Q58 / 155 · WBJEE 2022 · Mathematics

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

  • A
    $a=2, b=2$
  • Correct answer
    B
    $a=1 / 4, b=1$
  • C
    $a=-1, b=4$
  • D
    $a=2, b=4$

Solution

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, $b=4 a=1$

Hence, the correct answer is option: 'B'.

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