If Rolle's theorem holds for function f(x)=x^{3}-a x^{2}+b x-4, x ∈[1,2] with f ^{prime}left(f

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5. Continuity and Differentiation

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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

$(-5,-8)$

(JEE MAIN-2021)

Solution

$f (1)= f (2)$

$\Rightarrow 1-a+b-4=8-4 a+2 b-4$

$\Rightarrow 3 a-b=7$

Also $f ^{1}\left(\frac{4}{3}\right)=0$ (given)

$\Rightarrow\left(3 x ^{2}-2 ax + b \right)_{ x =\frac{4}{3}}=0$

$\Rightarrow \frac{16}{3}-\frac{8 a}{3}+b=0$

$\Rightarrow 8 a-3 b-16=0$ $…….(2)$

Solving $(1)$ and $(2)$

$a =5, b =8$

Standard 12

Mathematics

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