Solving P

Let the polynomial be $f(x) = ax^2 + bx + c$, where $a, b, c$ are non-negative integers as the degree is at most 2.

1. Condition 1: $f(0) = 0$ $f(0) = a(0)^2 + b(0) + c = c$. So, $c=0$. The polynomial is of the form $f(x) = ax^2 + bx$.

2. Condition 2: $\int_0^1 f(x)dx = 1$ $\int_0^1 (ax^2 + bx)dx = 1$ $\left[ \frac{ax^3}{3} + \frac{bx^2}{2} \right]_0^1 = 1$ $\frac{a}{3} + \frac{b}{2} = 1$ Multiplying by 6, we get the Diophantine equation: $2a + 3b = 6$.

3. Finding non-negative integer solutions for $(a,b)$:

   • If $b=0$, $2a = 6 \implies a=3$. Solution: $(3,0)$.
   • If $b=1$, $2a = 3 \implies a=3/2$. Not an integer.
   • If $b=2$, $2a = 0 \implies a=0$. Solution: $(0,2)$.
   • If $b > 2$, $3b > 6$, which means $2a = 6 - 3b < 0$, so $a$ would be negative. No more non-negative solutions.

There are two pairs of solutions: $(3,0)$ and $(0,2)$. These correspond to two polynomials: $f(x)=3x^2$ and $f(x)=2x$. The number of such polynomials is 2.

Matching P: The value is 2, which corresponds to item 2 in List-II. So, P $\to$ 2.

Solving Q

Let $f(x) = \sin(x^2) + \cos(x^2)$. We need to find the number of points in $[-\sqrt{13}, \sqrt{13}]$ where $f(x)$ is maximum.

1. Finding the maximum value: We can write $f(x)$ as: $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}}\sin(x^2) + \frac{1}{\sqrt{2}}\cos(x^2) \right) = \sqrt{2} \sin\left(x^2 + \frac{\pi}{4}\right)$ The maximum value of $f(x)$ is $\sqrt{2}$, which occurs when $\sin(x^2 + \frac{\pi}{4}) = 1$.

2. Condition for maximum: $x^2 + \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}, \quad n \in \mathbb{Z}$ $x^2 = 2n\pi + \frac{\pi}{4} = \frac{(8n+1)\pi}{4}$

3. Finding solutions in the interval: The interval for $x$ is $[-\sqrt{13}, \sqrt{13}]$, so $x^2$ must be in $[0, 13]$. We need to find integers $n$ such that $0 \le \frac{(8n+1)\pi}{4} \le 13$. Since $x^2 \ge 0$, we must have $8n+1 \ge 0$, so $n \ge -1/8$. Thus $n$ must be a non-negative integer ($n=0, 1, 2, ...$).

   • For $n=0$: $x^2 = \frac{\pi}{4} \approx \frac{3.14}{4} = 0.785$. This is in $[0, 13]$. So $x = \pm\sqrt{\frac{\pi}{4}}$ are two solutions.
   • For $n=1$: $x^2 = \frac{9\pi}{4} \approx \frac{9 \times 3.14}{4} = 7.065$. This is in $[0, 13]$. So $x = \pm\sqrt{\frac{9\[Pi]}{4}}$ are two more solutions.
   • For $n=2$: $x^2 = \frac{17\pi}{4} \approx \frac{17 \times 3.14}{4} = 13.345$. This is greater than 13. No solutions for $n \ge 2$.

There are a total of $2+2=4$ points where $f(x)$ attains its maximum value.

Matching Q: The value is 4, which corresponds to item 3 in List-II. So, Q $\to$ 3.

Solving R

Let $I = \int_{-2}^2 \frac{3x^2}{1 + e^x} dx$.

We use the property $\int_{-a}^a g(x) dx = \int_0^a (g(x) + g(-x)) dx$. Here, $g(x) = \frac{3x^2}{1+e^x}$. Then, $g(-x) = \frac{3(-x)^2}{1+e^{-x}} = \frac{3x^2}{1+1/e^x} = \frac{3x^2 e^x}{e^x+1}$.

Now, $g(x) + g(-x) = \frac{3x^2}{1+e^x} + \frac{3x^2 e^x}{1+e^x} = \frac{3x^2(1+e^x)}{1+e^x} = 3x^2$.

So, the integral becomes: $I = \int_0^2 3x^2 dx = \left[ x^3 \right]_0^2 = 2^3 - 0^3 = 8$

Matching R: The value is 8, which corresponds to item 1 in List-II. So, R $\to$ 1.

Solving S

Let the expression be $E = \frac{\int_{-1/2}^{1/2} \cos(2x)\log\left(\frac{1+x}{1-x}\right)dx}{\int_{0}^{1/2} \cos(2x)\log\left(\frac{1+x}{1-x}\right)dx}$.

Let's analyze the integrand in the numerator, $g(x) = \cos(2x)\log\left(\frac{1+x}{1-x}\right)$.

• $\cos(2x)$ is an even function, since $\cos(-2x) = \cos(2x)$.
• $\log\left(\frac{1+x}{1-x}\right)$ is an odd function, since $\log\left(\frac{1-x}{1+x}\right) = \log\left(\left(\frac{1+x}{1-x}\right)^{-1}\right) = -\log\left(\frac{1+x}{1-x}\right)$.

The product of an even function and an odd function is an odd function. So, $g(x)$ is an odd function.

The integral of an odd function over a symmetric interval $[-a, a]$ is zero. Therefore, the numerator is $\int_{-1/2}^{1/2} g(x) dx = 0$.

For the denominator, let's consider the integrand $g(x)$ on $(0, 1/2)$.

• For $x \in (0, 1/2)$, $2x \in (0, 1)$, so $\cos(2x) > 0$.
• For $x \in (0, 1/2)$, $1+x > 1-x > 0$, so $\frac{1+x}{1-x} > 1$, which means $\log\left(\frac{1+x}{1-x}\right) > 0$. Since the integrand is positive on $(0, 1/2)$, the denominator is a positive value, i.e., not zero.

Thus, the value of the expression is $E = \frac{0}{\text{non-zero value}} = 0$.

Matching S: The value is 0, which corresponds to item 4 in List-II. So, S $\to$ 4.

Conclusion

The final matching is:

• P $\to$ 2
• Q $\to$ 3
• R $\to$ 1
• S $\to$ 4

This corresponds to the option D: $P = 2,Q = 3,R = 1,S = 4$.