Question

If the prime sign (′) represents differentiation w.r.t.x and

ƒ′(x) = sin x + sin 4x . cosx then ƒ′$\left( 2x^{2} + \frac{\pi}{2} \right)$ at x = $\sqrt{\frac{\pi}{2}}$ is equal to –

• A. 0
• B. –1
• C. $- 2\sqrt{2\pi}$
• D. None of these

Answer 1

Answer: (C)

$- 2\sqrt{2\pi}$

Answer 2

ƒ′$\left( 2x^{2} + \frac{\pi}{2} \right)$ = $\frac{dƒ\left( 2x^{2} + \frac{\pi}{2} \right)}{dx}$ = $\frac{dƒ\left( 2x^{2} + \frac{\pi}{2} \right)}{d\left( 2x^{2} + \frac{\pi}{2} \right)}$

· $\frac{d\left( 2x^{2} + \frac{\pi}{2} \right)}{dx}$

Q ƒ′(x) = $\frac{dƒ(x)}{dx}$ = sin x + sin 4x . cos x, we get

$\frac{dƒ\left( 2x^{2} + \frac{\pi}{2} \right)}{d\left( 2x^{2} + \frac{\pi}{2} \right)}$ = sin$\left( 2x^{2} + \frac{\pi}{2} \right)$ +

sin 4$\left( 2x^{2} + \frac{\pi}{2} \right)$. cos$\left( 2x^{2} + \frac{\pi}{2} \right)$

= cos 2x² + sin 8x² . (–sin 2x²)

∴ ƒ′$\left( 2x^{2} + \frac{\pi}{2} \right)$ = (cos 2x² – sin 8x² . sin 2x²) . 4x

∴ at x = $\sqrt{\frac{\pi}{2}}$, ƒ′ $\left( 2x^{2} + \frac{\pi}{2} \right)$ = (cos π – sin 4π . sin π)

. 4 . $\sqrt{\frac{\pi}{2}}$ = $- 2\sqrt{2\pi}$.