Limit Continuity and Differentiability 7 Question 9
9. Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x)-f(y)| \leq 2|x-y|^{\frac{3}{2}}$, for all $x, y \in R$. If $f(0)=1$, then $\int _0^{1} f^{2}(x) d x$ is equal to
(2019 Main, 9 Jan II)
(a) 2
(b) $\frac{1}{2}$
(c) 1
(d) 0
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Answer:
Correct Answer: 9. (c)
Solution:
- We have, $x=3 \tan t$ and $y=3 \sec t$
Clearly, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{d}{d t}(3 \sec t)}{\frac{d}{d t}(3 \tan t)}$
$$ =\frac{3 \sec t \tan t}{3 \sec ^{2} t}=\frac{\tan t}{\sec t}=\sin t $$
and $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x} \quad \frac{d y}{d x}=\frac{d}{d t} \frac{d y}{d x} \cdot \frac{d t}{d x}$
$$ =\frac{\frac{d}{d t} \frac{d y}{d x}}{\frac{d x}{d t}}=\frac{\frac{d}{d t}(\sin t)}{\frac{d}{d t}(3 \tan t)}=\frac{\cos t}{3 \sec ^{2} t}=\frac{\cos ^{3} t}{3} $$
Now, $\frac{d^{2} y}{d x^{2}}$ at $t=\frac{\pi}{4}=\frac{\cos ^{3} \frac{\pi}{4}}{3}=\frac{1}{3(2 \sqrt{2})}=\frac{1}{6 \sqrt{2}}$
