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(B) For a probability density function (p.d.f.),the total area under the curve must be $1$. Thus,$\int_{1}^{3} f(x) dx = 1$. Substituting $f(x) = \fra

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$-9, 10$

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(B) For a probability density function (p.d.f.),the total area under the curve must be $1$. Thus,$\int_{1}^{3} f(x) dx = 1$. Substituting $f(x) = \frac{ax^2}{2} + bx$: $\int_{1}^{3} (\frac{ax^2}{2} + bx) dx = [\frac{ax^3}{6} + \frac{bx^2}{2}]_{1}^{3} = 1$. Evaluating at the limits: $(\frac{27a}{6} + \frac{9b}{2}) - (\frac{a}{6} + \frac{b}{2}) = 1$. $\frac{26a}{6} + \frac{8b}{2} = 1 \implies \frac{13a}{3} + 4b = 1 \implies 13a + 12b = 3$ (Equation $1$). Given $f(2) = 2$: $\frac{a(2)^2}{2} + b(2) = 2 \implies 2a + 2b = 2 \implies a + b = 1 \implies b = 1 - a$ (Equation $2$). Substitute Equation $2$ into Equation $1$: $13a + 12(1 - a) = 3$. $13a + 12 - 12a = 3$. $a = 3 - 12 = -9$. Using $b = 1 - a$: $b = 1 - (-9) = 10$. Thus,$a = -9$ and $b = 10$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

If a random variable $X$ has p.d.f. $f(x) = \begin{cases} \frac{ax^2}{2} + bx & , \text{if } 1 \leqslant x \leqslant 3 \\ 0 & , \text{otherwise} \end{cases}$ and $f(2) = 2$,then the values of $a$ and $b$ are,respectively

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