MCQ
Choose the correct answer: Area lying between the curves $y^2 = 4x$ and $y = 2x$ is:
- A$\frac23$
- ✓$\frac13$
- C$\frac14$
- D$\frac34.$
✓
Answer
Correct option: B.
$\frac13$
Equation of curve $($parabola$)$ is $y^2 = 4x ...(i)$
$\Rightarrow\text{y}=2\sqrt{\text{x}}=2\text{x}^{\frac12}...(\text{ii})$
Equation of another curve $($line$)$ is $y = 2x ...(iii)$ Solving $eq. (i)$ and $(iii),$
we get $x = 0$ or $x = 1$ and $y = 0$ or $y = 2$
Therefore, Points of intersections of circle $(i)$ and line $(ii)$ are $O(0, 0)$ and $A(1, 2).$
Now Area $\ce{OBAM} =$ Area bounded by parabola $(i)$ and $x-$axis $=\Bigg|\int\limits^1_0\text{ydx}\Bigg|$
$=\Bigg|\int\limits^1_02\text{x}^{\frac12}\text{dx}\Bigg|$
$= 2\frac{\Big(\text{x}^{\frac32}\Big)^1_0}{\frac32}$
$=\frac43(1-0)=\frac43\dots(\text{iv})$
Also, Area $\Delta\text{ OAM}=$ Area bounded by parabola $(iii)$ and $x-$axis
$=\Bigg|\int\limits^1_0\text{y dx}\Bigg|$
$=\Bigg|\int\limits^1_02\text{x dx}\Bigg|$
$=2\Big(\frac{\text{x}^2}{2}\Big)^1_0 $
$= (1 - 0) = 1 ...(v)$
Now Required shaded area $\ce{OBA} =$ Area $\ce{OBAM} -$ Area of $\Delta\text{ OAM}$
$=\frac43-1$
$=\frac{4-3}{3}$
$=\frac13\text{ sq. units}$
Therefore, option $(B)$ is correct.
$\Rightarrow\text{y}=2\sqrt{\text{x}}=2\text{x}^{\frac12}...(\text{ii})$
Equation of another curve $($line$)$ is $y = 2x ...(iii)$ Solving $eq. (i)$ and $(iii),$
we get $x = 0$ or $x = 1$ and $y = 0$ or $y = 2$
Therefore, Points of intersections of circle $(i)$ and line $(ii)$ are $O(0, 0)$ and $A(1, 2).$
Now Area $\ce{OBAM} =$ Area bounded by parabola $(i)$ and $x-$axis $=\Bigg|\int\limits^1_0\text{ydx}\Bigg|$
$=\Bigg|\int\limits^1_02\text{x}^{\frac12}\text{dx}\Bigg|$
$= 2\frac{\Big(\text{x}^{\frac32}\Big)^1_0}{\frac32}$
$=\frac43(1-0)=\frac43\dots(\text{iv})$
Also, Area $\Delta\text{ OAM}=$ Area bounded by parabola $(iii)$ and $x-$axis
$=\Bigg|\int\limits^1_0\text{y dx}\Bigg|$
$=\Bigg|\int\limits^1_02\text{x dx}\Bigg|$
$=2\Big(\frac{\text{x}^2}{2}\Big)^1_0 $
$= (1 - 0) = 1 ...(v)$
Now Required shaded area $\ce{OBA} =$ Area $\ce{OBAM} -$ Area of $\Delta\text{ OAM}$
$=\frac43-1$
$=\frac{4-3}{3}$
$=\frac13\text{ sq. units}$
Therefore, option $(B)$ is correct.
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