ISI2018-DCG-26

Question

The area of the region bounded by the curves $y=\sqrt x,$ $2y+3=x$ and $x$-axis in the first quadrant is

• A. $9$
• B. $\frac{27}{4}$
• C. $36$
• D. $18$

Comments

getting $\frac{28}{3}$?

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$\mathbf B$ is the right answer.

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@`JEET
Nope. I guess it should be A.

Answer 1

Here, The curve $y=\sqrt{x}$ and the line $2y+3=x$ will intersect each other. So

$\begin{align} y&=\sqrt{2y+3}\\\Rightarrow y^2-2y-3&=0\\\Rightarrow (y-3)(y+1)&=0\\\therefore y &= -1,~3\end{align}$

From the curve $y=\sqrt{x}$, we get $y$ is always non-negative i.e. $y\ge 0$. Therefore $y\ne -1$ so only $y=3$ is solution of the equations. Now putting $y=3$ to the equation $2y+3=x$, we obtain $x=9$.

Therefore, the curve and the line intersect at the point $(9,3)$. Here is shown the graph below.

 

 

Now, the line $2y+3=x\Rightarrow y = \frac{x-3}{2}$ cuts $x$-axis at the point $(3,0)$. $\therefore$ The required region is bounded the by points $(0,0), (3,0)\text{ and }(9,3)$.

 

$\begin{align} \therefore \mathrm{Area} &= \int_{0}^{9}\sqrt{x}~\mathrm{d}x-\int_{3}^{9}\frac{x-3}{2}~\mathrm{d}x\\&= \left[ \frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{9}-\left[ \frac{(x-3)^2}{4}\right]_{3}^{9}\\&=\left[ \frac{2}{3}(9)^{\frac{3}{2}} -0\right]-\left[ \frac{6^2}{4}-0 \right]\\&=18-9\\&=9  \end{align}$

 

So the correct answer is A.

Comments

Great!!