If from Lagrange's mean value theorem, we have f'(x_1)= f(b)-f(a)… - Vidyadip

MCQ

If from Lagrange's mean value theorem, we have
$\text{f}'(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then:

✓
$\text{a}<\text{x}_1\leq\text{b}$
B
$\text{a}\leq\text{x}_1<\text{b}$
C
$\text{a}<\text{x}_1<\text{b}$
D
$\text{a}\leq\text{x}_1\leq\text{b}$

✓

Answer

Correct option: A.

$\text{a}<\text{x}_1\leq\text{b}$

We have
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
In the Lagrange's mean value theorem, $\text{c}\in(\text{a},\text{b})$ such that $\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
So, if there is $x_1$ such that $\text{f}'(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then $\text{x}_1\in(\text{a},\text{b})$
$\Rightarrow\text{a}<\text{x}_1<\text{b}$

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