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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 2 · Set 1 · Q23
(a) Find the absolute maximum value of f (x)= cos x+ sin^2 x, x ∈[0, π]
(a)
Find the absolute maximum value of $\displaystyle \mathrm{f}(x)=\cos x+\sin ^{2} x, x \in[0, \pi]$
OR
(b)
If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.
Marking-scheme solution
$\displaystyle \mathrm{f}(x)=\cos x+\sin^{2} x, x \in[0, \pi]$
$\displaystyle \mathrm{f}^{\prime}(x)=-\sin x+2 \sin x \cos x$
$\displaystyle \mathrm{f}^{\prime}(x)=0 \Rightarrow \sin x(2 \cos x-1)=0$
$\displaystyle \Rightarrow \sin x=0$ or $\displaystyle \cos x=\dfrac{1}{2}$
$\displaystyle \therefore x=\dfrac{\pi}{3}$
Now, $\displaystyle \mathrm{f}(0)=1, \mathrm{f}\left(\dfrac{\pi}{3}\right)=\dfrac{5}{4}, \mathrm{f}(\pi)=-1$
$\displaystyle \therefore$ Absolute maximum value of $\displaystyle \mathrm{f}(x)$ is $\displaystyle \dfrac{5}{4}$.
$\displaystyle V=\dfrac{2}{3} \pi r^{3} \Rightarrow \dfrac{dV}{dt}=2 \pi r^{2} \dfrac{dr}{dt}=C$ (say)
$\displaystyle \Rightarrow \dfrac{dr}{dt}=\dfrac{C}{2 \pi r^{2}}$
Now, $\displaystyle S=3 \pi r^{2}$
$\displaystyle \Rightarrow \dfrac{dS}{dt}=6 \pi r \dfrac{dr}{dt}=6 \pi r \times \dfrac{C}{2 \pi r^{2}}=\dfrac{3C}{r}$
$\displaystyle \therefore S$ varies inversely as its radius.
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