CBSE 2026 · Region 2 · Set 3 · Q32 · 5 marks
Find the sub-interval of $\displaystyle \left(0, \frac{\pi}{2}\right)$ in which $\displaystyle \mathrm{f}(x)=\log (\sin x+\cos x)$ is increasing and decreasing.A rectangle of perimeter $\displaystyle 30$ 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 $\displaystyle \left(0, \frac{\pi}{2}\right)$ in which $\displaystyle \mathrm{f}(x)=\log (\sin x+\cos x)$ is increasing and decreasing.
A rectangle of perimeter $\displaystyle 30$ 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
$\displaystyle \mathrm{f}(x)=\log(\sin x+\cos x), x \in\left(0, \dfrac{\pi}{2}\right)$
$\displaystyle \Rightarrow \mathrm{f}^{\prime}(x)=\dfrac{\cos x-\sin x}{\sin x+\cos x}$
For critical points of $\displaystyle \mathrm{f}(x)$, put $\displaystyle \mathrm{f}^{\prime}(x)=0$
$\displaystyle \Rightarrow \dfrac{\cos x-\sin x}{\sin x+\cos x}=0 \Rightarrow \cos x=\sin x$
$\displaystyle \Rightarrow \tan x=1$
$\displaystyle \Rightarrow x=\dfrac{\pi}{4} \in\left(0, \dfrac{\pi}{2}\right)$
$\displaystyle \mathrm{f}^{\prime}(x)>0$ when $\displaystyle x \in\left(0, \dfrac{\pi}{4}\right) \Rightarrow \mathrm{f}(x)$ is increasing in $\displaystyle \left(0, \dfrac{\pi}{4}\right)$.
$\displaystyle \mathrm{f}^{\prime}(x)<0$ when $\displaystyle x \in\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right) \Rightarrow \mathrm{f}(x)$ is decreasing in $\displaystyle \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$.
Let the lengths of the sides of the rectangle be $\displaystyle x$ and $\displaystyle (15-x)$.
Let it be revolved around the side of length $\displaystyle (15-x)$, so that $\displaystyle x$ becomes the radius of the cylinder.
Volume of cylinder, $\displaystyle V=\pi x^{2}(15-x)=\pi\left(15 x^{2}-x^{3}\right)$
$\displaystyle \dfrac{dV}{dx}=\pi\left(30 x-3 x^{2}\right)$
For critical points, put $\displaystyle \dfrac{dV}{dx}=0$
$\displaystyle \Rightarrow x=10$ cm $\displaystyle (\because x \neq 0)$
Now $\displaystyle \dfrac{d^{2} V}{dx^{2}}=\pi(30-6 x)$ ; $\displaystyle \left.\dfrac{d^{2} V}{dx^{2}}\right]_{x=10\ \text{cm}}=-30\pi<0$
So, volume is maximum when $\displaystyle x=10$ cm.
Hence the dimensions of rectangle are $\displaystyle 10$ cm and $\displaystyle 5$ cm.
Application of DerivativesIncreasing and Decreasing FunctionsApplylong_answermedium
More from Application of Derivatives
- Overspeeding increases fuel consumption and decreases fuel economy as a result of tyre rolling friction and…2024 · asked 3×
- Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is…2024 · asked 3×
- A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed…2025 · asked 3×
- At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup…2026 · asked 3×
- A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they…2026 · asked 3×
- Sooraj's father wants to construct a rectangular garden using a brick wall on one side of the garden and wire…2023 · asked 3×
- If f(x)=a(x- cos x) is strictly decreasing in R, then ' a ' belongs to2023 · asked 3×
- A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better…2023 · asked 3×
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.