SATHEE: Limit Continuity and Differentiability 7 Question 9

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) $\frac{1}{2}$

(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 $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} \frac{d y}{d x} = \frac{d}{d t} \frac{d y}{d x} \cdot \frac{d t}{d x}$

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

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

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

Answer:

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Limit Continuity and Differentiability 7 Question 9

9. Let f be a differentiable function from R to R such that |f(x)−f(y)|≤2|x−y|32, for all x,y∈R. If f(0)=1, then ∫01f2(x)dx is equal to

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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