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:
- 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 $\sin x=1 / \sqrt{3}$
$$ \text { i.e. } \quad f \frac{1}{\sqrt{3}}=(\sqrt{3})^{2}=3 $$
