Discussion Forum

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 $\alpha^{2}$=

Answer:

Explanation:

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

    $\frac{64-\alpha^{2}}{8}=\frac{16+12\alpha^{2}}{12}$

 $\Rightarrow$      $192-3 \alpha^{2}=32+24 \alpha^{2}$

 $\Rightarrow$     $27 \alpha^{2}=160 \Rightarrow  \alpha^{2}= \frac{160}{27}$

 hence3, option(b) is correct