CBSE 2025 · Region 2 · Set 1 · Q38 · 4 marks
A small town is analyzing the pattern of a new street light installation. The lights are set up in such a way that the intensity of light at any point $\displaystyle x$ metres from the start of the street can be modelled by $\displaystyle \mathrm{f}(x)=\mathrm{e}^{x} \sin x$, where $\displaystyle x$ is in metres. Based on the above, answer the following :(i)Find the intervals on which the $\displaystyle \mathrm{f}(x)$ is increasing or decreasing, $\displaystyle x \in[0, \pi]$.(ii)Verify, whether each critical point when $\displaystyle x \in[0, \pi]$ is a point of local maximum or local minimum or a point of inflexion.
A small town is analyzing the pattern of a new street light installation. The lights are set up in such a way that the intensity of light at any point $\displaystyle x$ metres from the start of the street can be modelled by $\displaystyle \mathrm{f}(x)=\mathrm{e}^{x} \sin x$, where $\displaystyle x$ is in metres. Based on the above, answer the following :
(i)
Find the intervals on which the $\displaystyle \mathrm{f}(x)$ is increasing or decreasing, $\displaystyle x \in[0, \pi]$.
(ii)
Verify, whether each critical point when $\displaystyle x \in[0, \pi]$ is a point of local maximum or local minimum or a point of inflexion.
Marking-scheme solution
(i)
$\displaystyle f'(x) = e^{x}(\cos x + \sin x)$
For critical points, $\displaystyle f'(x) = 0$
$\displaystyle \Rightarrow \cos x + \sin x = 0$
$\displaystyle \Rightarrow \cos x = -\sin x$
For $\displaystyle x$ to be a critical point $\displaystyle x \in (0, \pi)$, hence, $\displaystyle x = \dfrac{3\pi}{4}$
For all $\displaystyle x \in \left[0, \dfrac{3\pi}{4}\right],\ f'(x) \geq 0$
Hence, $\displaystyle f$ is increasing in $\displaystyle \left[0, \dfrac{3\pi}{4}\right]$
For all $\displaystyle x \in \left[\dfrac{3\pi}{4}, \pi\right],\ f'(x) \leq 0$
Hence, $\displaystyle f$ is decreasing in $\displaystyle \left[\dfrac{3\pi}{4}, \pi\right]$
(ii)
$\displaystyle x = \dfrac{3\pi}{4}$ is a critical point
$\displaystyle f''(x) = e^{x}(\cos x - \sin x) + e^{x}(\cos x + \sin x)$
$\displaystyle = 2e^{x}\cos x$
$\displaystyle f''\left(\dfrac{3\pi}{4}\right) = -ve$
Hence, $\displaystyle \dfrac{3\pi}{4}$ is a point of local maximum.
Application of DerivativesIncreasing and Decreasing FunctionsApplycase_studymedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.