Question

Let $f\left(x\right)=e^{x^2+|x|}$ and $g\left(x\right)={(4{\cos }^{4}x-2\cos 2x-\frac{1}{2}\cos 4x-x^7)}^{1/7}.$ If domain of $f\left(x\right),g\left(x\right)$ is $[a,\infty ),R$ respectively and range of $f\left(x\right),g\left(x\right)$ is $[b,\infty ),R$ respectively, where $a={\sin }^{-1}\left(\sin 3\right)+{\sin }^{-1}\left(\sin 4\right)+{\sin }^{-1}\left(\sin 5\right),$ then which of the following is (are) CORRECT ?

• A. $a=-2$
• B. $b+a=-1$
• C. $f\left(gog\left(b\right)\right)=e^2$
• D. both $f\left(x\right)$ and $g\left(x\right)$ are invertible functions

Answer 1

Answer: (A), (B), (C)

$f\left(gog\left(b\right)\right)=e^2$


$a={\sin }^{-1}\left(\sin 3\right)+{\sin }^{-1}\left(\sin 4\right)+{\sin }^{-1}\left(\sin 5\right)$
$=(\pi -3)+(\pi -4)+(5-2\pi )=-2$

$\because x^2+|x|\in [0,\infty )\forall x\in [-2,\infty )$
$\Rightarrow e^{x^2+|x|}\in [1,\infty )$
$\Rightarrow f\left(x\right)\in [1,\infty )\Rightarrow b=1\therefore a+b=-1g\left(x\right)={({(1+\cos 2x)}^2-2\cos 2x-\frac{1}{2}(2{\cos }^{2}2x-1)-x^7)}^{1/7}={(1+2\cos 2x+{\cos }^{2}2x-2\cos 2x-{\cos }^{2}2x+\frac{1}{2}-x^7)}^{1/7}g\left(x\right)={(\frac{3}{2}-x^7)}^{1/7}$ $gog\left(x\right)={(\frac{3}{2}-{\left(g\left(x\right)\right)}^7)}^{1/7}={(\frac{3}{2}-(\frac{3}{2}-x^7))}^{1/7}={\left(x^7\right)}^{\frac{1}{7}}=x\Rightarrow f\left(gog\left(b\right)\right)=f\left(b\right)=e^{1+1}=e^2$

$\because f(-2)=f\left(2\right),$
$\Rightarrow f\left(x\right)$ is many-one function.
$\therefore f\left(x\right)$ is not an invertible function.