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JEE Advance - Mathematics (2012 - Paper 2 Offline - No. 9)

Let $$f\left( x \right) = {\left( {1 - x} \right)^2}\,\,{\sin ^2}\,\,x + {x^2}$$ for all $$x \in IR$$ and let
$$g\left( x \right) = \int\limits_1^x {\left( {{{2\left( {t - 1} \right)} \over {t + 1}} - In\,t} \right)f\left( t \right)dt} $$ for all $$x \in \left( {1,\,\infty } \right)$$.

Which of the following is true?

$$g$$ is increasing on $$\left( {1,\infty } \right)$$
$$g$$ is decreasing on $$\left( {1,\infty } \right)$$
$$g$$ is increasing on $$(1, 2)$$ and decreasing on $$\left( {2,\infty } \right)$$
$$g$$ is decreasing on $$(1, 2)$$ and increasing on $$\left( {2,\infty } \right)$$

Explanation

Given,

$$\begin{aligned} g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t \\ \text { and } f(x) & =(1-x)^2 \sin ^2 x+x^2 \\ \Rightarrow g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)\left((1-t)^2 \sin ^2 t+t^2\right) d t \end{aligned}$$

On differentiating the above equation w.r.t. $$x$$

$$\begin{aligned} & \Rightarrow g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \\ & \Rightarrow g^{\prime}(x)=\left(2-\frac{4}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \end{aligned}$$

Here $$(1-x)^2 \sin ^2 x+x^2>0$$ for all $$x \in \mathrm{R}$$

Now, draw the graph of $$y=2-\frac{4}{x+1}$$ and $$y=-\ln x$$

For $$x>0$$

IIT-JEE 2012 Paper 2 Offline Mathematics - Application of Derivatives Question 23 English Explanation 1

$$\text { Add the graph of } y=-\ln x \text { and } y=2-\frac{4}{x+1}$$

IIT-JEE 2012 Paper 2 Offline Mathematics - Application of Derivatives Question 23 English Explanation 2

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

Hence, $$g(x)$$ is decreasing function for $$x \in(1, \infty)$$

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