Let the maximum value of ( ^-1x)^2 + ( ^-1x)^2 for x [ - 32, 1 2 ] be mn ^2, where (m, n) = 1. Then m + n is equal to.

Given: The function is f(x) = ( ^-1x)^2 + ( ^-1x)^2 for x [ - 32, 1 2 ]. Find: The maximum value in the form mn ^2, and then compute m+n. Using the identity ^-1x + ^-1x = 2, let t = ^-1x. Then ^-1x = 2 - t. So the expression becomes g(t) = t^2 + ( 2 - t)^2. Expanding, g(t) = t^2 + ^24 - t + t^2 = 2t^2 - t + ^24. Now determine the interval for t: t [ ^-1 (- 32), ^-1 ( 1 2) ] = [ - 3, 4 ]. This is an upward-opening quadratic, so its maximum on the closed interval occurs at an endpoint. At t = 4, we get g ( 4) = 2 ( ^216) - ( 4) + ^24 = ^28. At t = - 3, we get g (- 3) = 2 ( ^29) - (- 3) + ^24. Thus, g (- 3) = 2 ^29 + ^23 + ^24 = 8 ^2 + 12 ^2 + 9 ^236 = 29 ^236. Comparing, 29 ^236 ^28. Therefore, the maximum value is 29 ^236. So, m = 29, n = 36. Hence, m+n = 65. Therefore, the required answer is 65.

Let the maximum value of ( ^-1x)^2 + ( ^-1x)^2 for x [ - 32, 1 2… — Solution | JEE Mathematics

NVAMediumJEE 2026Inverse Trigonometric Functions

Let the maximum value of $(\sin^{-1}x)^2 + (\cos^{-1}x)^2$ for $x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right]$ be $\frac{m}{n}\pi^2$ , where $\gcd(m, n) = 1$ . Then $m + n$ is equal to _____.

Answer

Correct answer:65

Step-by-step solution

Standard Method

Given: The function is

f(x) = (\sin^{-1}x)^2 + (\cos^{-1}x)^2

for

x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right].

Find: The maximum value in the form $\frac{m}{n}\pi^2$ , and then compute $m+n$ .

Using the identity

\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2},

let

t = \sin^{-1}x.

Then

\cos^{-1}x = \frac{\pi}{2} - t.

So the expression becomes

g(t) = t^2 + \left(\frac{\pi}{2} - t\right)^2.

Expanding,

g(t) = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}.

Now determine the interval for $t$ :

t \in \left[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right), \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] = \left[ -\frac{\pi}{3}, \frac{\pi}{4} \right].

This is an upward-opening quadratic, so its maximum on the closed interval occurs at an endpoint.

t = \frac{\pi}{4},

we get

g\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi^2}{16}\right) - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} = \frac{\pi^2}{8}.

t = -\frac{\pi}{3},

we get

g\left(-\frac{\pi}{3}\right) = 2\left(\frac{\pi^2}{9}\right) - \pi\left(-\frac{\pi}{3}\right) + \frac{\pi^2}{4}.

Thus,

g\left(-\frac{\pi}{3}\right) = \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \frac{8\pi^2 + 12\pi^2 + 9\pi^2}{36} = \frac{29\pi^2}{36}.

Comparing,

\frac{29\pi^2}{36} > \frac{\pi^2}{8}.

Therefore, the maximum value is $\frac{29\pi^2}{36}$ . So,

m = 29, \quad n = 36.

Hence,

m+n = 65.

Therefore, the required answer is $65$ .

Common mistakes

Taking the quadratic vertex as the maximum. This is wrong because $g(t)=2t^2-\pi t+\frac{\pi^2}{4}$ opens upward, so the vertex gives the minimum, not the maximum. Check the endpoints for the maximum on a closed interval.
Using the wrong range for $t=\sin^{-1}x$ . This is wrong because $t$ must be obtained by applying $\sin^{-1}$ to the given interval endpoints. Here, $t \in \left[-\frac{\pi}{3},\frac{\pi}{4}\right]$ , not a symmetric interval.
Forgetting the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ and treating both inverse trigonometric terms independently. This makes the expression harder than necessary. First rewrite everything in terms of a single variable.

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