Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2022 · 28 Jun · Shift 1 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2022 · 28 Jun · Shift 1 · Q28

Definite Integration question

2022 · 28 Jun · Shift 1 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral ∫01[−8x2+6x−1]dx\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx}0∫1​[−8x2+6x−1]dx is equal to :
  1. A
    −-− 1
  2. B
    −54{{ - 5} \over 4}4−5​
  3. C
    17−138{{\sqrt {17} - 13} \over 8}817​−13​
  4. D
    17−168{{\sqrt {17} - 16} \over 8}817​−16​
View written solutionFree

Correct answer: C

  1. We need to evaluate I=∫01[−8x2+6x−1]dx,I=\int_0^1 \left[ -8x^2+6x-1\right]dx,I=∫01​[−8x2+6x−1]dx, where [t][t][t] denotes the greatest integer function.

So the integrand is f(x)=[−8x2+6x−1].f(x)=\left[ -8x^2+6x-1\right].f(x)=[−8x2+6x−1].

  1. Let g(x)=−8x2+6x−1.g(x)=-8x^2+6x-1.g(x)=−8x2+6x−1. We first study the range of g(x)g(x)g(x) on [0,1][0,1][0,1].

This is a downward-opening parabola.

Its vertex is at x=−b2a=−62(−8)=38.x=\frac{-b}{2a}=\frac{-6}{2(-8)}=\frac{3}{8}.x=2a−b​=2(−8)−6​=83​. Then

=-\frac98+\frac94-1=\frac18.$$ Also, $$g(0)=-1, \qquad g(1)=-3.$$ Hence on $[0,1]$, the values of $g(x)$ lie between $-3$ and $\frac18$. Therefore, the possible values of $[g(x)]$ are $-3,-2,-1,0$. 3. Find where $g(x)=0,-1,-2$. ### (i) Solve $g(x)=0$ $$-8x^2+6x-1=0$$ $$8x^2-6x+1=0$$ $$x=\frac{6\pm\sqrt{36-32}}{16}=\frac{6\pm2}{16}$$ So, $$x=\frac14,\ \frac12.$$ Since the parabola opens downward, $g(x)\ge 0$ for $$x\in\left[\frac14,\frac12\right].$$ Thus, $$[g(x)]=0 \quad \text{for } x\in\left[\frac14,\frac12\right].$$ ### (ii) Solve $g(x)=-1$ $$-8x^2+6x-1=-1$$ $$-8x^2+6x=0$$ $$2x( -4x+3)=0$$ So, $$x=0,\ \frac34.$$ Thus, $$g(x)\ge -1 \text{ on } \left[0,\frac34\right],$$ and $g(x)<-1$ for $x>\frac34$. Combining with the previous result: - On $\left[0,\frac14\right)$ and $\left(\frac12,\frac34\right]$, we have $-1\le g(x)<0$, so $$[g(x)]=-1.$$ ### (iii) Solve $g(x)=-2$ $$-8x^2+6x-1=-2$$ $$-8x^2+6x+1=0$$ $$8x^2-6x-1=0$$ $$x=\frac{6\pm\sqrt{36+32}}{16}=\frac{6\pm\sqrt{68}}{16}=\frac{3\pm\sqrt{17}}{8}.$$ Only the root in $[0,1]$ is $$x=\frac{3+\sqrt{17}}{8}.$$ For $x\in\left(\frac34,\frac{3+\sqrt{17}}{8}\right)$, we have $$-2<g(x)<-1 \implies [g(x)]=-2.$$ For $x\in\left(\frac{3+\sqrt{17}}{8},1\right]$, we have $$-3\le g(x)<-2 \implies [g(x)]=-3.$$ 4. Now integrate piecewise: $$I=\int_0^{1/4}(-1)\,dx+\int_{1/4}^{1/2}0\,dx+\int_{1/2}^{3/4}(-1)\,dx+\int_{3/4}^{(3+\sqrt{17})/8}(-2)\,dx+\int_{(3+\sqrt{17})/8}^{1}(-3)\,dx.$$ Compute each part: $$\int_0^{1/4}(-1)dx=-\frac14,$$ $$\int_{1/4}^{1/2}0\,dx=0,$$ $$\int_{1/2}^{3/4}(-1)dx=-\frac14,$$ $$\int_{3/4}^{(3+\sqrt{17})/8}(-2)dx=-2\left(\frac{3+\sqrt{17}}{8}-\frac34\right),$$ $$\int_{(3+\sqrt{17})/8}^{1}(-3)dx=-3\left(1-\frac{3+\sqrt{17}}{8}\right).$$ So, $$I=-\frac12-2\left(\frac{3+\sqrt{17}-6}{8}\right)-3\left(\frac{8-3-\sqrt{17}}{8}\right).$$ That is, $$I=-\frac12-2\left(\frac{\sqrt{17}-3}{8}\right)-3\left(\frac{5-\sqrt{17}}{8}\right).$$ $$I=-\frac12-\frac{\sqrt{17}-3}{4}-\frac{15-3\sqrt{17}}{8}.$$ Take common denominator $8$: $$I=\frac{-4}{8}+\frac{-2\sqrt{17}+6}{8}+\frac{-15+3\sqrt{17}}{8}.$$ $$I=\frac{-13+\sqrt{17}}{8}.$$ Hence, $$I=\frac{\sqrt{17}-13}{8}.$$ 5. Comparing with the options, this is **Option C**.
PreviousNext

More from Definite Integration

  • Let f : R → R be a differentiable function such that f(4π​)=2​,f(2π​)=0 and f′(2π​)=1 and let g(x)=∫xπ/4​(f′(t)sect+tantsectf(t))dt…2022 · MCQ
  • Let f : R → R be a continuous function satisfying f(x) + f(x + k) = n, for all x ∈ R where k > 0 and n is a positive integer. If I1​=0∫4nk​f(x)dx and I2​=−k∫3k​f(x)dx, then :2022 · MCQ
  • The integral 0∫2π​​3+2sinx+cosx1​ dx is equal to :2022 · MCQ
  • If f(α)=1∫α​1+tlog10​t​dt,α>0, then f(e3)+f(e−3) is equal to :2022 · MCQ
  • If [t] denotes the greatest integer ≤t, then the value of ∫01​[2x−​3x2−5x+2​+1]dx is :2022 · MCQ
  • Let f:R→R be a function defined by : f(x)={max{t3−3t}t≤xx2+2x−6​;;​x≤225​ where [t] is the greatest integer…2022 · MCQ
  • ∫05​cos(π(x−[2x​]))dx, where [t] denotes greatest integer less than or equal to t, is equal to:2022 · MCQ
  • Let f be a real valued continuous function on [0, 1] and f(x)=x+0∫1​(x−t)f(t)dt. Then, which of the following points (x, y) lies on the curve y = f(x) ?2022 · MCQ