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?
- A \(g\) is increasing on \((1, \infty)\)
- B \(g\) is decreasing on \((1, \infty)\)
- C \(g\) is increasing on \((1,2)\) and decreasing on \((2, \infty)\)
- D \(g\) is decreasing on \((1,2)\) and increasing on \((2, \infty)\)
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 \((1, \infty)\).
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 \((1, \infty)\).
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