Question

Let f(x) = $\frac { \sin x } { \sqrt { 1 + \tan ^ { 2 } x } } - \frac { \cos x } { \sqrt { 1 + \cot ^ { 2 } x } }$ then range of f(x)is

• A. [-1, 0]
• B. [0, 1]
• C. [-1, 1]
• D. None

Answer 1

Answer: (C)

$-1, 1$

Answer 2

f(x) = $\frac { \sin x } { | \sec x | } - \frac { \cos x } { | \operatorname { cosec } x | }$

= sinx. |cosx| - cosx. |sinx|

Clearly domain of f(x) is R ~ $\left\{ n \pi , ( 2 n + 1 ) \frac { \pi } { 2 } \right\}$,

N ∈ I and period of f(x) is 2π.

f(x) = $\left\{ \begin{array} { l l } 0 & , x \in ( 0 , \pi / 2 ) \\ - \sin 2 x & , x \in ( \pi / 2 , \pi ) \\ 0 & , x \in ( \pi , 3 \pi / 2 ) \\ \sin 2 x & , x \in ( 3 \pi / 2,2 \pi ) \end{array} \right.$

⇒ Range of f(x) is [-1, 1].