Identify the correct order.

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

Question

(D) To determine the correct order,we evaluate the values of the inverse secant function:$1$. $\sec^{-1}(1) = 0$$2$. $\sec^{-1}(2) = \frac{\pi}{3} \ap

Vedclass · Accepted Answer

$sec^{-1}(1) < sec^{-1}(2) < sec^{-1}(-2) < sec^{-1}(-1)$

Vedclass · Answer

(D) To determine the correct order,we evaluate the values of the inverse secant function:$1$. $\sec^{-1}(1) = 0$$2$. $\sec^{-1}(2) = \frac{\pi}{3} \approx 1.047$$3$. $\sec^{-1}(-2) = \frac{2\pi}{3} \approx 2.094$$4$. $\sec^{-1}(-1) = \pi \approx 3.141$Comparing these values,we get $0 Thus,the correct order is $\sec^{-1}(1) < \sec^{-1}(2) < \sec^{-1}(-2) < \sec^{-1}(-1)$.

Similar Questions

Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $a=3, b=4$ and $A=\sin^{-1}\left(\frac{3}{4}\right)$,then the angle $B$ is (in $^{\circ}$)

$\sin^{-1}(\sin \frac{7\pi}{6}) = $

If $2 f(\sin x) + f(\cos x) = x$,then $f^{\prime}(x) = $

Find the principal values of $\sin ^{-1}\left(-\frac{1}{2}\right)$.

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module