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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
For the function $\text{f}(\text{x})=\text{x}+1\text{x},\text{x}\in[1,3],$ the value of c for the Lagrange's mean value theorem is:
  • A
    1
  • ✓
    $\sqrt3$
  • C
    2
  • D
    none of these
✓

Answer

Correct option: B.
$\sqrt3$
We have

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$

Clearly, f(x) is continuous on [1, 3] and derivable on (1, 3).

Thus, both the conditions of Lagrange's theorem is satisfied.

Concequently there exists $\text{c}\in(1,3)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$

Now, $\text{f}(\text{x})=\frac{\text{x}^2+1}{\text{x}}$

$\text{f}'(\text{x})=\frac{\text{x}^2-1}{\text{x}^2},\text{f}(1)=2,\text{f}(3)=\frac{10}{3}$

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{4}{6}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{2}{3}$

$\Rightarrow3\text{x}^2-3=2\text{x}^2$

$\Rightarrow\text{x}=\pm\sqrt3$

Thus, $\text{c}=\sqrt3\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}.$

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