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$\left( {2,\infty } \right)$
- ✓$\left( {0,1} \right) \cup \left( {2,\infty } \right)$
$f^{\prime}(x)=\left\{\begin{array}{ll}\frac{(x-2)}{x^{3}} & x<1 \\ \frac{-(x-2)}{x^{3}} & x \geq 1\end{array}\right.$
$x<1,$ 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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