List IList IIP. The number of polynomials f(x | vedclass

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(D) $(P)$ Let $f(x) = ax^2 + bx$,where $a, b \in \mathbb{W}$ (since $f(0)=0$).$\int_0^1 (ax^2 + bx) dx = \frac{a}{3} + \frac{b}{2} = 1 \implies 2a + 3

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$2 \quad 3 \quad 1 \quad 4$

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(D) $(P)$ Let $f(x) = ax^2 + bx$,where $a, b \in \mathbb{W}$ (since $f(0)=0$).$\int_0^1 (ax^2 + bx) dx = \frac{a}{3} + \frac{b}{2} = 1 \implies 2a + 3b = 6$.Possible non-negative integer solutions $(a, b)$ are $(3, 0)$ and $(0, 2)$.Thus,the number of such polynomials is $2$.$(Q)$ $f(x) = \sqrt{2} \sin(x^2 + \frac{\pi}{4})$.$f(x)$ is maximum when $x^2 + \frac{\pi}{4} = 2n\pi + \frac{\pi}{2} \implies x^2 = 2n\pi + \frac{\pi}{4}$.For $n=0$,$x^2 = \frac{\pi}{4} \approx 0.785 \in [0, 13)$.For $n=1$,$x^2 = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4} \approx 7.068 \in [0, 13)$.For $n=2$,$x^2 = 4\pi + \frac{\pi}{4} = \frac{17\pi}{4} \approx 13.35 
otin [0, 13)$.Since $x^2$ can take $2$ values for each $n$,and $x$ can be $\pm \sqrt{x^2}$,we check the interval $(-\sqrt{13}, \sqrt{13})$.For $n=0$,$x = \pm \sqrt{\pi/4} = \pm \sqrt{\pi}/2$ ($2$ points).For $n=1$,$x = \pm \sqrt{9\pi/4} = \pm 3\sqrt{\pi}/2$ ($2$ points).Total points $= 4$. Wait,checking the options,$Q$ matches $3$ ($4$ points).$(R)$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx = \int_0^2 3x^2 (\frac{1}{1+e^x} + \frac{1}{1+e^{-x}}) dx = \int_0^2 3x^2 (\frac{1+e^x}{1+e^x}) dx = \int_0^2 3x^2 dx = [x^3]_0^2 = 8$.$(S)$ The numerator is $\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx$. Since $\cos 2x$ is even and $\log(\frac{1+x}{1-x})$ is odd,the product is odd. Thus,the integral is $0$.

List $I$	List $II$
$P.$ The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$,satisfying $f(0)=0$ and $\int_0^1 f(x) dx=1$,is	$1.$ $8$
$Q.$ The number of points in the interval $(-\sqrt{13}, \sqrt{13})$ at which $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

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