Let f , be a function defined by f x = x - 1 2 + 1 , x ≥ 1 Statement - 1 : The set x : f x = f - 1 x = 1 , 2 Statement - 2 : f , is a bijection and f - 1 x = 1 + x - 1 , x ≥ 1

Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement- 1 .

Statement- 1 is true, Statement- 2 is true; Statement- 2 is NOT a correct explanation for Statement- 1

Statement- 1 is true, Statement- 2 is false

Statement- 1 is false, Statement- 2 is true

If f x + 2 f 1 x = 3 x , x ≠ 0 , and S = { x ∈ R : f ( x ) = f ( - x ) } ; then S ,

contains more than two elements.

Let f : 0 , 1 → R be defined by f x = b - x 1 - b x , where b is a constant such that 0 < b < 1 . Then

f ≠ f - 1 on 0 , 1 and f ' b = 1 f ' 0

f = f - 1 on 0 , 1 and f ' b = 1 f ' 0

f - 1 is differentiable on 0 , 1

For x ∈ R , f x = log 2 - sin x and g x = f f x , then

g is differentiable at x = 0 and g ' ( 0 ) = - sin ( log 2 )

g is not differentiable at x = 0 .

Let f x = - 1 , - 2 ≤ x < 0 x 2 - 1 , 0 ≤ x ≤ 2 and g ( x ) = | f ( x ) | + f ( | x | ) . Then, in the interval ( - 2 , 2 ) , g is

not differentiable at one point

not differentiable at two point

Find the domain of the function f x = - log x + 4 2 log 2 2 x - 1 3 + x

Given, f x = - log x + 4 2 log 2 2 x - 1 3 + x We know that in a , a ≥ 0 Here - log x + 4 2 log 2 2 x - 1 3 + x ≥ 0 On multiplying both the sides by - 1 , the sign of inequality will change. ⇒ log x + 4 2 log 2 2 x - 1 3 + x ≤ 0 We know that log a b ≤ m = b ≤ a m ; a > 1 b > a m ; 0 < a < 1 Case-I : When, 0 < x + 4 2 < 1 ⇒ - 4 < x < - 2 ⇒ log x + 4 2 log 2 2 x - 1 3 + x ≤ 0 ⇒ log 2 2 x - 1 3 + x ≥ x + 4 2 0 ⇒ log 2 x - 1 3 + x ≥ 1 Here as 2 > 1 (Base of logarithmic function) ⇒ 2 x - 1 3 + x ≥ 2 ⇒ 2 x - 1 3 + 2 - 2 ≥ 0 ⇒ 3 3 + x ≤ 0 (Number line, sign of ( x + 3 ) ] ⇒ x ∈ ( - ∞ , - 3 ) Intersection of x ∈ ( - 4 , - 2 ) and x ∈ ( - ∞ , - 3 ) is x ∈ ( - 4 , - 3 ) Case-II : When x + 4 2 > 1 ⇒ x > - 2 ⇒ x ∈ ( - 2 , ∞ ) ⇒ log x + 4 2 log 2 2 x - 1 3 + x ≤ 0 ⇒ log 2 2 x - 1 3 + x ≤ x + 4 2 0 ⇒ log 2 2 x - 1 3 + x ≤ 1 As 2 > 1 (Base of logarithmic function) ⇒ 2 x - 1 3 + x ≤ 2 ⇒ 2 x - 1 3 + x - 2 ≤ 0 ⇒ 7 3 + x ≥ 0 [Number line, sign of ( x + 3 ) ] ⇒ x ∈ ( - 3 , ∞ ) Intersection of x ∈ ( - 2 , ∞ ) and x ∈ ( - 3 , ∞ ) is x ∈ ( - 2 , ∞ ) Now, log is defined only when log 2 2 x - 1 3 + x > 0 and 2 x - 1 3 + x > 0 First solving for 2 x - 1 3 + x > 0 [Number line sign of 2 x - 1 3 + x ] So, x ∈ ( - ∞ , - 3 ) ∪ ( 1 2 , ∞ ) Now, solving for log 2 2 x - 1 3 + x > 0 As 2 > 1 (Base of logarithmic function) ⇒ 2 x - 1 3 + x > 1 ⇒ 2 x - 1 3 + x - 1 > 0 ⇒ x - 4 3 + x > 0 ⇒ x ∈ ( - ∞ , - 3 ) ∪ ( 4 , ∞ ) Taking intersection of both cases, we get x ∈ ( - ∞ , - 3 ) ∪ ( 4 , ∞ ) . . . i Union of Case-I, case-II and number i is ⇒ x ∈ ( - 4 , - 3 ) ∪ ( 4 , ∞ ) Hence, domain of f ( x ) is x ∈ ( - 4 , - 3 ) ∪ ( 4 , ∞ ) .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 8

HARD

JEE Main/Advance

IMPORTANT

Earn 100

The domain of the function $f (x) = \frac{1}{\sqrt{|x| - x}}$ is

73.68% studentsanswered this correctly

Important Questions on Functions

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Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 9

Let $f$ , be a function defined by $f (x) = {(x - 1)}^{2} + 1, (x \geq 1)$

Statement - $1 :$ The set $\{x : f (x) = f^{- 1} (x)\} = \{1, 2\}$
Statement - $2 : f$ , is a bijection and $f^{- 1} (x) = 1 + \sqrt{x - 1}, x \geq 1$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 10

If

f (x) + 2 f (\frac{1}{x}) = 3 x, x \neq 0

, and

S = {x \in R : f (x) = f (- x)}

; then

S

,

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 11

For

x \in R - {0, 1},

let

f_{1} (x) = \frac{1}{x}, f_{2} (x) = 1 - x

and

f_{3} (x) = \frac{1}{1 - x}

be three given functions. If a function,

J (x)

satisfies

(f_{2} \circ J \circ f_{1}) (x) = f_{3} (x)

, then

J (x)

is equal to

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 12

The number of functions

f

from

\{1, 2, 3, . . \dots, 20\}

, onto

\{1, 2, 3, \dots \dots \dots \dots, 20\}

such that

f (k)

is a multiple of

3

, wherever

k

is a multiple of

4

, is

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 13

Let

f : (0, 1) \to R

be defined by

f (x) = \frac{b - x}{1 - b x}

, where

b

is a constant such that

0 < b < 1

. Then

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Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 14

For $x \in R, f (x) = |\log 2 - \sin x|$ and $g (x) = f (f (x)),$ then

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-3>Q 15

Let $f (x) = \{\begin{cases} - 1, & - 2 \leq x < 0 \\ x^{2} - 1, & 0 \leq x \leq 2 \end{cases}$ and $g (x) = | f (x) | + f (| x |) .$ Then, in the interval $(- 2, 2), g$ is