Let f be a real valued function defined by f( | vedclass

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

Question

(A) For the function $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$ to be defined,the arguments of $

Vedclass · Accepted Answer

$[-4, 4]$

Vedclass · Answer

(A) For the function $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$ to be defined,the arguments of $\sin^{-1}$ and $\cos^{-1}$ must lie in the interval $[-1, 1]$.$1$. For $\sin^{-1} \left( \frac{1 - |x|}{3} \right)$,we have:$-1 \le \frac{1 - |x|}{3} \le 1$$-3 \le 1 - |x| \le 3$$-4 \le -|x| \le 2$$-2 \le |x| \le 4$Since $|x| \ge 0$,this implies $0 \le |x| \le 4$.$2$. For $\cos^{-1} \left( \frac{|x| - 3}{5} \right)$,we have:$-1 \le \frac{|x| - 3}{5} \le 1$$-5 \le |x| - 3 \le 5$$-2 \le |x| \le 8$Since $|x| \ge 0$,this implies $0 \le |x| \le 8$.Taking the intersection of both conditions:$0 \le |x| \le 4$This inequality is equivalent to $-4 \le x \le 4$.Thus,the domain of $f(x)$ is $[-4, 4]$.

Let $f$ be a real valued function defined by $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$. Then the domain of $f(x)$ is given by:

Similar Questions

The domain of the derivative of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatorname{Sin}^{-1}(x - 2)$ is

If $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

The domain of the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$ is

The number of real roots of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ is:

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the domain of $f(x) = \sec^{-1}(2[x] + 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module