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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?
Options:
Question: Which of the following is true?
Solution:
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Verified Answer
The correct answer is:
$g$ is decreasing on $(1, \infty)$
$g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$, $\therefore \quad g^{\prime}(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right] f(x)$
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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