JEE Mathematics Area Under The Curve MCQs Set 10

Answer: (a) \( (\pi-2)a^2 \) sq. units

Explanation:
1. The curves are \( y^2=\dfrac{a^2(a-x)}{x} \) and \( y^2=\dfrac{a^2x}{a-x} \); they meet at \( x=\dfrac a2 \) and both are symmetric about the \( x \)-axis.
2. The enclosed region has \( |y| \) bounded by the smaller of the two curves: \( A=2\left[\displaystyle\int_0^{a/2}a\sqrt{\tfrac{x}{a-x}}\,dx+\int_{a/2}^{a}a\sqrt{\tfrac{a-x}{x}}\,dx\right] \).
3. Each integral equals \( a^2\left(\tfrac\pi4-\tfrac12\right) \) (use \( x=a\sin^2t \)), so \( A=4a^2\left(\tfrac\pi4-\tfrac12\right)=(\pi-2)a^2 \).

Teacher's Note:
1. The substitution \( x=a\sin^2t \) turns \( \sqrt{x/(a-x)} \) into \( \tan t \).
2. Use symmetry about the x-axis and about the line x = a/2.

Answer: (c) \( \left(\frac{4-3\ln3}{2}\right) \)

Explanation:
1. For \( 0<x\le2 \), \( y=x \); for \( x>2 \), \( y=4-x \). Intersections with \( y=\dfrac3x \): \( x=\sqrt3 \) and \( x=3 \) (the point \( x=1 \) is not on the branch \( y=4-x \) for \( x>2 \)).
2. The curve \( y=2-|2-x| \) is above \( y=\dfrac3x \) between \( \sqrt3 \) and \( 3 \), so \( A=\displaystyle\int_{\sqrt3}^{2}\left(x-\dfrac3x\right)dx+\int_2^3\left(4-x-\dfrac3x\right)dx \).
3. \( =\left[\dfrac{x^2}2-3\ln x\right]_{\sqrt3}^2+\left[4x-\dfrac{x^2}{2}-3\ln x\right]_2^3=\dfrac{4-3\ln3}{2} \).

Teacher's Note:
1. Find the intersection points first, taking care with the two branches.
2. Add the areas piece by piece.

Answer: (d) i, iii and iv only

Explanation:
1. Case (i): \( \sin x-\sin^2x \) is \( \ge0 \) on \( [0,\pi] \), giving \( 2-\tfrac\pi2 \), and on \( [\pi,2\pi] \) it is \( \le0 \) with \( |\ldots|=|\sin x|+\sin^2x \), giving \( 2+\tfrac\pi2 \); the area is 4 per period, so 20 over \( 10\pi \) (5 periods). Case (iii): \( |\sin x|-\sin^3x \) integrates to \( \left(2-\tfrac43\right)+\left(2+\tfrac43\right)=4 \) per period; case (iv): \( \sin x-\sin^4x \) gives \( \left(2-\tfrac{3\pi}8\right)+\left(2+\tfrac{3\pi}8\right)=4 \) per period. So each of (i), (iii), (iv) is 20 over 5 periods.
2. Case (ii): the curves differ only where \( \sin x<0 \): \( |\sin x|-\sin x=2|\sin x| \), giving area \( 4 \) per period over 10 periods, i.e. 40.
3. So (i), (iii) and (iv) enclose equal areas (20 square units each) while (ii) encloses 40.

Teacher's Note:
1. Where sin x is negative, |sin x| and sin x are on opposite sides of the axis, doubling the gap.
2. Compute the area over one period and multiply by the number of periods.

Answer: (b) ii, iii and iv only

Explanation:
1. The vertical distance between the two curves at each \( x \) is \( p-q=|f-g|\ge0 \).
2. So the area is \( \int(p-q)\,dx=\int|p-q|\,dx=\int|f-g|\,dx \) (all three are the same quantity).
3. (i) is signed, so it can be smaller than the area (it vanishes when the curves swap places symmetrically).

Teacher's Note:
1. \( \max-\min=|f-g| \) is a useful identity.
2. Area requires a non-negative integrand.

Answer: (c) i, ii and iv only

Explanation:
1. On \( [0,1] \), \( e^{-x^2}\ge e^{-1} \), so \( S\ge\dfrac1e \) (i). Also \( x^2\le x \) gives \( e^{-x^2}\ge e^{-x} \), so \( S\ge\int_0^1e^{-x}dx=1-\dfrac1e \) (ii).
2. The exact value is \( S\approx0.7468 \). Statement (iii) claims \( S\le0.40 \), which is false.
3. For (iv): The exact value is \( S\approx0.7468 \) and the stated bound equals \( 0.7071+0.3679\times0.2929\approx0.815 \), so (iv) is true.

Teacher's Note:
1. Bound the integrand from above and below on sub-intervals.
2. A numerical value of S (about 0.747) lets you test each bound.

Answer: (c) 8

Explanation:
1. The disc \( x^2+y^2\le\pi^2 \) has area \( \pi\cdot\pi^2=\pi^3 \), and \( |x|\le\pi \) inside it.
2. The condition \( |y|\ge|\sin x| \) removes the region \( |y|<|\sin x| \) within the disc. Since \( |\sin x|\le\sqrt{\pi^2-x^2} \) for \( |x|\le\pi \), this region is the full strip under and above \( \sin x \): area \( =2\int_{-\pi}^{\pi}|\sin x|\,dx=2\times4=8 \).
3. So the required area is \( \pi^3-8 \) and \( A=8 \).

Teacher's Note:
1. Subtract the removed region from the total area of the disc.
2. \( \int_{-\pi}^{\pi}|\sin x|dx=4 \).

Answer: (c) 7

Explanation:
1. The function is even, so the area is \( 2\int_0^2\{x^2\}dx=2\left[\int_0^2x^2dx-\int_0^2\lfloor x^2\rfloor dx\right] \).
2. \( \int_0^2x^2dx=\dfrac83 \) and \( \int_0^2\lfloor x^2\rfloor dx=(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)=5-\sqrt2-\sqrt3 \).
3. Area \( =2\left(\sqrt2+\sqrt3-\dfrac73\right) \), so \( p=2,q=2,r=3 \) (or \( r,q \) swapped) and \( p+q+r=7 \).

Teacher's Note:
1. Write \( \{x^2\}=x^2-\lfloor x^2\rfloor \) and find where the floor changes value.
2. The floor of \( x^2 \) jumps at \( x=1,\sqrt2,\sqrt3 \) on \( [0,2] \).

Answer: (b) \( \dfrac12(1+e^{-\pi}) \) sq. units

Explanation:
1. \( S_0 \) is the first half-wave, \( 0\le x\le\pi \): \( S_0=\int_0^\pi e^{-x}\sin x\,dx \).
2. Using \( \int e^{-x}\sin x\,dx=-\dfrac{e^{-x}}2(\sin x+\cos x) \): \( S_0=\dfrac12\left(1+e^{-\pi}\right) \).

Teacher's Note:
1. Integrate by parts twice, or use the standard formula for \( \int e^{ax}\sin bx\,dx \).
2. The first half-wave is above the axis.

Answer: (b) \( e^{-\pi} \)

Explanation:
1. For \( S_n \) the interval is \( n\pi\le x\le(n+1)\pi \): \( S_n=\left|\int_{n\pi}^{(n+1)\pi}e^{-x}\sin x\,dx\right| \).
2. Substituting \( x=t+n\pi \): \( \sin x=(-1)^n\sin t \), so \( S_n=e^{-n\pi}S_0 \).
3. The common ratio is \( e^{-\pi} \).

Teacher's Note:
1. A shift by \( \pi \) multiplies the exponential factor by \( e^{-\pi} \) and flips the sign of the sine.
2. Each half-wave is a scaled copy of the first one.