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STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 6. Application of derivatives4 Marks
MCQ
Function $f\left( x \right) = \frac{{\left| {x - 1} \right|}}{{{x^2}}}$ is monotonic decreasing in
  • A
    $\left( { - \infty ,\infty } \right)$
  • B
    $(0,1)$
  • C
    $\left( {2,\infty } \right)$
  • ✓
    $\left( {0,1} \right) \cup \left( {2,\infty } \right)$
✓

Answer

Correct option: D.
$\left( {0,1} \right) \cup \left( {2,\infty } \right)$
d
$f(x)=\left\{\begin{array}{ll}\frac{-(x-1)}{x^{2}} & x<1 \\ \frac{(x-1)}{x^{2}} & x \geq 1\end{array}\right.$

$f^{\prime}(x)=\left\{\begin{array}{ll}\frac{(x-2)}{x^{3}} & x<1 \\ \frac{-(x-2)}{x^{3}} & x \geq 1\end{array}\right.$

$x<1,$ if $f^{\prime}(x)<0$ (for $f(x)$ to be monotonically decreasing $\Rightarrow \frac{(x-2)}{x^{3}}<0 \Rightarrow x \in(0,2)$

But $x<1 \Rightarrow x \in(0,1)$

For $x \geq 1,$ if $f^{\prime}(x)<0 \Rightarrow \frac{-(x-2)}{x^{3}}<0$

$\Rightarrow \frac{(x-2)}{x^{3}}>0 \Rightarrow x \in(-\infty, 0) \cup(2, \infty)$

But, $x \geq 1 \Rightarrow x \in(2, \infty)$

Hence, $x \in(0,1) \cup(2, \infty)$

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