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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
Choose the correct answers from the given four options : For the function $\text{f(x)}=\text{x}+\frac{1}{\text{x}},\text{x}\in[1,3],$ the value of $c$ for mean value theorem is :
  • A
    $1$
  • ✓
    $\sqrt{3}$
  • C
    $2$
  • D
    None of these
✓

Answer

Correct option: B.
$\sqrt{3}$
$\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ which is continuous and differentiable.
So, by mean value theorem there exists atleast one $\text{c}\in(1,3)$ such that
$\because\ \text{f}\ '\text{c}=\frac{\text{f(b)}-\text{f(a)}}{\text{b}-\text{a}}$
$\Rightarrow\ 1-\frac{1}{\text{c}^2}=\frac{\frac{10}{3}-2}{3-1}$
$\Rightarrow\ \frac{\text{c}^2-1}{\text{c}^2}=\frac{2}{3}$
$\Rightarrow\ 3(\text{c}^2-1)=2\text{c}^2$
$\Rightarrow\ 3\text{c}^2-2\text{c}^2=3$
$\Rightarrow\ \text{c}^2=3$
$\Rightarrow\ \text{c}=\sqrt{3}\in(1,3)$

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