Let -0xb t cos 4 t-a sin 4 t-t2 d t-a sin 4 x-x- then a and b can be

The correct option is B 1/4,1Since, ∫^x_0 (bt cos 4t a sin 4t/t^2)dt = a sin 4x/x Differentiating both sides w.r.t. x ∴bx cos 4x a sin 4x/x^2 = a{ 4x cos 4 ...

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Standard XII

Mathematics

Property 5

Let ∫0xb t co...

Question

Let $\int_{0}^{x} (\frac{b t c o s 4 t - a s i n 4 t}{t^{2}}) d t = \frac{a s i n 4 x}{x}$ , then a and b can be

\frac{1}{4}, 1

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2,2

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Solution

The correct option is A $\frac{1}{4}, 1$
Since, $\int_{0}^{x} (\frac{b t c o s 4 t - a s i n 4 t}{t^{2}}) d t = \frac{a s i n 4 x}{x}$
Differentiating both sides w.r.t. x
$∴ \frac{b x c o s 4 x - a s i n 4 x}{x^{2}} = \frac{a {4 x c o s 4 x - s i n 4 x}}{x^{2}}$
So b = 4a

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Let ∫x0(bt cos 4t−a sin 4tt2)dt=a sin 4xx, then a and b can be

The correct option is A 14,1Since, ∫x0(bt cos 4t−a sin 4tt2)dt=a sin 4xx Differentiating both sides w.r.t. x ∴bxcos4x−asin4xx2=a{4xcos4x−sin4x}x2 So b = 4a

Let $\int_{0}^{x} (\frac{b t c o s 4 t - a s i n 4 t}{t^{2}}) d t = \frac{a s i n 4 x}{x}$ , then a and b can be

The correct option is A $\frac{1}{4}, 1$
Since, $\int_{0}^{x} (\frac{b t c o s 4 t - a s i n 4 t}{t^{2}}) d t = \frac{a s i n 4 x}{x}$
Differentiating both sides w.r.t. x
$∴ \frac{b x c o s 4 x - a s i n 4 x}{x^{2}} = \frac{a {4 x c o s 4 x - s i n 4 x}}{x^{2}}$
So b = 4a