If f(x) = ax^3 + bx^2 + 11x - 6 for x in [1, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$,we have $f(1) = f(3)$.$a(1)^3 + b(1)^2 + 11(1) - 6 = a(3)^3 + b(3)^2 + 11(3) - 6$$a + b + 5 = 2

Vedclass · Accepted Answer

$1, -6$

Vedclass · Answer

(A) Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$,we have $f(1) = f(3)$.$a(1)^3 + b(1)^2 + 11(1) - 6 = a(3)^3 + b(3)^2 + 11(3) - 6$$a + b + 5 = 27a + 9b + 27$$26a + 8b + 22 = 0 \implies 13a + 4b + 11 = 0 \dots (i)$Now,$f'(x) = 3ax^2 + 2bx + 11$.Given $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,we substitute $x = 2 + \frac{1}{\sqrt{3}}$:$3a\left( 2 + \frac{1}{\sqrt{3}} \right)^2 + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$$3a\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$$3a\left( \frac{13}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$$13a + 4\sqrt{3}a + 4b + \frac{2b}{\sqrt{3}} + 11 = 0$Since $13a + 4b + 11 = 0$ from $(i)$,we have $4\sqrt{3}a + \frac{2b}{\sqrt{3}} = 0$.$4\sqrt{3}a = -\frac{2b}{\sqrt{3}} \implies 12a = -2b \implies b = -6a$.Substitute $b = -6a$ into $(i)$:$13a + 4(-6a) + 11 = 0$$13a - 24a + 11 = 0 \implies -11a = -11 \implies a = 1$.Then $b = -6(1) = -6$.

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

Similar Questions

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_e x$ on the interval $[1, 3]$ is

If $f$ is defined in $[1,3]$ by $f(x)=x^3+b x^2+a x$,such that $f(1)-f(3)=0$ and $f^{\prime}(c)=0$,where $c=2+\frac{1}{\sqrt{3}}$,then $(a, b)$ is equal to

Let $f$ be a function that is derivable on the interval $[0, 1]$. Then,which of the following statements is true?

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0$. Then:

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_{e}x$ on the interval $[1, 3]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module