Considering only the principal values of the | vedclass

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

Question

(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.This implies $-1 \leq \frac{x^{2}

Vedclass · Accepted Answer

$[-\frac{1}{4}, \infty)$

Vedclass · Answer

(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.This implies $-1 \leq \frac{x^{2}-4x+2}{x^{2}+3} \leq 1$.Since $x^{2}+3 > 0$ for all $x \in \mathbb{R}$,we can multiply by $(x^{2}+3)$:$-(x^{2}+3) \leq x^{2}-4x+2 \leq x^{2}+3$.First inequality: $-x^{2}-3 \leq x^{2}-4x+2 \implies 2x^{2}-4x+5 \geq 0$. The discriminant $D = (-4)^{2} - 4(2)(5) = 16 - 40 = -24  0$ for all $x \in \mathbb{R}$.Second inequality: $x^{2}-4x+2 \leq x^{2}+3 \implies -4x \leq 1 \implies x \geq -\frac{1}{4}$.Therefore,the domain is $[-\frac{1}{4}, \infty)$.

Considering only the principal values of the inverse trigonometric functions,the domain of the function $f(x) = \cos^{-1}\left(\frac{x^{2}-4x+2}{x^{2}+3}\right)$ is:

Similar Questions

The domain of $f(x) = \cos^{-1}[x]$ is,where $[x]$ denotes the greatest integer function.

The range of the real valued function $f(x) = \operatorname{Cos}^{-1}\left(\frac{3}{\sqrt{9x^2 - 12x + 22}}\right)$ is

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 range of $\tan^{-1} x$ is

If $y = \operatorname{cosec}^{-1}(x)$ and $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module