The domain of the real valued function f(x) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the function $f(x)$ to be defined,both parts must be defined.First,consider $\cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$. This is defined if

Vedclass · Accepted Answer

$\{-1, 1\}$

Vedclass · Answer

(D) For the function $f(x)$ to be defined,both parts must be defined.First,consider $\cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$. This is defined if $-1 \leq \frac{1 + x^2}{2 x} \leq 1$.This implies $\left| \frac{1 + x^2}{2 x} \right| \leq 1$,which means $|1 + x^2| \leq |2x|$.Since $1 + x^2 \geq 2|x|$ is only true when $|x| = 1$,we have $x = 1$ or $x = -1$.Now,check the first part $\sqrt{\cos(\sin x)}$.For $x = 1$,$\cos(\sin 1) > 0$ since $\sin 1 \approx 0.84$ radians,which is in the first quadrant.For $x = -1$,$\cos(\sin(-1)) = \cos(-\sin 1) = \cos(\sin 1) > 0$.Since both conditions are satisfied only at $x = 1$ and $x = -1$,the domain is $\{-1, 1\}$.

The domain of the real valued function $f(x) = \sqrt{\cos (\sin x)} + \cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$ is

Similar Questions

The range of the function $f(x) = \sqrt{9 - x^2}$ is

The domain of definition of the function $y(x)$ given by $2^x + 2^y = 2$ is

The domain of the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ is

Is it true that $x = e^{\log x}$ for all real $x$?

The range of the real valued function $f(x) = |x-2| + |x-3|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module