Solveeit Logo
Login

Question

Question: From mean value theorem f(2) – f(1) = (b – a).f '(x<sub>1</sub>), a < x<sub>1</sub> < b & f(x) = \(...

From mean value theorem f(2) – f(1) = (b – a).f '(x1),

a < x1 < b & f(x) = 1x\frac{1}{x} then x1 equal to

A

2aba+b\frac{2ab}{a + b}

B

b−ab+a\frac{b - a}{b + a}

C

ab\sqrt{ab}

D

a+b2\frac{a + b}{2}

Answer

ab\sqrt{ab}

Explanation

Solution

f '(x) = f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}

– 1x12=1b−1ab−a\frac{1}{x_{1}^{2}} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a} ̃ 1x12\frac{1}{x_{1}^{2}} = a−bab(b−a)\frac{a - b}{ab(b - a)}

– 1x12=−1ab\frac{1}{x_{1}^{2}} = - \frac{1}{ab} ̃ x12 = ab ̃ x1 = ab\sqrt{ab}