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

Area Under the Curves question

2023 · 29 Jan · Shift 1 · Q36
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. /Area Under the Curves
  5. /2023 · 29 Jan · Shift 1 · Q36

Area Under the Curves question

2023 · 29 Jan · Shift 1 · Q36

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Let [x][x][x] denote the greatest integer ≤x\le x≤x. Consider the function f(x)=max⁡{x2,1+[x]}f(x) = \max \left\{ {{x^2},1 + [x]} \right\}f(x)=max{x2,1+[x]}. Then the value of the integral ∫02f(x)dx\int\limits_0^2 {f(x)dx}0∫2​f(x)dx is
  1. A
    5+423{{5 + 4\sqrt 2 } \over 3}35+42​​
  2. B
    4+523{{4 + 5\sqrt 2 } \over 3}34+52​​
  3. C
    8+423{{8 + 4\sqrt 2 } \over 3}38+42​​
  4. D
    1+523{{1 + 5\sqrt 2 } \over 3}31+52​​
View written solutionFree

Correct answer: A

  1. We need to compute
I=∫02f(x) dx,f(x)=max⁡{x2, 1+[x]}.I=\int_0^2 f(x)\,dx, \qquad f(x)=\max\{x^2,\,1+[x]\}.I=∫02​f(x)dx,f(x)=max{x2,1+[x]}.
  1. Since the greatest integer function changes at integers, split the interval into:
  • 0≤x<10\le x<10≤x<1, where [x]=0[x]=0[x]=0
  • 1≤x<21\le x<21≤x<2, where [x]=1[x]=1[x]=1

So,

  • for 0≤x<10\le x<10≤x<1:
f(x)=max⁡{x2,1}f(x)=\max\{x^2,1\}f(x)=max{x2,1}

Since x2≤1x^2\le 1x2≤1 on [0,1][0,1][0,1], we get

f(x)=1.f(x)=1.f(x)=1.
  • for 1≤x<21\le x<21≤x<2:
f(x)=max⁡{x2,2}.f(x)=\max\{x^2,2\}.f(x)=max{x2,2}.

Now compare x2x^2x2 and 222:

x2=2  ⟺  x=2.x^2=2 \iff x=\sqrt{2}.x2=2⟺x=2​.

Thus,

  • for 1≤x≤21\le x\le \sqrt21≤x≤2​, x2≤2x^2\le 2x2≤2, so f(x)=2f(x)=2f(x)=2
  • for 2≤x≤2\sqrt2\le x\le 22​≤x≤2, x2≥2x^2\ge 2x2≥2, so f(x)=x2f(x)=x^2f(x)=x2

Hence,

I=∫011 dx+∫122 dx+∫22x2 dx.I=\int_0^1 1\,dx+\int_1^{\sqrt2}2\,dx+\int_{\sqrt2}^2 x^2\,dx.I=∫01​1dx+∫12​​2dx+∫2​2​x2dx.
  1. Evaluate each part:

First part:

∫011 dx=1.\int_0^1 1\,dx=1.∫01​1dx=1.

Second part:

∫122 dx=2(2−1)=22−2.\int_1^{\sqrt2}2\,dx=2(\sqrt2-1)=2\sqrt2-2.∫12​​2dx=2(2​−1)=22​−2.

Third part:

∫22x2 dx=[x33]22=83−(2)33=83−223.\int_{\sqrt2}^2 x^2\,dx=\left[\frac{x^3}{3}\right]_{\sqrt2}^2 =\frac{8}{3}-\frac{(\sqrt2)^3}{3} =\frac{8}{3}-\frac{2\sqrt2}{3}.∫2​2​x2dx=[3x3​]2​2​=38​−3(2​)3​=38​−322​​.
  1. Add them:
I=1+(22−2)+(83−223).I=1+(2\sqrt2-2)+\left(\frac{8}{3}-\frac{2\sqrt2}{3}\right).I=1+(22​−2)+(38​−322​​).

Simplify constants:

1−2+83=−1+83=53.1-2+\frac{8}{3}=-1+\frac{8}{3}=\frac{5}{3}.1−2+38​=−1+38​=35​.

Simplify the 2\sqrt22​ terms:

22−223=423.2\sqrt2-\frac{2\sqrt2}{3}=\frac{4\sqrt2}{3}.22​−322​​=342​​.

Therefore,

I=53+423=5+423.I=\frac{5}{3}+\frac{4\sqrt2}{3}=\frac{5+4\sqrt2}{3}.I=35​+342​​=35+42​​.
  1. Comparing with the options:
5+423\boxed{\frac{5+4\sqrt2}{3}}35+42​​​

which is Option A.

PreviousNext

More from Area Under the Curves

  • Let A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2​} and B={(x,y)∈R×R:0≤y≤min{2x,4−(x−1)2​}}. . Then…2023 · MCQ
  • The area of the region A={(x,y):∣cosx−sinx∣≤y≤sinx,0≤x≤2π​} is2023 · MCQ
  • Let α be the area of the larger region bounded by the curve y2=8x and the lines y=x and x=2, which lies in the first quadrant. Then the value of 3α is equal to ​.2023 · Numerical
  • Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2−px+45​p=0 are rational. Then the area of the region {(x,y):0≤y≤(x−q)2,0≤x≤q} is :2023 · MCQ
  • Let A be the area of the region {(x,y):y≥x2,y≥(1−x)2,y≤2x(1−x)}. Then 540 A is equal to :2023 · Numerical
  • Let for x∈R, f(x)=2x+∣x∣​ and g(x)={x,x2,​x<0x≥0​. Then area bounded by the curve y=(f∘g)(x) and the lines y=0,2y−x=15 is equal to ​…2023 · Numerical
  • Let the area of the region {(x,y):∣2x−1∣≤y≤​x2−x​,0≤x≤1} be A. Then (6 A+11)2 is equal to2023 · Numerical
  • Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1, R2. If max {R1, R2} = R2, then R1​R2​​ is equal to ​.2022 · Numerical