Question

If the function ƒ(x) = ax³ + bx² + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and ƒ ¢ $\left( 2 + \frac{1}{\sqrt{3}} \right)$ = 0, then values of a and b are respectively –

• A. 1, –6
• B. –2, 1
• C. –1, $\frac{1}{2}$
• D. –1, 6

Answer 1

Answer: (A)

1, –6

Answer 2

Here, ƒ(1) = ƒ(3) and ƒ ¢ $\left( 2 + \frac{1}{\sqrt{3}} \right)$ = 0

Ž a + b + 11 – 6 = 27a + 9b + 33 – 6

and 3a $\left( 2 + \frac{1}{\sqrt{3}} \right)^{2}$ + 2b $\left( 2 + \frac{1}{\sqrt{3}} \right)$ + 11 = 0

Ž 26 a + 8b = – 22

and 3a $\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right)$ + 4b + $\frac{2b}{\sqrt{3}}$ = –11

Ž 13a + 4b = – 11

and a (13 + 4$\sqrt{3}$) + b$\left( 4 + \frac{2}{\sqrt{3}} \right)$ = –11

Ž 13a + 4b = – 11 and $4\sqrt{3}$a + $\frac{2}{\sqrt{3}}$b = 0

Ž 13a + 4b = –11 and 12a + 2b = 0

Ž 13a + 4b = –11 and 6a + b = 0

Solving we get, a = 1, b = –6.