Indefinite Integration 3 Question 4
4. Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int _0^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int _0^{x} f(t) d t, 0 \leq x \leq 1 \quad$ and $f(0)=0$, then (2009) (a) $f \frac{1}{2}<\frac{1}{2}$ and $f \frac{1}{3}>\frac{1}{3}$
(b) $f \frac{1}{2}>\frac{1}{2}$ and $f \frac{1}{3}>\frac{1}{3}$
(c) $f \frac{1}{2}<\frac{1}{2}$ and $f \frac{1}{3}<\frac{1}{3}$
(d) $f \frac{1}{2}>\frac{1}{2}$ and $f \frac{1}{3}<\frac{1}{3}$
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Answer:
Correct Answer: 4. (c)
Solution:
- Given $\int _0^{x} \sqrt{1-{f^{\prime}(t) }^{2}} d t=\int _0^{x} f(t) d t, 0 \leq x \leq 1$
Differentiating both sides w.r.t. $x$ by using Leibnitz’s rule, we get
$$ \begin{aligned} & \quad \sqrt{1-{f^{\prime}(x) }^{2}}=f(x) \Rightarrow f^{\prime}(x)= \pm \sqrt{1-{f(x)}^{2}} \\ & \Rightarrow \quad \int \frac{f^{\prime}(x)}{\sqrt{1-{f(x)}^{2}}} d x= \pm \int d x \Rightarrow \sin ^{-1}{f(x)}= \pm x+c \\ & \text { Put } \quad x=0 \Rightarrow \sin ^{-1}{f(0)}=c \\ & \Rightarrow \quad c=\sin ^{-1}(0)=0 \\ & \qquad \quad[\because f(0)=0] \end{aligned} $$
$$ \begin{aligned} & \therefore \quad f(x)= \pm \sin x \\ & \text { but } \quad f(x) \geq 0, \forall x \in[0,1] \\ & \therefore \quad f(x)=\sin x \end{aligned} $$
$$ \begin{array}{ll} & \sin x<x, \forall x>0 \\ \therefore & \sin \frac{1}{2}<\frac{1}{2} \text { and } \sin \frac{1}{3}<\frac{1}{3} \\ \Rightarrow & f \frac{1}{2}<\frac{1}{2} \text { and } f \frac{1}{3}<\frac{1}{3} \end{array} $$
