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JEE Mains · Maths · STD 12 - 7.2 definite integral

વિધેય \(\mathrm{f}\) એ \([0,1]\) માં અનૃણ છે અને  \((0,1) \) પર દ્રીતીય વિકલનીય છે . જો \(\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int \limits_{0}^{x} f(t) \,d t\) \(0 \leq x \leq 1\) અને \(f(0)=0\) હોય તો  \(\lim \limits _{x \rightarrow 0} \frac{1}{x^{2}} \int \limits_{0}^{x} f(t)\, d t:\) ની કિમંત

  1. A  \(0\)
  2. B \(1\)
  3. C અસ્તિત્વ નથી 
  4. D  \(\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D)  \(\frac{1}{2}\)

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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