JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(g:(0, \infty) \rightarrow R\) be a differentiable function such that \(\int\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right)}{\left(e^{x}+1\right)^{2}}\right) d x=\frac{x g(x)}{e^{x}+1}+c\) for all \(x >0\), where \(c\) is an arbitrary constant. Then.
- A \(g\) is decreasing in \(\left(0, \frac{\pi}{4}\right)\)
- B \(g^{\prime}\) is increasing in \(\left(0, \frac{\pi}{4}\right)\)
- C \(g+g^{\prime}\) is increasing in \(\left(0, \frac{\pi}{2}\right)\)
- D \(g - g ^{\prime}\) is increasing in \(\left(0, \frac{\pi}{2}\right)\)
Answer & Solution
Correct Answer
(D) \(g - g ^{\prime}\) is increasing in \(\left(0, \frac{\pi}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\int\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right)}{\left(e^{x}+1\right)^{2}}\right) d x=\frac{x g(x)}{e^{x}+1}+c\) On differentiating both sides w.r.t. \(x\), we get…
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