Question

The range of the function
$f\left(x\right)=\sin \cos \left(ln\left(\frac{x^2+1}{x^2+e}\right)\right)$ is

• A. $[\sin \cos 1,1)$
• B. $[\sin \cos 1,\cos 1)$
• C. $[\sin \cos 1,\sin 1)$
• D. $(\sin \cos 1,\sin 1]$

Answer 1

Answer: (C)

$[\sin \cos 1,\sin 1)$

We have $1<e\Rightarrow x^2+1<x^2+e\Rightarrow \frac{x^2+1}{x^2+e}<1\Rightarrow ln\left(\frac{x^2+1}{x^2+e}\right)<0\Rightarrow \sin \cos$ $ln\left(\frac{x^2+1}{x^2+e}\right)<\sin 1$......(1).

Again we have,

$x^2\ge 0$ for $x^2=0$ we have $f\left(x\right)=\sin \cos 1$......($\because cos(-1)=cos1$)

So, for $x^2\ge 0$, $f\left(x\right)\ge \sin \cos 1$........(2).

From (1) and (2) we get range of $f\left(x\right)$ is $[\sin \cos 1,\sin 1)$.