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JEE Mains · Maths · STD 12 - 6. Application of derivatives

ધારો કે \(R -\{-1,1\}\) પર વ્યાખ્યાયિત વાસ્તવિક મૂલ્યવાળું વિધેય \('f'\) એ \(f(x)=3 \log _{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}\) મુજબ આપેલ છે. તો નીચેનામાંથી કયા અંતરાલોમાં વિધેય \(f ( x )\) વધે છે ?

  1. A \((-\infty,-1) \cup\left(\left[\frac{1}{2}, \infty\right)-\{1\}\right)\)
  2. B \((-\infty, \infty)-\{-1,1\}\)
  3. C \(\left(-1, \frac{1}{2}\right]\)
  4. D \(\left(-\infty, \frac{1}{2}\right]-\{-1\}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((-\infty,-1) \cup\left(\left[\frac{1}{2}, \infty\right)-\{1\}\right)\)

Step-by-step Solution

Detailed explanation

\(f(x)=3 \ell n(x-1)-3 \ell n(x+1)-\frac{2}{x-1}\) \(f^{\prime}(x)=\frac{3}{x-1}-\frac{3}{x+1}+\frac{2}{(x-1)^{2}}\) \(f^{\prime}(x)=\frac{4(2 x-1)}{(x-1)^{2}(x+1)}\) \(f^{\prime}(x) \geq 0\) \(\Rightarrow \quad x \in(-\infty,-1) \cup\left[\frac{1}{2}, 1\right) \cup(1, \infty)\)
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