CBSE 2026 · Region 2 · Set 2 · Q35 · 5 marks
Find the sub-interval of $\displaystyle (0, \pi)$ in which $\displaystyle \mathrm{f}(x)=\tan ^{-1}(\sin x-\cos x)$ is increasing and decreasing.A rectangle of perimeter $\displaystyle 24$ cm is revolved along one of its sides to sweep out a cylinder of maximum volume.
Find the dimensions of the rectangle.
Find the sub-interval of $\displaystyle (0, \pi)$ in which $\displaystyle \mathrm{f}(x)=\tan ^{-1}(\sin x-\cos x)$ is increasing and decreasing.
A rectangle of perimeter $\displaystyle 24$ cm is revolved along one of its sides to sweep out a cylinder of maximum volume.
Find the dimensions of the rectangle.
Marking-scheme solution
$\displaystyle \mathrm{f}(x)=\tan^{-1}(\sin x-\cos x), x \in(0, \pi)$
$\displaystyle \mathrm{f}^{\prime}(x)=\dfrac{\cos x+\sin x}{1+(\sin x-\cos x)^{2}}$
For critical points of $\displaystyle \mathrm{f}(x)$, put $\displaystyle \mathrm{f}^{\prime}(x)=0$
$\displaystyle \Rightarrow \cos x+\sin x=0$
getting $\displaystyle x=\dfrac{3\pi}{4} \in(0, \pi)$
$\displaystyle \mathrm{f}^{\prime}(x)>0$ when $\displaystyle \left(0, \dfrac{3\pi}{4}\right) \Rightarrow \mathrm{f}(x)$ is increasing when $\displaystyle x \in\left(0, \dfrac{3\pi}{4}\right)$.
$\displaystyle \mathrm{f}^{\prime}(x)<0$ when $\displaystyle \left(\dfrac{3\pi}{4}, \pi\right) \Rightarrow \mathrm{f}(x)$ is decreasing when $\displaystyle x \in\left(\dfrac{3\pi}{4}, \pi\right)$.
Let the lengths of the sides of the rectangle be $\displaystyle x$ and $\displaystyle (12-x)$.
Let it be revolved around the side of length $\displaystyle (12-x)$, so that $\displaystyle x$ becomes the radius of the cylinder.
Volume of cylinder, $\displaystyle V=\pi x^{2}(12-x)=\pi\left(12 x^{2}-x^{3}\right)$
$\displaystyle \dfrac{dV}{dx}=\pi\left(24 x-3 x^{2}\right)$
For critical points, put $\displaystyle \dfrac{dV}{dx}=0$
$\displaystyle \Rightarrow x=8$ cm $\displaystyle (\because x \neq 0)$
Now $\displaystyle \dfrac{d^{2} V}{dx^{2}}=\pi(24-6 x)$ ; $\displaystyle \left.\dfrac{d^{2} V}{dx^{2}}\right]_{x=8\ \text{cm}}=-24\pi<0$
So, volume is maximum when $\displaystyle x=8$ cm.
Hence the dimensions of rectangle are $\displaystyle 8$ cm and $\displaystyle 4$ cm.
Application of DerivativesIncreasing and Decreasing FunctionsApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.