Area Under Curves: JEE Main 2026 PYQs

Answer: 24

Key idea: Area between curves (hyperbola and line), Integration involving square roots

1. Convert hyperbola to standard form: $\frac{x^2}{9} - \frac{y^2}{16} = 1$ and line: $y = \frac{8x-24}{3}$
2. Find intersection points by substituting line equation into hyperbola: $x^2 - 8x + 15 = 0$ gives $x = 3, 5$
3. Points of intersection are $(3, 0)$ and $(5, \frac{16}{3})$. For the hyperbola (right branch): $y = \frac{4}{3}\sqrt{x^2-9}$
4. Set up area integral: $A = \int_3^5 \left[\frac{4}{3}\sqrt{x^2-9} - \frac{8x-24}{3}\right] dx$ (hyperbola is above the line)
5. Evaluate: $\int_3^5 (8x-24)dx = 16$ and $\int_3^5 \sqrt{x^2-9}dx = 10 - \frac{9}{2}\ln(3)$ using standard formula
6. Calculate: $A = \frac{1}{3}[4(10 - \frac{9}{2}\ln 3) - 16] = 8 - 6\ln 3$
7. Therefore: $3(A + 6\ln_e 3) = 3(8 - 6\ln 3 + 6\ln 3) = 3(8) = 24$

Answer: Option (2)

Key idea: Area bounded by curves involving absolute values and trigonometric functions

1. Step 1: Identify the boundaries: $y = \pi - |x|$, $y = |x \sin x|$, and $y = 0$. Note that $|x \sin x| = x \sin x$ on $[-\pi, \pi]$ and both functions are even.
2. Step 2: Use symmetry about the y-axis. Calculate area for $x \geq 0$ and double it.
3. Step 3: Find intersection of $\pi - x = x \sin x$ for $x \in [0, \pi]$. This occurs at $x = \frac{\pi}{2}$.
4. Step 4: For $x \in [0, \frac{\pi}{2}]$, the lower curve is $y = x \sin x$. For $x \in [\frac{\pi}{2}, \pi]$, the lower curve is $y = \pi - x$.
5. Step 5: Calculate $\int_0^{\pi/2} x \sin x \, dx$ using integration by parts: $[-x \cos x + \sin x]_0^{\pi/2} = 1$.
6. Step 6: Calculate $\int_{\pi/2}^{\pi} (\pi - x) \, dx = [\pi x - \frac{x^2}{2}]_{\pi/2}^{\pi} = \frac{\pi^2}{8}$.
7. Step 7: Total area = $2(1 + \frac{\pi^2}{8}) = 2 + \frac{\pi^2}{4}$.

Answer: Option (3)

Key idea: Area between two curves using integration

1. Step 1: Rewrite the curves as $x = -3y^2$ and $x = 1 - 4y^2$
2. Step 2: Find intersection points by setting $-3y^2 = 1 - 4y^2$, which gives $y^2 = 1$, so $y = \pm 1$
3. Step 3: Determine that $x = 1 - 4y^2$ is to the right of $x = -3y^2$ in the region of interest
4. Step 4: Set up the area integral: $A = \int_{-1}^{1} [(1 - 4y^2) - (-3y^2)] dy = \int_{-1}^{1} (1 - y^2) dy$
5. Step 5: Evaluate: $A = \left[y - \frac{y^3}{3}\right]_{-1}^{1} = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$

Answer: Option (2)

Key idea: Area bounded by curves using definite integration

1. Identify the region: $xy \leq 27$, $1 \leq y \leq x^2$. Find intersection points: $y=1$ and $y=x^2$ gives $x=1$; $y=x^2$ and $y=27/x$ gives $x=3$; $y=1$ and $y=27/x$ gives $x=27$
2. For $x \in [1,3]$: upper bound is $y=x^2$ (since $x^3 \leq 27$), lower bound is $y=1$
3. For $x \in [3,27]$: upper bound is $y=27/x$ (since $x^3 > 27$), lower bound is $y=1$
4. Calculate area: $A = \int_1^3 (x^2-1)dx + \int_3^{27} (27/x-1)dx$
5. First integral: $[x^3/3 - x]_1^3 = (9-3) - (1/3-1) = 20/3$
6. Second integral: $[27\ln x - x]_3^{27} = 27\ln(27) - 27\ln(3) - 24 = 54\ln(3) - 24$
7. Total area: $20/3 + 54\ln(3) - 24 = 54\log_e 3 - 52/3$

Answer: 16

Key idea: Functional equations, optimization of trigonometric functions, and area between curves

1. Solve the functional equation by substituting $x \to \frac{\pi}{2} - x$ to get $f\left(\frac{\pi}{2} - x\right) + 3f(x) = \cos x$, then solve the system to find $f(x) = \frac{3\cos x - \sin x}{8}$
2. Find maximum of $f(x)$ using $3\cos x - \sin x = \sqrt{10}\cos(x + \theta)$, giving $\alpha = \frac{\sqrt{10}}{8}$
3. Find intersection points of $x^2 = \beta x^3$: $x = 0$ and $x = \frac{1}{\beta}$
4. Calculate area: $A = \int_0^{1/\beta} (x^2 - \beta x^3)dx = \frac{1}{12\beta^3}$
5. Set $\frac{1}{12\beta^3} = \alpha^2 = \frac{10}{64} = \frac{5}{32}$ to get $\beta^3 = \frac{8}{15}$
6. Calculate $30\beta^3 = 30 \times \frac{8}{15} = 16$

Answer: Option (2)

Key idea: Area bounded by curves using integration

1. Identify the boundaries: $y \geq 0$, $x \leq 6-y$ (line), and $x \leq \frac{y^2+3}{4}$ (parabola)
2. Find intersection of line $x = 6-y$ and parabola $x = \frac{y^2+3}{4}$: Solve $6-y = \frac{y^2+3}{4}$ to get $y = 3$, $x = 3$
3. Determine boundaries: For $0 \leq y \leq 3$, parabola is the right boundary; for $3 \leq y \leq 6$, line is the right boundary
4. Calculate area: $A = \int_0^3 \frac{y^2+3}{4} dy + \int_3^6 (6-y) dy$
5. Evaluate first integral: $\frac{1}{4}[\frac{y^3}{3} + 3y]_0^3 = \frac{1}{4}(9+9) = \frac{9}{2}$
6. Evaluate second integral: $[6y - \frac{y^2}{2}]_3^6 = 18 - 13.5 = \frac{9}{2}$
7. Total area = $\frac{9}{2} + \frac{9}{2} = 9$

Answer: Option (1)

Key idea: Bijective Functions, Inverse Functions, and Area under Curves

1. Since $f: \{1,2,3,4\} \to \{1,e,e^2,e^3\}$ is strictly decreasing: $f(1)=e^3, f(2)=e^2, f(3)=e, f(4)=1$
2. Since $g: \{1,e,e^2,e^3\} \to \{\frac{1}{2},\frac{1}{3},\frac{1}{4}\}$ is strictly increasing: $g(e^2)=\frac{1}{2}$, so $g^{-1}(\frac{1}{2})=e^2$
3. From $f(2)=e^2$, we get $f^{-1}(e^2)=2$, thus $\phi(x)=[f^{-1}(g^{-1}(\frac{1}{2}))]^x = 2^x$
4. Set up the area integral: $A = \int_0^1 (2^x - x^2)\,dx$
5. Evaluate: $A = \left[\frac{2^x}{\ln 2}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{\ln 2} - \frac{1}{3}$
6. Simplify: $A = \frac{3 - \ln 2}{3\ln 2} = \frac{3 - \log_e(2)}{3\log_e(2)}$

Answer: Option (1)

Key idea: Area between curves using definite integration

1. Find intersection points by solving $x^2 - 8x = -x$, which gives $x^2 - 7x = 0$, so $x = 0$ or $x = 7$
2. Identify that the region has $y = x^2 - 8x$ as the lower boundary and $y = -x$ as the upper boundary
3. Set up the area integral: $A = \int_0^7 [(-x) - (x^2 - 8x)] dx = \int_0^7 (7x - x^2) dx$
4. Integrate: $A = \left[\frac{7x^2}{2} - \frac{x^3}{3}\right]_0^7$
5. Evaluate at $x = 7$: $A = \frac{343}{2} - \frac{343}{3} = \frac{1029 - 686}{6} = \frac{343}{6}$