Question

If $y= Cos^2 \frac {3x}{2}- Sin^2 \frac {3x}{2}$ ,then $ \frac {d^2y}{dx^2}$ is

• A. $9y$
• B. $-3\sqrt{1-y^2}$
• C. $3\sqrt{1-y^2}$
• D. $-9y$

Answer 1

Answer: (D)

Given, $y = \cos^{2} \frac{3x}{2} -\sin^{2} \frac{3x}{2} $
$ \Rightarrow y = \cos^{2} \frac{3x}{2} -\left(1 -\cos^{2} \frac{3x}{2}\right) $
$ \Rightarrow y = 2 \cos^{2} \frac{3x}{2} -1 $
$ \Rightarrow \frac{dy}{dx} = 2-2 \cos \frac{3x}{2} \left(- \sin \frac{3x}{2} \right) \left(\frac{3}{2}\right) $
$ \Rightarrow \frac{dy}{dx} =- 6 \cos \frac{3x}{2} \sin \frac{3x}{2} $
$ \Rightarrow \frac{d^{2}y}{dx^{2}} = - 6 \left[\cos \frac{3x}{2} \left(\cos \frac{3x}{2}\right) \frac{3}{2} - \sin \frac{3x}{2} \sin \frac{3x}{2} . \frac{3}{2}\right] $
$ \Rightarrow \frac{d^{2}y}{dx^{2}} = -9 \left[\cos^{2} \frac{3x}{2} -\sin^{2} \frac{3x}{2} \right] $
$ \Rightarrow \frac{d^{2}y}{dx^{2} } = - 9y $