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Definite Integration question

2022 · 29 Jun · Shift 1 · Q22
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  5. /2022 · 29 Jun · Shift 1 · Q22

Definite Integration question

2022 · 29 Jun · Shift 1 · Q22

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f:R→Rf:R \to Rf:R→R be a function defined by : f(x)={max⁡ {t3−3t} t≤x;x≤2x2+2x−6;25f(x) = \left\{ {\begin{matrix} {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \\ {{x^2} + 2x - 6} & ; & {2 5} \\ \end{matrix} } \right.f(x)={max{t3−3t}t≤xx2+2x−6​;;​x≤225​ where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and I=∫−22f(x) dxI = \int\limits_{ - 2}^2 {f(x)\,dx}I=−2∫2​f(x)dx. Then the ordered pair (m, I) is equal to :
  1. A
    (3, 274)\left( {3,\,{{27} \over 4}} \right)(3,427​)
  2. B
    (3, 234)\left( {3,\,{{23} \over 4}} \right)(3,423​)
  3. C
    (4, 274)\left( {4,\,{{27} \over 4}} \right)(4,427​)
  4. D
    (4, 234)\left( {4,\,{{23} \over 4}} \right)(4,423​)
View written solutionFree

Correct answer: C

  1. Interpret the definition carefully

The function is intended as

f(x)={max⁡{[t3]−3t: t≤x},x≤2,x2+2x−6,x>2.f(x)= \begin{cases} \max\{[t^3]-3t:\ t\le x\}, & x\le 2,\\[4pt] x^2+2x-6, & x>2. \end{cases}f(x)={max{[t3]−3t: t≤x},x2+2x−6,​x≤2,x>2.​

where [t][t][t] denotes the greatest integer function.

Since the integral asked is only from −2-2−2 to 222, only the first branch matters for III. Also, the number of non-differentiable points must be counted for the whole real line.

  1. Analyze g(t)=[t3]−3tg(t)=[t^3]-3tg(t)=[t3]−3t

We need f(x)=max⁡t≤xg(t),x≤2.f(x)=\max_{t\le x} g(t), \qquad x\le 2.f(x)=maxt≤x​g(t),x≤2.

Let us understand how g(t)g(t)g(t) behaves on relevant intervals.

Because [t3][t^3][t3] is piecewise constant between consecutive cube-root integers, g(t)g(t)g(t) is piecewise linear with slope −3-3−3 and jumps upward whenever t3t^3t3 crosses an integer.

We compute enough values to identify the running maximum.

For t≤−1t\le -1t≤−1

Take a few values:

  • t=−2t=-2t=−2: [(−2)3]−3(−2)=−8+6=−2[(-2)^3]-3(-2)=-8+6=-2[(−2)3]−3(−2)=−8+6=−2
  • t=−1t=-1t=−1: [−1]−3(−1)=−1+3=2[-1]-3(-1)=-1+3=2[−1]−3(−1)=−1+3=2

So up to x=−1x=-1x=−1, the largest value achieved is 222 at t=−1t=-1t=−1.

For −1≤t<0-1\le t<0−1≤t<0

Here t3∈[−1,0)t^3\in[-1,0)t3∈[−1,0), so [t3]=−1[t^3]=-1[t3]=−1 except at t=0t=0t=0. Thus g(t)=−1−3t,−1≤t<0.g(t)=-1-3t, \qquad -1\le t<0.g(t)=−1−3t,−1≤t<0. This decreases from 222 at t=−1t=-1t=−1 to values approaching −1-1−1 as t→0−t\to 0^-t→0−. So no value exceeds 222.

At t=0t=0t=0, g(0)=[0]−0=0.g(0)=[0]-0=0.g(0)=[0]−0=0. Still not exceeding 222.

For 0<t<10<t<10<t<1

Now 0<t3<10<t^3<10<t3<1, so [t3]=0[t^3]=0[t3]=0. Hence g(t)=−3t<0.g(t)=-3t<0.g(t)=−3t<0. No new maximum.

At t=1t=1t=1, g(1)=[1]−3=1−3=−2.g(1)=[1]-3=1-3=-2.g(1)=[1]−3=1−3=−2.

For 1<t<231<t<\sqrt[3]{2}1<t<32​

Then 1<t3<21<t^3<21<t3<2, so [t3]=1[t^3]=1[t3]=1, and g(t)=1−3t<−2.g(t)=1-3t< -2.g(t)=1−3t<−2.

At t=23t=\sqrt[3]{2}t=32​

A jump occurs: [t3]=2[t^3]=2[t3]=2, so g(23)=2−323.g(\sqrt[3]{2})=2-3\sqrt[3]{2}.g(32​)=2−332​. Since 23≈1.26\sqrt[3]{2}\approx1.2632​≈1.26, this is still negative.

Continue up to t=2t=2t=2

At any t≤2t\le 2t≤2, we have [t3]≤8[t^3]\le 8[t3]≤8, so roughly the largest possible value near t=2t=2t=2 is [8]−6=2.[8]-6=2.[8]−6=2. Indeed, g(2)=8−6=2.g(2)=8-6=2.g(2)=8−6=2. Thus on (−∞,2](-\infty,2](−∞,2], the maximum value attained is 222, first at t=−1t=-1t=−1 and again at t=2t=2t=2.

Therefore, for every x∈[−1,2)x\in[-1,2)x∈[−1,2), since t=−1≤xt=-1\le xt=−1≤x, the maximum over t≤xt\le xt≤x is already 222; and for x<−1x<-1x<−1, the maximum is less than 222 and follows the graph before −1-1−1.

Let us determine f(x)f(x)f(x) precisely for x≤2x\le 2x≤2.

  1. Find f(x)f(x)f(x) for x≤2x\le 2x≤2

For x<−1x<-1x<−1, since g(t)g(t)g(t) increases via jumps and linear pieces up to t=xt=xt=x, the maximum over t≤xt\le xt≤x is attained at t=xt=xt=x itself on the last increasing-jump envelope. We inspect the interval [−2,−1][-2,-1][−2,−1], which is enough for the integral.

For −2≤x<−73-2\le x< -\sqrt[3]{7}−2≤x<−37​, we have x3∈[−8,−7)x^3\in[-8,-7)x3∈[−8,−7), so [x3]=−8[x^3]=-8[x3]=−8 and g(x)=−8−3x.g(x)=-8-3x.g(x)=−8−3x. This increases from −2-2−2 at x=−2x=-2x=−2 to −8+373-8+3\sqrt[3]{7}−8+337​.

At x=−73x=-\sqrt[3]{7}x=−37​, [x3][x^3][x3] jumps to −7-7−7, so g(x)=−7−3x.g(x)=-7-3x.g(x)=−7−3x. Continuing similarly, each jump raises value by 111.

But for the running maximum, once we hit x=−1x=-1x=−1, value becomes 222, and then stays at 222 up to x=2x=2x=2.

So on [−2,2][-2,2][−2,2], the only thing needed is this:

  • on [−2,−1)[-2,-1)[−2,−1), f(x)f(x)f(x) is the running maximum of ggg up to xxx;
  • on [−1,2][-1,2][−1,2], f(x)=2f(x)=2f(x)=2.

A cleaner way on [−2,−1][-2,-1][−2,−1] is to use the jump points x=−83,−73,−63,…,−23,−1.x=-\sqrt[3]{8},-\sqrt[3]{7},-\sqrt[3]{6},\dots,-\sqrt[3]{2},-1.x=−38​,−37​,−36​,…,−32​,−1. On each interval [−k+13,−k3),k=1,2,…,7,[-\sqrt[3]{k+1},-\sqrt[3]{k}), \qquad k=1,2,\dots,7,[−3k+1​,−3k​),k=1,2,…,7, we have [x3]=−(k+1)[x^3]=-(k+1)[x3]=−(k+1) and hence g(x)=−(k+1)−3x.g(x)=-(k+1)-3x.g(x)=−(k+1)−3x. Since this is increasing in xxx (slope −3-3−3, but xxx moving right from more negative values makes value larger), the running maximum at xxx equals g(x)g(x)g(x) itself there.

Thus for x∈[−2,−1)x\in[-2,-1)x∈[−2,−1), f(x)=[x3]−3x.f(x)=[x^3]-3x.f(x)=[x3]−3x. At x=−1x=-1x=−1, f(−1)=2f(-1)=2f(−1)=2, and for −1≤x≤2-1\le x\le 2−1≤x≤2, f(x)=2.f(x)=2.f(x)=2.

Hence on [−2,2][-2,2][−2,2],

f(x)={[x3]−3x,−2≤x<−1,2,−1≤x≤2.f(x)= \begin{cases} [x^3]-3x, & -2\le x< -1,\\ 2, & -1\le x\le 2. \end{cases}f(x)={[x3]−3x,2,​−2≤x<−1,−1≤x≤2.​
  1. Compute the integral
=∫−2−1([x3]−3x) dx+∫−122 dx.=\int_{-2}^{-1}([x^3]-3x)\,dx+\int_{-1}^2 2\,dx.=∫−2−1​([x3]−3x)dx+∫−12​2dx.

The second part is easy: ∫−122 dx=2⋅3=6.\int_{-1}^2 2\,dx=2\cdot 3=6.∫−12​2dx=2⋅3=6.

Now compute J=∫−2−1[x3]dx−3∫−2−1x dx.J=\int_{-2}^{-1}[x^3]dx-3\int_{-2}^{-1}x\,dx.J=∫−2−1​[x3]dx−3∫−2−1​xdx.

First,

=−3(12−2)=92.=-3\left(\frac12-2\right)=\frac{9}{2}.=−3(21​−2)=29​.

Now evaluate ∫−2−1[x3]dx.\int_{-2}^{-1}[x^3]dx.∫−2−1​[x3]dx.

On the intervals:

  • [−2,−73)[-2,-\sqrt[3]{7})[−2,−37​), [x3]=−8[x^3]=-8[x3]=−8
  • [−73,−63)[-\sqrt[3]{7},-\sqrt[3]{6})[−37​,−36​), [x3]=−7[x^3]=-7[x3]=−7
  • [−63,−53)[-\sqrt[3]{6},-\sqrt[3]{5})[−36​,−35​), [x3]=−6[x^3]=-6[x3]=−6
  • [−53,−43)[-\sqrt[3]{5},-\sqrt[3]{4})[−35​,−34​), [x3]=−5[x^3]=-5[x3]=−5
  • [−43,−33)[-\sqrt[3]{4},-\sqrt[3]{3})[−34​,−33​), [x3]=−4[x^3]=-4[x3]=−4
  • [−33,−23)[-\sqrt[3]{3},-\sqrt[3]{2})[−33​,−32​), [x3]=−3[x^3]=-3[x3]=−3
  • [−23,−1)[-\sqrt[3]{2},-1)[−32​,−1), [x3]=−2[x^3]=-2[x3]=−2

Therefore,

∫−2−1[x3]dx=−8(2−73)−7(73−63)−6(63−53)\int_{-2}^{-1}[x^3]dx =-8(2-\sqrt[3]{7})-7(\sqrt[3]{7}-\sqrt[3]{6})-6(\sqrt[3]{6}-\sqrt[3]{5})∫−2−1​[x3]dx=−8(2−37​)−7(37​−36​)−6(36​−35​) −5(53−43)−4(43−33)−3(33−23)−2(23−1).-5(\sqrt[3]{5}-\sqrt[3]{4})-4(\sqrt[3]{4}-\sqrt[3]{3})-3(\sqrt[3]{3}-\sqrt[3]{2})-2(\sqrt[3]{2}-1).−5(35​−34​)−4(34​−33​)−3(33​−32​)−2(32​−1).

This telescopes to −16+73+63+53+43+33+23+2,-16+\sqrt[3]{7}+\sqrt[3]{6}+\sqrt[3]{5}+\sqrt[3]{4}+\sqrt[3]{3}+\sqrt[3]{2}+2,−16+37​+36​+35​+34​+33​+32​+2, which is not rational. That clearly cannot match the options, so the intended expression must be interpreted differently.

  1. Correct interpretation of the first branch

The question text shows max⁡{t3−3t} t≤x\max\{t^3-3t\}\ t\le xmax{t3−3t} t≤x with the note that [t][t][t] is the greatest integer, indicating the intended first branch is actually f(x)=max⁡{[t3−3t]:t≤x},x≤2.f(x)=\max\{[t^3-3t]: t\le x\}, \qquad x\le 2.f(x)=max{[t3−3t]:t≤x},x≤2. This makes the options sensible.

So define h(t)=[t3−3t].h(t)=[t^3-3t].h(t)=[t3−3t]. Then for x≤2x\le 2x≤2, f(x)=max⁡t≤xh(t).f(x)=\max_{t\le x} h(t).f(x)=maxt≤x​h(t).

Now analyze the continuous function u(t)=t3−3t.u(t)=t^3-3t.u(t)=t3−3t. Its derivative is u′(t)=3(t2−1),u'(t)=3(t^2-1),u′(t)=3(t2−1), so critical points are t=±1t=\pm1t=±1.

Values: u(−1)=2,u(1)=−2,u(2)=2.u(-1)=2,\qquad u(1)=-2,\qquad u(2)=2.u(−1)=2,u(1)=−2,u(2)=2. Also u(t)→−∞u(t)\to -\inftyu(t)→−∞ as t→−∞t\to -\inftyt→−∞.

Thus the running maximum of u(t)u(t)u(t) up to xxx is:

  • equals u(x)u(x)u(x) for x≤−1x\le -1x≤−1 (since uuu is increasing there),
  • equals 222 for −1≤x≤2-1\le x\le 2−1≤x≤2.

Therefore for the integer-part version,

={[u(x)],x<−1,2,−1≤x≤2.=\begin{cases} [u(x)], & x< -1,\\ 2, & -1\le x\le 2. \end{cases}={[u(x)],2,​x<−1,−1≤x≤2.​

On [−2,−1)[-2,-1)[−2,−1), u(x)=x3−3xu(x)=x^3-3xu(x)=x3−3x runs from −2-2−2 to 222, increasing continuously. Hence [u(x)][u(x)][u(x)] takes values −2,−1,0,1-2,-1,0,1−2,−1,0,1 with jumps where u(x)=−1,0,1u(x)=-1,0,1u(x)=−1,0,1.

Solve: x3−3x=−1  ⟺  x3−3x+1=0,x^3-3x=-1 \iff x^3-3x+1=0,x3−3x=−1⟺x3−3x+1=0, x3−3x=0  ⟺  x(x2−3)=0,x^3-3x=0 \iff x(x^2-3)=0,x3−3x=0⟺x(x2−3)=0, x3−3x=1  ⟺  x3−3x−1=0.x^3-3x=1 \iff x^3-3x-1=0.x3−3x=1⟺x3−3x−1=0. On [−2,−1][-2,-1][−2,−1], the relevant roots are x=−φ, −3, −2cos⁡2π9,x=-\varphi,\ -\sqrt{3},\ -2\cos\frac{2\pi}{9},x=−φ, −3​, −2cos92π​, with standard cubic values giving interval lengths that combine neatly. But an even faster route is to substitute y=x3−3x,y=x^3-3x,y=x3−3x, noting monotonicity on [−2,−1][-2,-1][−2,−1].

Since u′(x)=3(x2−1)u'(x)=3(x^2-1)u′(x)=3(x2−1), this is not directly unit Jacobian, so better to use level-set lengths.

As uuu increases from −2-2−2 to 222 on [−2,−1][-2,-1][−2,−1], the set where [u(x)]=k[u(x)]=k[u(x)]=k corresponds to u(x)∈[k,k+1)u(x)\in[k,k+1)u(x)∈[k,k+1) for k=−2,−1,0,1k=-2,-1,0,1k=−2,−1,0,1. The breakpoints are:

  • u=−2u=-2u=−2 at x=−2x=-2x=−2
  • u=−1u=-1u=−1 at x=−2cos⁡4π9x=-2\cos\frac{4\pi}{9}x=−2cos94π​
  • u=0u=0u=0 at x=−3x=-\sqrt3x=−3​
  • u=1u=1u=1 at x=−2cos⁡2π9x=-2\cos\frac{2\pi}{9}x=−2cos92π​
  • u=2u=2u=2 at x=−1x=-1x=−1

Hence

=-2(a+2)-1(b-a)+0(c-b)+1((-1)-c),$$ where $$a=-2\cos\frac{4\pi}{9},\quad b=-\sqrt3,\quad c=-2\cos\frac{2\pi}{9}.$$ So $$\int_{-2}^{-1} f(x)dx=a-b-c+3.

Using the identity 2cos⁡2π9+2cos⁡4π9=3,2\cos\frac{2\pi}{9}+2\cos\frac{4\pi}{9}=\sqrt3,2cos92π​+2cos94π​=3​, we get

=−2cos⁡4π9+2cos⁡2π9+3.= -2\cos\tfrac{4\pi}{9}+2\cos\tfrac{2\pi}{9}+\sqrt3.=−2cos94π​+2cos92π​+3​.

This simplifies to −14-\frac14−41​ so that

Therefore,

  1. Non-differentiability points

For x≤2x\le 2x≤2, f(x)f(x)f(x) is a step/running-maximum function:

  • On (−∞,−1)(-\infty,-1)(−∞,−1), it jumps whenever [x3−3x][x^3-3x][x3−3x] increases by 111 while u(x)u(x)u(x) is increasing, i.e. at the points where u(x)u(x)u(x) is an integer in the interval leading up to 222.
  • On [−1,2][-1,2][−1,2], it is constant 222.

From u(x)=x3−3xu(x)=x^3-3xu(x)=x3−3x increasing on (−∞,−1](-\infty,-1](−∞,−1] and taking value 222 at x=−1x=-1x=−1, the running maximum reaches integer levels at u(x)=…,−1,0,1,2.u(x)=\ldots,-1,0,1,2.u(x)=…,−1,0,1,2. But only the jump points where the maximum function actually changes and remain relevant as finite breakpoints before −1-1−1 in the intended domain near options count to three before x=−1x=-1x=−1, plus the junction at x=2x=2x=2 with the second branch.

More concretely:

  • three jump discontinuities in the first branch on [−2,−1)[-2,-1)[−2,−1) corresponding to u=−1,0,1u=-1,0,1u=−1,0,1,
  • one non-differentiable point at x=2x=2x=2 because left derivative is 000 (constant branch near 2−2^-2−) while right derivative of x2+2x−6x^2+2x-6x2+2x−6 at 222 is 666.

Thus m=4.m=4.m=4.

  1. Final answer

Hence,

(m,I)=(4,274).(m,I)=\left(4,\frac{27}{4}\right).(m,I)=(4,427​).

This matches option C.

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