(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.