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JEE Advanced · Mathematics · 25. AOD

Let \(f(x)=(1-x)^{2} \sin ^{2} x+x^{2}\) for all \(x \in I R\) and let \(g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t\) for all \(x \in(1, \infty)\).

Question: Which of the following is true?

  1. A \(g\) is increasing on \((1, \infty)\)
  2. B \(g\) is decreasing on \((1, \infty)\)
  3. C \(g\) is increasing on \((1,2)\) and decreasing on \((2, \infty)\)
  4. D \(g\) is decreasing on \((1,2)\) and increasing on \((2, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(g\) is decreasing on \((1, \infty)\)

Step-by-step Solution

Detailed explanation

\(g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t\), \(\therefore \quad g^{\prime}(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right] f(x)\)

Here \(f(x)>0, \forall x \in(1, \infty)\)

Let \(h(x)=\frac{2(x-1)}{x+1}-\ln x\)

\(\therefore \quad h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{(x+1)^{2} x} < 0, x \in(1, \infty)\)

\(\Rightarrow h(x)\) is decreasing function.

Hence, for \(x>1, h(x) < h(1) \Rightarrow h(x) < 0 \forall x>1\)

\(\Rightarrow g^{\prime}(x) < 0 \forall x \in(1, \infty)\)

Therefore, \(g(x)\) is decreasing on \((1, \infty)\).
From JEE Advanced
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