Question

If $y=a{\sin }^{2}x+b{\cos }^{2}x$, then $\frac{d^{2}y}{d{x}^2}$ at $x=\frac{x}{4}$ is:

• A. $2(b-a)$
• B. $\sqrt{2}(a-b)$
• C. $0$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

$0$

We have,

$y=a{\sin }^{2}x+b{\cos }^{2}x$

On differentiating w.r.t $x$, we get

$\frac{dy}{dx}=a\left(2\sin x\cos x\right)+b\left(2\cos x\right(-\sin x\left)\right)$

$\frac{dy}{dx}=a\left(\sin 2x\right)-b\left(\sin 2x\right)$

$\frac{dy}{dx}=(a-b)\sin 2x$

Again, differentiating w.r.t $x$, we get

$\frac{d^{2}y}{d{x}^2}=(a-b)\left(2\cos 2x\right)$

$\frac{d^{2}y}{d{x}^2}=2(a-b)\left(\cos 2x\right)$

On putting $x=\frac{\pi }{4}$, we get

$\frac{d^{2}y}{d{x}^2}=2(a-b)\left(\cos \frac{\pi }{2}\right)$

$\frac{d^{2}y}{d{x}^2}=2(a-b)\times 0=0$

Hence, this is the answer.