Indefinite Integration 3 Question 5

5. If $\int _{\sin x}^{1} t^{2} f(t) d t=1-\sin x, \forall x \in(0, \pi / 2)$, then $f \frac{1}{\sqrt{3}}$ is

(a) 3

(b) $\sqrt{3}$

(c) $1 / 3$

(d) None of these

(2005, 1M)

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Answer:

Correct Answer: 5. (a)

Solution:

  1. Since $\int _{\sin x}^{1} t^{2} f(t) d t=1-\sin x$, thus to find $f(x)$.

On differentiating both sides using Newton Leibnitz formula

i.e. $\quad \frac{d}{d x} \int _{\sin x}^{1} t^{2} f(t) d t=\frac{d}{d x}(1-\sin x)$

$\Rightarrow{1^{2} f(1) } \cdot(0)-\left(\sin ^{2} x\right) \cdot f(\sin x) \cdot \cos x=-\cos x$

$$ \Rightarrow \quad f(\sin x)=\frac{1}{\sin ^{2} x} $$

For $f \frac{1}{\sqrt{3}}$ is obtained when $\sin x=1 / \sqrt{3}$

$$ \text { i.e. } \quad f \frac{1}{\sqrt{3}}=(\sqrt{3})^{2}=3 $$



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