Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of is : [2025]

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Solution

Q.
Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$ is : [2025]

1 $22 π^{2}$

2 $24 π^{2}$

3 $18 π^{2}$

4 $31 π^{2}$

Ans.
(1)

$f (x) = 16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$

   $= 16 [(\sec^{- 1} x + {{cosec}^{- 1} x)}^{2} - 2 \sec^{- 1} x {cosec}^{- 1} x]$

   $= 16 [{(\frac{π}{2})}^{2} - 2 (\frac{π}{2} - {cosec}^{- 1} x) {cosec}^{- 1} x]$

$= 16 [\frac{π^{2}}{4} - π {cosec}^{- 1} x + 2 {({cosec}^{- 1} x)}^{2}]$

   $= 32 [{({cosec}^{- 1} x)}^{2} - \frac{π}{2} {cosec}^{- 1} x + \frac{π^{2}}{8}]$

$\Rightarrow f (x) = 32 [{({cosec}^{- 1} x - \frac{π}{4})}^{2} + \frac{π^{2}}{16}]$

f(x) is maximum when x = –1

$∴ f_{m a x} = 32 \times 10 \frac{π^{2}}{16}$

f(x) is minimum when $x = \sqrt{2}$

$∴ f_{\min} = 32 \times \frac{π^{2}}{16}$

$∴$ Required sum $= 32 \times 10 \frac{π^{2}}{16} + 32 \times \frac{π^{2}}{16} = 22 π^{2}$

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