JEE PYQ: Application Of Derivatives Question 16
Question 16 - 2021 (25 Feb Shift 1)
If Rolle’s theorem holds for the function $f(x) = x^3 - ax^2 + bx - 4$, $x \in [1, 2]$ with $f’\left(\frac{4}{3}\right) = 0$, then ordered pair $(a, b)$ is equal to:
(1) $(-5, 8)$
(2) $(5, 8)$
(3) $(5, -8)$
(4) $(-5, -8)$
Show Answer
Answer: (2)
Solution
$f(1) = f(2) \Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4 \Rightarrow 3a - b = 7$. $f’(4/3) = 0 \Rightarrow 3 \cdot \frac{16}{9} - \frac{8a}{3} + b = 0 \Rightarrow -8a + 3b = -16$. Solving: $a = 5, b = 8$.
BETA
