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

2021 · 25 Jul · Shift 1 · Q33
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. /2021 · 25 Jul · Shift 1 · Q33

Definite Integration question

2021 · 25 Jul · Shift 1 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f:[0,∞)→[0,∞)f:[0,\infty ) \to [0,\infty )f:[0,∞)→[0,∞) be defined as f(x)=∫0x[y]dyf(x) = \int_0^x {[y]dy}f(x)=∫0x​[y]dy where [x] is the greatest integer less than or equal to x. Which of the following is true?
  1. A
    f is continuous at every point in [0,∞)[0,\infty )[0,∞) and differentiable except at the integer points.
  2. B
    f is both continuous and differentiable except at the integer points in [0,∞)[0,\infty )[0,∞).
  3. C
    f is continuous everywhere except at the integer points in [0,∞)[0,\infty )[0,∞).
  4. D
    f is differentiable at every point in [0,∞)[0,\infty )[0,∞).
View written solutionFree

Correct answer: A

  1. Given function

We have

f(x)=∫0x[y] dy,f(x)=\int_0^x [y] \,dy,f(x)=∫0x​[y]dy,

where [y][y][y] denotes the greatest integer less than or equal to yyy.

We need to determine where fff is continuous and differentiable.

  1. Understand the integrand [y][y][y]

The function [y][y][y] is a step function:

  • For 0≤y<10 \le y < 10≤y<1, [y]=0[y]=0[y]=0
  • For 1≤y<21 \le y < 21≤y<2, [y]=1[y]=1[y]=1
  • For 2≤y<32 \le y < 32≤y<3, [y]=2[y]=2[y]=2
  • and so on.

So [y][y][y] has jump discontinuities at every integer point.

  1. Find an explicit form of f(x)f(x)f(x)

Let x∈[n,n+1)x \in [n,n+1)x∈[n,n+1) where nnn is a non-negative integer. Then

f(x)=∫010 dy+∫121 dy+∫232 dy+⋯+∫n−1n(n−1) dy+∫nxn dy.f(x)=\int_0^1 0\,dy+\int_1^2 1\,dy+\int_2^3 2\,dy+\cdots+\int_{n-1}^n (n-1)\,dy+\int_n^x n\,dy.f(x)=∫01​0dy+∫12​1dy+∫23​2dy+⋯+∫n−1n​(n−1)dy+∫nx​ndy.

Thus,

f(x)=0⋅1+1⋅1+2⋅1+⋯+(n−1)⋅1+n(x−n).f(x)=0\cdot 1+1\cdot 1+2\cdot 1+\cdots +(n-1)\cdot 1+n(x-n).f(x)=0⋅1+1⋅1+2⋅1+⋯+(n−1)⋅1+n(x−n).

Since

0+1+2+⋯+(n−1)=n(n−1)2,0+1+2+\cdots +(n-1)=\frac{n(n-1)}{2},0+1+2+⋯+(n−1)=2n(n−1)​,

we get

f(x)=n(n−1)2+n(x−n),x∈[n,n+1).f(x)=\frac{n(n-1)}{2}+n(x-n), \quad x\in[n,n+1).f(x)=2n(n−1)​+n(x−n),x∈[n,n+1).

So on each interval [n,n+1)[n,n+1)[n,n+1), fff is linear.

  1. Check continuity

Because fff is defined as an integral of a bounded function, it should be continuous. Let us verify at integer points.

Take an integer nnn.

For x∈[n−1,n)x \in [n-1,n)x∈[n−1,n),

f(x)=(n−1)(n−2)2+(n−1)(x−(n−1)).f(x)=\frac{(n-1)(n-2)}{2}+(n-1)(x-(n-1)).f(x)=2(n−1)(n−2)​+(n−1)(x−(n−1)).

So the left-hand limit at x=nx=nx=n is

lim⁡x→n−f(x)=(n−1)(n−2)2+(n−1)=n(n−1)2.\lim_{x\to n^-} f(x)=\frac{(n-1)(n-2)}{2}+(n-1)=\frac{n(n-1)}{2}.x→n−lim​f(x)=2(n−1)(n−2)​+(n−1)=2n(n−1)​.

For x=nx=nx=n itself, using the formula for interval [n,n+1)[n,n+1)[n,n+1),

f(n)=n(n−1)2+n(n−n)=n(n−1)2.f(n)=\frac{n(n-1)}{2}+n(n-n)=\frac{n(n-1)}{2}.f(n)=2n(n−1)​+n(n−n)=2n(n−1)​.

Hence,

lim⁡x→n−f(x)=f(n).\lim_{x\to n^-}f(x)=f(n).x→n−lim​f(x)=f(n).

Also,

lim⁡x→n+f(x)=n(n−1)2=f(n).\lim_{x\to n^+}f(x)=\frac{n(n-1)}{2}=f(n).x→n+lim​f(x)=2n(n−1)​=f(n).

Therefore fff is continuous at every integer point. Since it is clearly continuous on each open interval (n,n+1)(n,n+1)(n,n+1), we conclude:

f is continuous for all x∈[0,∞).f \text{ is continuous for all } x\in[0,\infty).f is continuous for all x∈[0,∞).
  1. Check differentiability

On each open interval (n,n+1)(n,n+1)(n,n+1),

f(x)=n(n−1)2+n(x−n),f(x)=\frac{n(n-1)}{2}+n(x-n),f(x)=2n(n−1)​+n(x−n),

so

f′(x)=n=[x].f'(x)=n=[x].f′(x)=n=[x].

Thus fff is differentiable at every non-integer point.

Now check at an integer n≥1n\ge 1n≥1:

  • Left derivative:
f−′(n)=n−1f'_-(n)=n-1f−′​(n)=n−1
  • Right derivative:
f+′(n)=nf'_+(n)=nf+′​(n)=n

Since

f−′(n)≠f+′(n),f'_-(n)\ne f'_+(n),f−′​(n)=f+′​(n),

fff is not differentiable at integer points.

At x=0x=0x=0, since domain is [0,∞)[0,\infty)[0,∞), right derivative exists and equals 000, but in standard school/JEE usage, when we say differentiable on an interval, interior points are considered and the issue is at positive integers where the jump occurs. Hence the intended statement is:

f is differentiable except at the integer points.f \text{ is differentiable except at the integer points.}f is differentiable except at the integer points.
  1. Evaluate options

  • A: fff is continuous at every point in [0,∞)[0,\infty)[0,∞) and differentiable except at the integer points.
    True.

  • B: fff is both continuous and differentiable except at the integer points.
    This says it is not continuous at integer points, which is false.

  • C: fff is continuous everywhere except at the integer points.
    False, because fff is continuous at integer points too.

  • D: fff is differentiable at every point in [0,∞)[0,\infty)[0,∞).
    False, since it is not differentiable at positive integers.

  1. Final answer

The correct option is

A\boxed{\text{A}}A​
  1. Comparison with stored answer

Stored correct answer: A
Derived answer: A

They agree.

PreviousNext

More from Definite Integration

  • If f(x)=⎩⎨⎧​0∫x​(5+∣1−t∣)dt,5x+1,​x>2x≤2​, then2021 · MCQ
  • The value of the integral −1∫1​log(x+x2+1​)dx is :2021 · MCQ
  • The value of 2​−1​∫2​1​​((x−1x+1​)2+(x+1x−1​)2−2)21​dx is :2021 · MCQ
  • If the value of the integral 0∫5​ex−[x]x+[x]​dx=αe−1+β, where α, β∈ R, 5 α + 6 β = 0, and [x] denotes the greatest integer less than or equal to x;…2021 · MCQ
  • The value of −2π​∫2π​​(1+πsinx1+sin2x​)dx is2021 · MCQ
  • The value of −π/2∫π/2​1+3xcos2x​dx is :2021 · MCQ
  • The value of n=1∑100​n−1∫n​ex−[x]dx, where [ x ] is the greatest integer ≤ x, is :2021 · MCQ
  • The value of the integral 0∫π​∣sin2x∣dx is ​.2021 · Numerical