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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
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:
  • A
    $\text{a}<\text{x}_1\leq\text{b}$
  • B
    $\text{a}\leq\text{x}_1<\text{b}$
  • ✓
    $\text{a}<\text{x}_1<\text{b}$
  • D
    $\text{a}\leq\text{x}_1\leq\text{b}$
✓

Answer

Correct option: C.
$\text{a}<\text{x}_1<\text{b}$
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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