If the function fx-ax3-bx2-11x-6- a-b- satisfies Rolle-s theorem in -1- 3- and f-2-1-3-0- then the value of a - b is

By Rolle’s theorem, f(1) = f(3) ⇒ a+b+11 6=27a +9b+33 6 ⇒ 13 a +4b +11 =0 Now, f'(x) = 3ax^2 + 2bx +11 Given, x=2+ 1√(3) nbsp; nbsp;is a root of f'(x)=0 Si ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Explanation of Intermediate Value Theorem

If the functi...

Question

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

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Solution

By Rolle’s theorem,
$f (1) = f (3)$
$\Rightarrow a + b + 11 - 6 = 27 a + 9 b + 33 - 6$
$\Rightarrow 13 a + 4 b + 11 = 0$

Now, $f^{'} (x) = 3 a x^{2} + 2 b x + 11$
Given, $x = 2 + \frac{1}{\sqrt{3}}$ is a root of $f^{'} (x) = 0$
Since coefficients are real, $x = 2 - \frac{1}{\sqrt{3}}$ is another root.
Product of roots $= \frac{11}{3 a} = \frac{11}{3}$
$\Rightarrow a = 1$
Sum of roots $= - \frac{2 b}{3} = 4$
$\Rightarrow b = - 6$
$∴ a - b = 1 + 6 = 7$

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If the function f(x)=ax3+bx2+11x−6,a,b∈R satisfies Rolle's theorem in [1,3] and f′(2+1√3)=0, then the value of (a−b) is

By Rolle’s theorem, f(1)=f(3) ⇒a+b+11−6=27a+9b+33−6 ⇒13a+4b+11=0 Now, f′(x)=3ax2+2bx+11 Given, x=2+1√3 is a root of f′(x)=0 Since coefficients are real, x=2−1√3 is another root. Product of roots =113a=113 ⇒a=1 Sum of roots =−2b3=4 ⇒b=−6 ∴a−b=1+6=7

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