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

If $A = \{x \in R : \sin^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$ and $B = \{y \in R : y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\}$,then which of the following is true?

The domain of ${\sin ^{ - 1}}({\log _3}x)$ is

If the domain of the function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}\right)$ is $[\alpha, \beta) \cup (\gamma, \delta]$,then $|3\alpha + 10(\beta + \gamma) + 21\delta|$ is equal to $.......$.

The range of $\left|\cos ^{-1} x\right|$ is . . . . . . .

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module