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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
For the function $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
  • A
    $x = 1$ is a point of maximum.
  • B
    $x = -1$ is a point of minimum.
  • C
    maximum value $ > $ minimum value.
  • ✓
    maximum value $ < $ minimum value.
✓

Answer

Correct option: D.
maximum value $ < $ minimum value.
Given, $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
lmplies that $\text{f}\ ''(\text{x})=\text{x}-\frac{1}{\text{x}}$
For a local maxima or a local minima, we must have $f\ '(x) = 0$
lmplies that $1-\frac{1}{\text{x}^{2}}=0$
lmplies that $\text{x}^{2}-1=0$
lmplies that $\text{x}^{2}=0$
lmplies that $\text{x}=\pm1$
Now, $\text{f}\ '(\text{x})=\frac{2}{\text{x}^{3}}$
lmplie that $\text{f}\ '(\text{1})=\frac{2}{\text{1}}=2<0$
Threrefore, $x = 1$ is a local minima.
Also, $f\ ''(1) - 2 < 0$
Threrefore, $x = -1$ is a local maxima.
The local minimum value is given by
$f(1) = 2$
The local maximum value is given by
$f(-1) = -2$
$\therefore$ maximum value $ < $ minimum value.

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