If $\int\limits_0^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}$ for all $x \neq 0$ then $a$ and $b$ are given by
Answer & explanation
Correct answer: option 1
We have,
$\int\limits_0^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}$
Differentiating both sides w.r. to $x$, we get
$\frac{b x \cos 4 x-a \sin 4 x}{x^2}=\frac{4 a x \cos 4 x-a \sin 4 x}{x^2}$ for all $x \neq 0 $
$\Rightarrow (b-4 a) x \cos 4 x=0$ for all $x \neq 0$
$\Rightarrow b-4 a=0 \Rightarrow b=4 a$
Clearly, $a=\frac{1}{4}$ and $b=1$ satisfy it.