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5. Continuity and Differentiation
medium
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
