Suppose,alpha is the minimum value of x^2+bx+ | vedclass

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(B) The minimum value of the quadratic expression $x^2+bx+5$ occurs at $x = -\frac{b}{2}$. Substituting this,we get $\alpha = (-\frac{b}{2})^2 + b(-\f

Vedclass · Accepted Answer

$16$

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(B) The minimum value of the quadratic expression $x^2+bx+5$ occurs at $x = -\frac{b}{2}$. Substituting this,we get $\alpha = (-\frac{b}{2})^2 + b(-\frac{b}{2}) + 5 = 5 - \frac{b^2}{4}$.The maximum value of the quadratic expression $-x^2+ax+5$ occurs at $x = \frac{a}{2}$. Substituting this,we get $\beta = -(\frac{a}{2})^2 + a(\frac{a}{2}) + 5 = 5 + \frac{a^2}{4}$.Given the inequality $x^2-10x+24 \leq 0$,we factor it as $(x-4)(x-6) \leq 0$,which implies $4 \leq x \leq 6$.Comparing this with the interval $[\alpha, \beta]$,we have $\alpha = 4$ and $\beta = 6$.Equating the expressions:$5 - \frac{b^2}{4} = 4$ $\Rightarrow \frac{b^2}{4} = 1$ $\Rightarrow b^2 = 4$.$5 + \frac{a^2}{4} = 6$ $\Rightarrow \frac{a^2}{4} = 1$ $\Rightarrow a^2 = 4$.Therefore,$a^2b^2 = 4 	imes 4 = 16$.

Suppose,$\alpha$ is the minimum value of $x^2+bx+5$ and $\beta$ is the maximum value of $-x^2+ax+5$. If $[\alpha, \beta]$ is the interval of maximum length for $x$ in which $x^2-10x+24 \leq 0$,then $a^2b^2$ is equal to

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