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

f x = x - 1 x 2 - 2 x + 3 Here, domain x ∈ R , Let y = x - 1 x 2 - 2 x + 3 ⇒ x 2 y - 2 x y + 3 y = x - 1 ⇒ x 2 y + x ( - 2 y - 1 ) + 3 y + 1 = 0 Here, x ∈ R so D ≥ 0 ( - 2 y - 1 ) 2 - 4 y ( 3 y + 1 ) ≥ 0 4 y 2 + 1 + 4 y - 12 y 2 - 4 y ≥ 0 - 8 y 2 + 1 ≥ 0 8 y 2 - 1 ≤ 0 y 2 - 1 8 ≤ 0 y 2 - 1 2 2 2 ≤ 0 y - 1 2 2 y + 1 2 2 ≤ 0 The range of f x is y ∈ - 1 2 2 , 1 2 2 .

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

f x = x - 1 x 2 - 2 x + 3 Here, domain x ∈ R , Let y = x - 1 x 2 - 2 x + 3 ⇒ x 2 y - 2 x y + 3 y = x - 1 ⇒ x 2 y + x ( - 2 y - 1 ) + 3 y + 1 = 0 Here, x ∈ R so D ≥ 0 ( - 2 y - 1 ) 2 - 4 y ( 3 y + 1 ) ≥ 0 4 y 2 + 1 + 4 y - 12 y 2 - 4 y ≥ 0 - 8 y 2 + 1 ≥ 0 8 y 2 - 1 ≤ 0 y 2 - 1 8 ≤ 0 y 2 - 1 2 2 2 ≤ 0 y - 1 2 2 y + 1 2 2 ≤ 0 The range of f x is y ∈ - 1 2 2 , 1 2 2 .

Find the range of f x = sin x 1 + tan 2 x - cos x 1 + cot 2 x .

f ( x ) = sin x 1 + tan 2 x - cos x 1 + cot 2 x f ( x ) = sinx secx - cosx cosecx Domain x ∈ R - ( 2 n + 1 ) π 2 , n π , where n ∈ I f ( x ) = ( sin x ) | cos x | - ( cos x ) | sin x | First, we'll check in [ 0 , 2 π ] interval f ( x ) = 0 x ∈ 0 , π 2 - 2 sin x cos x x ∈ π 2 , π 0 x ∈ π , 3 π 2 2 sin x cos x x ∈ 3 π 2 , 2 π So, f ( x ) = 0 x ∈ 0 , π 2 - sin 2 x x ∈ π 2 , π 0 x ∈ π , 3 π 2 sin 2 x x ∈ 3 π 2 , 2 π So, range of f ( x ) = [ - 1 , 1 ]

Find the range of f x = tan π x 2 - x 1 + sin cos x

f x = tan π [ x 2 - x ] 1 + sin cos x We know that for x = n π , n ∈ Z tan x = 0 Here, [ x 2 - x ] ∈ Z (greatest integer function always gives an integer) So, tan ( π [ x 2 - x ] ) = 0 f ( x ) = 0 Range = { 0 }

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 }