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Definite Integration question

2022 · 29 Jul · Shift 2 · Q27
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  5. /2022 · 29 Jul · Shift 2 · Q27

Definite Integration question

2022 · 29 Jul · Shift 2 · Q27

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If [t][t][t] denotes the greatest integer ≤t\leq t≤t, then the value of ∫01[2x−∣3x2−5x+2∣+1]dx\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x∫01​[2x−​3x2−5x+2​+1]dx is :
  1. A
    37+13−46\frac{\sqrt{37}+\sqrt{13}-4}{6}637​+13​−4​
  2. B
    37−13−46\frac{\sqrt{37}-\sqrt{13}-4}{6}637​−13​−4​
  3. C
    −37−13+46\frac{-\sqrt{37}-\sqrt{13}+4}{6}6−37​−13​+4​
  4. D
    −37+13+46\frac{-\sqrt{37}+\sqrt{13}+4}{6}6−37​+13​+4​
View written solutionFree

Correct answer: A

  1. We need to evaluate I=∫01[2x−∣3x2−5x+2∣+1]dx,I=\int_0^1 \left[2x-|3x^2-5x+2|+1\right]dx,I=∫01​[2x−∣3x2−5x+2∣+1]dx, where [t][t][t] denotes the greatest integer function.

So first define f(x)=2x−∣3x2−5x+2∣+1.f(x)=2x-|3x^2-5x+2|+1.f(x)=2x−∣3x2−5x+2∣+1. We must find the interval-wise value of [f(x)][f(x)][f(x)].

  1. Simplify the absolute value expression.

Factor: 3x2−5x+2=(3x−2)(x−1).3x^2-5x+2=(3x-2)(x-1).3x2−5x+2=(3x−2)(x−1). Its roots are x=23, 1.x=\frac23,\ 1.x=32​, 1.

Now on [0,1][0,1][0,1]:

  • for 0≤x<230\le x<\frac230≤x<32​, (3x−2)<0(3x-2)<0(3x−2)<0 and (x−1)<0(x-1)<0(x−1)<0, so product is positive,
  • for 23<x<1\frac23<x<132​<x<1, product is negative.

Hence

\begin{cases} 3x^2-5x+2, & 0\le x\le \frac23,\\[4pt] -(3x^2-5x+2)=-3x^2+5x-2, & \frac23\le x\le 1. \end{cases}$$ Therefore $$f(x)=2x-|3x^2-5x+2|+1= \begin{cases} 2x-(3x^2-5x+2)+1=-3x^2+7x-1, & 0\le x\le \frac23,\\[4pt] 2x-(-3x^2+5x-2)+1=3x^2-3x+3, & \frac23\le x\le 1. \end{cases}$$ --- 3. Determine possible integer values of $[f(x)]$. For $0\le x\le \frac23$, $$f_1(x)=-3x^2+7x-1.$$ We check where it crosses integers. Since $$f_1(0)=-1,\qquad f_1\left(\frac23\right)=-3\cdot\frac49+7\cdot\frac23-1=\frac73,$$ we expect values of the floor function to be $-1,0,1,2$. Solve successively: ### (i) $f_1(x)=0$ $$-3x^2+7x-1=0\iff 3x^2-7x+1=0.$$ Thus $$x=\frac{7\pm\sqrt{49-12}}{6}=\frac{7\pm\sqrt{37}}{6}.$$ Only the smaller root lies in $[0,1]$: $$\alpha=\frac{7-\sqrt{37}}{6}.$$ So on $[0,\alpha)$, $f_1(x)<0$, hence $[f(x)]=-1$. ### (ii) $f_1(x)=1$ $$-3x^2+7x-1=1\iff 3x^2-7x+2=0.$$ Thus $$x=\frac{7\pm\sqrt{49-24}}{6}=\frac{7\pm5}{6}.$$ Relevant root: $$x=\frac13.$$ So on $[\alpha,\frac13)$, $0\le f_1(x)<1$, hence $[f(x)]=0$. ### (iii) $f_1(x)=2$ $$-3x^2+7x-1=2\iff 3x^2-7x+3=0.$$ Thus $$x=\frac{7\pm\sqrt{49-36}}{6}=\frac{7\pm\sqrt{13}}{6}.$$ In $[0,\frac23]$, only $$\beta=\frac{7-\sqrt{13}}{6}$$ lies in the interval. So on $\left[\frac13,\beta\right)$, $1\le f_1(x)<2$, hence $[f(x)]=1$. And on $\left[\beta,\frac23\right]$, $2\le f_1(x)\le \frac73$, hence $[f(x)]=2$. --- 4. For $\frac23\le x\le 1$, $$f_2(x)=3x^2-3x+3.$$ Now $$f_2\left(\frac23\right)=\frac73,\qquad f_2(1)=3.$$ Also $$f_2(x)-2=3x^2-3x+1.$$ Its discriminant is $$(-3)^2-4\cdot 3\cdot 1=9-12=-3<0,$$ so $f_2(x)>2$ for all $x$. Further, for $x<1$, we have $f_2(x)<3$, and at $x=1$, $f_2(1)=3$. Thus on $\left[\frac23,1\right)$, $$[f(x)]=2,$$ and the single point $x=1$ does not affect the integral. So effectively, $[f(x)]=2$ on $\left[\beta,1\right]$. --- 5. Hence the floor function is $$[f(x)]= \begin{cases} -1, & 0\le x<\alpha,\\ 0, & \alpha\le x<\frac13,\\ 1, & \frac13\le x<\beta,\\ 2, & \beta\le x\le 1, \end{cases}$$ where $$\alpha=\frac{7-\sqrt{37}}{6},\qquad \beta=\frac{7-\sqrt{13}}{6}.$$ Therefore $$I=\int_0^{\alpha}(-1)dx+\int_{\alpha}^{1/3}0\,dx+\int_{1/3}^{\beta}1\,dx+\int_{\beta}^{1}2\,dx.$$ So $$I=-\alpha+\left(\beta-\frac13\right)+2(1-\beta).$$ Simplify: $$I=-\alpha+\beta-\frac13+2-2\beta =\frac53-\alpha-\beta.$$ Substitute $\alpha,\beta$: $$I=\frac53-\frac{7-\sqrt{37}}{6}-\frac{7-\sqrt{13}}{6}.$$ $$I=\frac{10-(14-\sqrt{37}-\sqrt{13})}{6} =\frac{\sqrt{37}+\sqrt{13}-4}{6}.$$ --- 6. Compare with the options: $$I=\frac{\sqrt{37}+\sqrt{13}-4}{6}.$$ This matches **Option A**. --- 7. Comparison with stored correct answer: - Derived answer: **A** - Stored correct answer: **A** They agree.
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