Question

# Let $$f: \displaystyle \left ( - \frac{\pi}{2} , \frac{\pi}{2} \right ) \rightarrow R$$ be given by $$f(x) = (\log (\sec x + \tan x))^3$$. Then

A
f(x) is an odd funtion
B
f(x) is a one-one function
C
f(x) is an onto function
D
f(x) is an even function

Solution

## The correct options are A $$f(x)$$ is an odd funtion B $$f(x)$$ is a one-one function C $$f(x)$$ is an onto function$$f(x) = (\log (\sec x + \tan x))^3 \forall x \in \displaystyle \left ( - \frac{\pi}{2} , \frac{\pi}{2} \right )$$$$f(-x) = -f(x)$$, hence $$f(x)$$ is odd functionLet $$g(x) =\sec x + \tan x \forall x \in \displaystyle \left ( - \frac{\pi}{2}, \frac{\pi}{2} \right )$$$$\Rightarrow g'(x) = \sec x (\sec x + \tan x) > 0 \forall x \in \displaystyle \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$$$$\Rightarrow g(x)$$ is one-one functionHence $$(\log_e(g(x)))^3$$ is one-one function.and $$g(x) \in (0, \alpha) \forall x \in \displaystyle \left ( - \frac{\pi}{2}, \frac{\pi}{2} \right )$$$$\Rightarrow \log(g(x)) \in R$$. Hence $$f(x)$$ is an onto function.Mathematics

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