If the real valued function f(x)=sin ^{-1}(x^ | vedclass

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

Question

(D) The function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is defined when both terms are defined.For $\sin ^{-1}(x^2-1)$, we require $-1 \leq x^2-1 \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is defined when both terms are defined.For $\sin ^{-1}(x^2-1)$, we require $-1 \leq x^2-1 \leq 1$, which implies $0 \leq x^2 \leq 2$, so $x \in [-\sqrt{2}, \sqrt{2}]$.For $\log _3(3^x-2)$, we require $3^x-2 > 0$, which implies $3^x > 2$, so $x > \log _3 2$.The domain of $f(x)$ is the intersection: $[-\sqrt{2}, \sqrt{2}] \cap (\log _3 2, \infty) = (\log _3 2, \sqrt{2}]$.The function is not defined for $x \in (-\infty, \log _3 2] \cup (\sqrt{2}, \infty)$.Comparing this with $(-\infty, a] \cup (b, \infty)$, we get $a = \log _3 2$ and $b = \sqrt{2}$.Thus, $3^a + b^2 = 3^{\log _3 2} + (\sqrt{2})^2 = 2 + 2 = 4$.

If the real valued function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is not defined for all $x \in(-\infty, a] \cup(b, \infty)$, then $3^a+b^2=$

Similar Questions

If the domain of the function $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$,then $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)$ is equal to $....$.

If $f:R \to S$ defined by $f(x) = \sin x - \sqrt{3} \cos x + 1$ is onto,then the interval of $S$ is

Let $f(x) = \frac{1}{2} - \tan \left(\frac{\pi x}{2}\right), -1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$, then the domain of $(f + g)$ is

Let $A = \{9, 10, 11, 12, 13\}$ and let $f: A \rightarrow N$ be defined by $f(n) = \text{the highest prime factor of } n$. Find the range of $f$.

The range of the real valued function $f(x) = \frac{x^2+x+1}{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module