The complete solution set of the inequality ( | vedclass

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

Question

(A) Let $y = \sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,so $y \in [0, \pi/2) \cup (\pi/2, \pi]$.Given inequality: $(y

Vedclass · Accepted Answer

$[\sec 2, \sec 1]$

Vedclass · Answer

(A) Let $y = \sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,so $y \in [0, \pi/2) \cup (\pi/2, \pi]$.Given inequality: $(y - 4)(y - 1)(y - 2) \ge 0$.Since $y \in [0, \pi] \approx [0, 3.14]$,the term $(y - 4)$ is always negative.Dividing by $(y - 4)$,the inequality reverses: $(y - 1)(y - 2) \le 0$.This holds for $y \in [1, 2]$.Since $y = \sec^{-1}x$,we have $1 \le \sec^{-1}x \le 2$.Applying the secant function (which is increasing on $[0, \pi/2)$ and $[1, \pi]$),we get $\sec 1 \le x \le \sec 2$.However,we must respect the domain of $\sec^{-1}x$. Since $1 \le y \le 2$,$x$ must be in the range of $\sec y$. For $y \in [1, 2]$,$x \in [\sec 1, \sec 2]$ is valid.

The complete solution set of the inequality $(\sec^{-1}x - 4)(\sec^{-1}x - 1)(\sec^{-1}x - 2) \ge 0$ is

Similar Questions

If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$,then for $x < 0$ the value of $500 x^4+270 x^2+997=$

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

If $\sinh ^{-1}(-\sqrt{3})+\cosh ^{-1}(2)=K$,then $\cosh K=$

The value of $\sin ^{-1}(\sin 100) + \cos ^{-1}(\cos 100) + \tan ^{-1}(\tan 100) + \cot ^{-1}(\cot 100)$ is:

If the equation $2 \operatorname{Cot}^{-1}(x^2+2x+k) = \pi - 3 \operatorname{Tan}^{-1}(x^2+2x+k)$ has two distinct real solutions,then all the values of $k$ lie in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module