Published by:
CGP EDU Academic Team
Published on: August 14, 2026
If the function ƒ(x) = ax 3 + bx 2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and ƒ ′ 
= 0, then values of a and b are respectively -
A
1, –6
B
–2, 1
C
–1, 
D
–1, 6
Text Solution
Verified by ExpertsThe correct answer is:
A
Here, ƒ(1) = ƒ(3) and ƒ ′ 
= 0
⇒ a + b + 11 – 6 = 27a + 9b + 33 – 6
and 3a 
+ 2b 
+ 11 = 0
⇒ 26 a + 8b = – 22
and 3a 
+ 4b + 
= –11
⇒ 13a + 4b = – 11
and a (13 + 4 
) + b 
= –11
⇒ 13a + 4b = – 11 and 
a + 
b = 0
⇒ 13a + 4b = –11 and 12a + 2b = 0
⇒ 13a + 4b = –11 and 6a + b = 0
Solving we get, a = 1, b = –6.
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