Polynomial Coefficients Definite Integrals and Function Maxima Matching List

To find the correct matching between List I and List II, we evaluate each item in List I individually:

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Evaluation of Entry P:

Let $f(x) = ax^2 + bx + c$, where $a, b, c$ are non-negative integers and the degree of $f(x)$ is $\le 2$.

Given $f(0) = 0$, we have: $c = 0 \implies f(x) = ax^2 + bx$

Now, using the given definite integral condition: $\int_0^1 f(x) \, dx = \int_0^1 (ax^2 + bx) \, dx = \left[ \frac{ax^3}{3} + \frac{bx^2}{2} \right]_0^1 = \frac{a}{3} + \frac{b}{2} = 1$

Multiplying by $6$ gives: $2a + 3b = 6$

Since $a$ and $b$ are non-negative integers:

1. If $b = 0 \implies 2a = 6 \implies a = 3$ (Valid: $f(x) = 3x^2$)
2. If $b = 1 \implies 2a = 3$ (No integer solution)
3. If $b = 2 \implies 2a = 0 \implies a = 0$ (Valid: $f(x) = 2x$)
4. If $b \ge 3 \implies 3b \ge 9 > 6$ (No solution)

Thus, there are 2 such polynomials ($f(x) = 3x^2$ and $f(x) = 2x$).

Therefore, P matches with 2.

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Evaluation of Entry Q:

Given the function: $f(x) = \sin(x^2) + \cos(x^2) = \sqrt{2} \sin\left(x^2 + \frac{\pi}{4}\right)$

For $x \in [-\sqrt{13}, \sqrt{13}]$, let $t = x^2$. Then $t \in [0, 13]$.

$f(x)$ attains its maximum value when: $\sin\left(t + \frac{\pi}{4}\right) = 1 \implies t + \frac{\pi}{4} = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$ $t = \frac{\pi}{4} + 2k\pi$

Finding all values of $t \in [0, 13]$:

• For $k = 0$: $t = \frac{\pi}{4} \approx 0.785 \in [0, 13]$
• For $k = 1$: $t = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \approx 7.068 \in [0, 13]$
• For $k = 2$: $t = \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} \approx 13.351 > 13$

For each valid positive value of $t = x^2$, $x = \pm \sqrt{t}$:

• $t = \frac{\pi}{4} \implies x = \pm \sqrt{\frac{\pi}{4}}$ (2 points)
• $t = \frac{9\pi}{4} \implies x = \pm \sqrt{\frac{9\pi}{4}}$ (2 points)

Total number of points in $[-\sqrt{13}, \sqrt{13}]$ is $2 + 2 = \mathbf{4}$.

Therefore, Q matches with 3.

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Evaluation of Entry R:

Let the given integral be: $I = \int_{-2}^2 \frac{3x^2}{1+e^x} \, dx \quad \text{--- (1)}$

Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$: $I = \int_{-2}^2 \frac{3(-x)^2}{1+e^{-x}} \, dx = \int_{-2}^2 \frac{3x^2 e^x}{e^x + 1} \, dx \quad \text{--- (2)}$

Adding equations (1) and (2): $2I = \int_{-2}^2 \frac{3x^2 (1+e^x)}{1+e^x} \, dx = \int_{-2}^2 3x^2 \, dx$ $2I = \left[ x^3 \right]_{-2}^2 = 8 - (-8) = 16 \implies I = \mathbf{8}$

Therefore, R matches with 1.

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Evaluation of Entry S:

Consider the function inside the integral: $g(x) = \cos 2x \log\left(\frac{1+x}{1-x}\right)$

Testing for symmetry: $g(-x) = \cos(-2x) \log\left(\frac{1-x}{1+x}\right) = \cos 2x \left(-\log\left(\frac{1+x}{1-x}\right)\right) = -g(x)$

Since $g(x)$ is an odd function, its integral over any symmetric interval $[-a, a]$ is zero: $\int_{-1/2}^{1/2} \cos 2x \log\left(\frac{1+x}{1-x}\right) \, dx = 0$

Since the denominator $\int_0^{1/2} \cos 2x \log\left(\frac{1+x}{1-x}\right) \, dx$ is non-zero (as the integrand is strictly positive on $(0, 1/2)$), the ratio equals: $\frac{0}{\text{non-zero value}} = \mathbf{0}$

Therefore, S matches with 4.

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Conclusion:

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

This corresponds to Option D.