TS EAMCET · Maths · Application of Derivatives
If the minimum value of \(f(x)=2 x^2+\alpha x+8\) is the same as the maximum value of \(g(x)=-3 x^2-4 x+\alpha^2\), then \(\alpha^2=\)
- A \(\frac{150}{27}\)
- B \(\frac{160}{27}\)
- C \(\frac{170}{27}\)
- D \(\frac{181}{27}\)
Answer & Solution
Correct Answer
(B) \(\frac{160}{27}\)
Step-by-step Solution
Detailed explanation
Since, the minimum value of \(f(x)=2 x^2+\alpha x+8\) is \(-\frac{\alpha^2-64}{8}=\frac{64-\alpha^2}{8}\) And the maximum value of \(g(x)=-3 x^2-4 x+\alpha^2\) is \(-\frac{12 \alpha^2-16}{-12}=\frac{16+12 \alpha^2}{12}\) Now, according to the question…
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