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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Let $\mathrm{f}$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int \limits_{0}^{x} f(t) \,d t$ $0 \leq x \leq 1$ and $f(0)=0$, then $\lim \limits _{x \rightarrow 0} \frac{1}{x^{2}} \int \limits_{0}^{x} f(t)\, d t:$
  • A
    equals $0$
  • B
    equals $1$
  • C
    does not exist
  • ✓
    equals $\frac{1}{2}$
✓

Answer

Correct option: D.
equals $\frac{1}{2}$
d
$\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t \quad 0 \leq x \leq 1$

differentiating both the sides

$\sqrt{1-\left(f^{\prime}(x)\right)^{2}}=f(x)$

$\Rightarrow 1-\left(f^{\prime}(x)\right)^{2}=f^{2}(x)$

$\frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$

$\sin ^{-1} f(x)=x+C$

$\because f(0)=0 \Rightarrow C=0 \Rightarrow f(x)=\sin x$

Now $\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t \,\mathrm{dt}}{x^{2}}\left(\frac{0}{0}\right)=\frac{1}{2}$

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