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TS EAMCET · Maths · Application of Derivatives

If the Rolle's theorem holds for the function \(f(x)=x^4+a x^3+b x\), in \(-1 \leq x \leq 1\), and \(f^{\prime}\left(\frac{1}{2}\right)=0\), then \(a b=\)

  1. A \(-4\)
  2. B \(-64\)
  3. C \(-1\)
  4. D \(-8\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-4\)

Step-by-step Solution

Detailed explanation

We have, \(f(x)=x^4+a x^3+b x,[-1,1]\) Since, RMVT is applicable on \(f(x)\) \(\therefore \quad f(-1)=f(1)\) \(\begin{aligned} \Rightarrow & & 1-a-b & =1+a+b \\ \Rightarrow & & a+b & =0\end{aligned}\) Also, we have \(f\left(\frac{1}{2}\right)=0\)ned}\( \)\therefore…