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

Limits Continuity and Differentiability question

2019 · 8 Apr · Shift 2 · Q43
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. /Limits Continuity and Differentiability
  5. /2019 · 8 Apr · Shift 2 · Q43

Limits Continuity and Differentiability question

2019 · 8 Apr · Shift 2 · Q43

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let ƒ : [–1,3] →\to→ R be defined as f(x)={∣x∣+[x],−1≤x<1x+∣x∣,1≤x<2x+[x],2≤x≤3f(x) = \left\{ {\begin{matrix} {\left| x \right| + \left[ x \right]} & , & { - 1 \le x \lt 1} \\ {x + \left| x \right|} & , & {1 \le x \lt 2} \\ {x + \left[ x \right]} & , & {2 \le x \le 3} \\ \end{matrix} } \right.f(x)=⎩⎨⎧​∣x∣+[x]x+∣x∣x+[x]​,,,​−1≤x<11≤x<22≤x≤3​ where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at:
  1. A
    only three points
  2. B
    four or more points
  3. C
    only two points
  4. D
    only one point
View written solutionFree

Correct answer: A

  1. Given function

We analyze

f(x)={∣x∣+[x],−1≤x<1x+∣x∣,1≤x<2x+[x],2≤x≤3f(x)= \begin{cases} |x|+[x], & -1\le x<1 \\ x+|x|, & 1\le x<2 \\ x+[x], & 2\le x\le 3 \end{cases}f(x)=⎩⎨⎧​∣x∣+[x],x+∣x∣,x+[x],​−1≤x<11≤x<22≤x≤3​

where [x][x][x] is the greatest integer function.

We need the points in [−1,3][-1,3][−1,3] where fff is discontinuous.

  1. Break the interval into simpler parts

The possible discontinuities can occur at:

  • points where the formula changes: x=1,2x=1,2x=1,2
  • points where [x][x][x] changes value: integers inside the relevant intervals, i.e. x=0,1,2,3x=0,1,2,3x=0,1,2,3
  • also check endpoint behavior if needed

Now simplify on each subinterval.

  1. For −1≤x<1-1\le x<1−1≤x<1, f(x)=∣x∣+[x]f(x)=|x|+[x]f(x)=∣x∣+[x]

(i) On −1≤x<0-1\le x<0−1≤x<0

Here ∣x∣=−x|x|=-x∣x∣=−x and [x]=−1[x]=-1[x]=−1. So

f(x)=−x−1f(x)=-x-1f(x)=−x−1

which is continuous on [−1,0)[-1,0)[−1,0).

(ii) At x=0x=0x=0

For 0≤x<10\le x<10≤x<1, we have ∣x∣=x|x|=x∣x∣=x and [x]=0[x]=0[x]=0, so

f(x)=xf(x)=xf(x)=x

Now check at x=0x=0x=0:

  • Left limit:
lim⁡x→0−f(x)=lim⁡x→0−(−x−1)=−1\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(-x-1)=-1x→0−lim​f(x)=x→0−lim​(−x−1)=−1
  • Right limit:
lim⁡x→0+f(x)=lim⁡x→0+x=0\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}x=0x→0+lim​f(x)=x→0+lim​x=0

Since left and right limits are different, fff is discontinuous at x=0x=0x=0.

(iii) On 0<x<10<x<10<x<1

f(x)=xf(x)=xf(x)=x

continuous there.

  1. At x=1x=1x=1

From the left (x<1x<1x<1), we use f(x)=∣x∣+[x]=x+0=xf(x)=|x|+[x]=x+0=xf(x)=∣x∣+[x]=x+0=x for 0≤x<10\le x<10≤x<1. Thus

lim⁡x→1−f(x)=1\lim_{x\to 1^-}f(x)=1x→1−lim​f(x)=1

At x=1x=1x=1 and for 1≤x<21\le x<21≤x<2, we use

f(x)=x+∣x∣=x+x=2xf(x)=x+|x|=x+x=2xf(x)=x+∣x∣=x+x=2x

so

f(1)=2,lim⁡x→1+f(x)=2f(1)=2, \qquad \lim_{x\to 1^+}f(x)=2f(1)=2,x→1+lim​f(x)=2

Since

lim⁡x→1−f(x)=1≠2=lim⁡x→1+f(x),\lim_{x\to 1^-}f(x)=1 \ne 2=\lim_{x\to 1^+}f(x),x→1−lim​f(x)=1=2=x→1+lim​f(x),

fff is discontinuous at x=1x=1x=1.

  1. For 1<x<21<x<21<x<2 Here x>0x>0x>0, so ∣x∣=x|x|=x∣x∣=x. Hence
f(x)=x+∣x∣=2xf(x)=x+|x|=2xf(x)=x+∣x∣=2x

which is continuous on (1,2)(1,2)(1,2).

  1. At x=2x=2x=2

From the left, for 1≤x<21\le x<21≤x<2, f(x)=2xf(x)=2xf(x)=2x, so

lim⁡x→2−f(x)=4\lim_{x\to 2^-}f(x)=4x→2−lim​f(x)=4

For 2≤x≤32\le x\le 32≤x≤3, we have

f(x)=x+[x]f(x)=x+[x]f(x)=x+[x]

Now for 2≤x<32\le x<32≤x<3, [x]=2[x]=2[x]=2, so

f(x)=x+2f(x)=x+2f(x)=x+2

Hence

lim⁡x→2+f(x)=4,f(2)=2+[2]=4\lim_{x\to 2^+}f(x)=4, \qquad f(2)=2+[2]=4x→2+lim​f(x)=4,f(2)=2+[2]=4

Thus fff is continuous at x=2x=2x=2.

  1. At x=3x=3x=3

For 2≤x<32\le x<32≤x<3, we have

f(x)=x+2f(x)=x+2f(x)=x+2

so

lim⁡x→3−f(x)=5\lim_{x\to 3^-}f(x)=5x→3−lim​f(x)=5

But

f(3)=3+[3]=6f(3)=3+[3]=6f(3)=3+[3]=6

Since

lim⁡x→3−f(x)≠f(3),\lim_{x\to 3^-}f(x)\ne f(3),x→3−lim​f(x)=f(3),

fff is discontinuous at x=3x=3x=3.

  1. Check endpoint x=−1x=-1x=−1

At x=−1x=-1x=−1,

f(−1)=∣−1∣+[−1]=1−1=0f(-1)=|-1|+[-1]=1-1=0f(−1)=∣−1∣+[−1]=1−1=0

For x→−1+x\to -1^+x→−1+, on [−1,0)[-1,0)[−1,0),

f(x)=−x−1→−(−1)−1=0f(x)=-x-1 \to -(-1)-1=0f(x)=−x−1→−(−1)−1=0

So fff is continuous at x=−1x=-1x=−1 (right-continuous endpoint check).

  1. List all discontinuity points

The discontinuities are at:

x=0,  1,  3x=0,\;1,\;3x=0,1,3

So there are exactly three points.

Therefore, the correct option is:

A: only three points\boxed{\text{A: only three points}}A: only three points​
PreviousNext

More from Limits Continuity and Differentiability

  • Let ƒ(x) = 15 – |x – 10|; x ∈ R. Then the set of all values of x, at which the function, g(x) = ƒ(ƒ(x)) is not differentiable, is :2019 · MCQ
  • If the function ƒ defined on , (6π​,3π​) by f(x)={cotx−12​cosxolimits−1​,k,​xe4π​x=4π​​…2019 · MCQ
  • If the function f(x)={a∣π−x∣+1,x≤5b∣x−π∣+3,x>5​ is continuous at x = 5, then the value of a – b is :-2019 · MCQ
  • If f(x)=[x]−[4x​],x ∈ 4 , where [x] denotes the greatest integer function, then2019 · MCQ
  • y→0lim​y41+1+y4​​−2​​2019 · MCQ
  • Let f : R → R be a function defined as f(x)=⎩⎨⎧​5a+bxb+5x30​;;;;​x≤11<x<33≤x<5x≥5​…2019 · MCQ
  • For each x ∈ R, let [x] be the greatest integer less than or equal to x. Then x→0−lim​∣x∣x([x]+∣x∣)sin[x]​ is…2019 · MCQ
  • If f(x)=⎩⎨⎧​xsin(p+1)x+sinx​qx23​x+x2​−x​​​,x<0,x=0,x>0​ is continuous at x =…2019 · MCQ