Find the range of f x = 1 2 - x - x , (where . represents fractional part of x )

Given, f x = 1 2 - x - x Domain of f x will be x ∈ R - Z Since, at all integers value x = 0 . We know that x + - x = 1 when, x ≠ Z ⇒ - x = 1 - x ⇒ f x = 1 2 1 - x - x ⇒ f x = 1 2 1 - x + 1 - x - 1 Let t = 1 - x ⇒ 1 1 - x = 1 t As 1 - x > 0 for all values of x ∈ R - Z So, applying AM-GM inequality, t + 1 2 t 2 ≥ t . 1 2 t ⇒ t + 1 2 t ≥ 2 2 ⇒ t + 1 2 t ≥ 2 ⇒ 1 - x + 1 2 1 - x ≥ 2 ⇒ 1 2 1 - x - x ≥ 2 - 1 ⇒ f x ≥ 2 - 1 Hence, range of f ( x ) is [ 2 - 1 , ∞ ) .

Find the range of f x = 1 2 - x - x , (where . represents fractional part of x )

Given, f x = 1 2 - x - x Domain of f x will be x ∈ R - Z Since, at all integers value x = 0 . We know that x + - x = 1 when, x ≠ Z ⇒ - x = 1 - x ⇒ f x = 1 2 1 - x - x ⇒ f x = 1 2 1 - x + 1 - x - 1 Let t = 1 - x ⇒ 1 1 - x = 1 t As 1 - x > 0 for all values of x ∈ R - Z So, applying AM-GM inequality, t + 1 2 t 2 ≥ t . 1 2 t ⇒ t + 1 2 t ≥ 2 2 ⇒ t + 1 2 t ≥ 2 ⇒ 1 - x + 1 2 1 - x ≥ 2 ⇒ 1 2 1 - x - x ≥ 2 - 1 ⇒ f x ≥ 2 - 1 Hence, range of f ( x ) is [ 2 - 1 , ∞ ) .

If f : R → R ; f x = x 2 + 1 - 3 x x 2 + 1 + x , then find the range of f x .

Given, f : R → R ; f x = x 2 + 1 - 3 x x 2 + 1 + x ⇒ f x = x 2 + 1 + x - 4 x x 2 + 1 + x ⇒ f x = 1 - 4 x x 2 + 1 + x For monotonicity, find f ' x ⇒ f ' x = - 4 x 2 + 1 + x - x x x 2 + 1 + 1 x 2 + 1 + x 2 = - 4 1 x 2 + 1 . x 2 + 1 + x 2 Which is always negative for all x ∈ R Hence, f x is always decreasing for all x ∈ R Now, x → - ∞ , lim x → ∞ f x = lim x → - ∞ 1 - 4 x x 2 + 1 + x = 1 - 4 - ∞ - ∞ 2 + 1 + - ∞ = a positive value. x → ∞ , lim x → ∞ f x = lim x → ∞ 1 - 4 x x 2 + 1 + x lim x → ∞ 1 - 4 x x 1 + 1 x 2 + 1 = 1 - 2 = - 1 Since, f x is a decreasing function, so at x → - ∞ , it will have maximum value So, the range of f x is [ - 1 , ∞ ) .

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 .

Find the domain and range of the following functions f x = tan - 1 x + - x + 2 - x + 1 x 2 , (where . denotes greatest integer function)

Given, f x = tan - 1 x + - x + 2 - x + 1 x 2 We know that, in t , t ≥ 0 Here, x + - x ≥ 0 and 2 - x ≥ 0 Now, x + - x = 0 , when x ∈ Z - 1 , when x ∈ R - Z So, x + - x ≥ 0 only when x ∈ Z . . . i Now, for 2 - x ≥ 0 ⇒ x ≤ 2 ⇒ x ∈ - 2 , 2 . . . i i Taking intersection of i and i i ⇒ x ∈ - 2 , - 1 , 0 , 1 , 2 For 1 x 2 , x ≠ 0 Here, the domain of f x is n ∈ - 2 , - 1 , 1 , 2 . Now, f - 2 = tan - 1 2 + - 2 + 2 - - 2 + 1 ( 2 ) 2 = tan - 1 0 + 0 + 1 4 f - 1 = tan - 1 1 + - 1 + 2 - ( - 1 ) + 1 ( - 1 ) 2 = 0 + 1 + 1 = 2 f 1 = tan - 1 1 + - 1 + 2 - 1 + 1 ( 1 ) 2 = 0 + 1 + 1 = 2 f 2 = tan - 1 2 + - 2 + 2 - 2 + 1 ( 2 ) 2 = 1 4 . Hence, range of f x is 1 4 , 2 .

If f ( x ) = sin 2 x + 4 sin x + 5 2 sin 2 x + 8 sin x + 8 , then range of f ( x ) is

f ( x ) = sin 2 x + 4 sin x + 5 2 sin 2 x + 8 sin x + 8 = 1 2 2 sin 2 x + 8 sin x + 10 2 sin 2 x + 8 sin x + 8 = 1 2 ( 2 sin 2 x + 8 sin x + 8 ) + 2 ( 2 sin 2 x + 8 sin x + 8 ) = 1 2 1 + 2 sin 2 x + 8 sin x + 8 = 1 2 1 + 1 sin 2 x + 4 sin x + 4 = 1 2 1 + 1 ( sin x + 2 ) 2 . The f ( x ) will be minimum when 1 ( sin x + 2 ) 2 is minimum and 1 ( sin x + 2 ) 2 is minimum when ( sin x + 2 ) 2 is maximum and it is maximum when sin x is maximum which is when sin x = 1 . ⇒ ( f ( x ) ) min = 1 2 1 + 1 ( 1 + 2 ) 2 = 1 2 1 + 1 9 = 5 9 . The f ( x ) will be maximum when 1 ( sin x + 2 ) 2 is maximum and 1 ( sin x + 2 ) 2 is maximum when ( sin x + 2 ) 2 is minimum and it is minimum when sin x is minimum, which is when sin x = - 1 . ⇒ f x max = 1 2 1 + 1 ( - 1 + 2 ) 2 = 1 2 ( 1 + 1 ) = 1 ⇒ ( f ( x ) ) max = 1 . Hence, range of f ( x ) is 5 9 , 1 .

Find range of the function f ( x ) = log 2 [ 3 x - [ x + [ x + [ x ] ] ] ] (where [ · ] is greatest integer function)

Given, f ( x ) = log 2 [ 3 x - [ x + [ x + [ x ] ] ] ] We know that, [ x + n ] = [ x ] + n where n ∈ Z So, here [ x + [ x ] ] = [ x ] + [ x ] = 2 [ x ] Also, [ x + [ x + [ x ] ] ] = [ x + 2 [ x ] ] = [ x ] + 2 [ x ] = 3 [ x ] now 3 x - x + x + x = 3 x - 3 x = 3 x - x = 3 x f x = log 2 3 x we know that 0 ≤ x < 1 for all x ∈ R ⇒ 0 ≤ 3 x < 3 3 x = 0 , 1 , 2 So, the greatest integer gives 0 , 1 , 2 ( 3 excluded) For log to be defined [ 3 x - [ x + [ x + [ x ] ] ] ] ≠ 0 So, range can be log 2 ( 1 ) and log 2 ( 2 ) = { 0 , 1 }

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 9

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Find the range of $f (x) = \frac{1}{2 \{- x\}} - \{x\}$ , (where $\{.\}$ represents fractional part of $x$ )

Important Questions on Functions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 10

If

f : R \to R; f (x) = \frac{\sqrt{x^{2} + 1} - 3 x}{\sqrt{x^{2} + 1} + x}

, then find the range of

f (x)

.

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 14

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

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-4>Q 15

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