QJEEAdv14Mac265: JEE Advanced 2014 Mathematics - Class 12 JEEAdv14Mac265 Solution: List - P. The number of polynomials with non-negative integer…

(P) Let $p(x)=a x^2+b x+c$

Given, $p(0)=0$

$\Rightarrow \quad c=0$

Also given $\int_0^1 f(x) d x=1$

$\begin{array}{lr}\Rightarrow & \int_0^1\left(a x^2+b x\right) d x=1 \\\Rightarrow & {\left[\frac{a x^3}{3}+\frac{b x^2}{2}\right]_0^1=1} \\\Rightarrow & \frac{a}{3}+\frac{b}{2}=1 \\\Rightarrow & 2 a+3 b=6\end{array}$

Here, $a$ and $b$ are non negative integer

$\therefore(a, b)=(3,0),(0,2)$

Hence, possible equation of $p(x)$ are $3 x^2$ and $2 x$ .

So, P Match with 2.

(Q) Given $f(x)=\sin x^2+\cos x^2$

Differentiate $f(x)$ w.r.t. $x$

$\begin{aligned}& \Rightarrow f^{\prime}(x)=2 x \cdot \cos x^2-2 x \cdot \sin x^2 \\& \Rightarrow f^{\prime}(x)=2 x\left(\cos x^2-\sin x^2\right)\end{aligned}$

Apply $f^{\prime}(x)=0$

$\begin{aligned}& \Rightarrow x=0, \tan x^2=1 \\& \Rightarrow x=0, \pm \sqrt{\frac{\pi}{4}}, \pm \sqrt{\frac{5 \pi}{4}}, \pm \sqrt{\frac{9 \pi}{4}}, \pm \sqrt{\frac{13 \pi}{4}}\end{aligned}$

$\begin{aligned}& \text { Differentiate } f^{\prime}(x) \text { w.r.t. } x \\& \Rightarrow f^{\prime \prime}(x)=2\left(\cos x^2-\sin x^2\right) \\& +2 x\left(-2 x \sin x^2-2 x \cos x^2\right) \\& \Rightarrow f^{\prime \prime}(x)=2\left(\cos x^2-\sin x^2\right) \\& -4 x^2\left(\sin x^2+\cos x^2\right) \\& \Rightarrow f^{\prime \prime}(x)<0 \text { at } x= \pm \sqrt{\frac{\pi}{4}}, \pm \sqrt{\frac{9 \pi}{4}} \\\end{aligned}$

So, $f(x)$ is maximum at four points. Hence, Q match with 3. (R) Let $\mathrm{I}=\int_{-2}^2 \frac{3 x^2}{1+e^x} d x\quad \text{... (i)}$

Apply the property $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$\begin{aligned}& \Rightarrow \quad \mathrm{I}=\int_{-2}^2 \frac{3(-x)^2}{1+e^{-x}} d x \\& \Rightarrow \quad \mathrm{I}=\int_{-2}^2 \frac{3 x^2 e^x}{e^x+1} d x \quad \text{.... (ii)}\end{aligned}$

Add equation (i) and (ii)

$\begin{aligned}& \Rightarrow \quad 2 \mathrm{I}=\int_{-2}^2 \frac{3 x^2}{e^x+1}\left[1+e^x\right] d x \\& \Rightarrow \quad \mathrm{I}=\frac{3}{2} \int_{-2}^2 x^2 \cdot d x \\& \Rightarrow \quad \mathrm{I}=\frac{3}{2}\left[\frac{x^3}{3}\right]_{-2}^2 \\& \Rightarrow \quad \mathrm{I}=16\end{aligned}$

(S) We know the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l}2 \int_0^a f(x) d x, \text { when } f(x) \text { is } \\\text { an even function } \\0, \text { when } f(x) \text { is an odd function }\end{array}\right.$

$\begin{aligned}& \text { Let } g(x)=\cos x \cdot \log \left(\frac{1+x}{1-x}\right) \\& \Rightarrow g(-x)=\cos (-x) \log \left(\frac{1-x}{1+x}\right) \\& \Rightarrow g(-x)=\cos x \cdot \log \left(\frac{1+x}{1-x}\right)^{-1} \\& \Rightarrow g(-x)=-\cos x \cdot \log \left(\frac{1+x}{1-x}\right)\end{aligned}$

$\Rightarrow g(-x)=-g(x)$

Hence, $g(x)$ is an odd function

So, $\int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x=0$

$\Rightarrow \frac{\frac{\int_{-1}^2}{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x}{\int_0^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x}=0$

Hence, $S$ match with 4.

Hint:

(i) Recall the property

$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

(ii) Recall the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l}2 \int_0^a f(x) d x, \text { when } f(x) \text { is an even function}\\0, \text { when } f(x) \text { is an oden function }\end{array}\right.$

(iii) At the point of local maxima, first derivative of the function is zero and second derivative of the function is negative.