JEE Areas Questions

Question 1 JEE Mains

Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+y=8$$ and y-axis that lies in the first quadrant. Let $$A_{2}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,y^{2}=x,x=2$$ and y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

$$\frac{2}{3}(2\sqrt{2}+1)$$

$$\frac{2}{3}(\sqrt{2}+1)$$

$$\frac{2}{3}(4\sqrt{2}+1)$$

$$\frac{2}{3}(3\sqrt{2}+1)$$

Solution

Given curves for $$A_1$$:

$$y = x^2 + 2, \quad x + y = 8 \implies y = 8 - x, \quad x = 0 \quad (\text{y-axis})$$

Finding the point of intersection of $$y = x^2 + 2$$ and $$y = 8 - x$$:

$$x^2 + 2 = 8 - x \implies x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0 \implies x = 2 \quad (\text{since } x \ge 0)$$

$$A_1 = \int_{0}^{2} \left[ (8 - x) - (x^2 + 2) \right] dx = \int_{0}^{2} (6 - x - x^2) dx$$

$$A_1 = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$

Given curves for $$A_2$$:

$$y = x^2 + 2, \quad y^2 = x \implies y = \sqrt{x}, \quad x = 2, \quad x = 0$$

$$A_2 = \int_{0}^{2} \left[ (x^2 + 2) - \sqrt{x} \right] dx$$

$$A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2} = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$$

$$A_1 - A_2 = \frac{22}{3} - \left( \frac{20}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1)$$

Question 1 JEE Mains

Let the area of the region $$\{(x, y) : x - 2y + 4 \geq 0, \; x + 2y^2 \geq 0, \; x + 4y^2 \leq 8, \; y \geq 0\}$$ be $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime numbers. Then $$m + n$$ is equal to _______.

Solution

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.

JEE Areas Questions

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