JEE Mathematics Area MCQs Set 01

Answer: (a) \( \int_\alpha^\beta f(x)\,dx \)

Explanation:
1. The area between a curve, the x-axis and the ordinates x = alpha and x = beta is the integral of y dx
2. Here y = f(x) is positive in the first quadrant

Teacher's Note:
1. Area under a curve from x = a to x = b is the integral of y dx.

Answer: (c) 13 sq. units

Explanation:
1. The region lies between x = -3 and x = 3 under y =
2. x - 2
3.
4. From -3 to 2, y = 2 - x gives 25/2. From 2 to 3, y = x - 2 gives 1/2
5. Area = 25/2 + 1/2 = 13

Teacher's Note:
1. Split at x = 2, where the modulus changes.

Answer: (d) \( 2\pi \)

Explanation:
1. y = sqrt(4 - x^2) is the upper half of the circle x^2 + y^2 = 4
2. The area is half of pi r^2 with r = 2, i.e. 2 pi

Teacher's Note:
1. Recognise the semicircle.

Answer: (b) \( \frac{16\sqrt2}{3} \) sq. units

Explanation:
1. By symmetry, area = 2 times the integral of 2 sqrt(x) from 0 to 2
2. = 4 times (2/3) x^(3/2) at 2 = (8/3)(2 sqrt2) = 16 sqrt2 / 3

Teacher's Note:
1. Use symmetry about the x-axis.

Answer: (a) 3

Explanation:
1. Area = 2 times the integral of sqrt(ky) from y = 0 to 3 = 4 sqrt(3k)
2. Setting 4 sqrt(3k) = 12 gives sqrt(3k) = 3, so k = 3

Teacher's Note:
1. Integrate with respect to y.

Answer: (c) \( \log_2e \)

Explanation:
1. The region extends to the left, x from -infinity to 0
2. Area = the integral of 2^x from -infinity to 0 = 1/ln 2 = log base 2 of e

Teacher's Note:
1. Improper integral on an infinite interval.

Answer: (b) \( a^2/6 \)

Explanation:
1. y = (sqrt a - sqrt x)^2 = a - 2 sqrt(a x) + x
2. Integral from 0 to a: a^2 - (4/3)a^2 + a^2/2 = a^2/6

Teacher's Note:
1. Expand the square, then integrate.

Answer: (d) 2 sin 1

Explanation:
1. The region is symmetric about the y-axis, with x from -1 to 1
2. Area = 2 times the integral of cos x from 0 to 1 = 2 sin 1

Teacher's Note:
1. Both cos 1 and sin 1 are positive here, so no sign worry.

Answer: (a) 9

Explanation:
1. The curve meets the line at x = 9 (y = 3)
2. Area under the curve from 0 to 9 = 18
3. Subtract the triangle under the line from x = 3 to 9: (1/2)(6)(3) = 9
4. Area = 18 - 9 = 9

Teacher's Note:
1. Area under the curve minus the triangle.

Answer: (c) 4 sq. units

Explanation:
1. The graphs meet at x = -1 and x = 2
2. The upper boundary minus the lower boundary is 2 + 2x on [-1, 0], 2 on [0, 1] and 4 - 2x on [1, 2]
3. Areas: 1 + 2 + 1 = 4

Teacher's Note:
1. Break into pieces where each modulus changes.

Answer: (d) 15 : 49

Explanation:
1. The parabolas meet at (0, 0) and (4, 4), enclosing area 16/3
2. Area for x from 0 to 1: integral of (2 sqrt x - x^2/4) = 4/3 - 1/12 = 5/4
3. The rest is 16/3 - 5/4 = 49/12
4. Ratio = 15 : 49

Teacher's Note:
1. Compute the part to the left of x = 1.

Answer: (b) \( \tfrac12 \)

Explanation:
1. Differentiate the area with respect to a: f(a) = a + (1/2) sin a + (a/2) cos a - (pi/2) sin a
2. At a = pi/2: pi/2 + 1/2 + 0 - pi/2 = 1/2

Teacher's Note:
1. Area function derivative gives the curve.

Answer: (a) \( |f(p)-f(q)| \)

Explanation:
1. Area = integral of c e^x from p to q = c(e^q - e^p) = f(q) - f(p)
2. Its magnitude is
3. f(p) - f(q)
4.

Teacher's Note:
1. The integral of f is f itself.

Answer: (c) \( 2\sqrt2 \)

Explanation:
1. Total area = 2 + 8(1/2 - 1/4) = 4, so each part is 2
2. The area from 2 to a is (a - 2) + 8(1/2 - 1/a) = a + 2 - 8/a
3. Setting it equal to 2 gives a = 8/a, so a = 2 sqrt2

Teacher's Note:
1. Set the half-area equal and solve.

Answer: (b) \( \frac{1}{12}(4\pi-3\sqrt3) \)

Explanation:
1. The centre is (-1, 0), so the first quadrant part has x from 0 to 1, where y = sqrt(1 - (x + 1)^2/4)
2. Put u = (x + 1)/2: area = 2 times the integral of sqrt(1 - u^2) from 1/2 to 1
3. = 2(pi/4 - sqrt3/8 - pi/12) = pi/3 - sqrt3/4 = (4 pi - 3 sqrt3)/12

Teacher's Note:
1. Shift the ellipse and integrate.