Question

If $ f\,(x)=\underset{y\to x}{\mathop{lim}}\,\,\frac{{{\sin }^{2}}y-{{\sin }^{2}}x}{{{y}^{2}}-{{x}^{2}}}, $ then $ \int{4x\,\,f(x)\,\,dx} $ =

• A. $ \cos \,\,2x+c $
• B. $ 2\,\cos \,\,2x+c $
• C. $ -\,\cos \,\,2x+c $
• D. $ -2\,\cos \,\,2x+c $

Answer 1

The Correct Option is C

$ f(x)=\underset{y\to x}{\mathop{\lim }}\,\,\,\,\frac{{{\sin }^{2}}\,y-\,{{\sin }^{2}}x}{{{y}^{2}}-{{x}^{2}}} $
$ \left[ \frac{0}{0}\,form \right] $
$ =\underset{y\to x}{\mathop{\lim }}\,\,\frac{2\sin \,y\,\cos \,y-0}{2y-0} $
$ =\frac{\sin \,2x}{2x} $
$ \therefore $ $ \int{4x\,\,f(x)\,\,dx=\int{4x\,\,\left( \frac{\sin 2x}{2x} \right)}\,\,dx} $
$ =2\int{\sin \,\,2x\,\,dx} $
$ =-\cos \,2x+c $