If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies conditions of Rolle's theorem in [1, 3] and $f'\left(2+\frac{1}{\sqrt{3}}\right)=0$, then values of a and b are respectively |
1, -6 –2, 1 -1, $\frac{1}{2}$ -1, 6 |
1, -6 |
f(1) = f(3) a + b + 11 – 6 = 27a + 9b + 27 ⇒ 26a + 8b + 22 = 0 ⇒ 13a + 4b + 11 = 0 Also $f'\left(2+\frac{1}{\sqrt{3}}\right)=0$ $\Rightarrow 3 a\left(2+\frac{1}{\sqrt{3}}\right)^2+2 b\left(2+\frac{1}{\sqrt{3}}\right)+11=0$ ⇒ a = 1, b = −6 |