Let the maximum value of ( sin - 1 x ) 2 + ( cos - 1 x ) 2 for x [ - 3 2 , ? 1 2 ] be m n 2 , where gcd ( m , n ) = 1 . Then m + n is equal to_____ [2026]

Question

Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]

Smart Achievers · Accepted Answer

(65)

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

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

$= \frac{π^{2}}{4} - 2 (\sin^{- 1} x) (\frac{π}{2} - \sin^{- 1} x)$

$= 2 {(\sin^{- 1} x - \frac{π}{4})}^{2} + \frac{π^{2}}{8}, where \sin^{- 1} x \in [\frac{- π}{3}, \frac{π}{4}]$

Then maximum value occurs at $\sin^{- 1} x = \frac{- π}{3}$

$Which is 2 {(\frac{π}{3} + \frac{π}{4})}^{2} + \frac{π^{2}}{8} = \frac{29 π^{2}}{36}$

$\Rightarrow m = 29 and n = 36$

$∴$ $m + n = 65$

Solution

Q.
Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let the maximum value of (sin-1x)2+(cos-1x)2 for x∈[-32,12] be mnπ2, where gcd(m,n)=1. Then m+n is equal to_____ [2026]

Ans. (65) (sin-1x)2+(cos-1x)2 =(sin-1x+cos-1x)2-2sin-1xcos-1x =π24-2(sin-1x)(π2-sin-1x) =2(sin-1x-π4)2+π28, where sin-1x∈[-π3,π4] Then maximum value occurs at sin-1x=-π3 Which is 2(π3+π4)2+π28=29π236 ⇒m=29 and n=36 ∴ m+n=65

Want Us to Email you About Special Offers & Updates?

Q.
Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]