JEE Mathematics Area Under The Curve MCQs Set 04

Answer: (a) \( \dfrac2{\sqrt3} \)

Explanation:
1. Tangent: \( y=(1+\alpha^2)-2\alpha x \), with intercepts \( \dfrac{1+\alpha^2}{2\alpha} \) and \( 1+\alpha^2 \).
2. Area of OPQ \( =\dfrac{(1+\alpha^2)^2}{4\alpha} \), minimum at \( \alpha=\dfrac1{\sqrt3} \), giving \( \dfrac{4\sqrt3}{9} \).
3. The area under the curve is \( \displaystyle\int_0^1(1-x^2)dx=\dfrac23 \).
4. \( k=\dfrac{4\sqrt3/9}{2/3}=\dfrac{2}{\sqrt3} \).

Teacher's Note:
1. Write the triangle area as a function of \( \alpha \) and set its derivative to zero.
2. Compare the minimum with the area under the curve.

Answer: (a) \( 4A+8\cos a=7 \)

Explanation:
1. \( y=\sin x\,(2\cos x-\sqrt3) \) is zero first at \( \cos x=\dfrac{\sqrt3}2 \), i.e. \( a=\dfrac\pi6 \), and positive before it.
2. \( A=\displaystyle\int_0^{\pi/6}(\sin2x-\sqrt3\sin x)dx=\left[-\dfrac{\cos2x}2+\sqrt3\cos x\right]_0^{\pi/6}=\dfrac74-\sqrt3 \).
3. \( 4A+8\cos a=7-4\sqrt3+8\cdot\dfrac{\sqrt3}2=7 \).

Teacher's Note:
1. Find the first zero by factorising the expression.
2. Test each option by substituting \( a=\pi/6 \) and the value of A.

Answer: (a) \( \dfrac1{24} \)

Explanation:
1. Passing through the origin: \( c=0 \). Slope 1 at the origin: \( b=1 \). Through (1, 2): \( a=1 \). So \( y=x^2+x \).
2. The minimum is at \( x=-\dfrac12 \) (ordinate means the vertical line \( x=-\dfrac12 \)).
3. Area between the curve and the line \( y=x \) from \( x=-\dfrac12 \) to 0: \( \displaystyle\int_{-1/2}^0\left[(x^2+x)-x\right]dx=\int_{-1/2}^0x^2dx=\dfrac1{24} \).

Teacher's Note:
1. A tangent at a point of the curve gives two conditions: the point lies on the curve and the slopes match.
2. The ordinate at the minimum is the vertical line through the vertex.

Answer: (a) \( \sqrt2-1 \)

Explanation:
1. The general solution is \( y=\sin x+Ce^x \). Boundedness as \( x\to\infty \) forces \( C=0 \), so \( f(x)=\sin x \).
2. \( \sin x \) and \( \cos x \) meet at \( x=\dfrac\pi4 \).
3. Area \( =\displaystyle\int_0^{\pi/4}(\cos x-\sin x)dx=\sqrt2-1 \).

Teacher's Note:
1. The exponential term \( Ce^x \) is unbounded, so the bounded solution has \( C=0 \).
2. Always find where two curves meet before setting the limits.

Answer: (a) \( \dfrac{16\sqrt2}3 \)

Explanation:
1. \( f(x)=(x-1)^2+4(x-1)=x^2+2x-3 \), so \( f(x-1)=x^2-4 \).
2. It meets \( y=-x^2 \) where \( x^2-4=-x^2 \), i.e. \( x=\pm\sqrt2 \).
3. Area \( =\displaystyle\int_{-\sqrt2}^{\sqrt2}\left[-x^2-(x^2-4)\right]dx=2\left[4x-\dfrac{2x^3}3\right]_0^{\sqrt2}=\dfrac{16\sqrt2}3 \).

Teacher's Note:
1. Replace x by \( x-1 \) in the known formula for f.
2. The two parabolas are symmetric about the y-axis.

Answer: (a) 2009

Explanation:
1. The curves meet at \( x=0 \) and \( x=1 \), where \( x^{1/2009}\ge x^{2009} \).
2. Area \( =\displaystyle\int_0^1\left(x^{1/2009}-x^{2009}\right)dx=\dfrac{2009}{2010}-\dfrac1{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005} \).
3. \( p+q=1004+1005=2009 \).

Teacher's Note:
1. Use \( \int_0^1x^ndx=\dfrac1{n+1} \).
2. Reduce the fraction to lowest terms before adding p and q.

Answer: (c) 1 : 2

Explanation:
1. \( 4x^2=3x \) gives \( x=0,\tfrac34 \). \( A_1=\displaystyle\int_0^{3/4}(3x-4x^2)dx=\dfrac{27}{32}-\dfrac{9}{16}=\dfrac9{32} \).
2. The region for g is symmetric about the y-axis (replace x by \( -x \)), so \( A_2=2A_1 \).
3. \( A_1:A_2=1:2 \).

Teacher's Note:
1. The modulus makes the region symmetric about the y-axis.
2. Compute one side and double it.

Answer: (c) \( \dfrac{\pi^2}4 \)

Explanation:
1. For \( x\in\left[\tfrac{3\pi}2,2\pi\right] \): \( \cos^{-1}(\sin x)=\dfrac{5\pi}2-x \) and \( \sin^{-1}(\cos x)=x-\dfrac{3\pi}2 \).
2. So \( f(x)=4\pi-2x \), a straight line falling from \( \pi \) at \( x=\tfrac{3\pi}2 \) to 0 at \( x=2\pi \).
3. The area is a right triangle: \( \dfrac12\cdot\dfrac\pi2\cdot\pi=\dfrac{\pi^2}4 \).

Teacher's Note:
1. Reduce inverse trigonometric expressions to linear functions on the given interval.
2. Then the area is just a triangle.

Answer: (a) \( \dfrac{9\pi}2-12 \)

Explanation:
1. The circle is \( x^2+(y-3)^2=9 \) (area \( 9\pi \)); the parabola \( y=\dfrac{x^2}3 \) meets it at \( (\pm3,3) \).
2. Area of the part of the disc above the parabola: \( \displaystyle\int_{-3}^3\left[3+\sqrt{9-x^2}-\dfrac{x^2}3\right]dx=18+\dfrac{9\pi}2-6=12+\dfrac{9\pi}2 \).
3. The required part (below the parabola) is \( 9\pi-12-\dfrac{9\pi}2=\dfrac{9\pi}2-12 \).

Teacher's Note:
1. Find the total area of the disc and subtract the part that is not wanted.
2. The condition \( 3y\le x^2 \) describes the region on and below the parabola.

Answer: (a) \( \dfrac{\pi^3}2 \)

Explanation:
1. The disc has area \( \pi\cdot\pi^2=\pi^3 \).
2. The curve \( y=\sin x \) is odd, so it is symmetric about the origin and divides the disc into two congruent parts (rotate by \( 180^\circ \)).
3. Each part has area \( \dfrac{\pi^3}2 \).

Teacher's Note:
1. Look for symmetry before integrating.
2. An odd function through the centre of a disc splits it into equal halves.

Answer: (a) 1

Explanation:
1. The area between the curves is \( \dfrac13 \). The curves are mirror images in \( y=x \), so the line \( y=x \) divides the area into two equal parts.
2. The triangle formed by \( y=x \), \( y=0 \) and \( x+y=2 \) has vertices (0, 0), (2, 0) and (1, 1).
3. Area \( =\dfrac12\cdot2\cdot1=1 \).

Teacher's Note:
1. The curves \( y=x^2 \) and \( y=\sqrt x \) are inverse functions, so they are reflections in the line \( y=x \).
2. That reflection explains why \( y=x \) bisects the region.

Answer: (b) is many-one onto

Explanation:
1. \( f'(x)=3x^2-16x+20=(3x-10)(x-2) \), so \( x_1=2,\ x_2=\dfrac{10}3 \).
2. \( f(2)=3 \) and \( f\!\left(\dfrac{10}3\right)=\dfrac{49}{27} \): both positive, so the product is positive, and f is not one-one.
3. A cubic is onto \( \mathbb R \). Also \( f=(x-1)(x^2-7x+13) \) has only one real root.
4. Hence f is many-one onto.

Teacher's Note:
1. A cubic is one-one only when it has no turning points, i.e. its derivative has no real roots of sign change.
2. Compare the values at the two turning points to count roots.

Answer: (a) \( \dfrac{65}{12} \)

Explanation:
1. \( f(0)=-13 \) and the only real root is \( x=1 \).
2. On \( (0,1) \), f is negative: area \( =-\displaystyle\int_0^1f(x)dx \).
3. \( \int_0^1f\,dx=\dfrac14-\dfrac83+10-13=-\dfrac{65}{12} \), so the area is \( \dfrac{65}{12} \).

Teacher's Note:
1. Find where the curve crosses the axes to fix the limits.
2. Take the absolute value of an integral over a region below the axis.

Answer: (b) \( \dfrac2x-x \)

Explanation:
1. Tangents meeting on the y-axis: \( f-xf'=g-xg' \), so \( (f-g)=cx \).
2. Normals meeting on the x-axis: \( ff'=gg' \), so \( f^2-g^2=k \). Hence \( f+g=\dfrac k{cx} \).
3. \( f=\dfrac12\left(cx+\dfrac k{cx}\right) \), \( g=\dfrac12\left(\dfrac k{cx}-cx\right) \). With \( f(1)=1 \) and \( g(2)=3 \): \( c=-2,\ k=-8 \).
4. So \( f(x)=\dfrac2x-x \).

Teacher's Note:
1. The tangent meets the y-axis at height \( f-xf' \) and the normal meets the x-axis at \( x+ff' \).
2. Equal intercepts give two simple differential conditions.

Answer: (d) \( \dfrac2x+x \)

Explanation:
1. Equal intercepts of the tangents on the y-axis give \( f-g=cx \); normals meeting on the x-axis give \( f^2-g^2=k \). Using \( f(1)=1,\ g(2)=3 \) one gets \( c=-2,\ k=-8 \) and \( f=\dfrac2x-x \).
2. Then \( g=f-cx=f+2x=\dfrac2x+x \).
3. Check: \( g(2)=1+2=3 \).

Teacher's Note:
1. Once f is known, g follows from \( g=f-cx \).
2. Always verify the given point on the final curve.

Answer: (d) none of these

Explanation:
1. Tangent and normal conditions give \( f-g=cx \) and \( f^2-g^2=k \); with \( f(1)=1,\ g(2)=3 \) this yields \( f=\dfrac2x-x \) and \( g=\dfrac2x+x \).
2. \( f(x)=g(x) \) gives \( -x=x \), so \( x=0 \), which is not in the domain.
3. Hence there is no positive integral solution.

Teacher's Note:
1. Equate the two expressions and check the answer against the domain.
2. The curves never meet because they differ by 2x.

Answer: (a) I and III only

Explanation:
1. \( \text{Area}(R)=\displaystyle\int_0^c(cx-x^2)dx=\dfrac{c^3}2-\dfrac{c^3}3=\dfrac{c^3}6 \). So (I) is true and (II) false.
2. \( \text{Area}(T)=\dfrac12\cdot c\cdot c^2=\dfrac{c^3}2 \).
3. \( \dfrac{\text{Area}(T)}{\text{Area}(R)}=3 \) for every c, so the limit is 3: (III) true, (IV) false.

Teacher's Note:
1. Both areas are multiples of \( c^3 \), so the ratio is a constant.
2. A limit of a constant is the constant itself.

Answer: (a) I, III and IV only

Explanation:
1. \( f(x)=1-\dfrac2{x^2+1} \) has range \( [-1,1) \) (value 1 is never reached), so f is not onto: (I) does not hold.
2. \( f'(x)=\dfrac{4x}{(x^2+1)^2} \), so (II) holds. The minimum \( -1 \) is attained at \( x=0 \): (III) does not hold.
3. Area between \( y=1 \) and the curve \( =\displaystyle\int_{-\infty}^\infty\dfrac2{x^2+1}dx=2\pi\ne\pi \): (IV) does not hold.

Teacher's Note:
1. Rewrite the function as \( 1-\dfrac2{x^2+1} \) to read off its range, monotonicity and area.
2. The integral of \( \dfrac1{1+x^2} \) over the whole line is \( \pi \).

Answer: (a) I and III only

Explanation:
1. On \( [0,\tfrac\pi2) \) the values lie in \( (0,1] \); on \( [\tfrac\pi2,\pi) \) they lie in \( \left[0,\tfrac{\pi^2}4\right) \). Union: \( \left[0,\tfrac{\pi^2}4\right) \), so (I) is true.
2. At \( x=\pi \), the left limit is \( \dfrac{\pi^2}4 \) but \( f(\pi)=f(0)=1 \), so f is discontinuous at multiples of \( \pi \): (II) is false.
3. Area over one period: \( \displaystyle\int_0^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi}\left(\tfrac\pi2-x\right)^2dx=1+\dfrac{\pi^3}{24} \). Over \( 2n \) periods: \( 2n\left(1+\dfrac{\pi^3}{24}\right) \), so (III) is true.

Teacher's Note:
1. Check continuity at the joining points, including the periodic repetition.
2. The area over several periods is the number of periods times the area of one period.

Answer: (b) 2

Explanation:
1. \( y^2\le x^2-x^6 \), so \( |y|\le|x|\sqrt{1-x^4} \) for \( |x|\le1 \).
2. Area \( =4\displaystyle\int_0^1x\sqrt{1-x^4}\,dx \). With \( u=x^2 \): \( =2\int_0^1\sqrt{1-u^2}\,du=2\cdot\dfrac\pi4=\dfrac\pi2\approx1.57 \).
3. The least integer greater than or equal to 1.57 is 2.

Teacher's Note:
1. The substitution \( u=x^2 \) reduces the integral to a quarter circle area.
2. The ceiling of 1.57 is 2.

Answer: (c) 7

Explanation:
1. Put \( y=0 \): \( f(1)=f(x)\cdot1-1-x+2 \), so \( f(x)=x+f(1)-1 \). With \( f(0)=1 \): \( f(x)=x+1 \) (this satisfies the relation).
2. The curves \( y=x+1 \) and \( y=x^2+1 \) meet at \( x=0,1 \).
3. Area \( =\displaystyle\int_0^1(x-x^2)dx=\dfrac16 \), so \( p+q=1+6=7 \).

Teacher's Note:
1. Substituting y = 0 gives f directly in terms of f(1).
2. Check that the function found satisfies the original relation.