Find the domain and range of the following functions f ( x ) = cos - 1 log [ x ] | x | x , where [ . ] denotes the greatest integer function.

Let f ( x ) = cos - 1 log [ x ] | x | x Now, log is defined only when [ x ] > 0 , & [ x ] ≠ 1 and | x | x > 0 ⇒ x ∈ [ 1 , ∞ ) & x ∈ ( - ∞ , 1 ) ∪ [ 2 , ∞ ) & x ∈ 0 , ∞ Intersection of all three is x ∈ [ 2 , ∞ ) Hence, Domain of f ( x ) is x ∈ [ 2 , ∞ ) For ∈ [ 2 , ∞ ) , | x | x = x x = 1 f ( x ) = cos - 1 log [ x ] | x | x = cos - 1 log [ x ] 1 ⇒ cos - 1 ( 0 ) = π 2 Hence, range of f ( x ) is π 2

Find the domain and range of the following functions f x = log 1 2 log 2 x 2 + 4 x + 5 where . denotes the greatest integer function

Given, f x = log 1 2 log 2 x 2 + 4 x + 5 We know that in a , a ≥ 0 So, here log 1 2 log 2 x 2 + 4 x + 5 ≥ 0 We know that log a b ≥ m = b ≥ a m ; a > 1 b ≤ a m ; 0 < a < 1 As 0 < 1 2 < 1 ⇒ log 1 2 log 2 x 2 + 4 x + 5 ≥ 0 ⇒ log 2 x 2 + 4 x + 5 ≤ 1 2 0 ⇒ log 2 x 2 + 4 x + 5 ≤ 1 As, 2 > 1 [Base of logarithm] ⇒ x 2 + 4 x + 5 ≤ 2 We know that if [ t ] ≤ 2 then t must be such that t < 3 So, x 2 + 4 x + 5 < 3 ⇒ x 2 + 4 x + 2 < 0 . Find roots of the quadratic equation using Shree Dharacharya method ⇒ x = - 4 ± 16 - 8 2 = - 4 ± 8 2 = - 2 ± 2 ⇒ x = - 2 + 2 or x = - 2 - 2 ⇒ x ∈ - 2 - 2 , 0 - 2 + 2 . Now, log is defined only when log 2 x 2 + 4 x + 5 > 0 and x 2 + 4 x + 5 > 0 As 2 > 1 [Base logarithm] ⇒ x 2 + 4 x + 5 > 2 0 ⇒ x 2 + 4 x + 5 > 1 ⇒ x 2 + 4 x + 5 ≥ 2 As, t > 1 when t ≥ 2 ⇒ x 2 + 4 x + 3 ≥ 0 ⇒ x 2 + 3 x + x + 3 ≥ 0 ⇒ x x + 3 + 1 x + 3 ≥ 0 ⇒ x + 1 x + 3 ≥ 0 ⇒ x ∈ - ∞ , - 3 ∪ [ - 1 , ∞ ) . Now for ⇒ x 2 + 4 x + 5 ≥ 1 As t > 0 when t ≥ 1 ⇒ x 2 + 4 x + 4 ≥ 0 ⇒ x + 2 2 ≥ 0 ⇒ x ∈ - ∞ , ∞ . Now taking the intersection of x ∈ - 2 - 2 , 2 - 2 , x ∈ ( - ∞ , - 3 ] ∪ [ - 1 , ∞ ) and x ∈ - ∞ , ∞ Hence, x ∈ ( - 2 - 2 , - 3 ] ∪ [ - 1 , 2 - 2 ) . As from above, we have seen that 0 < log 2 x 2 + 4 x + 5 ≤ 1 ⇒ 2 0 < x 2 + 4 x + 5 ≤ 2 1 ⇒ 1 < x 2 + 4 x + 5 ≤ 2 . As x 2 + 4 x + 5 is an integer and it lies between ( 1 , 2 ] So, x 2 + 4 x + 5 must be equal to 2 So, x 2 + 4 x + 5 = 2 So, f x = log 1 2 log 2 x 2 + 4 x + 5 = log 1 2 log 2 2 ⇒ log 1 2 1 = 0 ⇒ Range of f x is 0 .

Find the domain and range of the following functions f x = sin - 1 log 2 x 2 2 , where . denotes greatest integer function

f x = sin - 1 log 2 x 2 2 , where . denotes greatest integer function We know that, the domain of sin - 1 t is t ∈ - 1 , 1 Therefore, here - 1 ≤ log 2 x 2 2 ≤ 1 We know that, t ≥ - 1 ⇒ t ≥ - 1 And t ≤ 1 ⇒ t < 2 So here, - 1 ≤ log 2 x 2 2 < 2 When, log 2 x 2 2 ≥ - 1 We know that, log b a ≥ m = a ≥ b m , b > 1 a ≤ b m , 0 < b < 1 As here 2 > 1 [base of logarithm] ⇒ log 2 x 2 2 ≥ - 1 ⇒ x 2 2 ≥ 1 2 ⇒ x 2 ≥ 1 ⇒ x 2 - 1 ≥ 0 ⇒ x - 1 x + 1 ≥ 0 ⇒ x ∈ - ∞ , - 1 ∪ [ 1 , ∞ ) When log 2 x 2 2 < 2 ⇒ x 2 2 < 2 2 ⇒ x 2 < 8 ⇒ x 2 - 8 < 0 ⇒ x - 8 x + 8 < 0 ⇒ x ∈ - 8 , 8 Taking intersection of x ∈ ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) and x ∈ - 8 , 8 ⇒ x ∈ ( - 8 , - 1 ] ∪ [ 1 , 8 ) . Hence, the domain of f x is x ∈ ( - 8 , - 1 ] ∪ [ 1 , 8 ) . Here, the argument of sin - 1 will be an integer which are - 1 , 0 , 1 as log 2 x 2 2 will give - 1 , 0 , 1 as integers for x in the domain of f x So, range of f ( x ) = sin - 1 - 1 , sin - 1 0 , sin - 1 1 = - π 2 , 0 , π 2 . Hence, range of f x is - π 2 , 0 , π 2 .

If f ( x ) = x , x < 1 x 2 , 1 ≤ x ≤ 4 , 8 x , x > 4 then find f - 1 ( x ) .

Given, f x = x , x < 1 x 2 , 1 ≤ x ≤ 4 8 x , x > 4 Let f ( x ) = y , in order to find inverse of f ( x ) We take, x = f - 1 ( y ) For x < 1 , given f ( x ) = x ⇒ y = x ⇒ x = y , here y < 1 ⇒ f - 1 ( y ) = y , y < 1 For 1 ≤ x ≤ 4 , given f ( x ) = x 2 ⇒ y = x 2 , 1 ≤ x ≤ 4 ⇒ x = y ⇒ f - 1 ( y ) = y As, 1 ≤ x ≤ 4 ⇒ 1 ≤ y ≤ 4 ⇒ 1 ≤ y ≤ 16 So, f - 1 ( y ) = y , 1 ≤ y ≤ 16 For x > 4 , given f ( x ) = 8 x ⇒ y = 8 x , x > 4 ⇒ y 2 = 64 x ⇒ x = y 2 64 , y 2 64 > 4 ⇒ x = y 2 64 , y 2 > 64 · 4 Now, y 2 > 256 ⇒ y 2 - 256 > 0 ⇒ ( y - 16 ) ( y + 16 ) > 0 ⇒ y ∈ ( - ∞ , - 16 ) ∪ ( 16 , ∞ ) y cannot be negative since, y = 8 x and x > 0 for x > 4 Hence, y ∈ ( 16 , ∞ ) or y > 16 So, x = y 2 64 , y > 16 ⇒ f - 1 y = y 2 64 , y > 16 So, f - 1 y = y , y , y 2 64 , y < 1 1 ≤ y ≤ 16 y > 16 Hence, f - 1 x = x , x , x 2 64 , x < 1 1 ≤ x ≤ 16 x > 16

f x = lim n → ∞ x sin n x sin n x + 1 Find domain and range of f x , where n ∈ N .

Given, f x = lim n → ∞ x sin n x sin n x + 1 f x is defined for sin n x + 1 ≠ 0 ⇒ sin n x ≠ - 1 ⇒ x ≠ 2 k π - π 2 , k ∈ Z So, domain of f x = R - 2 k π - π 2 , k ∈ Z We know that range of sin x ∈ - 1 , 1 but here sin x ∈ ( - 1 , 1 ] Case I : If sin x ∈ [ 0 , 1 ) If sin x ∈ [ 0 , 1 ) ⇒ lim n → ∞ sin n x = 0 ⇒ f ( x ) = 0 Case II : If sin x ∈ - 1 , 0 , for this lim n → ∞ sin n x = 0 ⇒ f ( x ) = 0 . Case III : If sin x = 1 ⇒ x = 2 k π + π 2 , k ∈ Z Now, lim n → ∞ sin n x = 1 ⇒ f ( x ) = 1 2 2 k π + π 2 , k ∈ Z ⇒ f ( x ) = k π + π 4 , k ∈ Z So, the range of f ( x ) is 0 ∪ k π + π 4 , k ∈ Z .

Let f : R + → R satisfies the equation f ( x y ) = e x y - x - y e y f ( x ) + e x f ( y ) ∀ x , y ∈ R + If f ' ( 1 ) = e , then find f ( x )

Given f ( x y ) = e x y - x - y e y f ( x ) + e x f ( y ) ∀ x , y ∈ R + Put x = y = 1 we get f 1 = e - 1 e f 1 + e f 1 ⇒ f 1 = 0 Using the definition of differentiability f ' x = lim h → 0 f x + h - f x h = lim h → 0 f x 1 + h x - f x h = lim h → 0 e x 1 + h x - x - 1 + h x e 1 + h x f x + e x f 1 + h x - f x h = lim h → 0 e h f x + e h - 1 + x - h x + f 1 + h x - f x h = lim h → 0 e h - 1 f x + e h - 1 + x - h x f 1 + h x - f 1 h = lim h → 0 e h - 1 f x h + lim h → 0 e h - 1 + x - h x f 1 + h x - f 1 h x . x = f x + e x - 1 f ' 1 x ⇒ f ' x = f x + e x x Now f x e x - f ' x e x e 2 x = 1 x ⇒ d d x f x e x = 1 x Integrating in both sides with respect to x ∫ d d x f x e x = ∫ 1 x d x ⇒ f x e x = log e x + c ∵ f 1 = 0 ⇒ c = 0 Therefore, f x = x log e x .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4

Embibe Experts Solutions for Chapter: Functions, Exercise 4: Exercise-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Functions, Exercise 4: Exercise-4

Attempt the free practice questions on Chapter 24: Functions, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Functions, Exercise 4: Exercise-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 11

Find the domain and range of the following functions $f (x) = \cos^{- 1} \sqrt{\log_{[x]} \frac{| x |}{x}},$ where $[.]$ denotes the greatest integer function.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 12

Find the domain and range of the following functions $f (x) = \sqrt{\log_{\frac{1}{2}} \log_{2} [x^{2} + 4 x + 5]}$ where $[.]$ denotes the greatest integer function

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 13

Find the domain and range of the following functions $f (x) = \sin^{- 1} [\log_{2} (\frac{x^{2}}{2})],$ where $[.]$ denotes greatest integer function

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 24

Suppose $X$ and $Y$ are two sets and $f : X \to Y$ is a function. For a subset $A$ of $X,$ define $f (A)$ to be the subset $\{f (a) : a \in A\}$ of $Y .$ For a subset $B$ of $Y,$ define $f^{- 1} (B)$ to be the subset $\{x \in X : f (x) \in B\}$ of $X .$ Then prove the followings
$(i)$ Statement $" f^{- 1} (f (A)) = A$ for every $A \subset X "$ is false
$(ii)$ Statement $" f^{- 1} (f (A)) = A$ for every $A \subset X$ if only if $f (X) = Y^{''}$ is false
$(iii)$ Statement $" f (f^{- 1} (B)) = B$ for every $B \subset Y "$ is false
$(iv)$ Statement $" f (f^{- 1} (B)) = B$ for every $B \subset Y$ if only if $f (X) = Y^{''}$ is true

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 27

Let $f (x) = \log_{2} \log_{3} \log_{4} \log_{5} (\sin x + a^{2}) .$ Find the set of values of $a$ for which domain of $f (x)$ is $R .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 28

If $f (x) = [\begin{matrix} x, & x < 1 \\ x^{2}, & 1 \leq x \leq 4, \\ 8 \sqrt{x}, & x > 4 \end{matrix}$ then find $f^{- 1} (x) .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 29

$f (x) = \lim_{n \to \infty} [\frac{x \sin^{n} x}{\sin^{n} x + 1}]$ Find domain and range of $f (x)$ , where $n \in N$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 30

Let $f : R^{+} \to R$ satisfies the equation $f (x y) = e^{x y - x - y} (e^{y} f (x) + e^{x} f (y)) \forall x, y \in R^{+}$
If $f^{'} (1) = e,$ then find $f (x)$

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