JEE Mathematics Application of Integrals MCQs Set 08

Answer: (b) \(\dfrac x{\sqrt{1+x^2}}\)

Explanation:
1. \(\displaystyle\int_1^bf(x)dx=\sqrt{b^2+1}-\sqrt2\).
2. Differentiating with respect to \(b\): \(f(b)=\dfrac b{\sqrt{b^2+1}}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(8\) sq. units

Explanation:
1. For \([x]=n\) (\(n=1,2,3,4\)) we need \([y]=\pm n\).
2. Each value gives two unit squares (\(y\in[n,n+1)\) and \(y\in[-n,-n+1)\)), so \(4\times2=8\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac{\sqrt3}2\)

Explanation:
1. The point \((2\sqrt3,-1)\) lies on the line, since \(\sqrt3\cdot2\sqrt3-7=-1\). So the locus is the perpendicular to the line through that point.
2. Its equation is \(x+\sqrt3y=\sqrt3\), with intercepts \(\sqrt3\) and \(1\); the area is \(\dfrac12\sqrt3\cdot1=\dfrac{\sqrt3}2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(2\ln2\)

Explanation:
1. For \(0<x<\frac\pi2\), \(\tan x,\cot x>0\), so \(f=(\tan x+\cot x)-|\tan x-\cot x|=2\min(\tan x,\cot x)\).
2. By symmetry about \(\dfrac\pi4\): area \(=2\displaystyle\int_0^{\pi/4}2\tan x\,dx=4\cdot\dfrac12\ln2=2\ln2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(0\)

Explanation:
1. \(x^2-2mx-4=0\) gives \(x_2-x_1=\sqrt{4m^2+16}\).
2. The area between a parabola \(y=\frac{x^2}2\) and a chord is \(\dfrac{(x_2-x_1)^3}{12}\), smallest when \(m=0\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(3\pi a^2\) sq. units

Explanation:
1. The curve is symmetric about the \(x\)-axis: area \(=2\displaystyle\int_0^{2a}\dfrac{x^{3/2}}{\sqrt{2a-x}}dx\).
2. With \(x=2a\sin^2\theta\) the integral is \(2\cdot4a^2\displaystyle\int_0^{\pi/2}2\sin^4\theta\,d\theta=2\cdot4a^2\cdot\dfrac{3\pi}8=3\pi a^2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac2e\)

Explanation:
1. For \(0\le x\le1\), \(xe^x\ge xe^{-x}\), so the area is \(\displaystyle\int_0^1x\left(e^x-e^{-x}\right)dx\).
2. \(=\left[xe^x-e^x+xe^{-x}+e^{-x}\right]_0^1=\dfrac2e\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(2\) sq. units

Explanation:
1. Let \(u=x+y-1\), \(v=2x+y-1\); the region is \(|u|+|v|\le1\), a square of diagonals \(2\) and \(2\), area \(2\).
2. The Jacobian of \((u,v)\) with respect to \((x,y)\) is \(\left|1\cdot1-1\cdot2\right|=1\), so the area is unchanged.

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(\dfrac{59}4\) sq. units

Explanation:
1. Compare the three expressions on \([-2,4]\) (real cube root); the maximum is the upper envelope and is non-negative throughout.
2. Integrating the envelope piece by piece (checked numerically) gives \(14.75=\dfrac{59}4\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(\displaystyle\int_0^{\sqrt2-1}\dfrac{4t}{(1+t^2)\sqrt{1-t^2}}dt\)

Explanation:
1. With \(t=\tan\dfrac x2\), the limits are \(t=0\) to \(t=\tan\dfrac\pi8=\sqrt2-1\).
2. The difference of the two roots reduces to \(\dfrac{4t}{(1+t^2)\sqrt{1-t^2}}\cdot\dfrac{dt}{1}\) after \(dx=\dfrac{2dt}{1+t^2}\); numerical evaluation of both sides gives \(0.3312\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac43\)

Explanation:
1. Area \(=\displaystyle\int_0^1(x^2-x^3)dx+\int_1^p(x^3-x^2)dx=\dfrac1{12}+\left(\dfrac{p^4}4-\dfrac{p^3}3+\dfrac1{12}\right)\).
2. Setting this to \(\dfrac16\) gives \(\dfrac{p^4}4=\dfrac{p^3}3\), so \(p=\dfrac43\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \((\pi-2):(3\pi+2)\)

Explanation:
1. The distance of the line from the centre is \(\sqrt2\), so the chord subtends \(90^\circ\) at the centre.
2. Smaller part \(=\dfrac14\pi\cdot4-\dfrac12\cdot2\cdot2=\pi-2\); greater part \(=4\pi-(\pi-2)=3\pi+2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac16\)

Explanation:
1. The area between a parabola \(y=x^2+\dots\) and a chord is \(\dfrac{(x_2-x_1)^3}6\), which grows with the horizontal extent.
2. For a chord of length \(1\) the horizontal extent is greatest (\(=1\)) when the chord is horizontal, giving area \(\dfrac16\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(e^2-e-\dfrac76\)

Explanation:
1. The curves are \(y=e^x+\dfrac{\sin x}x\) and \(y=\dfrac{\sin x}x+\dfrac{x^2}2\); their difference is \(e^x-\dfrac{x^2}2>0\).
2. Area \(=\displaystyle\int_1^2\left(e^x-\dfrac{x^2}2\right)dx=e^2-e-\dfrac{8-1}6=e^2-e-\dfrac76\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(1\)

Explanation:
1. \(x-[x]=\{x\}\) and \(-x-[-x]=\{-x\}\); for non-integer \(x\) the minimum is \(\min(t,1-t)\) with \(t=\{x\}\).
2. Over each unit interval \(\displaystyle\int_0^1\min(t,1-t)dt=\dfrac14\); four unit intervals give \(1\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(\dfrac{10}3\)

Explanation:
1. On each unit interval \(\displaystyle\int_0^1\left(\sqrt t-t^2\right)dt=\dfrac23-\dfrac13=\dfrac13\).
2. Ten unit intervals give \(\dfrac{10}3\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(2\)

Explanation:
1. The functional equation with \(f'(1)=2\) gives \(f(x)=x^2\).
2. The curves meet at \(x=\pm1\): area \(=\displaystyle\int_{-1}^1\left(\dfrac2{1+x^2}-x^2\right)dx=\pi-\dfrac23\), so \(a=2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(0\)

Explanation:
1. \(f'(x)=3k\left(x^2-\dfrac43\right)\) and \(f(0)=0\) give \(f=k(x^3-4x)\); \(f(1)=-2\) gives \(k=\dfrac23\). So \(f\) is an odd function.
2. The graph is symmetric about the origin, the centre of the circle, so the two parts are congruent and the difference is \(0\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(7\)

Explanation:
1. \(\displaystyle\int_0^2\{x^2\}dx=\dfrac83-\sum_{n=1}^3n\left(\sqrt{n+1}-\sqrt n\right)\ \ (\text{for }x\in[0,2])=\sqrt2+\sqrt3-\dfrac73\).
2. Doubling for the symmetric half gives \(p=2\), \(q=2\), \(r=3\), so \(p+q+r=7\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\left(\ln\left(3+2\sqrt2\right)-\dfrac\pi2\right)\) sq. units

Explanation:
1. The curves meet at \(x=\sqrt2\) where both equal \(\dfrac\pi4\); on \([1,\sqrt2]\), \(\operatorname{cosec}^{-1}x\ge\sec^{-1}x\).
2. Area \(=\displaystyle\int_1^{\sqrt2}\left(\operatorname{cosec}^{-1}x-\sec^{-1}x\right)dx=\ln\left(3+2\sqrt2\right)-\dfrac\pi2\approx0.192\) (positive).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(\dfrac{8a^2}{15}\) sq. units

Explanation:
1. The loop lies for \(0\le x\le a\), symmetric about the \(x\)-axis: area \(=\dfrac2{\sqrt a}\displaystyle\int_0^ax\sqrt{a-x}\,dx\).
2. \(\displaystyle\int_0^ax\sqrt{a-x}\,dx=\dfrac{4a^{5/2}}{15}\), so the area is \(\dfrac{8a^2}{15}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \((\pi-2)a^2\)

Explanation:
1. Take \(a=1\): \(y^2=\dfrac{1-x}x\) and \(y^2=\dfrac x{1-x}\) meet where \(x=\dfrac12\), \(y=\pm1\).
2. The enclosed area is \(2\displaystyle\int_0^1\min\left(\sqrt{\tfrac{x}{1-x}},\sqrt{\tfrac{1-x}x}\right)dx=\pi-2\) (numerically \(1.1416\)).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac{4-3\ln3}2\) sq. units

Explanation:
1. The tent \(y=x\) (\(x\le2\)), \(y=4-x\) (\(x\ge2\)) meets \(y=\dfrac3x\) at \(x=\sqrt3\) and \(x=3\).
2. Area \(=\displaystyle\int_{\sqrt3}^2\left(x-\dfrac3x\right)dx+\int_2^3\left(4-x-\dfrac3x\right)dx=\dfrac{4-3\ln3}2\approx0.352\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(a+b-2c-2d\)

Explanation:
1. On \([0,3]\), \(f\ge g\): area \(=\displaystyle\int_0^3f-\int_0^3g=(a-c)-d\).
2. On \([3,5]\), \(g\ge f\): area \(=\displaystyle\int_3^5g-\int_3^5f=(b-d)-c\).
3. Total \(=a+b-2c-2d\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) I, II and IV only

Explanation:
1. \(S=\displaystyle\int_0^1e^{-x^2}dx\approx0.7468\). Since \(e^{-x^2}\ge e^{-1}\) on \([0,1]\), I holds; \(0.7468>1-\dfrac1e\approx0.632\), so II holds.
2. III would need \(S\le0.40\), which is false. IV: the bound is about \(0.885\), true (use \(e^{-x^2}\le1\) on \([0,\frac1{\sqrt2}]\) and \(\le e^{-1/2}\) on \([\frac1{\sqrt2},1]\)).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(8\)

Explanation:
1. The disc has area \(\pi\cdot\pi^2=\pi^3\). The region excluded is \(|y|<|\sin x|\) for \(|x|\le\pi\) (it lies inside the disc).
2. \(A=\displaystyle\int_{-\pi}^{\pi}2|\sin x|dx=2\cdot4=8\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(2\)

Explanation:
1. The square has diagonals \(2\) along both axes, area \(2\). The circle has radius \(\dfrac1{\sqrt3}<\dfrac1{\sqrt2}\) (the inradius), so it lies inside.
2. Area \(=2-\dfrac\pi3\), so \(p=2\) and \(\dfrac\pi{p+1}=\dfrac\pi3\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

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

Explanation:
1. \(S_0=\displaystyle\int_0^\pi e^{-x}\sin x\,dx=\left[-\dfrac{e^{-x}(\sin x+\cos x)}2\right]_0^\pi\).
2. \(=\dfrac{e^{-\pi}+1}2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

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

Explanation:
1. \(S_n=\displaystyle\int_{n\pi}^{(n+1)\pi}\left|e^{-x}\sin x\right|dx\); substituting \(x=t+\pi\) shows \(S_{n+1}=e^{-\pi}S_n\).
2. So the common ratio is \(e^{-\pi}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.