If the function f(x) = ax3 + bx2 + 11x - 6 satisfies the conditions of Rolle's theorem in [1, 3] and f' ( 2 + (1/√3) ) = 0 , then a + b =

Question

If the function f(x) = ax3 + bx2 + 11x - 6 satisfies the conditions of Rolle's theorem in [1, 3] and f' ( 2 + (1/√3) ) = 0 , then a + b =  (A) -5 (B) -3 (C) 4 (D) 7

Tardigrade Admin · Accepted Answer

Given, f(x)=a x3+b x2+11 x-6 satisfies the Rolle's theorem in ⌊ 1,3 rfloor . So, f(1)=0 and f(3)=0 a+b+5=0 dots(i) and 27 a+9 b+27=0 ⇒ 9 a+3 b+9=0 ⇒ 3 a+b+3 =0 dots(ii) From Eq. (i), we get a+b=-5

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

-5

-3

4

7

Given, $f (x) = a x^{3} + b x^{2} + 11 x - 6$
satisfies the Rolle's theorem in $⌊ 1, 3 ⌋ .$
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 $a + b = - 5$

Q. If the function f(x)=ax3+bx2+11x−6 satisfies the conditions of Rolle's theorem in [1,3] and f′(2+3​1​)=0, then a+b=

-5

-3

4

7

Given, f(x)=ax3+bx2+11x−6 satisfies the Rolle's theorem in ⌊1,3⌋. So, f(1)=0 and f(3)=0 a+b+5=0…(i) and 27a+9b+27=0 ⇒9a+3b+9=0 ⇒3a+b+3=0…(ii) From Eq. (i), we get a+b=−5

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

Given, $f (x) = a x^{3} + b x^{2} + 11 x - 6$
satisfies the Rolle's theorem in $⌊ 1, 3 ⌋ .$
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 $a + b = - 5$