Question

If $f\left(x\right)=x+\sin x;g\left(x\right)=e^{-x};u=\sqrt{c+1}-\sqrt{c};v=\sqrt{c}-\sqrt{c-1};(c>1),$ then

• A. $f\circ g\left(u\right)<f\circ g\left(v\right)$
• B. $g\circ f\left(u\right)<g\circ f\left(v\right)$
• C. $g\circ f\left(u\right)>g\circ f\left(v\right)$
• D. $f\circ g\left(u\right)>f\circ g\left(v\right)$

Answer 1

The correct option is C $g\circ f\left(u\right)>g\circ f\left(v\right)$

$u=\sqrt{c+1}-\sqrt{c}$
$u=\frac{1}{\sqrt{c+1}+\sqrt{c}}$
and $v=\frac{1}{\sqrt{c-1}+\sqrt{c}}$
Clearly, $u<v$ and they lie in $(0,1)$
$f^{\prime }\left(x\right)=1+\cos x$
For $x\in (0,1)$
$f^{\prime }\left(x\right)>0$, so $f\left(x\right)$ is increasing .
Therefore,
$f\left(u\right)<f\left(v\right)$
We know that $g\left(x\right)$ is decreasing function, so
$g\circ f\left(u\right)>g\circ f\left(v\right)$