JEE Mathematics Area Under The Curve MCQs Set 05

Answer: (b) \( \left(\frac{\pi^3}6-\frac{\pi^2}2+\frac43\right) \) sq unit

Explanation:
1. \( u=\sin^{-1}|\sin x| \) equals x on [0, π/2] and is symmetric, giving 4 identical pieces over [0, 2π].
2. On [0, π/2] the curves \( u \) and \( u^2 \) meet at u = 0 and u = 1.
3. One piece: \( \int_0^1(u-u^2)du+\int_1^{\pi/2}(u^2-u)du=\frac16+\left[\frac{u^3}3-\frac{u^2}2\right]_1^{\pi/2}=\frac13+\frac{\pi^3}{24}-\frac{\pi^2}8 \).
4. Total \( =4\times\text{one piece}=\frac43+\frac{\pi^3}6-\frac{\pi^2}2 \).

Teacher's Note:
1. Reduce to one quarter-period using the symmetry of sin⁻¹|sin x|.
2. The curves cross at u = 1, so split the integral there.

Answer: (a) \( \frac7{120} \) sq unit

Explanation:
1. \( y'=8x^3-2x=0 \) gives x = 0, ±1/2; the minima are at x = ±1/2.
2. For |x| < 1/√2, y < 0, so the area is \( 2\int_0^{1/2}(x^2-2x^4)dx=2\left[\frac{x^3}3-\frac{2x^5}5\right]_0^{1/2} \).
3. \( =2\left(\frac1{24}-\frac1{80}\right)=\frac{7}{120} \).

Teacher's Note:
1. Find the minima first; they set the limits of integration.
2. The curve lies below the x-axis between the minima, so take the absolute value.

Answer: (b) 15 sq unit

Explanation:
1. \( [x-3]=[x]-3 \). On [−2, −1) the value is |−5| = 5, on [−1, 0) it is 4, on [0, 1) it is 3, on [1, 2) it is 2 and on [2, 3) it is 1.
2. Each strip has width 1, so the area is 5 + 4 + 3 + 2 + 1 = 15.

Teacher's Note:
1. Because 3 is an integer, [x − 3] = [x] − 3.
2. The graph is a staircase of unit-wide rectangles.

Answer: (b) 2 sq unit

Explanation:
1. The curves meet where |x| − 1 = 1 − |x|, i.e. x = ±1, at (±1, 0).
2. The region is a square with vertices (±1, 0), (0, ±1), whose diagonals have length 2.
3. Area \( =\frac12\cdot2\cdot2=2 \).

Teacher's Note:
1. Sketch the two V-shapes; together they form a rotated square.
2. Area of a rhombus = half the product of the diagonals.

Answer: (c) \( \frac{4-3\ln3}2 \) sq unit

Explanation:
1. The first graph is y = x for x ≤ 2 and y = 4 − x for x > 2. It meets y = 3/x at \( x=\sqrt3 \) (on y = x) and at x = 3 (on y = 4 − x).
2. Area \( =\int_{\sqrt3}^2\left(x-\frac3x\right)dx+\int_2^3\left(4-x-\frac3x\right)dx \).
3. \( =\left(\frac12-3\ln2+\frac32\ln3\right)+\left(\frac32-3\ln3+3\ln2\right)=2-\frac32\ln3=\frac{4-3\ln3}2 \).

Teacher's Note:
1. Split the graph at x = 2 where the modulus changes.
2. For x < 0 the graphs do not enclose any region.

Answer: (c) \( (e+e^{-1}-2) \) sq unit

Explanation:
1. The curves meet at x = 0; for 0 ≤ x ≤ 1, \( e^x\ge e^{-x} \).
2. Area \( =\int_0^1(e^x-e^{-x})dx=\left[e^x+e^{-x}\right]_0^1=e+e^{-1}-2 \).

Teacher's Note:
1. Find where the curves intersect (x = 0).
2. The upper curve is e^x for x > 0.

Answer: (d) \( 12\sqrt3 \) sq unit

Explanation:
1. \( 5x^2=2x^2+9\Rightarrow x=\pm\sqrt3 \).
2. Area \( =\int_{-\sqrt3}^{\sqrt3}(9-3x^2)dx=2\left[9x-x^3\right]_0^{\sqrt3}=2(9\sqrt3-3\sqrt3)=12\sqrt3 \).

Teacher's Note:
1. Equate the two curves to find the limits.
2. The parabola y = 2x² + 9 lies above y = 5x² between the intersections.

Answer: (c) \( \frac12\left(1-\frac1{\sqrt e}\right) \)

Explanation:
1. \( y'=e^{-x^2}(1-2x^2)=0 \) gives the maximum at \( x=\frac1{\sqrt2} \).
2. Area \( =\int_0^{1/\sqrt2}xe^{-x^2}dx=\left[-\frac12e^{-x^2}\right]_0^{1/\sqrt2}=\frac12\left(1-e^{-1/2}\right) \).

Teacher's Note:
1. Locate the maximum ordinate by setting the derivative to zero.
2. Integrate by the substitution u = x².

Answer: (b) −2

Explanation:
1. The curves meet where \( x-x^2=mx \): x = 0 and x = 1 − m.
2. The area between a parabola and a chord is \( \int_0^{1-m}(x-x^2-mx)dx=\frac{(1-m)^3}6 \) (in absolute value).
3. \( \frac{|1-m|^3}6=\frac92\Rightarrow|1-m|=3 \), so m = −2 or m = 4. Only m = −2 is listed.

Teacher's Note:
1. Integrate (1 − m)x − x² between the intersection points.
2. Take the absolute value, since the limit 1 − m may be negative.

Answer: (c) \( 6\pi \)

Explanation:
1. Complete the squares: \( 4(x-1)^2+(y+2)^2=12 \), i.e. \( \frac{(x-1)^2}3+\frac{(y+2)^2}{12}=1 \).
2. This is an ellipse with semi-axes \( \sqrt3 \) and \( \sqrt{12} \).
3. Area \( =\pi\sqrt3\sqrt{12}=6\pi \).

Teacher's Note:
1. Rewrite the equation as an ellipse in standard form.
2. Area of an ellipse = π × (semi-axis a) × (semi-axis b).

Answer: (a) 1

Explanation:
1. max{|x|, |y|} = 1 is a square with vertices (±1, ±1) and area 4.
2. \( \pi(x^2+y^2)=3 \) is a circle of radius \( \sqrt{3/\pi}\approx0.98<1 \), which lies completely inside the square; its area is \( \pi\cdot\frac3\pi=3 \).
3. The required area is 4 − 3 = 1.

Teacher's Note:
1. The circle's radius is less than 1, so it fits inside the square.
2. Subtract the area of the circle from the square.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The region \( |x-y|\le2,\ |x+y|\le2 \) is a square with vertices (±2, 0) and (0, ±2); its diagonal is 4, so the area is \( \frac{4^2}2=8 \): A is true.
2. R is also true, but it concerns a different square (side 4) and does not explain A.

Teacher's Note:
1. Draw the lines x ± y = ±2 to see the rotated square.
2. Area of a square = diagonal²/2.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With f(0) = 0 and f′(±2) = 0 we get \( f(x)=ax^3-12ax \), which is an odd function, so its graph is symmetric about the origin.
2. The circle is also symmetric about the origin, so the curve divides the circle into two regions that map onto each other by the half-turn about the origin: equal areas, ratio 1 : 1. R explains A.

Teacher's Note:
1. An odd function has point symmetry about the origin.
2. A half-turn about the origin swaps the two parts of the circle.