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

(A) For the function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ to be defined,the expressions under the square roots must be non-negative: $1$. $x^2 - 144 \

Vedclass · Accepted Answer

$[-15, -12] \cup [12, 15]$

Vedclass · Answer

(A) For the function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ to be defined,the expressions under the square roots must be non-negative: $1$. $x^2 - 144 \geq 0$ $\Rightarrow x^2 \geq 144$ $\Rightarrow |x| \geq 12$. $2$. $9 - \sqrt{x^2 - 144} \geq 0 \Rightarrow 9 \geq \sqrt{x^2 - 144}$. Squaring both sides,we get $81 \geq x^2 - 144$ $\Rightarrow x^2 \leq 225$ $\Rightarrow |x| \leq 15$. Combining both conditions,we have $12 \leq |x| \leq 15$. This implies $x \in [-15, -12] \cup [12, 15]$.

The domain of the real valued function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ is

Similar Questions

The domain of the real valued function $f(x) = \log_2 \log_3 \log_5(x^2 - 5x + 11)$ is

If $f(x) = \tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$ is a real-valued function, then the range of $f$ is:

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

The domain of the real valued function $f(x) = \frac{\sqrt{2-x} + \sqrt{1+x}}{\sqrt{x+3}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module