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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
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B
f(x) is a one-one function
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C
f(x) is an onto function
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D
f(x) is an even function
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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 function
Let $$g(x) =\sec  x + \tan  x  \forall  x \in \displaystyle \left ( - \frac{\pi}{2}, \frac{\pi}{2} \right )$$
$$\Rightarrow
 g'(x) = \sec  x (\sec  x + \tan  x) > 0 \forall x \in \displaystyle
\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$$
$$\Rightarrow g(x)$$ is one-one function
Hence $$(\log_e(g(x)))^3$$ 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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