Relations and Functions — MATHS STD 11 Science — Question

2. $\Big[-1,\frac{1}{3}\Big]$

Solution:

We know that $-1\leq\cos\text{x}\leq1$ for all $\text{x}\in\text{R}$

Now,

$-1\leq\cos\text{x}\leq1$

$\Rightarrow-1\leq\cos\text{x}\leq1$

$\Rightarrow-2\leq-2\cos\text{x}\leq2$

$\Rightarrow-1\leq1-2\cos\text{x}\leq3$ (Adding 1 ro each term)

But,

$\cos\text{x}\neq\frac{1}{2}$

$\Rightarrow1-2\cos\text{x}\in\big[-1,3\big]-\{0\}$

$\Rightarrow\frac{1}{1-2\cos\text{x}}\in\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

$\therefore\ \text{Range of }\text{f(x)}=\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as $\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$