Find the range of f x = e x x + 1 , x ≥ 0 .

f x = e x x + 1 , x ≥ 0 For x ≥ 0 , e x ≥ 1 (also increasing) [ x + 1 ] is also positive and for x ≥ 0 ⇒ e x ≥ [ x + 1 ] ⇒ e x [ x + 1 ] ≥ 1 So, range of f ( x ) = [ 1 , ∞ ) .

Find the range of f x = sin x + cos x , where · denotes the greatest integer function.

f x = sin x + cos x | sin x | ∈ [ 0 , 1 ] | cos x | ∈ [ 0 , 1 ] and sin x + cos x ∈ [ - 2 , 2 ] If | sin x | = 0 therefore | cos x | = 1 and | cos x | = 0 therefore | sin x | = 1 So, minimum value of | sin x | + | cos x | = 1 and maximum value of | sin x | + | cos x | = 2 | sin x | + | cos x | ∈ [ 1 , 2 ) So, [ | sin x | + | cos x | ] = 1 f ( x ) = 1 So, range = { 1 } .

Find the range of f x = - x 2 + 4 x - 3 + sin π 2 sin π 2 x - 1 .

f x = - x 2 + 4 x - 3 + sin π 2 sin π 2 x - 1 Let f ( x ) = g ( x ) + h ( x ) So, the domain of f ( x ) is the intersection of domain of g ( x ) and h ( x ) . g ( x ) = - x 2 + 4 x - 3 So, - x 2 + 4 x - 3 ≥ 0 x 2 - 4 x + 3 ≤ 0 ( x - 3 ) ( x - 1 ) ≤ 0 x ∈ [ 1 , 3 ] h ( x ) = sin π 2 sin π 2 ( x - 1 ) So, sin π 2 sin π 2 ( x - 1 ) ≥ 0 but the maximum value of sin θ i s 1 , So, 0 ≤ sin π 2 sin π 2 ( x - 1 ) ≤ 1 sin - 1 ( 0 ) ≤ π 2 sin π 2 ( x - 1 ) ≤ sin - 1 ( 1 ) 0 ≤ π 2 sin π 2 ( x - 1 ) ≤ π 2 0 ≤ sin π 2 ( x - 1 ) ≤ 1 sin - 1 ( 0 ) ≤ π 2 ( x - 1 ) ≤ sin - 1 ( 1 ) 0 ≤ π 2 ( x - 1 ) ≤ π 2 0 ≤ x - 1 ≤ 1 1 ≤ x ≤ 2 So, x ∈ [ 1 , 2 ] . . . ( ii ) So, the domain of f ( x ) = Domain of g ( x ) ∩ domain of h ( x ) . Domain of f ( x ) ∈ [ 1 , 2 ] For x ∈ 1 , 2 ⇒ 0 ≤ - x 2 + 4 x - 3 ≤ 1 . . . i And 0 ≤ sin π 2 sin π 2 x - 1 ≤ 1 . . . ii ∵ Both g x and h x are monotonically increasing function ∴ 0 ≤ - x 2 + 4 x - 3 + sin π 2 sin π 2 x - 1 ≤ 2 ⇒ 0 ≤ f x ≤ 2 So, range of f ( x ) = [ 0 , 2 ] .

Find the domain and range of f x = log e cos x + 1 2 , where · denotes the greatest integer function.

Given that f x = log e cos x + 1 2 Logarithmic function is defined when cos x + 1 2 > 0 ⇒ cos | x | + 1 2 ≥ 1 ⇒ cos | x | ≥ 1 2 ⇒ Domain = 2 n π - π 3 , 2 n π + π 3 , where n ∈ I Now, we know that - 1 ≤ cos | x | ≤ 1 ⇒ - 1 2 ≤ cos | x | + 1 2 ≤ 3 2 ⇒ cos | x | + 1 2 = - 1 , 0 , 1 ⇒ log e cos | x | + 1 2 = 0 Range = { 0 }

Find the domain and range of f x = sin - 1 log e x + log e sin - 1 x , where · denotes the greatest integer function.

Given that f x = sin - 1 log e x + log e sin - 1 x Let f x = f 1 x + f 2 x Now for f 1 x , Inverse of sine function is defined. ⇒ - 1 ≤ log e [ x ] ≤ 1 ⇒ 1 e ≤ [ x ] ≤ e ⇒ [ x ] = 1 , 2 ⇒ x ∈ [ 1 , 3 ) Now for f 2 x , Logarithmic function is defined. ⇒ sin - 1 [ x ] > 0 ⇒ [ x ] = 1 ⇒ x ∈ [ 1 , 2 ) ⇒ Domain of function f x is [ 1 , 2 ) . Now, we know, 1 ≤ x < 2 [ x ] = 1 f x = sin - 1 log e 1 + log e sin - 1 1 = sin - 1 0 + log e π 2 = log e π 2 Range = log e π 2

Find the domain and range of f x = log sin - 1 x 2 + 3 x + 2 , where · denotes the greatest integer function.

Given that f x = log e sin - 1 x 2 + 3 x + 2 ⇒ sin - 1 x 2 + 3 x + 2 > 0 We know , - π 2 ≤ sin - 1 x ≤ π 2 ∴ 0 < sin - 1 x 2 + 3 x + 2 ≤ π 2 ⇒ 0 < x 2 + 3 x + 2 ≤ 1 ⇒ 0 < x 2 + 3 x + 2 ≤ 1 ⇒ x 2 + 3 x + 2 > 0 ⇒ x + 1 x + 2 > 0 x ∈ - ∞ , - 2 ∪ - 1 , ∞ And x 2 + 3 x + 2 ≤ 1 x 2 + 3 x + 1 ≤ 0 x ∈ - 3 - 5 2 , 5 - 3 2 ⇒ Domain of f x is - 3 - 5 2 , - 2 ∪ - 1 , 5 - 3 2 . Now, x 2 + 3 x + 2 > 0 ⇒ 0 < sin - 1 x 2 + 3 x + 2 ≤ π 2 ⇒ log e sin - 1 x 2 + 3 x + 2 ≤ log e π 2 ⇒ log e sin - 1 x 2 + 3 x + 2 ≤ log e π 2 ≤ 0 ∴ Range of f x = { . . . - 3 , - 2 , - 1 , 0 }

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.3

Amit M Agarwal Solutions for Chapter: Functions, Exercise 3: Target Exercise 3.3

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 3: Target Exercise 3.3

Attempt the practice questions on Chapter 3: Functions, Exercise 3: Target Exercise 3.3 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.