WBJEE · Maths · Definite Integration
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 \(a=2, b=2\)
- B \(a=1 / 4, b=1\)
- C \(a=-1, b=4\)
- D \(a=2, b=4\)
Answer & Solution
Correct Answer
(B) \(a=1 / 4, b=1\)
Step-by-step Solution
Detailed explanation
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, \(b=4 a=1\)
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