If for some \alpha > 0, the area of the region \ (x,y):|x + \alpha | \le y \le 2 - |x|\ is equal to 3 \over 2 , then the area of the region \ (x,y):0 \le y \le x + 2\alpha ,\,|x| \le 1\ is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2022 · 30 Jun · Shift 1 · Q36 with Solution

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Correct answer: 8-4\SQRT{3}

Let $R_1=\{(x,y): |x+\alpha|\le y\le 2-|x|\}.$ We are given that the area of $R_1$ is $\dfrac32$ for some $\alpha>0$ .

We must then find the area of $R_2=\{(x,y):0\le y\le x+2\alpha,\, |x|\le 1\}.$

1. Interpreting the first region

The first region lies between the curves $y=|x+\alpha| \quad \text{and} \quad y=2-|x|.$ So its area is $A_1=\int (\text{upper} - \text{lower})\,dx$ over those $x$ for which $|x+\alpha|\le 2-|x|.$

Thus we first solve $|x+\alpha|+|x|\le 2.$

Now the vertical height is $(2-|x|)-|x+\alpha|=2-(|x|+|x+\alpha|).$ Using the same identity,

=(2-\alpha)-2\left|x+\frac\alpha2\right|.$$ So the area is $$A_1=\int_{-1}^{1-\alpha}\left[(2-\alpha)-2\left|x+\frac\alpha2\right|\right]dx.$$ This is the area of an isosceles triangle with: - base length $$\left(1-\alpha\right)-(-1)=2-\alpha,$$ - maximum height at $x=-\alpha/2$ equal to $$2-\alpha.$$ Therefore, $$A_1=\frac12(2-\alpha)(2-\alpha)=\frac{(2-\alpha)^2}{2}.$$ Given $A_1=\dfrac32$, we get $$\frac{(2-\alpha)^2}{2}=\frac32.$$ So $$(2-\alpha)^2=3.$$ Since $\alpha>0$ and also $\alpha\le 2$, we must have $$2-\alpha=\sqrt3,$$ thus $$\alpha=2-\sqrt3.$$ --- ## 2. Area of the second region Now consider $$R_2=\{(x,y):0\le y\le x+2\alpha,\, |x|\le 1\}.$$ Since $|x|\le 1$, we have $x\in[-1,1]$. Its area is $$A_2=\int_{-1}^{1}(x+2\alpha)\,dx,$$ provided $x+2\alpha\ge 0$ on this interval. Now $$2\alpha=2(2-\sqrt3)=4-2\sqrt3>1,$$ so $$x+2\alpha\ge -1+2\alpha = -1+4-2\sqrt3=3-2\sqrt3>0.$$ Thus the whole strip contributes. Hence, $$A_2=\int_{-1}^{1}(x+2\alpha)\,dx =\int_{-1}^{1}x\,dx+2\alpha\int_{-1}^{1}dx.$$ Now, $$\int_{-1}^{1}x\,dx=0, \qquad \int_{-1}^{1}dx=2.$$ So $$A_2=0+2\alpha\cdot 2=4\alpha.$$ Substitute $\alpha=2-\sqrt3$: $$A_2=4(2-\sqrt3)=8-4\sqrt3.$$ --- ## 3. Comparison with stored answer Our derived answer is $$8-4\sqrt3,$$ which is approximately $1.072$, not $4$. So the stored correct answer appears to be incorrect.

More from Area Under the Curves

1. Interpreting the first region

The first region lies between the curves

y=|x+\alpha| \quad \text{and} \quad y=2-|x|.

So its area is

A_1=\int (\text{upper} - \text{lower})\,dx

over those

x

for which

|x+\alpha|\le 2-|x|.

Thus we first solve

|x+\alpha|+|x|\le 2.

A standard identity is:

|x+\alpha|+|x|=2\left|x+\frac\alpha2\right|+\alpha.

Hence the inequality becomes

2\left|x+\frac\alpha2\right|+\alpha\le 2,

\left|x+\frac\alpha2\right|\le 1-\frac\alpha2.

Therefore this region exists only if

\alpha\le 2

, and the allowed interval is

x\in\left[-1,-\alpha+1\right].

Now the vertical height is

(2-|x|)-|x+\alpha|=2-(|x|+|x+\alpha|).

Using the same identity,

=(2-\alpha)-2\left|x+\frac\alpha2\right|.$$ So the area is $$A_1=\int_{-1}^{1-\alpha}\left[(2-\alpha)-2\left|x+\frac\alpha2\right|\right]dx.$$ This is the area of an isosceles triangle with: - base length $$\left(1-\alpha\right)-(-1)=2-\alpha,$$ - maximum height at $x=-\alpha/2$ equal to $$2-\alpha.$$ Therefore, $$A_1=\frac12(2-\alpha)(2-\alpha)=\frac{(2-\alpha)^2}{2}.$$ Given $A_1=\dfrac32$, we get $$\frac{(2-\alpha)^2}{2}=\frac32.$$ So $$(2-\alpha)^2=3.$$ Since $\alpha>0$ and also $\alpha\le 2$, we must have $$2-\alpha=\sqrt3,$$ thus $$\alpha=2-\sqrt3.$$ --- ## 2. Area of the second region Now consider $$R_2=\{(x,y):0\le y\le x+2\alpha,\, |x|\le 1\}.$$ Since $|x|\le 1$, we have $x\in[-1,1]$. Its area is $$A_2=\int_{-1}^{1}(x+2\alpha)\,dx,$$ provided $x+2\alpha\ge 0$ on this interval. Now $$2\alpha=2(2-\sqrt3)=4-2\sqrt3>1,$$ so $$x+2\alpha\ge -1+2\alpha = -1+4-2\sqrt3=3-2\sqrt3>0.$$ Thus the whole strip contributes. Hence, $$A_2=\int_{-1}^{1}(x+2\alpha)\,dx =\int_{-1}^{1}x\,dx+2\alpha\int_{-1}^{1}dx.$$ Now, $$\int_{-1}^{1}x\,dx=0, \qquad \int_{-1}^{1}dx=2.$$ So $$A_2=0+2\alpha\cdot 2=4\alpha.$$ Substitute $\alpha=2-\sqrt3$: $$A_2=4(2-\sqrt3)=8-4\sqrt3.$$ --- ## 3. Comparison with stored answer Our derived answer is $$8-4\sqrt3,$$ which is approximately $1.072$, not $4$. So the stored correct answer appears to be incorrect.

Area Under the Curves question

Area Under the Curves question

1. Interpreting the first region

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Area Under the Curves question

Area Under the Curves question

1. Interpreting the first region

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