Let A = { z ∈ C : 1 ≤ | z − ( 1 + i ) ≤ 2 } and B = { z ∈ A : | z − ( 1 − i ) | = 1 } . Then, B :

A = { z ∈ c : 1 ≤ | z − ( 1 + i ) | ≤ 2 } | z − ( 1 + i ) | ≥ 1 and | z − ( 1 + i ) | ≤ 2 and also | z − ( 1 − i ) | = 1 hence B has infinite set.

Let R be a relation defined on the set of natural numbers N as follows: R={(x,y):x∈N,y∈N,2x+y=41} . Find the domain and range of the relation R . Also, verify whether R is reflexive, symmetric, and transitive.

This is a Long Answer Type Question as classified in NCERT Exemplar Sol: Given?? that x∈ N, y∈N and 2x + y=41 ∴ Domain of R = {1, 2, 3, 4, 5, …, 20} and Range = {39, 37, 35, 33, 31, …, 1} Here, (3, 3) ∉ R as 2× 3 + 3 ≠ 41 So, R is not reflexive. R is not symmetric as (2, 37) ∈ R but (37, 2) ∉ R R is not transitive as (11, 19) ∈ R and (19, 3) ∈ R but (11, 3) ∉ R. Hence, R is neither reflexive, nor symmetric and nor transitive.

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

Q: The surface area of a balloon of spherical shape being inflated increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

Surface area, S = 4pr2 ? d s d t = 4 ? . 2 r d r d t = 8 ? d r d t = c o n s t a n t = k ( s a y ) ? d s d t = k ? s = k t + c ? 4 ? r 2 = k t + c Initially t = 0, r = 3 c = 36 p When t = 5, r = 7, k = 32p When t = 9, r = r, r = 9

Q: Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random are found to be 1 red and 1 black. If the probability that both balls come from Bag A is 6 1 1 , then n is equal to ________.

6 1 1 = Required probability After solving, we get n = 4

Q: The number of distinct real roots of the equation x5 (x3−x2−x+1) + x(3x3−4x2−2x+4)−1=0 is …………..

x8−x7−x6+x5+3x4−4x3−2x2+4x−1=0 ⇒x7 (x−1)−x5 (x−1)+3x3 (x−1)−x (x2−1)+2x (1−x)+ (x−1)=0 ⇒ (x−1) (x2−1) (x5+3x−1)=0∴x=±1 are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or 1. ∴ 3 real roots.

Q: Let f : R -> R be a continuous function such that f(3x) – f(x) =. If f(8) = 7, then f(14) is equal to:

f (3x)- f (x) = x Replace x→x3⇒f (x)−f (x3)=x3 Again replace x→x3⇒f (x3)−f (x32)−f (x32)=x32 ⇒f (3x)−f (0)=3x2putting x=83⇒f (8)−f (0)=4∴f (0)=3 Also putting x = 143 in f (3x) – 3 = 3x2⇒ F (14) – 3 = 7 f (14) = 10

Q: Let O be the origin and A be the point z1 = 1 + 2i. If B is the point z2, Re(z2) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

z2−0 (1+2i)−0=|OBOA|eiπ4⇒z2= (1+2i) (1+i)=−1+3i∴arg z2=π−tan−13 and |z2|=10 z1−2z2=3−4i∴arg (z1−2z2)=−tan−143⇒|z1−2z2|=|2+4i+1−3i|=10

Q: If the system of linear equations. 8x + y + 4z = -=2 x + y + z = 0 λx 3y = μ has infinitely many solutions, then the distance of the point (λ,μ−12) from the plane 8x + y + 4z + 2 = 0 is

Δ=|81411λ−30|=12−3λ So for λ = 4, it is having infinitely many solutions. Δx=|−214011μ−30| = 6 3μ=0⇒−6−3μ=0 For μ=−2 distance of (4, −2, −12) from 8x + y + 4z + 2= 0 |32−2−2+264+1+16|=103 units

Q: Let A be a 2 × 2 matrix with det(A) = 1 and det ((A+I)(Adj(A)+I))=4. Then the sum of the diagonal elements of A can be:

| (A + I) (adj A + I)| = 4 |A adj A + A + Adj A + I| = 4 | (A)I + A + adj A + I|= 4|A| = 1 |A + adj A| = 4 A= [abcd]⇒adj A= [a−b−cd]⇒| (a+d)00 (a+d)|=4⇒a+d=±2

Updated on Jul 17, 2025 12:12 IST

By Vishal Baghel, Executive Content Operations

Table of contents

Relations and Function Question and Answers
JEE Mains 2022
JEE Mains Solutions 2022,24th june , Maths, first shift

Relations and Function Question and Answers

Q.1. If $A = {1, 2, 3, 4}$ , define relations on $A$ which have properties of being:

(a) Reflexive, transitive but not symmetric

(b) Symmetric but neither reflexive nor transitive

(c) Reflexive, symmetric and transitive.

Sol:

$\begin{array}{l} G i v e n t h a t : A = {1, 2, 3} \\ a n d R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} \\ H e r e, 1 R 1, 2 R 2 a n d 3 R 3, S o, R i s r e f l e x i v e . \\ 1 R 2 b u t 2 R 1 o r 2 R 3 b u t 3 R 2, \\ S o, R i s n o t s y m m e t r i c . \\ 1 R 1 a n d 1 R 2 \Rightarrow 1 R 3, S o, R i s t r a n s i t i v e . \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

Q.2. Let $R$ be a relation defined on the set of natural numbers N as follows:

$R = {(x, y) : x \in N, y \in N, 2 x + y = 41}$ . Find the domain and range of the relation $R$ . Also, verify whether $R$ is reflexive, symmetric, and transitive.

Sol:

$\begin{array}{l} G i v e n t h a t x \in N, y \in N a n d 2 x + y = 41 \\ ∴ D o m a i n o f R = {1, 2, 3, 4, 5, \dots, 20} \\ a n d R a n g e = {39, 37, 35, 33, 31, \dots, 1} \\ H e r e, (3, 3) \notin R \\ a s 2 \times 3 + 3 \neq 41 \\ S o, R i s n o t r e f l e x i v e . \\ R i s n o t s y m m e t r i c a s (2, 37) \in R b u t (37, 2) \notin R \\ R i s n o t t r a n s i t i v e a s (11, 19) \in R a n d (19, 3) \in R \\ b u t (11, 3) \notin R . \\ H e n c e, R i s n e i t h e r r e f l e x i v e, n o r s y m m e t r i c a n d n o r t r a n s i t i v e . \end{array}$

Q.3. Given $A = {2, 3, 4}, B = {2, 5, 6, 7}$ , construct an example of each of the following:

(a) An injective mapping from $A$ to $B$

(b) A mapping from $A$ to $B$ which is not injective

(c) A mapping from $B$ to $A$ .

Sol:

$\begin{array}{l} H e r e, A = {2, 3, 4} a n d B = {2, 5, 6, 7} \\ (i) L e t f : A \to B b e t h e m a p p i n g f r o m A t o B \\ f = {(x, y) : y = x + 3} \\ ∴ f = {(2, 5), (3, 6), (4, 7)} w h i c h i s a n i n j e c t i v e m a p p i n g . \\ (i i) L e t g : A \to B b e t h e m a p p i n g f r o m A \to B s u c h t h a t \\ ∴ g = {(2, 5), (3, 5), (4, 2)} w h i c h i s n o t a n i n j e c t i v e m a p p i n g . \\ (i i i) L e t h : B \to A b e t h e m a p p i n g f r o m B t o A \\ h = {(y, x) : x = y - 2} \\ ∴ h = {(5, 3), (6, 4), (7, 3)} w h i c h i s t h e m a p p i n g f r o m B t o A . \end{array}$

Q.4. Give an example of a map:

(i) Which is one-one but not onto

(ii) Which is not one-one but onto

(iii) Which is neither one-one nor onto.

Sol: $\begin{array}{l} \end{array}$

Commonly asked questions

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.

$I. {(x, y) : xisaperson, y isthemonthofx}$ .

$II. {(a, b) : aisaperson, bisanancestorof a}$ .

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

Sol:

$\begin{array}{l} (i) G i v e n s e t o f o r d e r e d p a i r i s {(x, y) : x i s a p e r s o n, y i s t h e m o t h e r o f x} . \end{array}$ $\begin{array}{l} I t r e p r e s e n t s a f u n c t i o n . \end{array}$

$\begin{array}{l} H e r e, t h e i m a g e o f d i s t i n c t e l e m e n t s o f x u n d e r f a r e n o t d i s t i n c t, s o i t i s n o t a n i n j e c t i v e b u t i t i s a s u r j e c t i v e . \end{array}$

$\begin{array}{l} (i i) S e t o f o r d e r e d p a i r s = {(a, b) : a i s a p e r s o n, b i s a n a n c e s t o r o f a} \end{array}$

$\begin{array}{l} H e r e, e a c h e l e m e n t o f d o m a i n d o e s n o t h a v e a u n i q u e i m a g e . \end{array}$

$\begin{array}{l} S o, i t d o e s n o t r e p r e s e n t f u n c t i o n . \end{array}$

If $A = {1, 2, 3, 4}$ , define relations on $A$ which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric and transitive.

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

Sol:

$G i v e n t h a t :$
$\begin{array}{l} A = {1, 2, 3} \\ a n d R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} \\ H e r e, 1 R 1, 2 R 2 a n d 3 R 3, S o, R i s r e f l e x i v e . \\ 1 R 2 b u t 2 R 1 o r 2 R 3 b u t 3 R 2, \\ S o, R i s n o t s y m m e t r i c . \\ 1 R 1 a n d 1 R 2 \Rightarrow 1 R 3, S o, R i s t r a n s i t i v e . \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

Let $R$ be a relation defined on the set of natural numbers N as follows:

$R = {(x, y) : x \in N, y \in N, 2 x + y = 41}$ . Find the domain and range of the relation $R$ . Also, verify whether $R$ is reflexive, symmetric, and transitive.

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

Sol:

$\begin{array}{l} G i v e n ? ? t h a t x \in N, y \in N a n d 2 x + y = 41 \\ ∴ D o m a i n o f R = {1, 2, 3, 4, 5, \dots, 20} \\ a n d R a n g e = {39, 37, 35, 33, 31, \dots, 1} \\ H e r e, (3, 3) \notin R \\ a s 2 \times 3 + 3 \neq 41 \\ S o, R i s n o t r e f l e x i v e . \\ R i s n o t s y m m e t r i c a s (2, 37) \in R b u t (37, 2) \notin R \\ R i s n o t t r a n s i t i v e a s (11, 19) \in R a n d (19, 3) \in R \\ b u t (11, 3) \notin R . \\ H e n c e, R i s n e i t h e r r e f l e x i v e, n o r s y m m e t r i c a n d n o r t r a n s i t i v e . \end{array}$

Given $A = {2, 3, 4}, B = {2, 5, 6, 7}$ , construct an example of each of the following:

(a) An injective mapping from $A$ to $B$

(b) A mapping from $A$ to $B$ which is not injective

(c) A mapping from $B$ to $A$ .

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

Sol:

Give an example of a map:

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.

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

Sol: $\begin{array}{l} \end{array}$

Let $A = R - {3}, B = R - {1}$ . Let $f : A \to B$ be defined by $f (x) = \frac{x - 2}{x - 3} \forall x \in A$ . Then Show that $f$ is bijective.

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

Sol:

$\begin{array}{l} H e r e, A \in R - {3}, B = R - {1} \\ G i v e n t h a t f : A \to B d e f i n e d b y f (x) = \frac{x - 2}{x - 3} \forall x \in A . \\ L e t x_{1}, x_{2} \in f (x) \\ ∴ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{1} - 3) = (x_{2} - 2) (x_{2} - 3) \\ \Rightarrow - x_{1} = - x_{2} \Rightarrow x_{1} = x_{2} \\ S o, i t i s i n j e c t i v e f u n c t i o n . \\ N o w, L e t y = \frac{x - 2}{x - 3} \\ \Rightarrow x y - 3 y = x - 2 \Rightarrow x y - x = 3 y - 2 \\ \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1} \\ f (x) = \frac{x - 2}{x - 3} = \frac{\frac{3 y - 2}{y - 1} - 2}{\frac{3 y - 2}{y - 1} - 3} \Rightarrow \frac{3 y - 2 - 2 y + 2}{3 y - 2 - 3 y + 3} \Rightarrow y \\ \Rightarrow f (x) = y \in B . \\ S o, f (x) i s s u r j e c t i v e f u n c t i o n . \\ H e n c e, f (x) i s b i j e c t i v e f u n c t i o n . \end{array}$

Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

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

Sol:

$\begin{array}{l} (i) G i v e n t h a t - 1 \leq x \leq 1 \\ L e t x_{1}, x_{2} \in f (x) \\ f (x_{1}) = \frac{1}{x_{1}} a n d f (x_{2}) = \frac{1}{x_{2}} \\ f (x_{1}) = f (x_{2}) \Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}} \Rightarrow x_{1} = x_{2} \\ S o, f (x) i s o n e - o n e f u n c t i o n . \\ L e t f (x) = y = \frac{x}{2} \Rightarrow x = 2 y \\ F o r y = 1, x = 2 \notin [- 1, 1] \\ S o, f (x) i s n o t o n t o . H e n c e, f (x) i s n o t b i j e c t i v e f u n c t i o n . \\ (i i) H e r e, g (x) = | x | \\ g (x_{1}) = g (x_{2}) \Rightarrow | x_{1} | = | x_{2} | \Rightarrow x_{1} = \pm x_{2} \\ S o, g (x) i s n o t o n e - o n e f u n c t i o n . \\ L e t g (x) = y = | x | \Rightarrow x = \pm y \notin A \forall y \in A \\ S o, g (x) i s n o t o n t o f u n c t i o n . \\ H e n c e, g (x) i s n o t b i j e c t i v e f u n c t i o n . \\ (i i i) H e r e, h (x) = x | x | \\ h (x_{1}) = h (x_{2}) \\ \Rightarrow x_{1} | x_{1} | = x_{2} | x_{2} | \Rightarrow x_{1} = x_{2} \\ S o, h (x) i s o n e - o n e f u n c t i o n . \\ N o w, l e t h (x) = y = x | x | = x^{2} o r - x^{2} \\ \Rightarrow &thi \end{array}$

Each of the following defines a relation on $N$ :

(i) $x$ is greater than $y, x, y \in N$

(ii) $x + y = 10, x, y \in N$

(iii) $x y, i s s q u a r e o f a n i n t e g e r x, y \in N$

(iv) $x + 4 y = 10, x, y \in N$ .

Determine which of the above relations are reflexive, symmetric, and transitive.

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

Sol:

Let $A = {1, 2, 3, \dots, 9}$ and $R$ be the relation in $A \times A$ defined by $(a, b) R (c, d)$ if $a + d = b + c$ for $(a, b), (c, d) i n A \times A$ . Prove that $R$ is an equivalence relation and also obtain the equivalent class $[(2, 5)]$ .

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

Sol:

$\begin{array}{l} H e r e, A = {1, 2, 3, \dots, 9} \\ a n d R \to A \times A d e f i n e d b y (a, b) R (c, d) \Rightarrow a + d = b + c \\ \forall (a, b), (c, d) \in A \times A \\ F o r r e f l e x i v e : (a, b) R (a, b) = a + b = b + a \forall a, b \in A w h i c h i s t r u e . S o, R i s r e f l e x i v e . \\ F o r s y m m e t r i c : (a, b) R (c, d) = (c, d) R (a, b) \\ L . H . S . a + d = b + c \\ R . H . S . c + b = d + a \\ L . H . S . = R . H . S . S o, R i s s y m m e t r i c . \\ F o r t r a n s i t i v e : (a, b) R (c, d) = (c, d) R (e, f) \Leftrightarrow (a, b) R (e, f) \\ \Rightarrow a + d = b + c a n d c + f = d + e \\ \Rightarrow a + d = b + c a n d d + e = c + f \\ \Rightarrow (a + d) - (d + e) = (b + c) - (c + f) \\ \Rightarrow a - e = b - f \\ \Rightarrow a + f = b + e \\ \Rightarrow (a, b) R (e, f) \\ S o, R i s t r a n s i t i v e . \\ H e n c e, R i s a n e q u i v a l e n c e r e l a t i o n . \\ E q u i v a l e n t c l a s s o f {(2, 5)} i s {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} \end{array}$

Using the definition, prove that the function $f : A \to B$ is invertible if and only if $f$ is both one-one and onto.

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

Sol:

$\begin{array}{l} A f u n c t i o n f : X \to Y i s s a i d t o b e i n v e r t i b l e i f t h e r e e x i s t s a \\ f u n c t i o n g : Y \to X s u c h t h a t g o f = I_{X} a n d f o g = I_{Y} a n d t h e n t h e \\ i n v e r s e o f f i s d e n o t e d b y f^{- 1} . \\ A f u n c t i o n f : X \to Y i s s a i d t o b e i n v e r t i b l e i f f f i s a b i j e c t i v e f u n c t i o n . \end{array}$

Functions $f, g : R \to R$ are defined, respectively, by $f (x) = x^{2} + 3 x + 1$ , $g (x) = 2 x - 3$ , find

(i) $f \circ g$

(ii) $g \circ f$

(iii) $f \circ f$

(iv) $g \circ g$

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

Sol:

$\begin{array}{l} (i) f o g \Rightarrow f [(g (x)] = [g] {(x)}^{2} + 3 [(g (x)] + 1 \\ = {(2 x - 3)}^{2} + 3 (2 x - 3) + 1 \\ = 4 x^{2} + 9 - 12 x + 6 x - 9 + 1 = 4 x^{2} - 6 x + 1 \\ (i i) g o f \Rightarrow g [f (x)] = 2 [x^{2} + 3 x + 1] - 3 \\ = 2 x^{2} + 6 x + 2 - 3 = 2 x^{2} + 6 x + 1 \\ (i i i) f o f \Rightarrow f [(f (x)] = {[f (x)]}^{2} + 3 [f (x)] + 1 \\ = ({x^{2} + 3 x + 1)}^{2} + 3 (x^{2} + 3 x + 1) + 1 \\ = x^{4} + 9 x^{2} + 1 + 6 x^{3} + 6 x + 2 x^{2} + 3 x^{2} + 9 x + 3 + 1 \\ = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 \\ (i v) g o g \Rightarrow g [g (x)] = 2 [g (x)] - 3 = 2 (2 x - 3) - 3 = 4 x - 6 - 3 = 4 x - 9 \end{array}$

Let $*$ be the binary operation defined on $Q$ . Find which of the following binary operations are commutative:

(i) $a * b = a - b \forall a, b \in Q$

(ii) $a * b = a^{2} + b^{2} \forall a, b \in Q$

(iii) $a * b = a + a b \forall a, b \in Q$

(iv) $a * b = {(a - b)}^{2} \forall a, b \in Q$ .

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

Sol:

$\begin{array}{l} (i) a * b = a - b \in Q a, b \in Q, \\ S o, * i s b i n a r y o p e r a t i o n . \\ a * b = a - b a n d b * a = b - a \forall a, b \in Q \\ a - b \neq b - a \\ S o, * i s n o t c o m m u t a t i v e . \\ (i i) a * b = a^{2} + b^{2} \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = b * a \\ \Rightarrow a^{2} + b^{2} = b^{2} + a^{2} \forall a, b \in Q \\ W h i c h i s t r u e . S o, * i s c o m m u t a t i v e . \\ (i i i) a * b = a + a b \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = a + a b a n d b * a = b + b a \\ a + a b \neq b + b a \Rightarrow a * b \neq b * a \forall a, b \in Q . \\ S o, * i s n o t c o m m u t a t i v e . \\ (i v) a * b = {(a - b)}^{2} \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = {(a - b)}^{2} a n d b * a = {(b - a)}^{2} \\ a * b = b * a \Rightarrow {(a - b)}^{2} = {(b - a)}^{2} \forall a, b \in Q . \\ S o, * i s c o m m u t a t i v e . \end{array}$

Let $*$ be binary operation defined on $R$ by $a * b = 1 + a b, \forall a, b \in R$ . Then the operation $*$ is:

(i) Commutative but not associative

(ii) Associative but not commutative

(iii) Neither commutative nor associative

(iv) Both commutative and associative.

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

Sol:

$\begin{array}{l} (i) : G i v e n t h a t \\ a * b = 1 + a b \forall a, b \in R \\ a n d b * a = 1 + b a \forall a, b \in R \\ a * b = b * a = 1 + a b \\ S o, * i s c o m m u t a t i v e . \\ N o w a * (b * c) = (a * b) * c \forall a, b, c \in R \\ L . H . S . a * (b * c) = a * (1 + b c) = 1 + a (1 + b c) = 1 + a + a b c \\ R . H . S . (a * b) * c = (1 + a b) * c = 1 + (1 + a b) . c = 1 + c + a b c \\ S o, * i s n o t a s s o c i a t i v e . \\ H e n c e, * i s c o m m u t a t i v e b u t n o t a s s o c i a t i v e . \end{array}$

Let $A = {a, b, c}$ and the relation $R$ be defined on $A$ as follows:

$R = {(a, a), (b, c), (a, b)}$ .

Then, write minimum number of ordered pairs to be added in $R$ to make $R$ reflexive and transitive.

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

Sol: R = { (a, a), (b, c), (a, b)}.

To make R as reflexive we must add (b, b) and (c, c) to R. Also, to make R as transitive we must add

(a, c) to R.

So, minimum number of ordered pair is to be added are (b, b), (c, c), (a, c).

Kindly consider the following

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

Sol:

Let $f, g : R \to R$ be defined by $f (x) = 2 x + 1$ and $g (x) = x^{2} - 2$ , $\forall x \in R$ .

Respectively then, Find $g \circ f$ .

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

Sol:

$\begin{array}{l} f (x) = 2 x + 1 a n d g (x) = x^{2} - 2 \\ ∴ g o f = g [f (x)] \\ = {[2 x + 1]}^{2} - 2 \\ = 4 x^{2} + 4 x + 1 - 2 \\ = 4 x^{2} + 4 x - 1 \\ H e n c e, g o f = 4 x^{2} + 4 x - 1 . \end{array}$

Let $f : R \to R$ be the function defined by $f (x) = 2 x - 3, \forall x \in R$ Write $f^{- 1}$ .

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

Sol:

$\begin{array}{l} \begin{array}{l} f (x) = 2 x - 3 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} L e t f (x) = y = 2 x - 3 \end{array} \\ \Rightarrow y + 3 = 2 x \Rightarrow x = \frac{y ? + 3}{2} \\ ∴ f^{- 1} (y) = \frac{y + 3}{2} 0 r f^{- 1} (x) = \frac{x + 3}{2} \end{array}$

If $A = {a, b, c, d}$ and the function $f = {(a, b), (b, d), (c, a), (d, c)}$ , write $f^{- 1}$

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

Sol:

$\begin{array}{l} G i v e n t h a t, A = {a, b, c, d} \end{array}$

$\begin{array}{l} a n d f = {(a, b), (b, d), (c, a), (d, c)} \end{array}$

$\begin{array}{l} f^{- 1} = {(b, a), (d, b), (a, c), (c, d)} \end{array}$

If $f : R \to R$ is defined by $f (x) = x^{2} - 3 x + 2$ , write $f (f (x))$ .

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

Sol:

$\begin{array}{l} G i v e n t h a t \end{array}$ $\begin{array}{l} f (x) = x^{2} - 3 x + 2 \end{array}$ $\begin{array}{l} f {f (x)} = f (x^{2} - 3 x + 2) \end{array}$

$\begin{array}{l} = {(x^{2} - 3 x + 2)}^{2} - 3 (x^{2} - 3 x + 2) + 2 \end{array}$

$\begin{array}{l} = x^{4} + 9 x^{2} + 4 - 6 x^{3} - 12 x + 4 x^{2} - 3 x^{2} + 9 x - 6 + 2 \end{array}$

$\begin{array}{l} = x^{4} + 10 x^{2} - 6 x^{3} - 3 x \end{array}$

$\begin{array}{l} H e n c e f {f (x)} = x^{4} - 6 x^{3} + 10 x^{2} - 3 x \end{array}$

Is $g = {(1, 1), (2, 3), (3, 5), (4, 7)}$ a function? If $g$ is described by $g (x) = α x + β$ , then what value should be assigned to $α$ and $β$ .

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

Sol:

Given that, g = { (1, 1), (2, 3), (3, 5), (4, 7)}.

Here, each element of domain has unique image. So, g is a function.

Now given that, g (x ) = $α$ x + $β$

For (1,1) g (1) = $α$ + $β$

$α$ + $β$ = 1… (i)

For (2,3) g (2) = 2 $α$ + $β$

2 $α$ + $β$ = 3 … (ii)

From Equations. (i) and (ii),

2 (1 - $β$ ) + $β$ = 3

2 - 2 $β$ + $β$ = 3

2 - $β$ = 3

$β$ = - 1

If $β$ = - 1, then $α$ = 2

$∴$ $α$ = 2, $β$ = - 1

If the mappings $f$ and $g$ are given by

$f = {(1, 2), (3, 5), (4, 1)}$ and $g = {(2, 3), (5, 1), (1, 3)}$ , write $f ? g$ .

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

Sol:

$G i v e n t h a t,$
$\begin{array}{l} f = {(1, 2), (3, 5), (4, 1)} \end{array}$

$a n d$
$\begin{array}{l} g = {(2, 3), (5, 1), (1, 3)} \end{array}$

$\begin{array}{l} N o w, f o g (2) = f {g (2)} = f (3) = 5 \end{array}$

$\begin{array}{l} f o g (5) = f {g (5)} = f (1) = 2 \end{array}$

$\begin{array}{l} f o g (1) = f {g (1)} = f (3) = 5 \end{array}$

$\begin{array}{l} f o g = {(2, 5), (5, 2), (1, 5)} \end{array}$

Let $C$ be the set of complex numbers. Prove that the mapping $f : C \to R$ given by $f (z) = ∣ z ∣, \forall z \in C$ , is neither one-one nor onto.

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

Sol:

$\begin{array}{l} f (z) = | z | \forall z \in C \\ f (1) = | 1 | = 1 \\ f (- 1) = | - 1 | = 1 \\ f (1) = f (- 1) \\ B u t 1 \neq - 1 \\ T h e r e f o r e, i t i s n o t o n e - o n o . \\ N o w, l e t f (z) = y = | z | . \\ H e r e, t h e r e i s n o; p r e - i m a g e o f n e g a t i v e n u m b e r s . \\ H e n c e, i t i s n o t o n t o . \end{array}$

Let the function $f : R \to R$ be defined by $f (x) = c o s x, \forall x \in R .$ Show that $f$ is neither one-one nor onto.

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

Sol:

$\begin{array}{l} G i v e n f u n c t i o n, f (x) = c o s x \forall x \in R \\ L e t [- \frac{π}{2}, \frac{π}{2}] \in f (x) \\ f (- \frac{π}{2}) = c o s (- \frac{π}{2}) = c o s \frac{π}{2} = 0 \\ c o s (\frac{π}{2}) = c o s \frac{π}{2} = 0 \\ B u t - \frac{π}{2} \neq \frac{π}{2} \\ ∴, f (x) i s n o t o n e - o n e . \\ N o w, f (x) = c o s x, \forall x \in R i s n o t o n t o a s t h e r e i s n o p r e - i m a g e f o r a n y r e a l n u m b e r . \\ W h i c h d o e s n o t b e l o n g t o t h e i n t e r v a l s [- 1, 1], t h e r a n g e o f c o s x . \end{array}$

Let $X = {1, 2, 3}$ and $Y = {4, 5}$ . Find whether the following subsets of $X \times Y$ are functions from $X$ to $Y$ or not:

$I. f = {(1, 4), (1, 5), (2, 4), (3, 5)}$

$II. g = {(1, 4), (2, 4), (3, 4)}$

$III. h = {(1, 4), (2, 5), (3, 5)}$

$IV. k = {(1, 4), (2, 5)}$ .

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

Sol:

$G i v e n t h a t,$

$\begin{array}{l} X = {1, 2, 3} a n d Y = {4, 5} \end{array}$

$\begin{array}{l} ∴ X \times ? Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} \end{array}$

$\begin{array}{l} (i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} \end{array}$

$\begin{array}{l} f i s n o t a f u n c t i o n b e c a u s e f h a s n o u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i i) g = {(1, 4), (2, 4), (3, 4)} \end{array}$

$\begin{array}{l} S i n c e, g i s a f u n c t i o n a s e a c h e l e m e n t o f t h e d o m a i n h a s a u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i i i) h = {(1, 4), (2, 5), (3, 5)} \end{array}$

$\begin{array}{l} I t i s a f u n c t i o n a s e a c h e l e m e n t o f t h e d o m a i n h a s a u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i v) k = {(1, 4), (2, 5)} \end{array}$

$\begin{array}{l} k i s n o t a f u n c t i o n a s 3 d o e s n o t h a v e a n y i m a g e u n d e r t h e m a p p i n g . \end{array}$

If functions $f : A \to B$ and $g : B \to A$ satisfy $g ? f = I_{A}$ , then show that $f$ is one-one and $g$ is onto.

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

Sol:

$\begin{array}{l} \end{array} L e t x_{1}, x_{2} \in g o f$

$g o f {f (x_{1})} = g o f {f (x_{2})}$

$\Rightarrow g (x_{1}) = g (x_{2}) \begin{array}{l} \end{array}$

$∴$
$\begin{array}{l} x_{1} = x_{2} \end{array}$

$\begin{array}{l} H e n c e, f i s o n e - o n e a n d g i s o n t o . \end{array}$

Let $f : R \to R$ be the function defined by $f (x) = \frac{1}{2 - c o s x}, \forall x \in R$ . Then, find the range of $f$ .

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

Sol:

$\begin{array}{l} f (x) \frac{1}{2 - c o s x}, \forall x \in R \\ R a n g e o f c o s x i s [- 1, 1] \\ L e t f (x) = y = \frac{1}{2 - c o s x} \\ \Rightarrow 2 y - y c o s x = 1 \Rightarrow y c o s x = 2 y - 1 \\ \Rightarrow c o s x = \frac{2 y - 1}{y} = 2 - \frac{1}{y} \\ N o w - 1 \leq c o s x \leq 1 \\ \Rightarrow - 1 \leq 2 - \frac{1}{y} \leq 1 \Rightarrow - 1 - 2 \leq - \frac{1}{y} \leq 1 - 2 \\ \Rightarrow - 3 \leq - \frac{1}{y} \leq - 1 \Rightarrow 3 \geq \frac{1}{y} \geq 1 \Rightarrow \frac{1}{3} \leq y \leq 1 \\ ∴ t h e r a n g e o f f = [\frac{1}{3}, 1] . \end{array}$

Let $n$ be a fixed positive integer. Define a relation $R$ in $Z$ as follows: $\forall a, b \in Z, a R b$ if and only if $a - b$ is divisible by $n$ . Show that $R$ is an equivalence relation.

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

Sol: $\begin{array}{l} \begin{array}{l} \begin{array}{l} \begin{array}{l} \end{array} \end{array} \end{array} \end{array}$

Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $a R b$ if $a$ is congruent to $b \forall a, b \in T$ . Then $R$ is:

(A) Reflexive but not transitive

(B) Transitive but not symmetric

(C) Equivalence

(D) None of these.

This is a Objective Type Question as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

$\begin{array}{l} I f a = b \forall a, b \in T \\ t h e n a R a \Rightarrow a ≅ a w h i c h i s t r u e f o r a l l a \in T \\ S o, R i s r e f l e x i v e \\ N o w, a R b a n d b R a . \\ i . e ., a ≅ b a n d b ≅ a w h i c h i s t r u e f o r a l l a \in T \\ R i s s y m m e t r i c . \\ L e t a R b a n d b R c . \\ \Rightarrow a ≅ b a n d b ≅ a \Rightarrow a ≅ c \forall a, b, c \in T \\ S o, R i s t r a n s i t i v e . \\ H e n c e, R i s e q u i v a l e n c e r e l a t i o n . \\ S o, t h e c o r r e c t a n s w e r i s (c) . \end{array}$

Consider the non-empty set consisting of children in a family and a relation $R$ defined as $a R b$ if $a$ is the brother of $b$ . Then $R$ is:

(A) Symmetric but not transitive

(B) Transitive but not symmetric

(C) Neither symmetric nor transitive

(D) Both symmetric and transitive.

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, a R b \Rightarrow a i s a b r o t h e r o f b . \\ a R a \Rightarrow a i s a b r o t h e r o f a w h i c h i s n o t t r u e . \\ S o, R i s n o t r e f l e x i v e . \\ a R b \Rightarrow a i s a b r o t h e r o f b . \\ b R a \Rightarrow w h i c h i s n o t t r u e b e c a u s e b m a y b e s i s t e r o f a . \\ \Rightarrow a R b \neq b R a \\ S o, R i s n o t s y m m e t r i c . \\ N o w, a R b, b R c \Rightarrow a R c \\ \Rightarrow a i s t h e b r o t h e r o f b a n d b i s t h e b r o t h e r o f c . \\ ∴ a i s a l s o t h e b r o t h e r o f c . \end{array}$

The maximum number of equivalence relations on the set $A = {1, 2, 3}$ are

(A) 1

(B) 2

(C) 3

(D) 5

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, A = {1, 2, 3} \\ T h e n u m b e r o f e q u i v a l e n c e r e l a t i o n s a r e a s f o l l o w s : \\ R_{1} = {(1, 1), (1, 2), (2, 1), (2, 3), (1, 3)} \\ R_{2} = {(2, 2), (1, 3), (3, 1), (3, 2), (1, 2)} \\ R_{3} = {(3, 3), (1, 2), (2, 3), (1, 3), (3, 2)} \\ H e n c e, c o r r e c t a n s w e r i s (d) \end{array}$

If a relation $R$ on the set ${1, 2, 3}$ be defined by $R = {(1, 2)}$ , then $R$ is

(A) Reflexive

(B) Transitive

(C) Symmetric

(D) None of these.

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : R = {(1, 2)} \\ a R a, ? ? s o i t i s n o t r e f l e x i v e . \\ a R b b u t b R a, s o i t i s n o t s y m m e t y r i c . \\ a R b a n d b R c \Rightarrow a R c w h i c h i s t r u e . \\ S o, R i s t r a n s i t i v e . \\ H e n c e, c o r r e c t a n s w e r i s (b) . \end{array}$

Let us define a relation $R$ in $R$ as $a R b$ if $a \geq b$ . Then $R$ is:

(A) An equivalence relation

(B) Reflexive, transitive but not symmetric

(C) Symmetric, transitive but not reflexive

(D) Neither transitive nor reflexive but symmetric

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, a R b i f a \geq b \\ \Rightarrow a R a \Rightarrow a \geq a w h i c h i s t r u e, s o i t i s r e f l e x i v e . \\ L e t a R b \Rightarrow a \geq b, b u t b \geq a, s o b R a \\ R i s n o t s y m m e t r i c . \\ N o w, a \geq b, b \geq c, \Rightarrow a \geq c, w h i c h i s t r u e . \\ S o, R i s t r a n s i t i v e . \\ H e n c e, c o r r e c t a n s w e r i s (b) . \end{array}$

Let $A = {1, 2, 3}$ and consider the relation

$R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}$ .

Then $R$ is

(A) Reflexive but not symmetric

(B) Reflexive but not transitive

(C) Symmetric and transitive

(D) Neither symmetric, nor transitive

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

The identity element for the binary operation $*$ defined on $Q ~ {0}$ as $a * b = \frac{a b}{2} \forall a, b \in Q ~ {0}$ is

(A) 1

(B) 0

(C) 2

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a * b = \frac{a b}{2} \forall a, b \in Q - {0} \\ L e t e b e t h e i d e n t i t y e l e m e n t \\ ∴ a * e = \frac{a e}{2} = a \Rightarrow e = 2 \\ H e n c e, t h e c o r r e c t a n s w e r i s (c) . \end{array}$

If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is

(A) 720

(B) 120

(C) 0

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f A a n d B s e t s h a v e m a n d n e l e m e n t s r e s p e c t i v e l y, \\ t h e n t h e n u m b e r o f o n e - o n e a n d o n t o m a p p i n g f r o m A t o B i s \\ n! i f m = n \\ a n d 0 i f m \neq n \\ H e r e, m = 5 a n d n = 6 \\ 5 \neq 6 \\ S o, N u m b e r o f m a p p i n g = 0 \\ H e n c e, t h e c o r r e c t a n s w e r i s (c) \end{array}$

Let $A = {1, 2, 3, \dots, n}$ and $B = {a, b}$ . Then the number of surjections from $A$ into $B$ is

(A) ⁿp₂

(B) $2^{n} - 2$

(C) $2^{n} - 1$

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Let $f : R \to R$ be defined by $f (x) = \frac{1}{x} \forall x \in R$ . Then $f$ is

(A) One-one

(B) Onto

(C) Bijective

(D) $f$ is not defined

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{1}{x} \\ P u t t i n g x = 0 ∴ f (x) = \frac{1}{0} = \infty \\ S o, f (x) i s n o t d e f i n e d . \\ H e n c e, t h e c o r r e c t a n s w e r i s (d) . \end{array}$

Let $f : R \to R$ be defined by $f (x) = 3 x^{2} - 5$ and $g : R \to R$ by $g (x) = \frac{X}{x^{2} + 1}$ . Then $g ? f$ is

(A) $\frac{3 x^{2} - 5}{9 x^{4} - 30 x^{2} + 26}$

(B) $\frac{3 x^{2} - 5}{9 x^{4} - 6 x^{2} + 26}$

(C) $\frac{3 x^{2}}{x^{4} + 2 x^{2} - 4}$

(D) $\frac{3 x^{2}}{9 x^{4} + 30 x^{2} - 2}$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, f (x) = 3 x^{2} - 5 a n d g (x) = \frac{x}{x^{2} + 1} \\ ∴ g o f = g o f (x) = g [3 x^{2} - 5] \\ = \frac{3 x^{2} - 5}{(3 x^{2} - 5)^{2} + 1} = \frac{3 x^{2} - 5}{9 x^{4} + 25 - 30 x^{2} + 1} \\ ∴ g o f = \frac{3 x^{2} - 5}{9 x^{4} + 25 - 30 x^{2} + 26} \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

Which of the following functions from $Z$ into $Z$ are bijections?

(A) $f (x) = x^{3}$

(B) $f (x) = x + 2$

(C) $f (x) = 2 x + 1$

(D) $f (x) = x^{2} + 1$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f : Z \to Z \\ L e t x_{1}, x_{2} \in f (x) \Rightarrow f (x_{1}) = x_{1} + 2, f (x_{2}) = x_{2} + 2 \\ f (x_{1}) = f (x_{2}) \Rightarrow x_{1} + 2 = x_{2} + 2 \Rightarrow x_{1} = x_{2} \\ S o, f (x) i s o n e - o n e f u n c t i o n . \\ N o w, l e t y = x + 2 ∴ x = y - 2 \in Z \forall y \in Z \\ S o, f (x) i s o n t o f u n c t i o n . \\ ∴ f (x) i s b i j e c t i v e f u n c t i o n . \\ H e n c e, t h e c o r r e c t a n s w e r i s (b) . \end{array}$

Let $f : R \to R$ be the functions defined by $f (x) = x^{3} + 5$ . Then $f^{- 1} (x)$ is

(A) ${(x + 5)}^{\frac{1}{3}}$

(B) ${(x - 5)}^{\frac{1}{3}}$

(C) ${(5 - x)}^{\frac{1}{3}}$

(D) $5 - x$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Let $f : A \to B$ and $g : B \to C$ be the bijective functions. Then ${(g \circ f)}^{- 1}$ is

(A) $f^{- 1} \circ g^{- 1}$

(B) $f \circ g$

(C) $g^{- 1} \circ f^{- 1}$

(D) $g \circ f$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Let $f : R - {\frac{3}{5}} \to R$ be defined by $f (x) = \frac{3 x + 2}{5 x - 3}$ . Then

(A) $f^{- 1} (x) = f (x)$

(B) $f^{- 1} (x) = - f (x)$

(C) ( $f \circ f$ )x=-x

(D) $f^{- 1} (x) = \frac{1}{19} f (x)$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{3 x + 2}{5 x - 3} \forall x \neq \frac{3}{5} \\ L e t y = \frac{3 x + 2}{5 x - 3} \\ \Rightarrow y (5 x - 3) = 3 x + 2 \\ \Rightarrow 5 x y - 3 y = 3 x + 2 \\ \Rightarrow 5 x y - 3 x = 2 + 3 y \\ \Rightarrow x (5 y - 3) = 2 + 3 y \\ \Rightarrow x = \frac{3 y + 2}{5 y - 3} \\ \Rightarrow f^{- 1} (x) = \frac{3 x + 2}{5 x - 3} \\ \Rightarrow f^{- 1} (x) = f (x) \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

Let $f : [0, 1] \to [0, 1]$ be defined by $f (x) = {\begin{matrix} x, & if x is rational \\ 1 - x, & if x is irrational \end{matrix}$

Then $(f \circ f) x i s$

(A) Constant

(B) $1 + x$

(C) $x$

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f : [0, 1] \to [0, 1] \\ ∴ f = f^{- 1} \\ S o, (f o f) x = x (i d e n t i t y e l e m e n t) \\ H e n c e, ? ? ? t h e c o r r e c t a n s w e r i s (c) . \end{array}$

Let $f : [2, \infty) \to R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . Then the range of $f$ is

(A) $?$

(B) $[1, \infty)$

(C) $[4, \infty)$

(D) $[5, \infty)$

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Let $f : N \to R$ be the function defined by $f (x) = \frac{2 x - 1}{2}$ and $g : Q \to R$ be another function defined by $g (x) = x + 2$ . Then $($ $(g \circ f)$
$\frac{3}{2}$ is

(A) 1

(B) 1

(C) $\frac{7}{2}$

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

Let $f : R \to R$ be defined by:

Then $f (- 1) + f (2) + f (4)$ is

(A) 9

(B) 14

(C) 5

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ f (x) = {\begin{cases} 2 x : x > 3 \\ x^{2} : 1 < x \leq 3 \\ 3 x : x \leq 1 \end{cases}} \\ ∴ f (- 1) + f (2) + f (4) = 3 (- 1) + {(2)}^{2} + 2 (4) = - 3 + 4 + 8 = 9 \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

Let $f : R \to R$ be given by $f (x) = t a n x$ . Then $f^{- 1} (1)$ is

(A) $\frac{π}{4}$

(B) ${n π + \frac{π}{4} : n \in ℤ}$

(C) Does not exist

(D) None of these

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

Let the relation $R$ be defined in $ℕ$ by $a R b$ if $2 a + 3 b = 30$ . Then $R =________.$

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

Sol:

Let the relation $R$ be defined on the set $A = {1, 2, 3, 4, 5}$ by $R = {(a, b) : ∣ a^{2} - b^{2} ∣ < 8}$ . Then $R$ is given by ________

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

Sol:

Let $f = {(1, 2), (3, 5), (4, 1)}$ and $g = {(2, 3), (5, 1), (1, 3)}$ . Then $g \circ f =______$ and $f \circ g =________.$

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

Sol:

$\begin{array}{l} \end{array} G i v e n t h a t, f = {(1, 2), (3, 5), (4, 1)} a n d g = {(2, 3), (5, 1), (1, 3)}$

$g o f (1) = g {f (1)} = g (2) = 3$

$g o f (3) = g {f (3)} = g (5) = 1$

$g o f (4) = g {f (4)} = g (1) = 3$

$\begin{array}{l} ∴ g o f = {(1, 3), (3, 1), (4, 3)} \end{array}$

$\begin{array}{l} \end{array} f o g (2) = f {g (2)} = f (3) = 5$

$f o g (5) = f {g (5)} = f (1) = 2$

$f o g (1) = f {g (1)} = f (3) = 5$

$∴ f o g = {(2, 5), (5, 2), (1, 5)}$

Kindly consider the following

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

If $f (x) = (4 - {(x - 7)}^{3}}$ , then $f^{- 1} (x) =________.$

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

Sol:

$\begin{array}{l} f (x) = [4 - {(x - 7)}^{3}] \\ L e t y = [4 - {(x - 7)}^{3}] \\ \Rightarrow {(x - 7)}^{3} = 4 - y \\ \Rightarrow x - 7 = {(4 - y)}^{\frac{1}{3}} \Rightarrow x = {(4 - y)}^{\frac{1}{3}} + 7 \\ H e n c e, f^{- 1} (x) = {(4 - y)}^{\frac{1}{3}} + 7 \end{array}$

State True or False for the statements in each of the Exercises 6 to 14:

Let $R = {(3, 1), (1, 3), (3, 3)}$ be a relation defined on the set $A = {1, 2, 3}$ . Then $R$ is symmetric, transitive but not reflexive.

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

Sol:

Let $f : R \to R$ be the function defined by $f (x) = s i n (3 x + 2), \forall x \in R$ . Then $f$ is invertible.

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

Sol:

$\begin{array}{l} G i v e n t h a t, f (x) = s i n (3 x + 2), \forall x \in R \\ f (x) i s n o t o n e - o n e f u n c t i o n . \\ H e n c e, t h e s t a t e m e n t i s f a l s e . \end{array}$

Every relation which is symmetric and transitive is also reflexive.

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

Sol:

An integer $m$ is said to be related to another integer $n$ if $m$ is a integral multiple of $n$ . This relation in $Z$ is reflexive, symmetric, and transitive.

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

Sol:

$\begin{array}{l} H e r e, m = k n (w h e r e k i s a n i n t e g e r) \\ I f k = 1 m = n, s o z i s r e f l e x i v e . \\ C l e a r l y z i s n o t s y m m e t r i c b u t z i s t r a n s i t i v e . \\ H e n c e, t h e s t a t e m e n t i s f a l s e . \end{array}$

Let $A = {0, 1}$ and $N$ be the set of natural numbers. Then the mapping $f : N \to A$ defined by $f (2 n - 1) = 0, f (2 n) = 1, \forall n \in N$ , is onto.

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

Sol:

The relation $R$ on the set $A = {1, 2, 3}$ defined as $R = {(1, 1), (1, 2), (2, 1), (3, 3)}$ is reflexive, symmetric, and transitive.

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

Sol:

The composition of functions is commutative.

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

Sol:

The composition of functions is associative.

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

Sol:

$\begin{array}{l} L e t f (x) = 2 x, g (x) = x - 1 a n d h (x) = 2 x + 3 \\ f o {g o h (x)} = f o {g (2 x + 3)} \\ = f (2 x + 3 - 1) = f (2 x + 2) = 2 (2 x + 2) = 4 x + 4 \\ a n d (f o g) o h (x) = (f o g) {h (x)} \\ = f o g (2 x + 3) \\ = f (2 x + 3 - 1) = f (2 x + 2) = 2 (2 x + 2) = 4 x + 4 \\ ∴ f o {g o h (x)} = (f o g) o h (x) = 4 x + 4 \\ H e n c e, t h e s t a t e m e n t i s T r u e . \end{array}$

Every function is invertible.

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

Sol:

$\begin{array}{l} O n l y b i j e c t i v e f u n c t i o n s a r e i n v e r t i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

A binary operation on a set has always the identity element.

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

Sol:

$\begin{array}{l} ' +' i s a b i n a r y o p e r a t i o n o n t h e s e t N b u t i t h a s n o i d e n t i t y e l e m e n t . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

JEE Mains 2022

Commonly asked questions

The number of distinct real roots of the equation x⁵ $(x^{3} - x^{2} - x + 1)$ + $x (3 x^{3} - 4 x^{2} - 2 x + 4) - 1 = 0$ is …………..

$x^{8} - x^{7} - x^{6} + x^{5} + 3 x^{4} - 4 x^{3} - 2 x^{2} + 4 x - 1 = 0$

$\Rightarrow x^{7} (x - 1) - x^{5} (x - 1) + 3 x^{3} (x - 1) - x (x^{2} - 1) + 2 x (1 - x) + (x - 1) = 0$

$\Rightarrow (x - 1) (x^{2} - 1) (x^{5} + 3 x - 1) = 0 ∴ x = \pm 1$ are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or 1.

$∴$ 3 real roots.

Let f : R -> R be a continuous function such that f(3x) – f(x) =. If f(8) = 7, then f(14) is equal to:

f (3x)- f (x) = x

Replace $x \to \frac{x}{3} \Rightarrow f (x) - f (\frac{x}{3}) = \frac{x}{3}$

Again replace $x \to \frac{x}{3} \Rightarrow f (\frac{x}{3}) - f (\frac{x}{3^{2}}) - f (\frac{x}{3^{2}}) = \frac{x}{3^{2}}$

$\Rightarrow f (3 x) - f (0) = \frac{3 x}{2} p u t t i n g x = \frac{8}{3} \Rightarrow f (8) - f (0) = 4 ∴ f (0) = 3$

Also putting x = $\frac{14}{3}$ in f (3x) – 3 = $\frac{3 x}{2} \Rightarrow$ F (14) – 3 = 7 f (14) = 10

Let O be the origin and A be the point z₁ = 1 + 2i. If B is the point z₂, Re(z₂) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

$\frac{z_{2} - 0}{(1 + 2 i) - 0} = | \frac{O B}{O A} | e^{\frac{i π}{4}} \Rightarrow z_{2} = (1 + 2 i) (1 + i) = - 1 + 3 i ∴ a r g z_{2} = π - t a n^{- 1} 3 a n d | z_{2} | = \sqrt[]{10}$

$z_{1} - 2 z_{2} = 3 - 4 i ∴ a r g (z_{1} - 2 z_{2}) = - t a n^{- 1} \frac{4}{3} \Rightarrow | z_{1} - 2 z_{2} | = | 2 + 4 i + 1 - 3 i | = \sqrt[]{10}$

If the system of linear equations.

8x + y + 4z = -=2

x + y + z = 0

$λ x$ 3y = $μ$

has infinitely many solutions, then the distance of the point $(λ, μ - \frac{1}{2})$ from the plane 8x + y + 4z + 2 = 0 is

$Δ = | \begin{matrix} 8 & 1 & 4 \\ 1 & 1 \\ λ & - 3 & 0 \end{matrix} | = 12 - 3 λ$

So for $λ$ = 4, it is having infinitely many solutions. $Δ_{x} = | \begin{matrix} - 2 & 1 & 4 \\ 0 & 1 & 1 \\ μ & - 3 & 0 \end{matrix} |$ = 6 $3 μ = 0 \Rightarrow - 6 - 3 μ = 0$

For $μ = - 2$ distance of $(4, - 2, - \frac{1}{2})$ from 8x + y + 4z + 2= 0 $| \frac{32 - 2 - 2 + 2}{\sqrt[]{64 + 1 + 16}} | = \frac{10}{3}$ units

Let A be a 2 × 2 matrix with det(A) = 1 and det $((A + I) (A d j (A) + I)) = 4 .$ Then the sum of the diagonal elements of A can be:

| (A + I) (adj A + I)| = 4 |A adj A + A + Adj A + I| = 4 | (A)I + A + adj A + I|= 4|A| = 1

|A + adj A| = 4

$A = [\begin{matrix} a & b \\ c & d \end{matrix}] \Rightarrow a d j A = [\begin{matrix} a & - b \\ - c & d \end{matrix}] \Rightarrow | \begin{matrix} (a + d) & 0 \\ 0 & (a + d) \end{matrix} | = 4 \Rightarrow a + d = \pm 2$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$

a = 5

Consider two G.P.’s. 2, 2², 2³, …… and 4, 4², 4³, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to:

Given G.P’s 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${(2)}^{\frac{225}{8}} \Rightarrow {(2, 2^{2}, 2^{3}, . . . .)}^{\frac{1}{60 + n}} = {(2)}^{\frac{225}{8}} \Rightarrow n = \frac{57}{8}, 20 s o n = 20$

$∴ \sum_{k = 1}^{n} k (n - k) 20 \times \frac{20 \times 21}{2} - \frac{20 \times 21 \times 41}{6} = 1330$

If the function f(x) = ${\frac{l o g_{e} (1 - x + x^{2}) + l o g_{e} (1 + x + x^{2})}{\begin{matrix} s e c x - c o s x & k \end{matrix}}}, x \in (\frac{- π}{2}, \frac{π}{2}) - {0}$ is continuous at x = 0, then k is equal to:

$f (x) = {\begin{cases} \frac{l o g_{e} (1 - x + x^{2}) + l o g_{e} (1 + x + x^{2})}{s e c x - c o s x}, x \in (\frac{- π}{2}, \frac{π}{2}) - {0} \\ k, x = 0 \end{cases}$ for continuity at x = 0

$\underset{x \to 0}{l i m} f (x) = k ∴ k = \underset{x \to 0}{l i m} \frac{l o g_{e} (1 + x^{2} + x^{4})}{s e c x - c o s x} (\frac{0}{0} f o r m) = \underset{x \to 0}{l i m} \frac{c o s x l o g_{e} (1 + x^{2} + x^{4})}{s i n^{2} x} = 1$

If f(x) = ${\begin{matrix} x + a, x \leq 0 & | x - 4 |, x > 0 \end{matrix} a n d g (x) = {\begin{matrix} x + 1, x < 0 & {(x - 4)}^{2} + b, x \geq 0 \end{matrix}$ are continuous on R, then (gof)(2) + (fog)(2) is equal to:

$f (x) = {\begin{matrix} x + a, & x \leq 0 \\ | x - 4 |, & x > 0 \end{matrix} a n d g (x) = {\begin{matrix} x + 1, & x < 0 \\ {(x - 4)}^{2} + b, & x \geq 0 \end{matrix}$

$?$ f (x) and g (x) are continuous on R $∴$ a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (2) = g (2) + f (-1) = -11 + 3 = -8

Let f(x) = ${\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 & - 2 x + l o g_{2} (b^{2} - 4), x > 1 \end{matrix} .$ Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

$f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, & x \leq 1 \\ - 2 x + l o g_{2} (b^{2} - 4), & x > 1 \end{matrix}$

If f (x) has maximum value at x = 1 then

$f (1) \leq f (1) \Rightarrow - 2 + l o g_{2} (b^{2} - 4) \leq 1 - 1 + 10 - 7$

$\Rightarrow l o g_{2} (b^{2} - 4) \leq 5 \Rightarrow 0 < b^{2} - 4 \leq 32$

$\Rightarrow b^{2} - 4 > 0 \Rightarrow b \in (- \infty, - 2) \cup (2, \infty)$ ……. (i)

$A n d b^{2} - 4 \leq 32 \Rightarrow b \in [- 6, 6]$ ……. (ii)

From (i) and (ii) we get $b \in [- 6, - 2) \cup (2, 6]$

If a = $\underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{2 n}{n^{2} + k^{2}}$ and f(x) = $\sqrt[]{\frac{1 - c o s x}{1 + c o s x}}$ , x $\in$ (0, 1), then:

$a = \underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{2 n}{n^{2} + k^{2}} = \underset{n \to \infty}{l i m} \frac{1}{n} \sum_{k = 1}^{n} \frac{2}{1 + {(\frac{k}{n})}^{2}}$

$∴ a = \int_{0}^{1} \frac{2}{1 + x^{2}} d x = 2 t a n^{- 1} x]_{0}^{1} = \frac{π}{2}$

$f' (\frac{a}{2}) = \sqrt[]{2} f (\frac{a}{2})$

$I f \frac{d y}{d x}$ + 2y tan x = sin x, 0 < x < $\frac{π}{2}$ and y $(\frac{π}{3})$ = 0, then the maximum value of y(x) is:

$\frac{d y}{d x} + 2 y t a n x = s i n x, ∴ I . F . e^{\int 2 t a n x d x} = s e c^{2} x$

= cos x – 2 cos² x= $- 2 {(c o s x - \frac{1}{4})}^{2} + \frac{1}{8} ∴ y_{m a x} = \frac{1}{8}$

A point P moves so that the sum of squares of its distances from the points (1, 2) and (2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A , B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to:

Let point P : (h, k)

Therefore according to question, ${(h - 1)}^{2} + {(k - 2)}^{2} + {(h + 2)}^{2} + {(k - 1)}^{2} = 14$

$∴$ locus of P (h, k) is $x^{2} + y^{2} + x - 3 y - 2 = 0$

Now intersection with x – axis are $x^{2} + x - 2 = 0 \Rightarrow x = - 2, 1$

Now intersection with y – axis are $y^{2} - 3 y - 2 = 0 \Rightarrow y = \frac{3 \pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2} (| x_{1} | + | x_{2} |) (| y_{1} | + | y_{2} |) = \frac{1}{2} \times 3 \times \sqrt[]{17} = \frac{3 \sqrt[]{17}}{2}$

Let the tangent drawn to the parabola y²= 24x at the point (α, β) is perpendicular to the line 2x + 2y = 5. Then the normal to the hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ at the point ( α+ 4, β+ 4) does NOT pass through the point:

Any tangent to y² = 24x at (α, β) is βy = 12 (x + α) therefore Slope = $\frac{12}{β}$

and perpendicular to 2x + 2y = 5 =>12 =β and α= 6 Hence hyperbola is $\frac{x^{2}}{6^{2}} - \frac{y^{2}}{1 2^{2}}$ = 1 and normal is drawn at (10, 16)

therefore equation of normal $\frac{36 x}{10} + \frac{144 y}{16} = 36 + 144 \Rightarrow \frac{x}{50} + \frac{y}{20} = 1$ This does not pass through (15, 13) out of given option.

The length of the perpendicular from the point (1, 2, 5) on the line passing through (1, 2, 4) and parallel to the line x + y – z = 0 = x – 2y + 3z – 5 is:

he line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

$\vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 1 & - 2 & 3 \end{matrix} | = (1, 4, - 3)$ Equation of line through P(1, 2, 4) and parallel to $\vec{b} \frac{x - 1}{1} = \frac{y - 2}{- 4} = \frac{z - 4}{- 3}$

Let $N \equiv (λ + 1, - 4 λ + 2, - 3 λ + 4) \bar{Q N} = (λ, - 4 λ + 4, - 3 λ - 1)$

$\bar{Q N}$ is perpendicular to $\vec{b} \Rightarrow (λ, - 4 λ + 4, - 3 λ - 1) . (1, 4, - 3) = 0 \Rightarrow λ = \frac{1}{2} .$

Hence $\bar{Q N} (\frac{1}{2}, 2, \frac{- 5}{2}) a n d \bar{| Q N |} = \frac{21}{2}$

Let $\vec{a} = α \hat{i} + \hat{j} - \hat{k} a n d \vec{b} = 2 \hat{i} + \hat{j} - α \hat{k}, α > 0 .$ If the projection of $\vec{a} \times \vec{b}$ on the vector $- \hat{i} + 2 \hat{j} - 2 \hat{k}$ is 30, then equal to:

Given : $\vec{a} = (α, 1, - 1) a n d \vec{b} = (2, 1, - α) \vec{c} = \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 1 & - 1 \\ 2 & 1 & - α \end{matrix} |$

$= (- α + 1) \hat{i} + (α^{2} - 2) \hat{j} + (α - 2) \hat{k}$ Projection of $\vec{c}$ on $\vec{d} = - \hat{i} + 2 \hat{j} - 2 \hat{k} = | \vec{c} . \frac{\vec{d}}{| \vec{d} |} | = 30 {G i v e n}$

$\Rightarrow | \frac{α - 1 - 4 + 2 α^{2} - 2 α + 4}{\sqrt[]{1 + 4 + 4}} | = 30$

On solving $α = \frac{- 13}{2}$ (Rejected as > 0) and = 7

The mean and variance of a binomial distribution are and $\frac{α}{3}$ respectively. If P(X = 1) = $\frac{4}{243}$ then P(X = 4 or 5) is equal to:

Given, mean = np = . and variance = npq = $\frac{α}{3} \Rightarrow q = \frac{1}{3} a n d p = \frac{2}{3}$

$P (X = 1) = n p^{1} q^{n - 1} = \frac{4}{243} \Rightarrow n {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{n - 1} = \frac{4}{243} \Rightarrow n = 6$

$P (X = 4 o r 5) =^{6} C_{4} {(\frac{2}{3})}^{4} {(\frac{1}{3})}^{2} +^{6} C_{5} {(\frac{2}{3})}^{5} {(\frac{1}{3})}^{1} = \frac{16}{27}$

Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁) = $\frac{2 + 3 p}{6}$ , P(E₃) = $\frac{2 - p}{8}$ and P(E3) = $\frac{1 - p}{2} .$ If the maximum and minimum values of p are p₁ and p₂, then (p₁ + p₂) is equal to:

$0 \leq \frac{2 + 3 p}{6} \leq 1 \Rightarrow p \in [- \frac{2}{3}, \frac{4}{3}], 0 \leq \frac{2 - p}{8} \leq 1 \Rightarrow p \in [- 6, 2] a n d 0 \leq \frac{1 - p}{2} \leq 1 \Rightarrow p \in [- 1, 1]$

$0 < P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 10 \leq \frac{13}{12} - \frac{p}{8} \leq 1 \Rightarrow p \in [\frac{2}{3}, \frac{26}{3}]$ Taking intersection to all $p \in [\frac{2}{3}, 1]$

$∴ p_{1} + p_{2} = \frac{5}{3}$

Let S = ${θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then $n (S) + \sum_{θ \in S}^{} (s e c (\frac{π}{4} + 2 θ) c o s e c (\frac{π}{4} + 2 θ))$ is equal to:

$S = {θ \in [0, 2 π] : 8^{2 s i n^{2} x} + 8^{2 c o s^{2} x} = 16}$

Now apply AM $\geq G M$ for $\frac{8^{2 s i n^{2} x} + 8^{2 c o s^{2} x}}{2} \leq {(8^{2 s i n^{2} x + 2 c o s^{2} x})}^{\frac{1}{2}} \Rightarrow 8^{2 s i n^{2} x} = 8^{2 c o s^{2} x}$

$s i n^{2} θ = c o s^{2} θ$ $∴ θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

$= 4 + [c o s e c (\frac{π}{2} + π) + c o s e c (\frac{π}{2} + 3 π) + c o s e c (\frac{π}{2} + 5 π) + c o s e c (\frac{π}{2} + 7 π)]$

$= 4 - 2 (4) = - 4$

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8})$ is equal to:

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8}) t a n (2 t a n^{- 1} \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \times \frac{1}{8}} + s e c^{- 1} \frac{\sqrt[]{5}}{2})$

$= t a n (t a n^{- 1} \frac{3}{4} + t a n^{- 1} \frac{1}{2}) = t a n (t a n^{- 1} \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}})$

$= t a n (t a n^{- 1} \frac{\frac{5}{4}}{\frac{5}{8}}) = 2$

The statement $(\sim (p \Leftrightarrow \sim q)) \land q i s :$

(D)

$\sim (p \Leftrightarrow \sim q) \land q = (p \Leftrightarrow \sim q) \land q i s :$

$∴ (\sim (P \Leftrightarrow \sim Q)) \land$ q is equivalent to $(p \Rightarrow q) \land$ p

If for some q, q, r $\in$ R, not at all have same sign, one of the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$ is also a root of the equation x² + 2x – 8 = 0, then $\frac{q^{2} + r^{2}}{p^{2}}$ is equal to………….

Let and are the roots of $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$

$∴ α + β > 0 a n d α β > 0$ Also, it has a common root with x² + 2x – 8 = 0

$∴$ The common root between above two equations is 4.

$\Rightarrow 16 (p^{2} + q^{2}) - 8 q (p + r) + q^{2} + r^{2} = 0 \Rightarrow (16 p^{2} - 8 p q + q^{2}) + (16 q^{2} - 8 q r + r^{2}) = 0$

$\Rightarrow {(4 p - q)}^{2} + {(4 q - r)}^{2} = 0 \Rightarrow q = 4 p a n d r = 16 p ∴ \frac{q^{2} + r^{2}}{p^{2}} = \frac{16 p^{2} + 256 p^{2}}{p^{2}} = 272$

The number of 5-digit natural numbers, such that the product of their digits is 36, is………

Factors of 36 = 2².3².1

Five-digit combinations can be

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers $\frac{5! 5!}{2! 2!} + \frac{5!}{2! 2!} + \frac{5!}{2! 2!} + \frac{5!}{3!} + \frac{5!}{2!} + \frac{5!}{3! 2!} = (30 \times 3) + 20 + 60 + 10 = 180 .$

The series of positive multiples of 3 is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21, 24, 27}, . . . . . .$ Then the sum of the elements in the 11th set is equal to…………..

Given series ${3 \times 1}, {3 \times 2, 3 \times 3, 3 \times 4}, {3 \times 5, 3 \times 6, 3 \times 7, 3 \times 8, 3 \times 9} . . . . . . . . .$

$∴$ 11th set will have 1 + (10)2 = 21 terms

Also up to 10th set total 3 × k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$∴ S e t 11 = {3 \times 101, 3 \times 102, . . . . . . 3 \times 121}$ $∴$ Sum of elements = 3 × (101 + 102 + ….+121)

$= \frac{3 \times 222 \times 21}{2} = 6993 .$

If the coefficients of x and x² in the expansion of (1 + x)^p (1 – x)^q, p, q 15, are -3 and -5 respectively, then the coefficient of x³ is equal to…………………

Coefficient of x in ${(1 + x)}^{p} {(1 - x)}^{q} = -^{p} {C_{0}}^{q} C_{1} +^{p} {C_{1}}^{q} C_{0} = - 3 \Rightarrow p - q = 3$

Coefficient of x² in ${(1 + x)}^{p} {(1 - x)}^{q} =^{p} {C_{0}}^{q} C_{2} -^{p} {C_{1}}^{q} C_{1} -^{p} {C_{2}}^{q} C_{0} = - 5$

$\Rightarrow \frac{q (q - 1)}{2} - p q + \frac{p (p - 1)}{2} = - 5 \Rightarrow \frac{q (q - 1)}{2} - (q - 3) q + \frac{(q - 3) (q - 4)}{2} = - 5 \Rightarrow q = 11, p = 8$

Coefficient of x³ in ${(1 + x)}^{8} {(1 - x)}^{11} = -^{11} C_{3} +^{8} {C_{1}}^{11} C_{2} -^{8} {C_{2}}^{11} C_{1} +^{8} C_{3} = 23$

If $n (2 n + 1) \int_{0}^{1} {(1 - x^{n})}^{2 n} d x = 1177 \int_{0}^{1} {(1 - x^{n})}^{2 n + 1}$ dx, then n $\in$ N is equal to………….

$\int_{0}^{1} 1 . {(1 - x^{n})}^{2 n + 1} d x$ using by parts we get,

$(2 n^{2} + n + 1) \int_{0}^{1} {(1 - x^{n})}^{2 n + 1} d x = 1177 \int_{0}^{1} {(1 - x^{n})}^{2 n + 1} d x$

$\Rightarrow 2 n^{2} + n + 1 = 1177 \Rightarrow n = 24 o r - \frac{49}{2} ∴ n = 24$

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then is equal to………….

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), $- \frac{1}{2} < a < 2,$ then p is equal to…………

Slope of AH = $\frac{a + 2}{1}$ slope of BC = $- \frac{1}{p} ∴ p = a + 2$ $∴ C (\frac{18 p - 30}{p + 1}, \frac{15 p - 33}{p + 1})$

slope of HC = $\frac{16 p - p^{2} - 31}{16 p - 32}$

slope of BC × slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

Let the function f(x) = 2x² – log_e x, x > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y² = 4ax at a point P on it passes through the point (8a, 8a – 1) but does not pass through the point $(- \frac{1}{a}, 0) .$ If the equation of the normal at P is $\frac{x}{α} + \frac{γ}{β} = 1,$ then a + b is equal to…………….

$f' (x) = \frac{4 x^{2} ? 1}{x}$ so f (x) is decreasing in $(0, \frac{1}{2}) a n d ? ? (\frac{1}{2}, ?)$ $? a = \frac{1}{2}$

Tangent at y² = 2x is y = mx + $\frac{1}{2 m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2} o r ? ? \frac{1}{4}$

So tangent may be $y = \frac{1}{2} x + 1 ? ? o r ? ? y = \frac{1}{4} x + 2 ? ? ? b u t ? ? y = \frac{1}{2} x + 1$ passes through (-2, 0) so rejected.

Equation of normal $\frac{x}{9} + \frac{y}{36} = 1$

Let Q and R be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $\sqrt[]{26}$ from the point P(4, 2, 7). Then the surface of the area of the triangle PQR is……………

Let PT perpendicular to QR

$\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2} = λ \Rightarrow T (2 λ - 1, 3 λ - 2, 2 λ + 1)$ therefore

$2 (2 λ - 5) + 3 (3 λ - 4) + 2 (2 λ - 6) = 0 \Rightarrow λ = 2$

$T (3, 4, 5) ∴ P T = \sqrt[]{1 + 4 + 4} = 3 ∴ Q T = \sqrt[]{26 - 9} = \sqrt[]{17}$

$∴ Δ P Q R = \frac{1}{2} \times 2 \sqrt[]{17} \times 3 = 3 \sqrt[]{17}$

Therefore square of $a r (Δ P Q R)$ = 153.

JEE Mains Solutions 2022,24th june , Maths, first shift

Commonly asked questions

Let A = ${z \in C : 1 \leq | z - (1 + i) \leq 2}$ and B = ${z \in A : | z - (1 - i) | = 1}$ . Then, B :

$A = {z \in c : 1 \leq | z - (1 + i) | \leq 2}$

| z - (1 + i) | \geq 1

and $| z - (1 + i) | \leq 2$

and also $| z - (1 - i) | = 1$

hence B has infinite set.

The remainder when 3²⁰²²is divided by 5 is :

(3 * 3 * 3 * ………2022 times) ¸ 5

Remainder = (-1) (-1) (-1) ….1011 times)

Remainder = -1 + 5 = 4

The surface area of a balloon of spherical shape being inflated increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

Surface area, S = 4pr²

$? \frac{d s}{d t} = 4 ? . 2 r \frac{d r}{d t} = 8 ? \frac{d r}{d t} = c o n s t a n t = k (s a y)$

$? \frac{d s}{d t} = k ? s = k t + c$

$? 4 ? r^{2} = k t + c$

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11},$ then n is equal to ________.

$\frac{6}{11}$ = Required probability

After solving, we get n = 4

Let x²+ y² + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x² at (2, 4). Then A + C is equal to

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

The number of values of a for which the system of equations :

x + y + z = a

ax + 2ay + 3z = -1

x + 3ay + 5z = 4

is inconsistent, is

For inconsistent, $Δ = 0, | \begin{matrix} 1 & 1 & 1 \\ α & 2 a & 3 \\ 1 & 3 α & 5 \end{matrix} | = 0$

a = 1

$∴ Δ_{1} = | \begin{matrix} 1 & 1 & 1 \\ - 1 & 2 & 3 \\ 4 & 3 & 5 \end{matrix} | = 1 (1) + 1 (12 + 5) + 1 (- 3 - 8) = 7 \neq 0$

If the sum of the squares of the reciprocals of the roots a and b of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 {(α^{3} + β^{3})}^{2}$ is equal to :

$α + β = - \frac{λ}{3}, α β = - \frac{1}{3}$

$∴ \frac{1}{α^{2}} + \frac{1}{β^{2}} = 15 \Rightarrow λ = \pm 3$

$∴ 6 {(α^{3} + β^{3})}^{2} = 6 {({(α + β)}^{3} - 3 α β (α + β))}^{2} = 6 \times {(- (1 + 1))}^{2} = 24$

The set of all values of k for which ${(t a n^{- 1} x)}^{3} + {(c o t^{- 1} x)}^{3} = k π^{3}, x \in R$ , is the interval :

${(t a n^{- 1} x)}^{3} + {(c o t^{- 1} x)}^{3} = k π^{3}, x \in R$

$\Rightarrow (t a n^{- 1} x + c o t^{- 1} x) ({(t a n^{- 1} x + c o t^{- 1} x)}^{2} - 3 t a n^{- 1} x c o t^{- 1} x) = k π^{3}$

$\Rightarrow \frac{1}{3} \leq k < \frac{7}{8}$

Let S = ${\sqrt[]{n} : 1 \leq n \leq 50 a n d n i s o d d}$ .

Let $a \in S a n d A = [\begin{matrix} 1 & 0 & a \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix}] .$ If $\sum_{a \in S}^{} d e t (a d j A)$ = 100l, then l is equal to :

S = ${\sqrt[]{n} : 1 \leq n \leq 50 & n = o d d}$

= ${\sqrt[]{1}, \sqrt[]{3}, \sqrt[]{5}, . . . . . ., \sqrt[]{49} (25 t e r m s)}$

$| A | = | \begin{matrix} 1 & 0 & 4 \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix} |$ = 1 + a²

^{$∴ \sum_{a \in S}^{} d e t (a d j A) = 100 λ \Rightarrow \sum {(1 + a^{2})}^{2} = 100 λ \Rightarrow λ = 221$}

For the function f(x) = 4 log_e(x – 1) – 2x² + 4x + 5, x > 1, which one of the following is NOT correct?

f (x) = 4log_e (x – 1) -2x² + 4x + 5, x > 1

$(A) f' (x) = \frac{4}{x - 1} - 4 x + 4$

$\Rightarrow f' (x) > 0 \Rightarrow \frac{x (- x + 2)}{x - 1} > 0$

option (A) is correct.

(B) f (x) = -1, has two solution

option (B) is correct

$(C) f " (x) = - \frac{4}{{(x - 1)}^{2}} - 4$

$f' (e) - f " (2) = \frac{4 e (2 - e)}{e - 1} + 8 > 0$

option (C) is not correct

(D) f (e) = 4log_e (e – 1) -2 (e² – 2e + 1) + 7 > 0

f (e + 1) = 4 – 2 (e + 1)² + 4 (e + 1) +5+ < 0

option (D) is correct

If the tangent at the point (x₁, y₁) on the curve y = x³ + 3x² + 5 passes through the origin, then (x₁, y₁) does NOT lie on the curve:

f(x) = $| 2 x^{2} + 3 x - 2 | + s i n x c o s x$

$= | (2 x - 1) (x + 2) | + s i n x c o s x$

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

$f' (x) = {\begin{matrix} - 4 x - 3 + c o s 2 x, & 0 < x < \frac{1}{2} \\ 4 x + 3 + c o s 2 x, & \frac{1}{2} \leq x < 1 \end{matrix}$

f(1) = 3 + sin 1 cos 1 and $f (\frac{1}{2}) = s i n \frac{1}{2} c o s \frac{1}{2}$

$∴ f (1) + f (\frac{1}{2}) = 3 + \frac{s i n 2}{2} + \frac{s i n 1}{2} = 3 + \frac{1}{2} (s i n 1 + s i n 2)$

$= 3 + \frac{s i n 1}{2} (1 + 2 c o s 1)$

The sum of absolute maximum and absolute minimum values of the function f(x) = |2x² + 3x + 2| + sin x cos x in the interval [0, 1] is :

f(x) = $| 2 x^{2} + 3 x ? 2 | + s i n x c o s x$

$= | (2 x ? 1) (x + 2) | + s i n x c o s x$

$f (x) = {\begin{matrix} ? 2 x^{2} ? 3 x + 2 + s i n x c o s x, & 0 < x < \frac{1}{2} \\ 2 x^{2} + 3 x ? 2 + s i n x c o s x, & \frac{1}{2} ? x < 1 \end{matrix}$

$f' (x) = {\begin{matrix} ? 4 x ? 3 + c o s 2 x, & 0 < x < \frac{1}{2} \\ 4 x + 3 + c o s 2 x, & \frac{1}{2} ? x < 1 \end{matrix}$

f(1) = 3 + sin 1 cos 1 and $f (\frac{1}{2}) = s i n \frac{1}{2} c o s \frac{1}{2}$

$? f (1) + f (\frac{1}{2}) = 3 + \frac{s i n 2}{2} + \frac{s i n 1}{2} = 3 + \frac{1}{2} (s i n 1 + s i n 2)$

$= 3 + \frac{s i n 1}{2} (1 + 2 c o s 1)$

If ${a_{i}}_{i = 1}^{n},$ where n is an even integer, is an arithmetic progression with common difference 1, and $\sum_{i = 1}^{n} a_{i} = 192,$ then n is equal to :

$\sum_{i = 1}^{n} a_{i} = 192 \Rightarrow a_{1} + a_{2} + a_{3} + . . . . . . + a_{n} = 192$

$∴ n (a_{1} + a_{n}) = 384 . . . . . . . . . . . . . . (i)$

and $\sum_{i = 1}^{n / 2} a_{2 i} = 120 \Rightarrow a_{2} + a_{4} + a_{6} + . . . . . . + a_{n} = 120$

n (a₂ + a_n) = 480 ……………… (ii)

n (a₂ – a₁) = 96

$∴ n = 96$

If x = x(y) is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then x(e) is equal to :

$y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}$

$\frac{d x}{d y} - \frac{2}{y} x = y^{3} (1 + y) e^{y} . . . . . . . . . . (i)$

Equation (i) is in linear form, so I.F. = y^-2

$\begin{array}{l} ∴ x y^{- 2} = y e^{y} + c \Rightarrow 0 = e + c \Rightarrow c = - e \\ ∴ x = e^{3} (e^{e} - 1) \end{array}$

Let lx – 2y = µ be a tangent to the hyperbola $a^{2} x^{2} - y^{2} = b^{2} .$ Then ${(\frac{λ}{a})}^{2} - {(\frac{μ}{b})}^{2}$ is equal to

$λ x - 2 y = μ . . . . . . . . (i) a n d \frac{x^{2}}{(\frac{b^{2}}{a^{2}})} - \frac{y^{2}}{b^{2}} = 1 . . . . . . . . . (i i)$

Let (x₁, y₁) be a point on the curve equation of tangent of eq (ii)

$y = m x \pm \sqrt[]{\frac{b^{2}}{a^{2}} m^{2} - b^{2}} . . . . . . . . . . (i i i)$

From (i) and (ii) are identical

$∴ m = \frac{λ}{2} a n d \frac{b^{2}}{a^{2}} m^{2} - b^{2} = \frac{μ^{2}}{4}$

$\Rightarrow \frac{b^{2}}{a^{2}} \times \frac{λ^{2}}{4} - b^{2} = \frac{μ^{2}}{4}$

$\Rightarrow {(\frac{λ}{a})}^{2} - {(\frac{μ}{b})}^{2} = 4$

Let $\hat{a}, \hat{b}$ and be unit vectors. If $\hat{c}$ be vector such that the angle between $\hat{a} a n d \vec{c} i s \frac{π}{12},$ and $\hat{b} = \vec{c} + 2 (\vec{c} \times \hat{a}),$ then ${| 6 \vec{c} |}^{2}$ is equal to :

$| \hat{a} | = 1, | \hat{b} | = 1$

${| \hat{b} |}^{2} = {| \vec{c} + 2 (\vec{c} \times \hat{a}) |}^{2}$

$\Rightarrow 1 = {| \vec{c} |}^{2} + 4 {| \vec{c} |}^{2} {(\frac{\sqrt[]{3} - 1}{2 \sqrt[]{2}})}^{2}$

${| \vec{c} |}^{2} [1 + 4 \times \frac{{(\sqrt[]{3} - 1)}^{2}}{8}] = {| \vec{c} |}^{2} (3 - \sqrt[]{3})$

$∴ {| \vec{c} |}^{2} = \frac{1}{3 - \sqrt[]{3}} = \frac{3 + \sqrt[]{3}}{6}$

$\Rightarrow {| 6 \vec{c} |}^{2} = 6 (3 + \sqrt[]{3})$

If a random variable X follows the Binomial distribution B (33, p) such that 3P (X = 0) = P (X = 1) then the value of $\frac{P (X = 15)}{P (X = 18)} - \frac{P (X = 16)}{P (X = 17)}$ is equal to :L

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

$\Rightarrow^{33} C_{0} p^{0} q^{33} =^{33} C_{1} p q^{32}$

$\Rightarrow p = \frac{1}{12}, q = \frac{11}{12} \Rightarrow \frac{q}{p} = 11$

………. (i)

Subtracting, (ii) – (i), we get 1320

The domain of the function f(x) = $\frac{c o s^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{l o g_{e} (x^{2} - 3 x + 2)}$ is :

$\frac{x^{2} - 5 x + 6}{x^{2} - 9} \geq - 1 & \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$

$\frac{2 x + 1}{x + 3} \geq 0, x \neq - 3 & \frac{1}{x + 3} \geq 0, x \neq - 3 \Rightarrow x > - 3$

$x \in [- \frac{1}{2}, \infty] . . . . . . . . . . . (i)$

$x^{2} - 3 x + 2 > 0 a n d x^{2} - 3 x + 1 \neq 0$

(x – 2) (x – 1) > 0 and $x \neq \frac{3 \pm \sqrt[]{5}}{2}$

$x \in (- \infty, 1) \cup (2, \infty) - {\frac{3 \pm \sqrt[]{5}}{2}}$

From (i) and (ii) $x \in [- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 \pm \sqrt[]{5}}{2}}$

Let S = ${θ \in [- π, π] - {\pm \frac{π}{2}} : s i n θ t a n θ + t a n θ = s i n 2 θ}$ . If $\sum_{θ \in S}^{} c o s 2 θ,$ then T = n(S) is equal to

$\frac{s i n θ s i n θ}{c o s θ} + \frac{s i n θ}{c o s θ} = 2 s i n θ c o s θ$

$\Rightarrow \frac{s i n θ}{c o s θ} [s i n θ + 1] = 2 s i n θ c o s θ$

$∴$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$

$θ = \frac{3 π}{2}, θ = \frac{π}{6}, \frac{5 π}{6}$

$θ = \frac{3 π}{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{π}{3} + c o s \frac{5 π}{3} = 4$

$∴$ T + n(s) = 4 + 5 = 9

The number of choices for $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ , such that $(p Δ q) \Rightarrow ((p Δ ~ q) \lor (~ p) Δ q)$ is a tautology, is :

$(p Δ q) \Rightarrow (p Δ \sim q) \lor (\sim p Δ q)$

Case I

When $Δ$ is same as $\lor .$

Then $(p Δ \sim q) \lor (- p Δ q)$ becomes

$(p \lor \sim q) \lor (\sim p \lor q)$ which is always true, so x becomes tautology.

Case II

When $Δ$ is same as $\lor$

Then $(p \land q) \Rightarrow (p \land \sim q) \lor (\sim p \land q)$ becomes $p \land q$ is T, then $(p \land \sim q) \lor (\sim p \land q)$ is false, so x cannot be tautology.

Case III

When $Δ$ is same as $\Rightarrow$

Then $(p \Rightarrow \sim q) \lor (- p \Rightarrow q)$ is same as $(\sim p \lor \sim q) \lor (p \lor q)$ which is true, so x becomes tautology.

Case IV

When $Δ$ is same as $\Leftrightarrow$

Then $(p \Leftrightarrow q) \Rightarrow (p \Leftrightarrow \sim q) \lor (\sim p \Leftrightarrow q)$

$p \Leftrightarrow q$ is true when p and q have same truth values $p \Leftrightarrow \sim q a n d \sim p \Leftrightarrow q$ both are false. Hence x cannot be tautology.

The number of one-one functions f:{a, b, c, d} ® {0, 1, 2, ……., 10} such that 2f (a) – f(b) + 3f(c) + f(d) = 0 is ___________.

f(b) = 2f(a) + 3f(c) + f(d)

Value of f(c) Value of f(a) Number of functions

1 7

2 5

0 3 3

4 2

0 6

1 2 2

3 1

2 0 3

1 1

3 0 1

Total number of function 31

In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is __________

Let x = correct answer, y = incorrect answer

$∴ 3 x - 2 y = 5, x + y \leq 5, x, y \in w$

only possible (x, y) is (3, 2)

$∴$ Required number of ways =

Let $A (\frac{3}{\sqrt[]{a}}, \sqrt[]{a}), a > 0$ , be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D (3 cos q, a sin q) is a point in the fourth quadrant such that the maximum area of $Δ A C D$ is 12 square units, then a is equal to _________.

$B (- \frac{3}{\sqrt[]{a}}, \sqrt[]{a}) a n d c (- \frac{3}{\sqrt[]{a}}, - \sqrt[]{a})$

$∴ A r e a o f Δ A C D = \frac{1}{2} | \begin{matrix} \frac{3}{\sqrt[]{a}} & \sqrt[]{a} & 1 \\ \frac{- 3}{\sqrt[]{a}} & - \sqrt[]{a} & 1 \\ 3 c o s θ & a s i n θ & 1 \end{matrix} |$ = 12

$\Rightarrow Δ = 3 \sqrt[]{a} | c o s θ + s i n θ | = 12$

$∴ Δ m a x = 3 \sqrt[]{a} . \sqrt[]{2} = 12 \Rightarrow a = 8$

Let a line having direction ratios 1, -4, 2 intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the points A and B. Then (AB)² is equal to _________.

$\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1} = r_{1}$

$A (3 r_{1} + 7, 1 - r_{1}, - 2 + r_{1})$

and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1} = r_{2}$

$B (2 r_{2}, 7 + 3 r_{2}, r_{2})$

$A / q, \frac{3 r_{1} - 2 r_{2} + 7}{1} = \frac{3 r_{2} + r_{1} + 6}{4} = \frac{r_{1} - r_{2} - 2}{2}$

$\Rightarrow r_{1} = - 5, r_{2} = - 3$

$∴ A (- 8, 6, - 7) a n d B (- 6, - 2, - 3)$

AB² = 84

$f (x) = {\begin{cases} | 2 x^{2} - 3 x - 7 |, x \leq - 1 \\ [4 x^{2} - 1], - 1 < x < 1 \\ | x + 1 | + | x - 2 |, x \geq 1 \end{cases}$

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x = \pm \frac{1}{2}, \pm \frac{1}{\sqrt[]{2}}, \pm \frac{\sqrt[]{3}}{2}$

The number of points where the function f(x) = ${\begin{cases} | 2 x^{2} - 3 x - 7 | i f x \leq - 1 \\ [4 x^{2} - 1] i f - 1 < x < 1 \\ | x + 1 | + | x - 2 | i f x \geq 1 \end{cases}$ [t] denotes the greatest integer $\leq$ t is discontinuous is __________-.

$y = 2 | x^{2} - \frac{3}{2} x - \frac{7}{2} |$

$= 2 | {(x - \frac{3}{4})}^{2} - \frac{65}{16} |$

Let f (θ) = $s i n θ + \int_{- \frac{π}{2}}^{\frac{π}{2}} (s i n θ + t c o s θ) f (t) d t .$ Then the value of $| \int_{0}^{\frac{π}{2}} f (θ) d θ |$ is

$f (θ) = s i n θ + \int_{- \frac{π}{2}}^{\frac{π}{2}} (s i n θ + t c o s θ) f (t) d t$

$= s i n θ + s i n θ \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t + c o s θ \int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t$

$= (1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t) s i n θ + (\int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t) c o s θ$

f(q) = a sin q + b cos q

$∴ a = 1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t = 1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} (a s i n t + b c o s t) d t$

$∴ a = 2 b + 1 . . . . . . . . . . (i)$

$b = \int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t = \int_{- \frac{π}{2}}^{\frac{π}{2}} (a s i n t + b c o s t) d t$

Solving (i) & (ii) we get a = $- \frac{1}{3}, b = - \frac{2}{3}$

$∴ | \int_{0}^{\frac{π}{2}} f (θ) d θ | = 1$

Let $\underset{0 \leq x \leq 2}{M a x} {\frac{9 - x^{2}}{5 - x}} = α a n d \underset{0 \leq x \leq 2}{M i n} = {\frac{9 - x^{2}}{5 - x}} = β$

If $\int_{β - \frac{8}{3}}^{2 α - 1} M a x {\frac{9 - x^{2}}{5 - x}, x} d x = α_{1} + α_{2} l o g_{e} (\frac{8}{15})$ then $α_{1} + α_{2}$ is equal to __________.

Let f(x) = $\frac{x^{2} - 9}{x - 5}$

$f' (x) = \frac{2 x (x - 5) - (x^{2} - 9)}{{(x - 5)}^{2}}$

$= \frac{x^{2} - 10 x + 9}{{(x - 5)}^{2}} = \frac{(x - 1) (x - 9)}{{(x - 5)}^{2}}$

$α = f (1) = 2, β = {f (0), f (2)} = \frac{5}{3}$

$∴ \int_{- 1}^{3} m a x {\frac{x^{2} - 9}{x - 5}, x} d x + {[\frac{x^{2}}{2}]}_{9 / 5}^{3}$

a₁ = 18, a₂ = 16

a₁ + a₂ = 34

If two tangents drawn from a point (a, b) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4x are such that the slope of one tangent is four times the other, then the value of ${(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2}$ equals __________.

$\frac{x^{2}}{\frac{1}{25}} + \frac{y^{2}}{\frac{1}{4}} = 1$

$25 α^{2} + 4 β^{2} = 1 . . . . . . . . . . (i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$ passes

though (a, b)

$α m^{2} - β m + 1 = 0$

$m_{1} + m_{2} = \frac{β}{α}, 4 m_{1}^{2} = \frac{1}{α}$

$∴ f r o m (i) & (i i)$

25(a² + a) = 1 …………(iii)

$∴ {(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2} = 2929$

$∴ a = 1 + 0 + b {(s i n - \frac{π}{2})}^{\frac{π}{2}} + 2 b + 1$

Let S be the region bonded by the curves y = x³ and y² = x. The curve y = 2|x| divides S into two regions of areas R₁, and R₂ _________.

If max {R₁, R₂} = R₂, then $\frac{R_{2}}{R_{1}}$ is equal to _________.

$R_{2} = \int_{0}^{1} (x - x^{3}) d x = {(\frac{x^{2}}{2} - \frac{x^{4}}{4})}_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

R_{1} = \int_{0}^{\frac{1}{4}} (\sqrt[]{x} - 2 x) d x = {(\frac{2 x^{3 / 2}}{3} - x^{2})}_{0}^{\frac{1}{4}} = \frac{1}{12} - \frac{1}{16}

= \frac{4 - 3}{48} = \frac{1}{48}

R_{1} + R_{2} = {\int_{0}^{1} (\sqrt[]{x} - x^{3}) d x = (\frac{2 x^{3 / 2}}{3} - \frac{x^{4}}{4})}_{0}^{1} = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}

\frac{R_{1} + R_{2}}{R_{1}} = 1 + \frac{R_{2}}{R_{1}} = \frac{\frac{5}{12}}{\frac{1}{48}} = 20

∴ \frac{R_{2}}{R_{1}} = 19

If the shortest distance between the lines $\vec{r} = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j}) a n d \vec{r} = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\sqrt[]{\frac{2}{3}}$ , then the integral value of ‘a’ is equal to __________.

${\vec{b}}_{1} \times {\vec{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - a & 0 \\ 1 & - 1 & 1 \end{matrix} |$ = $- a \hat{i} - \hat{j} + (a - 1) \hat{k}$

${\vec{a}}_{1} - {\vec{a}}_{2} = - \hat{i} + \hat{j} + \hat{k}$

Shortest distance = $| \frac{({\vec{a}}_{1} - {\vec{a}}_{2}) . ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$

$= \frac{2 (a - 1)}{\sqrt[]{a^{2} + 1 + {(a + 1)}^{2}}} = \sqrt[]{\frac{2}{3}} = \frac{4 {(a - 1)}^{2}}{a^{2} + 1 + {(a - 1)}^{2}} = \frac{2}{3}$

$12 {(a - 1)}^{2} = 2 (a^{2} + 1) + 2 {(a - 1)}^{2}$

$10 {(a - 1)}^{2} = 2 (a^{2} + 1)$

$5 a^{2} - 10 a + 5 = a^{2} + 1 \Rightarrow 4 a^{2} - 10 a + 4 = 0$

$2 a^{2} - 5 a + 2 = 0$

$(2 a - 1) (a - 2) = 0$

$∴ a = \frac{1}{2}, 2$

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