JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If Rolle's theorem holds for the function \(f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]\) with \(f ^{\prime}\left(\frac{4}{3}\right)=0,\) then ordered pair \(( a , b )\) is equal to
- A \((5,8)\)
- B \((-5,8)\)
- C \((5,-8)\)
- D \((-5,-8)\)
Answer & Solution
Correct Answer
(A) \((5,8)\)
Step-by-step Solution
Detailed explanation
\(f (1)= f (2)\) \(\Rightarrow 1-a+b-4=8-4 a+2 b-4\) \(\Rightarrow 3 a-b=7\) Also \(f ^{1}\left(\frac{4}{3}\right)=0\) (given) \(\Rightarrow\left(3 x ^{2}-2 ax + b \right)_{ x =\frac{4}{3}}=0\) \(\Rightarrow \frac{16}{3}-\frac{8 a}{3}+b=0\) \(\Rightarrow 8 a-3 b-16=0\)…
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