The range of f(x)=sqrt{frac{a-|x|}{(a+1)-|x|} | vedclass

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

Question

(C) Given function is $f(x) = \sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a > 0)$.Since $f(x) \geq 0$ for all $x$ in the domain,let $y^2 = \frac{a-|x|}{(a+1)-|x|

Vedclass · Accepted Answer

$\left[0, \sqrt{\frac{a}{a+1}}\right] \cup (1, \infty)$

Vedclass · Answer

(C) Given function is $f(x) = \sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a > 0)$.Since $f(x) \geq 0$ for all $x$ in the domain,let $y^2 = \frac{a-|x|}{(a+1)-|x|}$ where $y \geq 0$.Then $y^2((a+1)-|x|) = a-|x|$.$y^2(a+1) - y^2|x| = a - |x|$.$|x|(1 - y^2) = a - y^2(a+1)$.$|x| = \frac{a - y^2(a+1)}{1 - y^2} = \frac{y^2(a+1) - a}{y^2 - 1}$.Since $|x| \geq 0$,we have $\frac{y^2(a+1) - a}{y^2 - 1} \geq 0$.Solving the inequality using the critical points $y^2 = \frac{a}{a+1}$ and $y^2 = 1$,we get $y^2 \in [0, \frac{a}{a+1}] \cup (1, \infty)$.Since $f(x) = y$,the range is $[0, \sqrt{\frac{a}{a+1}}] \cup (1, \infty)$.

The range of $f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a>0)$ is

Similar Questions

If $f: R \to R$,then the range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

Domain of $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ 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

If $f(x) = \frac{2x-3}{(x-2)(x-3)}$ is a real-valued function,then the value that $f(x)$ does not take is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module