If the minimum value of f(x)=2x^2+alpha x+8 i | vedclass

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(B) The minimum value of a quadratic function $f(x)=ax^2+bx+c$ (where $a>0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $f(x)=2x^2+\alp

Vedclass · Accepted Answer

$\frac{160}{27}$

Vedclass · Answer

(B) The minimum value of a quadratic function $f(x)=ax^2+bx+c$ (where $a>0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $f(x)=2x^2+\alpha x+8$,we have $a=2, b=\alpha, c=8$. The minimum value is $\frac{4(2)(8)-\alpha^2}{4(2)} = \frac{64-\alpha^2}{8}$.
The maximum value of a quadratic function $g(x)=ax^2+bx+c$ (where $a < 0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $g(x)=-3x^2-4x+\alpha^2$,we have $a=-3, b=-4, c=\alpha^2$. The maximum value is $\frac{4(-3)(\alpha^2)-(-4)^2}{4(-3)} = \frac{-12\alpha^2-16}{-12} = \frac{12\alpha^2+16}{12}$.
Equating the two values:
$\frac{64-\alpha^2}{8} = \frac{12\alpha^2+16}{12}$
Multiplying both sides by $24$:
$3(64-\alpha^2) = 2(12\alpha^2+16)$
$192-3\alpha^2 = 24\alpha^2+32$
$160 = 27\alpha^2$
$\alpha^2 = \frac{160}{27}$.
Thus,option $(B)$ is correct.

If the minimum value of $f(x)=2x^2+\alpha x+8$ is the same as the maximum value of $g(x)=-3x^2-4x+\alpha^2$,then $\alpha^2=$

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