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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

यदि \(x \in\left(0, \frac{1}{4}\right)\) के लिए \(\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right)\) का अवकलन
\(\sqrt{x} \cdot g(x)\) है, तो \(g(x)\) बराबर है:

  1. A \(\frac{3}{{1 + 9{x^3}}}\)
  2. B \(\frac{9}{{1 + 9{x^3}}}\)
  3. C \(\frac{{3x\sqrt x }}{{1 - 9{x^3}}}\)
  4. D \(\;\frac{3}{{1 - 9{x^3}}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{9}{{1 + 9{x^3}}}\)

Step-by-step Solution

Detailed explanation

Let \(F\left( x \right) = {\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)\) where \(x \in \left( {0,\frac{1}{4}} \right)\).…
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