Question

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 ${\alpha }^2=$

• A. $\frac{150}{27}$
• B. $\frac{160}{27}$
• C. $\frac{170}{27}$
• D. $\frac{181}{27}$

Answer 1

Answer: (B)

Since, the minimum value of
$f(x)=2x^2+\alpha x+8$ is $-\frac{{\alpha }^2-64}{8}=\frac{64-{\alpha }^2}{8}$
And the maximum value of $g(x)=-3x^2-4x+{\alpha }^2$
is $-\frac{12{\alpha }^2-16}{-12}=\frac{16+12{\alpha }^2}{12}$
Now, according to the question
$\frac{64-{\alpha }^2}{8}=\frac{16+12{\alpha }^2}{12}$
$\Rightarrow 192-3{\alpha }^2=32+24{\alpha }^2$
$\Rightarrow 27{\alpha }^2=160$
$\Rightarrow {\alpha }^2=\frac{160}{27}$