Let F: R rightarrow R be a thrice differentia | vedclass

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(A-D) $1.$ Given $f(x) = x F(x)$,then $f'(x) = F(x) + x F'(x)$.
At $x=1$,$f'(1) = F(1) + 1 \cdot F'(1) = 0 + F'(1) = F'(1)$.

Vedclass · Accepted Answer

$(ABC, CD)$

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(A-D) $1.$ Given $f(x) = x F(x)$,then $f'(x) = F(x) + x F'(x)$.
At $x=1$,$f'(1) = F(1) + 1 \cdot F'(1) = 0 + F'(1) = F'(1)$. Since $F'(x) < 0$,$f'(1) < 0$. Thus $(A)$ is correct.
For $(B)$,$f(2) = 2 F(2)$. Since $F(1)=0$ and $F'(x) < 0$ on $(1/2, 3)$,$F(x)$ is strictly decreasing. Thus $F(2) < F(1) = 0$,so $f(2) < 0$. Thus $(B)$ is correct.
For $(C)$ and $(D)$,$f'(x) = F(x) + x F'(x)$. Since $F(x) < 0$ and $F'(x) < 0$ for $x \in (1, 3)$,$f'(x) < 0$ for all $x \in (1, 3)$. Thus $f'(x) 
eq 0$. Thus $(C)$ is correct.
Correct options for $1$: $(A, B, C)$.
$2.$ $\int_1^3 f(x) dx = \int_1^3 x F(x) dx$. Using integration by parts: $\left[ \frac{x^2}{2} F(x) ight]_1^3 - \int_1^3 \frac{x^2}{2} F'(x) dx = \frac{9}{2}(-4) - 0 - \frac{1}{2}(-12) = -18 + 6 = -12$. Thus $(D)$ is correct.
For $(A)$ and $(C)$,$\int_1^3 x^3 F''(x) dx = [x^3 F'(x)]_1^3 - 3 \int_1^3 x^2 F'(x) dx = 27 F'(3) - F'(1) - 3(-12) = 40$.
$27 F'(3) - F'(1) + 36 = 40 \Rightarrow 27 F'(3) - F'(1) = 4$.
Since $f'(x) = F(x) + x F'(x)$,$f'(3) = F(3) + 3 F'(3) = -4 + 3 F'(3)$ and $f'(1) = F(1) + F'(1) = F'(1)$.
$9 f'(3) - f'(1) = 9(-4 + 3 F'(3)) - F'(1) = -36 + 27 F'(3) - F'(1) = -36 + 4 = -32$.
So $9 f'(3) - f'(1) + 32 = 0$. Thus $(C)$ is correct.
Correct options for $2$: $(C, D)$.

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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