Solve qn 48 and 49 pls

Question

Answer 1

(48) Given f(x) = 3sin^-1x - 2cos^-1x

\(\therefore\) f(-x) = 3sin^-1(-x) - 2cos^-1(-x)

 = -3sin^-1x - 2(π - cos^-1x)

(\(\because\) sin^-1x + -sin^-1x and cos^-1(x) = π - cos^-1x)

= -(3sin^-1x - 2 cos^-1x) - 2π

 = -f(x) - 2π

\(\because\) f(-x) \(\neq\) f(x) and f(-x) \(\neq\) - f(x)

\(\therefore\) f(x) is neither even nor off function.

(49) Correct option is (B) \(\frac{\pi^3}{32}\)

Let f(x) = (sin^-1x)³ + (cos^-1x)³

 = (sin^-1x + cos^-1x)((sin^-1x)² + (cos^-1x)² - sin^-1x cos^-1x)

(\(\because\) a³ + b³ = (a + b) (a² + b² - ab))

= \(\frac{\pi}2((sin^{-1}x+cos^{-1}x)^2-3sin^{-1}xcos^{-1}x)\)

(\(\because sin^{-1}x+cos^{-1}x=\frac{\pi}2\) and a² + b² = (a + b)² - 2ab)

= \(\frac{\pi}2((\frac{\pi}2)^2-3sin^{-1}xcos^{-1}x)\) (\(\because sin^{-1}x+cos^{-1}x = \frac{\pi}2\))

 = \(\frac{\pi}2(\frac{\pi^2}4-3sin^{-1}x(\frac{\pi}2-sin^{-1}x))\) 

= \(\frac{\pi}2(\frac{\pi^2}4-\frac{3\pi}4sin^{-1}x+3(sin^{-1}x)^2)\)

 = \(\frac{\pi}8(\pi^2-6\pi sin^{-1}x + 12(sin^{-1}x)^2)\) 

 = \(\frac{12\pi}8((sin^{-1}x)^2-\frac{\pi}2sin^{-1}x+\frac{\pi^2}{12})\) 

 = \(\frac{3\pi}2((sin^{-1}x-\frac{\pi}4)^2+\frac{\pi^2}{12}-\frac{\pi^2}{16})\) 

 = \(\frac{3\pi}2((sin^{-1}x-\frac{\pi}4)^2+\frac{\pi^2}{48})\) 

\(\therefore\) (sin^-1x)³ + (cos^-1x)³ = \(\frac{3\pi}2((sin^{-1}x-\frac{\pi}4)^2+\frac{\pi^2}{48})\)

\(\therefore\) Minimum value of (sin^-1x)³ + (cos^-1x)³ occurs when

sin^-1x = \(\frac{\pi}4\) and minimum value is \(\frac{3\pi}2\times\frac{\pi^2}{48}\) = \(\frac{\pi^3}{32}\)