AP EAMCET · Maths · Application of Derivatives
If the function \(f(x)=a x^3+b x^2+26 x-24\) satisfies the conditions of Rolle's theorem in \([2,4]\) and \(f^{\prime}\left(3+\frac{1}{\sqrt{3}}\right)=0\), then the value of \(a b\) is equal to
- A \(-9\)
- B \(9\)
- C \(-3\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(-9\)
Step-by-step Solution
Detailed explanation
\(f(x)=a x^3+b x^2+26 x-24\) ...(i) on \([2,4]\) \(\because f(x)\) satisfies the Role's theorem \(\begin{array}{ll}\Rightarrow & f(2)=f(4) \\ \Rightarrow & a(2)^3+b\left(2^2\right)+26(2)-24\end{array}\) \(=a(4)^3+b(4)^2+26(4)-24\) \(\Rightarrow \quad 8 a+4 b+28=64 a+16 b+80\)…
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