Let ∫0x(bt cos 4t−asin4tt2) dt=asin4xx then a and b are given by

∵∫0x(btcos4t−asin4tt2) dt=asin4xxDiffer.both sides w.r.t ‘x’bxcos4x−asin4xx2=a(4xcos4x−sin4x)x2On comparing b=4a∵a=14,b=1

Let ∫0x(bt cos 4t−asin4tt2) dt=asin4xx then a and b are given by

∵∫0x(btcos4t−asin4tt2) dt=asin4xxDiffer.both sides w.r.t ‘x’bxcos4x−asin4xx2=a(4xcos4x−sin4x)x2On comparing b=4a∵a=14,b=1

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Let $\int_{0}^{x} (\frac{b t \cos 4 t - a \sin 4 t}{t^{2}}) d t = \frac{a \sin 4 x}{x}$ then $a$ and $b$ are given by

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$a = 2, b = 2$

$a = \frac{1}{4}, b = 1$

$a = - 1, b = 4$

$a = 2, b = 4$

answer is A.

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Detailed Solution

$∵ \int_{0}^{x} (\frac{b t \cos 4 t - a \sin 4 t}{t^{2}}) d t = \frac{a \sin 4 x}{x}$
Differ.both sides w.r.t ‘x’
$\frac{b x \cos 4 x - a \sin 4 x}{x^{2}} = \frac{a (4 x \cos 4 x - \sin 4 x)}{x^{2}}$
On comparing $b = 4 a$
$∵ a = \frac{1}{4}, b = 1$