Limit Continuity And Differentiability

Limit Continuity And Differentiability Ques 140

Let $f$ be a differentiable function from $\mathbb{R}$ to $\mathbb{R}$ such that $|f(x)-f(y)| \leq 2|x-y|^{\frac{3}{2}}$, for all $x, y \in \mathbb{R}$. If $f(0)=1$, then $\int _0^{1} f^{2}(x) , dx$ is equal to

(2019 Main, 9 Jan II)

(b) $\frac{1}{2}$

Show Answer

Answer:

Correct Answer: 140.(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 $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x} \left( \frac{d y}{d x} \right)=\frac{d}{d t} \left( \frac{d y}{d x} \right) \cdot \frac{d t}{d x}$

$$ =\frac{\frac{d}{d t} \frac{d y}{d x}}{\frac{d x}{d t}}=\frac{\frac{d}{d t}(\sin t)}{\frac{d}{d t}(3 \tan t)}=\frac{\cos t}{3 \sec ^{2} t}=\frac{1}{3} \cos t \cos^{2} t $$

Now, $\frac{d^{2} y}{d x^{2}}$ at $t=\frac{\pi}{4}=\frac{\cos ^{3} \frac{\pi}{4}}{3}=\frac{1}{3 \cdot 2 \sqrt{2}}=\frac{1}{6 \sqrt{2}}$