Area Ques 25
Question
- The area (in sq units) of region described by $(x, y) y^{2} \leq 2 x$ and $y \geq 4 x-1$ is
(2015 JEE Main)
(a) $\frac{7}{32}$
(b) $\frac{5}{64}$
(c) $\frac{15}{64}$
(d) $\frac{9}{32}$
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Answer:
Correct Answer: 25.(d)
Solution:
Formula:
- Given region is $\left\{(x, y): y^{2} \leq 2 x\right.$ and $\left.y \geq 4 x-1\right\}$
$y^{2} \leq 2 x$ represents a region inside the parabola
$ y^{2}=2 x $ $\quad$ …….(i)
and $y \geq 4 x-1$ represents a region above the line
$ y=4 x-1 $\quad$ …….(ii)
The point of intersection of the curves (i) and (ii) is
$ \begin{alignedat} & (4 x-1)^{2}=2 x \Rightarrow 16 x^{2}-8 x+1=2 x \\ \Rightarrow \quad & 16 x^{2}-10 x+11=0 \Rightarrow \quad x=\frac{1}{2}, \frac{1}{8} \end{aligned} $
So, the points where these curves intersect are $(\frac{1}{2}, 1)$ and $(\frac{1}{8}, \frac{1}{2})$.
$\therefore$ Required area $=\int _{-1 / 2}^{1} (\frac{y+1}{4}-\frac{y^{2}}{2}) d y$
$ \begin{alignedat} & =\frac{1}{4} (\frac{y^{2}}{2}+y){ } _{-1 / 2}^{-1}-\frac{1}{6}\left(y^{3}\right) _{-1 / 2}^{1} \\ & =\frac{1}{4} [(\frac{1}{2}+1)-(\frac{1}{8}-\frac{1}{2})]-\frac{1}{6} (1+\frac{1}{8} )\\ & =\frac{1}{4} (\frac{3}{2}+\frac{3}{8})-\frac{1}{6} (\frac{9}{8}) \\ & =\frac{1}{4} \times \frac{15}{8}-\frac{3}{16}=\frac{3}{32} \text { sq units } \end{aligned} $
BETA
