JEE Main 2025 Jan 22 Shift 1, Mathematics Q5: Inverse Trigonometric Functions

JEE Main2025Jan 22, Shift 1Mathematics

Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of $16 ((sec^{- 1} x)^{2} + (csc^{- 1} x)^{2})$ is :

A
$24 π^{2}$
B
$18 π^{2}$
C
$31 π^{2}$
D
$22 π^{2}$

Solution

Use the identity $sec^{- 1} x + csc^{- 1} x = \frac{π}{2}$ . Let $a = sec^{- 1} x \in [0, π] ∖ {π /2}$ ; then $csc^{- 1} x = \frac{π}{2} - a$ .

$f (a) = 16 [a^{2} + (\frac{π}{2} - a)^{2}] = 16 [2 a^{2} - π a + \frac{π ^{2}}{4}]$ .

This quadratic in $a$ has its minimum at $a = \frac{π}{4}$ , where $f = 16 [\frac{2 π ^{2}}{16} - \frac{π ^{2}}{4} + \frac{π ^{2}}{4}] = 2 π^{2}$ .

Maximum on the allowed interval occurs at the endpoint $a = π$ (which gives the larger value than $a = 0$ ): $f (π) = 16 [2 π^{2} - π^{2} + \frac{π ^{2}}{4}] = 16 \cdot \frac{5 π ^{2}}{4} = 20 π^{2}$ .

Sum $= 20 π^{2} + 2 π^{2} = 22 π^{2}$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q5: Inverse Trigonometric Functions

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