JEE Mains · Maths · STD 12 - 7.2 definite integral
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
- D equals \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) equals \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t \quad 0 \leq x \leq 1\) differentiating both the sides \(\sqrt{1-\left(f^{\prime}(x)\right)^{2}}=f(x)\) \(\Rightarrow 1-\left(f^{\prime}(x)\right)^{2}=f^{2}(x)\)…
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