Relations and Functions

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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) $[5, \infty)$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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) $g \circ f$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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) $5 - x$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

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

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

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

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