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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना फलन \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\left(1+\mathrm{x}^4\right)^{1 / 4}}\) द्वारा परिभाषित है तथा \(\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x}))))\) है। तो \(18 \int_0^{\sqrt{2 \sqrt{5}}} x^2 g(x) d x\) = ...........

  1. A \(33\)
  2. B \(36\)
  3. C \(42\)
  4. D \(39\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(39\)

Step-by-step Solution

Detailed explanation

\( f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}} \) \( f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\)…
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