Let \(f:\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\to R\) be given by \(f(x)=(\log(\sec x+\tan x))^3\), then which is not true:
f(x) is an even function
Let \(g(x)=\log(\sec x+\tan x)\). Then \(g(-x)=\log(\sec x-\tan x)=\log\left(\dfrac{1}{\sec x+\tan x}\right)=-\log(\sec x+\tan x)=-g(x)\), so \(g\) is an odd function.
Since \(f(x)=(g(x))^3\) and the cube of an odd function is odd, \(f(-x)=(g(-x))^3=(-g(x))^3=-f(x)\), so f(x) is odd — statement (A) is true.
Differentiating, \(g'(x)=\sec x>0\) on \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), so g is strictly increasing. Since \(h(y)=y^3\) is strictly increasing on R, \(f=h\circ g\) is strictly increasing, hence one-one — statement (B) is true.
As \(x\to \dfrac{\pi}{2}^{-}\), \(\sec x+\tan x\to\infty\) so \(g(x)\to\infty\) and \(f(x)\to\infty\); as \(x\to -\dfrac{\pi}{2}^{+}\), \(f(x)\to-\infty\). Being continuous and strictly increasing from \(-\infty\) to \(\infty\), f takes every real value, so f is onto — statement (C) is true.
Since f is a non-zero odd function, \(f(-x)=-f(x)\ne f(x)\) in general, so f cannot be an even function. Statement (D) is the one that is NOT true.
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