Question

Let $f\left(x\right)=\frac{2}{x+1}$; $g\left(x\right)=\cos x$ and $h\left(x\right)=\sqrt{x+3}$ then the range of the composite function $\left(fogoh\right)$ is

• A. $R^{+}$
• B. $R-\left\{0\right\}$
• C. $[1,\infty )$
• D. $R^{+}-\left\{1\right\}$

Answer 1

Answer: (C)

$[1,\infty )$

Given, $f\left(x\right)=\frac{2}{x+1}$; $g\left(x\right)=\cos x$ and $h\left(x\right)=\sqrt{x+3}$

So, $g\left(h\right(x\left)\right)$ $=\cos \left(\sqrt{x+3}\right)$
Now, $f\left(g\right(h\left(x\right)\left)\right)$ $=\frac{2}{\cos \left(\sqrt{x+3}\right)+1}$

Also, we know that the range of $cos\theta$ is $[-1,1]$

So, $-1\le \cos \left(\sqrt{x+3}\right)\le 1$

Adding $1$ on both the sides

So, $0\le \cos \left(\sqrt{x+3}\right)+1\le 2$

$⟹\infty \ge \frac{1}{\cos \left(\sqrt{x+3}\right)+1}\ge \frac{1}{2}$

Multiply whole inequality with 2

$⟹\infty \ge \frac{2}{\cos \left(\sqrt{x+3}\right)+1}\ge 1$

$⟹\infty \ge f\left(g\right(h\left)\right))\ge 1$

Therefore the range of $\left(fogoh\right)$ is $[1,\infty )$.