Question

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

• A. $fog\left(u\right)\ge fog\left(v\right)$
• B. $fog\left(u\right)\le fog\left(v\right)$
• C. $fog\left(u\right)>fog\left(v\right)$
• D. $fog\left(u\right)<fog\left(v\right)$

Answer 1

Answer: (C)

$fog\left(u\right)>fog\left(v\right)$

$f\left(g\right(x\left)\right)=e^{-x}+\sin \left(e^{-x}\right)$

Let $h\left(x\right)=fog\left(x\right)=e^{-x}+\sin \left(e^{-x}\right)$

$\Rightarrow h^{\prime }\left(x\right)=-e^{-x}+\cos \left(e^{-x}\right)\cdot (-e^{-x})=-e^{-x}(1+\cos \left(e^{-x}\right))1+\cos \left(t\right)\ge 0,\forall tand{e}^{-x}>0,\forall x\Rightarrow h^{\prime }\left(x\right)<,\forall x$

$\Rightarrow fog\left(x\right)$ is a decreasing function.

Now, $(\sqrt{c+1}-\sqrt{c})<(\sqrt{c}-\sqrt{c-1})\Rightarrow u<v\Rightarrow fog\left(u\right)>fog\left(v\right)$

Because $fog\left(x\right)$ is a decreasing function

Hence, $fog\left(u\right)>fog\left(v\right)$