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Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}-1,(0 < x < \pi / 4)$. Then a and $b$ are given by
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The correct answer is:
$a=1 / 4, b=1$
Let $g(x)=\lim _{c \rightarrow 0} \int_E^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}-1$
$\lim _{x \rightarrow 0} g(x)=0=4 a-1 \quad \Rightarrow a=1 / 4
$$
$$
g^{\prime}(x)=\frac{b x \cos 4 x-a \sin 4 x}{x^2}$
Comparing, $b=4 a=1$
$\lim _{x \rightarrow 0} g(x)=0=4 a-1 \quad \Rightarrow a=1 / 4
$$
$$
g^{\prime}(x)=\frac{b x \cos 4 x-a \sin 4 x}{x^2}$
Comparing, $b=4 a=1$
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