Range of the function f(x) = cot^{-1} left( l | vedclass

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

Question

(B) Let $g(x) = 5x^2 - 8x + 4$.Completing the square,$g(x) = 5(x^2 - \frac{8}{5}x) + 4 = 5(x - \frac{4}{5})^2 + 4 - 5(\frac{16}{25}) = 5(x - \frac{4}{

Vedclass · Accepted Answer

$[\frac{\pi}{4}, \pi)$

Vedclass · Answer

(B) Let $g(x) = 5x^2 - 8x + 4$.Completing the square,$g(x) = 5(x^2 - \frac{8}{5}x) + 4 = 5(x - \frac{4}{5})^2 + 4 - 5(\frac{16}{25}) = 5(x - \frac{4}{5})^2 + \frac{4}{5}$.Since $5(x - \frac{4}{5})^2 \geq 0$,the range of $g(x)$ is $[\frac{4}{5}, \infty)$.Let $u = \log_{4/5}(g(x))$. Since the base $4/5 For $t \in [\frac{4}{5}, \infty)$,$\log_{4/5}(t) \in (\log_{4/5}(\infty), \log_{4/5}(\frac{4}{5})] = (-\infty, 1]$.Now,$f(x) = \cot^{-1}(u)$ where $u \in (-\infty, 1]$.Since $\cot^{-1}(u)$ is a strictly decreasing function,the range is $[\cot^{-1}(1), \cot^{-1}(-\infty)) = [\frac{\pi}{4}, \pi)$.

Range of the function $f(x) = \cot^{-1} \left( \log_{4/5} (5x^2 - 8x + 4) \right)$ is:

Similar Questions

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the domain of $f(x) = \sec^{-1}(2[x] + 1)$ is:

The number of real roots of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ is:

The domain of the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty, -a] \cup [a, \infty)$. Then $a$ is equal to

The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1} \left( \frac{4x+2[x]}{3} \right)$ is $[\alpha, \beta]$, then $12(\alpha + \beta)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module