CBSE 2024 · Region 3 · Set 1 · Q27 · 3 marks
Find the intervals in which the function $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\log \mathrm{x}}{\mathrm{x}}$ is strictly increasing or strictly decreasing.Find the absolute maximum and absolute minimum values of the function f given by $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}}$, on the interval $\displaystyle [1,2]$.
Find the intervals in which the function $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\log \mathrm{x}}{\mathrm{x}}$ is strictly increasing or strictly decreasing.
Find the absolute maximum and absolute minimum values of the function f given by $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}}$, on the interval $\displaystyle [1,2]$.
Marking-scheme solution
$$\mathrm{f}(\mathrm{x})=\frac{\log \mathrm{x}}{\mathrm{x}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}} ; \mathrm{x}>0for strictly increasing/decreasing, put $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=e$ for strictly increasing, $\displaystyle \mathrm{x} \in(0, e)$ and for strictly decreasing $\displaystyle \mathrm{x} \in(e, \infty)$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}} ; \mathrm{x} \in[1,2]
& \Rightarrow \mathrm{f}^{I}(\mathrm{x})=\frac{1}{2}-\frac{2}{\mathrm{x}^{2}}
\end{aligned}for absolute maximum / minimum, put $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=0$
$\displaystyle \Rightarrow \mathrm{x}^{2}=4 \Rightarrow \mathrm{x}=2$
Now, $\displaystyle \mathrm{f}(1)=\frac{5}{2}$ and $\displaystyle \mathrm{f}(2)=2$
∴ absolute maximum value $\displaystyle =\frac{5}{2}$ and absolute minimum value $\displaystyle =2$
Application of DerivativesIncreasing and Decreasing FunctionsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.