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Mathematics · 2022 · 4 marks
CBSE 2022 · Region 2 · Set 2 · Q11
(a) Using integration, find the area of the smaller region enclosed by the curve 4 x^2+4 y^2=9 and the line 2 x +2 y =3.
(a)
Using integration, find the area of the smaller region enclosed by the curve $\displaystyle 4 \mathrm{x}^{2}+4 \mathrm{y}^{2}=9$ and the line $\displaystyle 2 \mathrm{x}+2 \mathrm{y}=3$.
OR
(b)
If the area of the region bounded by the curve $\displaystyle \mathrm{y}^{2}=4 \mathrm{ax}$ and the line $\displaystyle \mathrm{x}=4 \mathrm{a}$ is $\displaystyle \frac{256}{3}$ sq. units, then using integration, find the value of a, where a > 0.
Marking-scheme solution
(a)
Clearly point of intersection are $\displaystyle \left(\dfrac{3}{2},\,0\right)$ & $\displaystyle \left(0,\,\dfrac{3}{2}\right)$
Required area $\displaystyle = \displaystyle\int_{0}^{3/2}\sqrt{\dfrac{9}{4}-x^{2}}\,dx-\int_{0}^{3/2}\left(\dfrac{3}{2}-x\right)dx$
\[=\left.\dfrac{x}{2}\sqrt{\dfrac{9}{4}-x^{2}}+\dfrac{9}{8}\sin^{-1}\dfrac{2x}{3}\right|_{0}^{3/2}+\left.\dfrac{\left(\dfrac{3}{2}-x\right)^{2}}{2}\right|_{0}^{3/2}
\]
\[=\dfrac{9\pi}{16}-\dfrac{9}{8}
\]
Or(b) Given area $\displaystyle =\dfrac{256}{3}$
Area of Shaded region $\displaystyle = 2\displaystyle\int_{0}^{4a}\sqrt{4ax}\;dx$
\[=\left.8\sqrt{a}\,\dfrac{x^{3/2}}{3}\right|_{0}^{4a}
\]
\[=\dfrac{64a^{2}}{3}
\]
\[\dfrac{64a^{2}}{3}=\dfrac{256}{3}
\]
$\displaystyle \Rightarrow a^{2}=4$ gives $\displaystyle a=2$ (as $\displaystyle a>0$)
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CBSE Class 12 Mathematics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.