Indefinite Integration 3 Question 1
1. If $\int _0^{x} f(t) d t=x^{2}+\int _x^{1} t^{2} f(t) d t$, then $f^{\prime} \frac{1}{2}$ is
(a) $\frac{24}{25}$
(b) $\frac{18}{25}$
(c) $\frac{6}{25}$
(d) $\frac{4}{5}$
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Answer:
Correct Answer: 1. (a)
Solution:
- Given, $\int _0^{x} f(t) d t=x^{2}+\int _x^{1} t^{2} f(t) d t$
On differentiating both sides, w.r.t. ’ $x$ ‘, we get
$$ f(x)=2 x+0-x^{2} f(x) $$
$$ \begin{aligned} & \because \frac{d}{d x} \int _{\varphi(x)}^{\psi(x)} f(t) d t=f(\Psi(x)) \frac{d}{d x} \Psi(x)-f(\varphi(x)) \frac{d}{d x} \varphi(x) \\ & \Rightarrow \quad\left(1+x^{2}\right) f(x)=2 x \Rightarrow \quad f(x)=\frac{2 x}{1+x^{2}} \end{aligned} $$
On differentiating w.r.t. ’ $x$ ’ we get
$$ \begin{aligned} f^{\prime}(x) & =\frac{\left(1+x^{2}\right)(2)-(2 x)(0+2 x)}{\left(1+x^{2}\right)^{2}} \\ & =\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2-2 x^{2}}{\left(1+x^{2}\right)^{2}} \end{aligned} $$
$$ \therefore f^{\prime} \frac{1}{2}=\frac{2-2 \frac{1}{2}^{2}}{1+\frac{1}{2}^{2^{2}}}=\frac{2-2 \frac{1}{4}}{1+\frac{1}{4}^{2}}=\frac{2-\frac{1}{2}}{\frac{5}{4}^{2}}=\frac{\frac{3}{2}}{\frac{25}{16}}=\frac{24}{25} $$
