Question

If the function $f(x) = ax^3 + bx^2 + 11x - 6$ satisfies the conditions of Rolle's theorem in $[1, 3]$ and $f' \left( 2 + \frac{1}{\sqrt{3}} \right) = 0 $, then $a + b = $

• A. -5
• B. -3
• C. 4
• D. 7

Answer 1

Answer: (A)

Given, $f(x)=a x^{3}+b x^{2}+11 x-6$
satisfies the Rolle's theorem in $\lfloor 1,3\rfloor .$
So, $f(1)=0$ and $f(3)=0$
$a+b+5=0 \dots$(i)
and $ 27 a+9 b+27=0$
$\Rightarrow 9 a+3 b+9=0 $
$ \Rightarrow 3 a+b+3 =0 \dots$(ii)
From Eq. (i), we get $a+b=-5$