JEE Main 2025 Jan 23 Shift 1, Mathematics Q21: Area as Definite Integral

JEE Main2025Jan 23, Shift 1Mathematics

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = ∣ x - 1∣$ is $\frac{1}{4} (bπ + c)$ , $b, c \in N$ , then $b + c$ is equal to ________

Your answer

Solution

The line $y = ∣ x - 1∣$ is a V-shape with vertex $(1, 0)$ . It meets the circle $x^{2} + y^{2} = 25$ at $x = 4$ (on $y = x - 1$ ) and $x = - 3$ (on $y = 1 - x$ ).

So the V-curve cuts the disk between $(- 3, 4)$ and $(4, 3)$ . The smaller region lies above the V and inside the circle.

Area of the smaller region $= \int_{- 3}^{4} 25 - x^{2} d x - (area of the V below the upper boundary)$ . Using the standard antiderivative $\int 25 - x^{2} d x = \frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}$ evaluated from $- 3$ to $4$ , then subtracting the V triangles, the smaller area works out to $\frac{25 π}{4} - \frac{1}{2}$ .

Larger area $= 25 π - (\frac{25 π}{4} - \frac{1}{2}) = \frac{75 π}{4} + \frac{1}{2} = \frac{1}{4} (75 π + 2)$ .

Hence $b = 75$ , $c = 2$ , and $b + c = 77$ .

Concept behind this question

Area as Definite IntegralNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 1, Mathematics Q21: Area as Definite Integral

Concept behind this question

More previous year questions on Area as Definite Integral

Practice more Mathematics