ \Rightarrow 3 x^{2} - 8 x - 3 = 0 \Rightarrow 3 x^{2} - 9 x + x - 3 = 0 \\ \Rightarrow 3 x (x - 3) + 1 (x - 3) = 0 \Rightarrow (x - 3) (3 x + 1) = 0 \\ ∴ x = 3, \frac{- 1}{3} \\ P u t t i n g x = 3 i n e q (i), w e g e t \\ y = 4 {(3)}^{2} + 2 (3) - 8 = 36 + 6 - 8 = 34 \\ S o, t h e r e q u i r e d point i s (3, 34) \\ N o w f o r x = \frac{- 1}{3} \\ y = 4 {(\frac{- 1}{3})}^{2} + 2 (\frac{- 1}{3}) - 8 = 4 \times \frac{1}{9} - \frac{2}{3} - 8 \\ = \frac{4}{9} - \frac{2}{3} - 8 = \frac{4 - 6 - 72}{9} = \frac{- 74}{9} \\ ∴ O t h e r r e q u i r e d point i s (\frac{- 1}{3}, \frac{- 74}{9}) . \\ H e n c e, t h e r e \end{array}$

The equation of the normal to the curve y = tanx at (0, 0) is ____.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

The values of a for which the function f(x) = sinx – ax + b increases on R are __.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} W e h a v e f (x) = s i n x - a x + b \Rightarrow f^{'} (x) = c o s x - a \\ F o r increasing t h e f u n c t i o n f^{'} (x) > 0 \\ ∴ c o s x - a > 0 \\ Since c o s x \in [- 1, 1] \\ ∴ a < - 1 \Rightarrow a \in (- \infty, - 1) \\ H e n c e, t h e v a l u e o f a \in (- \infty, - 1) . \end{array}$

The function f(x) = $\frac{2 x^{2} + 1}{x^{4}}$ , x > 0 decreases in the interval ___.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = \frac{2 x^{2} - 1}{x^{4}} \\ \Rightarrow f^{'} (x) = \frac{x^{4} (4 x) - (2 x^{2} - 1) . 4 x^{3}}{x^{8}} \\ \Rightarrow f^{'} (x) = \frac{4 x^{5} - (2 x^{2} - 1) . 4 x^{3}}{x^{8}} = \frac{4 x^{3} [x^{2} - 2 x^{2} + 1]}{x^{8}} = \frac{4 [- x^{2} + 1]}{x^{5}} \\ F o r d e creasing t h e f u n c t i o n f^{'} (x) < 0 \\ ∴ \frac{4 (- x^{2} + 1)}{x^{5}} < 0 \Rightarrow - x^{2} + 1 < 0 \Rightarrow x^{2} > 1 \\ ∴ x > \pm 1 \Rightarrow x \in (1, \infty) \\ H e n c e, t h e r e q u i r e d i n t e r v a l i s (1, \infty) . \end{array}$

The least value of the function f(x) = ax + ( $\frac{b}{x}$ ), where a > 0, b > 0, x > 0, is ___.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

26th June 2022 (First Shift)

Commonly asked questions

Let f(x) = $\frac{x - 1}{x + 1}, x \in R - {0, - 1, 1} .$ If $f^{n + 1} (x) = f (f^{n} (x)) f o r a l l n \in N,$ then f⁶ (6) + f⁷ (7) is equal a to

$f (x) = \frac{x - 1}{x + 1} \Rightarrow f (f (x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = - \frac{1}{x}$

$f^{3} (x) = - \frac{x + 1}{x - 1} \Rightarrow f^{4} (x) = \frac{\frac{x - 1}{x + 1} + 1}{\frac{x - 1}{x + 1} - 1} = x$

$S o, f^{6} (6) + f^{7} (7) = f^{2} (6) + f^{3} (7)$

= $- \frac{1}{6} - \frac{7 + 1}{7 - 1} = - \frac{9}{6} = - \frac{3}{2}$

Let A be a 3 × 3 invertible matrix. If $| a d j (24 A) | = | a d j (3 a d j (2 A)) |, t h e n {| A |}^{2}$ is equal to

$| a d j (24 A) | = | a d j (3 a d j (2 A)) |$

$\Rightarrow {| 24 A |}^{2} = {| 3 a d j (2 A) |}^{2}$

$2 4^{6} {| A |}^{2} = 3^{6} . {(2^{3})}^{4} {| A |}^{4}$

${| A |}^{2} = \frac{2 4^{6}}{3^{6} . 2^{12}} = \frac{2^{18} . 3^{6}}{3^{6} . 2^{12}} = 2^{6}$

The ordered pair (a, b), for which the system of linear equations

3x – 2y + z = b

5x – 8y + 9z = 3

2x + y + az = -1

has no solution, is

System of equation can be written as

$(\begin{matrix} 3 & - 2 & 1 \\ 5 & - 8 & 9 \\ 2 & 1 & a \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} b \\ 3 \\ - 1 \end{array})$

$(\begin{matrix} 3 & - 2 & 1 \\ 15 & - 24 & 27 \\ 6 & 3 & 3 a \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} b \\ 9 \\ - 3 \end{array})$

$R_{3} - 2 R_{1}, R_{2} - 5 R_{1}$

for no solution

3a + 9 = 0 but $\frac{3}{2} - \frac{9 b}{2} \neq 0$

$\Rightarrow a = - 3 \Rightarrow b \neq \frac{1}{3}$

The remainder when (2021)²⁰²³ is divided by 7 is

$2021 \equiv - 2$ mod (7)

$\Rightarrow {(2021)}^{2023} \equiv {(- 2)}^{2023} m o d (7)$ …… (i)

Now, ${(- 2)}^{3} \equiv - 1 m o d (7)$

$\Rightarrow {(- 2)}^{2023} \equiv (- 2) m o d (7) \equiv 5 m o d (7)$ ……. (ii)

(i) & (ii)

$\Rightarrow {(2021)}^{2023} \equiv 5 m o d (7)$

$∴$ Remainder = 5

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$ is equal to

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$

let $c o s^{- 1} x = \frac{π}{4} + θ$

= $\underset{θ \to \infty}{l i m} \frac{2 s i n θ}{- 2 t a n θ} (1 - t a n θ) = -$ $1$

Let f, g : R → R be two real valued function defined as f(x) = ${\begin{matrix} - | x + 3 |, & x < 0 \\ e^{x}, & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} x^{2} + k_{1} x, & x < 0 \\ 4 x + k_{2}, & x \geq 0 \end{matrix}$ , where k₁ and k₂ are real constants. If (gof) is differentiable at x = 0, then (gof)(-4) is equal to

GOF is differentiable at x = 0

So R.H.D = L.H.D.

$\frac{d}{d x} (4 e^{x} + k_{2}) = \frac{d}{d x} ({(- | x + 3 |)}^{2} - k_{1} | x + 3 |)$

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

= 2 (2e⁴ – 1)

The sum of the absolute minimum and the absolute maximum values of the friction f(x) = $| 3 x - x^{2} + 2 | - x$ in the interval [1, 2] is

$f (x) = | x^{2} - 3 x - 2 | - x$

$= | (x - \frac{3 - \sqrt[]{17}}{2}) (x - \frac{3 + \sqrt[]{17}}{2}) | - x$

$\Rightarrow f (x) = [\begin{matrix} x^{2} - 4 x - 2 - 1 \leq x \leq \frac{3 - \sqrt[]{17}}{2} & - x^{2} + 2 x + 2 \frac{3 - \sqrt[]{17}}{2} < x \leq 2 \end{matrix}]$

absolute minimum $f (\frac{3 - \sqrt[]{17}}{2}) = \frac{- 3 + \sqrt[]{17}}{2}$

absolute maximum = 3

$∴ s u m 3 + \frac{- 3 + \sqrt[]{17}}{2} = \frac{3 + \sqrt[]{17}}{2}$

Let S be the set of all the natural numbers, for which the line $\frac{x}{a} + \frac{y}{b} = 2$ is a tangent to the curve ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$ at the point (a, b), ab Then

${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$

$\Rightarrow \frac{n}{a} {(\frac{x}{a})}^{n - 1} + \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{b}{a} {(\frac{b x}{a y})}^{n - 1}$

$\frac{d y}{d x_{(a, b)}} = - \frac{b}{a}$

So line always touches the given curve.

The area bounded by the curve y = $| x^{2} - 9 |$ and the line y = 3 is :

$| x^{2} - 9 | = 3$

$\Rightarrow x = \pm 2 \sqrt[]{3}, \pm \sqrt[]{6}$

Required area = A

$\frac{A}{2} = \int_{0}^{\sqrt[]{6}} (9 - x^{2} - 3) d x + \int_{0}^{3} (\sqrt[]{9 + y} - \sqrt[]{9 - y}) d y$

$A = 16 \sqrt[]{6} + 32 \sqrt[]{3} - 72 = 8 [2 \sqrt[]{6} + 4 \sqrt[]{3} - 9]$

Note : No option in the question paper is correct.

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $Δ P Q R$ is

RM = $| \frac{3 + 7 - 5}{\sqrt[]{2}} | = \frac{5}{\sqrt[]{2}}$

$l s i n 60 ° = \frac{5}{\sqrt[]{2}} \Rightarrow l = 5 \sqrt[]{\frac{2}{3}}$

$∴ A r e a o f Δ P Q R = \frac{3}{4} l^{2} =$ =25/2√3

Let C be a circle passing through the points A(2, 1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to

Equation of perpendicular bisector of AB is

$y - \frac{3}{2} = - \frac{1}{5} (x - \frac{5}{2}) \Rightarrow x + 5 y = 10$

Solving it with equation of given circle,

${(x - 5)}^{2} + {(\frac{10 - x}{5} - 1)}^{2} = \frac{13}{2}$

$\Rightarrow x - 5 = \pm \frac{5}{2} \Rightarrow x = \frac{5}{2} o r \frac{15}{2}$

But

$x \neq \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$(\frac{15}{2}, \frac{1}{2})$

Now,

$r^{2} = {(\frac{15}{2} - 2)}^{2} + {(\frac{1}{2} + 1)}^{2} = \frac{65}{2}$

Let the normal at the point P on the parabola y² = 6x pass through the point (5, 8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is

Let P (at², 2 at) where

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

T : yt = x + at² so point Q is

$(- a, a t - \frac{a}{t})$

N : y = -tx + 2at + at³ passes through (5, -8)

$- 8 = - 5 t + 3 t + \frac{3}{2} t^{3}$

⇒ $3 t^{3} - 4 t + 16 = 0$

⇒ t = -2

So ordinate of point Q is $- \frac{9}{4}$

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

$? l_{1} a n d l_{2}$ are perpendicular, so

$3 \times 1 + (- 2) (\frac{α}{2}) + 0 \times 2 = 0$

a = 3

Now angle between $l_{2} a n d l_{3}$ ,

$c o s θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt[]{1 + \frac{α^{2}}{4} + 4} . \sqrt[]{9 + 4 + 16}}$

$\Rightarrow c o s θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = c o s^{- 1} (\frac{4}{29}) = s e c^{- 1} (\frac{29}{4})$

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane in B(a, b, c), then

Consider the equation of plane,

$P : (2 x + 3 y + z + 20) + λ (x - 3 y + 5 z - 8) = 0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So, $4 + 2 λ + 9 - 9 λ + 1 + 5 λ = 0$

0 $λ = 7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$(2, - \frac{1}{2}, 2)$

In plane P is (a, b, c) then

$\frac{a - 2}{1} = \frac{b + \frac{1}{2}}{- 2} = \frac{c - 2}{4}$

and $(\frac{a + 2}{2}) - 2 (\frac{b - \frac{1}{2}}{2}) + 4 (\frac{c + 2}{2}) = 4$

clearly

$a = \frac{4}{3}, b = \frac{5}{6} a n d c = - \frac{2}{3}$

So, a : b : c = 8 : 5 : 4

If $\vec{a} . \vec{b} = 1, \vec{b} . \vec{c} = 2 \vec{c} . \vec{a} = 3,$ then the value of $[\vec{a} \times (\vec{b} \times \vec{c}), \vec{b} \times (\vec{c} \times \vec{a}), \vec{c} \times (\vec{b} \times \vec{a})] i s :$

$\vec{a} \times (\vec{b} \times \vec{c}) = 3 \vec{b} - \vec{c} = \vec{u}$

$\vec{b} \times (\vec{c} \times \vec{a}) = \vec{c} - 2 \vec{a} = \vec{v}$

$\vec{c} \times (\vec{b} \times \vec{a}) = 3 \vec{b} - 2 \vec{a} = \vec{w}$

$∴ \vec{u} + \vec{v} = \vec{w}$

so vectors

$\vec{u}, \vec{v} a n d \vec{w}$

are coplanar, hence their Scalar triple product will be zero.

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)

6x = 5 = 0 $x = \frac{5}{6}$

P (atmost 2H)

$P (O H, 5 T) + P (1 H, 4 T) + P (2 H, 3 T)$

$= \frac{1}{6^{5}} (1 + 25 + 250) = \frac{276}{6^{5}} = \frac{4^{6}}{6^{4}}$

The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to

$\bar{x} = 6 = \frac{a + b + 8 + 5 + 10}{5} \Rightarrow a + b = 7$ …… (i)

and

$σ^{2} = \frac{a^{2} + b^{2} + 8^{2} + 5^{2} + 1 0^{2}}{5} - 6^{2} = 6.8$

a² + b² = 25 ……. (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M = \frac{1}{5} (3 + 2 + 2 + 1 + 4) = \frac{12}{5}$

25M = 60

Let f(x) = $2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1] . I f [a, b]$ is the range of the function f, then 4a – b is equal to

$f (x) = 2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10 \forall x \in [- 1, 1]$

$\Rightarrow f' (x) = - \frac{2}{\sqrt[]{1 - x^{2}}} - \frac{4}{1 + x^{2}} - 6 x - 2 < 0 \forall x \in [- 1, 1]$

So, f (x) is decreasing function and range of f (x) is

$[f (1), f (- 1)],$ which is $[π + 5, 5 π + 9]$

Now 4a – b = 4 ( + 5) (5 + 9) = 11 - π

Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to

Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

Case – II If $Δ \equiv \nabla \equiv \lor$

$(p \lor q) ? \to ((p \lor q) \lor r) \equiv$ tautology

then $(p \lor q) \lor r \equiv (p Δ r) \lor q$

Case – III If $Δ \equiv v, \nabla \equiv \land$

then $(p \land q) \to {(p \lor q) \land r}$

Not a tautology

Case – IV If $Δ \equiv \land, \nabla \equiv \lor$

$(p \land q) \to {(p \land q) \lor r}$

Not a tautology

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is

$x^{4} ? 3 x^{3} ? 2 x^{2} + 3 x + 1 = 0$

$? x = \pm 1$

and let, are roots of x2 – 3x – 1 = 0

$? ? + ? = 3 ? ? = ? 1$

$? 1^{3} + {(? 1)}^{3} + ?^{3} + ?^{3}$

= 27 + 3 (3) = 36

There are ten boys B₁, B₂,….., B₁₀ and five girls G₁, G₂,….., G₅ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B₁ and B₂ together should not be the members of a group, is………….

Boys (10) Girls (5)

(3) (3)

B₁ & B₂ should not be selected together

Total number of ways

= (56 + 56) × 10 = 1120

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence

$\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

Let f(x) = $m a x {| x + 1 |, | x + 2 |, . . . . . ., | x + 5 |} . T h e n \int_{- 6}^{0} f (x) d x$ is equal to

$∴ \int_{- 6}^{0} f (x) d x = 2 \times \frac{1}{2} (2 + 5) \times 3 = 21$

Let the solution curve y = y(x) of the differential equation (4 + x²)dy – 2x(x² + 3y + 4)dx = 0 pass through the origin. They y(2) is equal to

$(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$

$\Rightarrow \frac{d y}{d x} = (\frac{6 x}{x^{2} + 4}) y + 2 x$

$e^{- 3 l n (x^{2} + 4)} = \frac{1}{{(x^{2} + 4)}^{3}}$

$\frac{y}{{(x^{2} + 4)}^{3}} = \int \frac{2 x}{{(x^{2} + 4)}^{3}} d x + c$

$\Rightarrow y = - \frac{1}{2} (x^{2} + 4) + c {(x^{2} + 4)}^{3}$

When x = 0, y = 0 gives

$c = \frac{1}{32}$

So, for x = 2, y = 12

If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to

$(s i n 10 ° . s i n 50 ° . s i n 70 °) . (s i n 10 ° . s i n 20 ° . s i n 40 °)$

$= (\frac{1}{4} s i n 30 °) . [\frac{1}{2} s i n 10 ° (c o s 20 ° - c o s 60 °)]$

$= \frac{1}{32} [s i n 30 ° - s i n 10 ° - s i n 10 °]$

$\frac{1}{64} - \frac{1}{16} s i n 10 °$

Clearly $α = \frac{1}{64}$

Hence 16 + a^-1= 80

Let A = ${n \in N : H . C . F . (n, 45) = 1}$ and Let B = ${2 k : k \in {1, 2, . . . . . ., 100}} .$ Then the sum of all the elements of $A \cup B i s . . . . . . . . . .$

Sum of all elements of $A \cap B = 2$ [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$= 2 [\frac{100 \times 101}{2} - 3 (\frac{33 \times 34}{2}) - 5 (\frac{20 \times 21}{2}) + 15 (\frac{6 \times 7}{2})]$

= 10100 – 3366 – 2100 + 630

= 5264

The value of the integral $\frac{48}{π^{4}} \int_{0}^{π} (\frac{3 π x^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s^{2} x} d x$ is equal to

$l = \frac{48}{π^{4}} \int_{0}^{π} [{(\frac{π}{2} - x)}^{3} - \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{s i n x d x}{1 + c o s^{2} x}$

Using

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

we get

$l = \frac{48}{π^{4}} \int_{0}^{π} [- {(\frac{π}{2} - x)}^{3} + \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{s i n x d x}{1 + c o s^{2} x}$

Adding these two equations, we get

$\Rightarrow l = \frac{12}{π} {[- t a n^{- 1} (c o s x)]}_{0}^{π} = \frac{12}{π} . \frac{π}{2} = 6$

Let $A = \sum_{i = 1}^{10} \sum_{j = 1}^{10} m i n {i, j} a n d B = \sum_{i = j}^{10} \sum_{j = 1}^{10} m a x {i, j} .$ Then A + B is equal to

Each element of ordered pair (i, j) is either present in A or in B.

So, A + B = Sum of all elements of all ordered pairs {i, j} for $1 \leq i \leq 10$ and $1 \leq j \leq 10$

= 20 (1 + 2 + 3 + … + 10) = 1100

Let S = (0, 2) – ${\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}} .$ Let y = y(x), $x \in S,$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}, y (\frac{π}{4}) = \frac{1}{2} .$ If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $y = \sqrt[]{2} s i n x i s \frac{k π}{12},$ then k is equal to

$\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}$

$d y = \frac{s e c^{2} x d x}{{(1 + t a n x)}^{2}}$

$\Rightarrow y = - \frac{1}{1 + t a n x} + c$

When

$x = \frac{π}{4}, y = \frac{1}{2}$ gives c = 1

$x + \frac{π}{4} = \frac{5 π}{6} o r \frac{13 π}{6} \Rightarrow x = \frac{7 π}{12} o r \frac{23 π}{12}$

sum of all solutions =

$π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{42 π}{12}$

Hence k = 42

Maths NCERT Exemplar Solutions Class 12th Chapter Six Exam

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