(A) Given $f(x) = \frac{1}{x}$.Then $f'(x) = -\frac{1}{x^2}$.According to the Mean Value Theorem,$f'(x_1) = \frac{f(b) - f(a)}{b - a}$.Substituting th

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(A) Given $f(x) = \frac{1}{x}$.Then $f'(x) = -\frac{1}{x^2}$.According to the Mean Value Theorem,$f'(x_1) = \frac{f(b) - f(a)}{b - a}$.Substituting the values,we get $-\frac{1}{x_1^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}$.Simplifying the right side: $\frac{\frac{a - b}{ab}}{b - a} = \frac{-(b - a)}{ab(b - a)} = -\frac{1}{ab}$.Thus,$-\frac{1}{x_1^2} = -\frac{1}{ab}$,which implies $x_1^2 = ab$.Since $a < x_1 < b$,we have $x_1 = \sqrt{ab}$.

Class 12 Mathematics Continuity and Differentiability Rolle’s theorem, Lagrange's mean value theorem

Easy

From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

A
$\sqrt{ab}$
B
$\frac{a + b}{2}$
C
$\frac{2ab}{a + b}$
D
$\frac{b - a}{b + a}$

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Continuity and Differentiability

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Rolle’s theorem, Lagrange's mean value theorem

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

DifficultAIEEE 2002

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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 $b$.

Medium

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Verify the Mean Value Theorem for the function $f(x) = x^{3} - 5x^{2} - 3x$ in the interval $[1, 3]$. Find all $c \in (1, 3)$ such that $f^{\prime}(c) = \frac{f(3) - f(1)}{3 - 1}$.

Difficult

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For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

EasyAP EAMCET 2024

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Let $f(x) = \log(1 + x^2)$ and $A$ be a constant such that $\frac{|f(x) - f(y)|}{|x - y|} \leq A$ for all real $x, y$ where $x \neq y$. Then,the least possible value of $A$ is

DifficultKVPY 2017

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From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

Similar Questions

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

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 $b$.

Verify the Mean Value Theorem for the function $f(x) = x^{3} - 5x^{2} - 3x$ in the interval $[1, 3]$. Find all $c \in (1, 3)$ such that $f^{\prime}(c) = \frac{f(3) - f(1)}{3 - 1}$.

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