Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1, R2. If max R1, R2 = R2, then R_2 \over R_1 is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q42 with Solution

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Correct answer: 19

Identify the region $S$

The curves are:

$y=x^3$
$y^2=x \Rightarrow x=y^2$

It is easier to describe both in terms of $y$ :

From $y=x^3$ , we get $x=y^{1/3}$
From $y^2=x$ , we have $x=y^2$

Their intersection points satisfy $y=x^3,\quad x=y^2$ Substitute $x=y^2$ into $y=x^3$ : $y=(y^2)^3=y^6$ $y^6-y=0$ $y(y^5-1)=0$ So real solutions are: $y=0,1$ Hence intersection points are $(0,0)$ and $(1,1)$ .

For $0\le y\le 1$ , $y^2 \le y^{1/3}$ so the bounded region $S$ lies between $x=y^2 \quad \text{and} \quad x=y^{1/3}, \qquad 0\le y\le 1.$

Thus total area of $S$ is $A_S=\int_0^1 \left(y^{1/3}-y^2\right)dy$

=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}.$$ --- 2. **Understand the dividing curve $y=2|x|$** Since $S$ lies in the first quadrant, $|x|=x$. So inside $S$, the dividing curve is simply $$y=2x \quad \Rightarrow \quad x=\frac y2.$$ We compare $x=\frac y2$ with the left and right boundaries of $S$: - left boundary: $x=y^2$ - right boundary: $x=y^{1/3}$ For $0<y<1$, $$y^2<\frac y2<y^{1/3}$$ (because $2y^2<y$ i.e. $y<\tfrac12$ is not always true, so let us check intersections carefully). We find where the line meets the boundaries. ### With $x=y^2$ $$\frac y2=y^2$$ $$y\left(y-\frac12\right)=0$$ So intersections at $$y=0,\frac12.$$ ### With $x=y^{1/3}$ $$\frac y2=y^{1/3}$$ For $y>0$, $$\frac{y^{2/3}}{2}=1 \Rightarrow y^{2/3}=2 \Rightarrow y=2^{3/2}>1$$ which is outside the region $0\le y\le1$. So within $S$, the line only meets the parabola $x=y^2$ at $y=0$ and $y=\tfrac12$. Now compare positions: - For $0\le y<\frac12$, we have $y^2<\frac y2<y^{1/3}$, so the line passes through the region. - For $\frac12<y\le1$, we have $\frac y2<y^2$, so the line lies to the **left** of the region and does not cut it. Therefore the line cuts off a small region for $0\le y\le \frac12$ between $x=y^2$ and $x=\frac y2$. --- 3. **Compute the smaller area** The area between $x=y^2$ and $x=\frac y2$ from $y=0$ to $y=\frac12$ is $$R_1=\int_0^{1/2}\left(\frac y2-y^2\right)dy$$ $$=\left[\frac{y^2}{4}-\frac{y^3}{3}\right]_0^{1/2}$$ $$=\frac{1}{16}-\frac{1}{24}$$ $$=\frac{1}{48}.$$ This is the smaller part. Hence the larger part is $$R_2=A_S-R_1=\frac{5}{12}-\frac{1}{48}$$ $$=\frac{20}{48}-\frac{1}{48}=\frac{19}{48}.$$ --- 4. **Find the ratio** $$\frac{R_2}{R_1}=\frac{19/48}{1/48}=19.$$ So, $$\boxed{\frac{R_2}{R_1}=19}.$$ --- 5. **Compare with stored answer** Stored correct answer: $19$ Our derived answer is also $19$, so it agrees.

More from Area Under the Curves

View written solutionFree

Correct answer: 19

Identify the region $S$

The curves are:

$y=x^3$
$y^2=x \Rightarrow x=y^2$

It is easier to describe both in terms of $y$ :

From $y=x^3$ , we get $x=y^{1/3}$
From $y^2=x$ , we have $x=y^2$

For $0\le y\le 1$ , $y^2 \le y^{1/3}$ so the bounded region $S$ lies between $x=y^2 \quad \text{and} \quad x=y^{1/3}, \qquad 0\le y\le 1.$

Thus total area of $S$ is $A_S=\int_0^1 \left(y^{1/3}-y^2\right)dy$

=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}.$$ --- 2. **Understand the dividing curve $y=2|x|$** Since $S$ lies in the first quadrant, $|x|=x$. So inside $S$, the dividing curve is simply $$y=2x \quad \Rightarrow \quad x=\frac y2.$$ We compare $x=\frac y2$ with the left and right boundaries of $S$: - left boundary: $x=y^2$ - right boundary: $x=y^{1/3}$ For $0<y<1$, $$y^2<\frac y2<y^{1/3}$$ (because $2y^2<y$ i.e. $y<\tfrac12$ is not always true, so let us check intersections carefully). We find where the line meets the boundaries. ### With $x=y^2$ $$\frac y2=y^2$$ $$y\left(y-\frac12\right)=0$$ So intersections at $$y=0,\frac12.$$ ### With $x=y^{1/3}$ $$\frac y2=y^{1/3}$$ For $y>0$, $$\frac{y^{2/3}}{2}=1 \Rightarrow y^{2/3}=2 \Rightarrow y=2^{3/2}>1$$ which is outside the region $0\le y\le1$. So within $S$, the line only meets the parabola $x=y^2$ at $y=0$ and $y=\tfrac12$. Now compare positions: - For $0\le y<\frac12$, we have $y^2<\frac y2<y^{1/3}$, so the line passes through the region. - For $\frac12<y\le1$, we have $\frac y2<y^2$, so the line lies to the **left** of the region and does not cut it. Therefore the line cuts off a small region for $0\le y\le \frac12$ between $x=y^2$ and $x=\frac y2$. --- 3. **Compute the smaller area** The area between $x=y^2$ and $x=\frac y2$ from $y=0$ to $y=\frac12$ is $$R_1=\int_0^{1/2}\left(\frac y2-y^2\right)dy$$ $$=\left[\frac{y^2}{4}-\frac{y^3}{3}\right]_0^{1/2}$$ $$=\frac{1}{16}-\frac{1}{24}$$ $$=\frac{1}{48}.$$ This is the smaller part. Hence the larger part is $$R_2=A_S-R_1=\frac{5}{12}-\frac{1}{48}$$ $$=\frac{20}{48}-\frac{1}{48}=\frac{19}{48}.$$ --- 4. **Find the ratio** $$\frac{R_2}{R_1}=\frac{19/48}{1/48}=19.$$ So, $$\boxed{\frac{R_2}{R_1}=19}.$$ --- 5. **Compare with stored answer** Stored correct answer: $19$ Our derived answer is also $19$, so it agrees.

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