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

Answer: (C)

$-\,\cos \,\,2x+c$

Answer 2

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