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Limits Continuity and Differentiability question

2023 · 11 Apr · Shift 1 · Q34
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  5. /2023 · 11 Apr · Shift 1 · Q34

Limits Continuity and Differentiability question

2023 · 11 Apr · Shift 1 · Q34

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x)=[x2−x]+∣−x+[x]∣f(x)=\left[x^{2}-x\right]+|-x+[x]|f(x)=[x2−x]+∣−x+[x]∣, where x∈Rx \in \mathbb{R}x∈R and [t][t][t] denotes the greatest integer less than or equal to ttt. Then, fff is :
  1. A
    continuous at x=0x=0x=0, but not continuous at x=1x=1x=1
  2. B
    continuous at x=0x=0x=0 and x=1x=1x=1
  3. C
    continuous at x=1x=1x=1, but not continuous at x=0x=0x=0
  4. D
    not continuous at x=0x=0x=0 and x=1x=1x=1
View written solutionFree

Correct answer: C

  1. Given function

We have

f(x)=⌊x2−x⌋+∣−x+⌊x⌋∣,f(x)=\lfloor x^2-x\rfloor+\left|-x+\lfloor x\rfloor\right|,f(x)=⌊x2−x⌋+∣−x+⌊x⌋∣,

where ⌊t⌋\lfloor t\rfloor⌊t⌋ is the greatest integer function.

We need to check continuity at x=0x=0x=0 and x=1x=1x=1.

  1. Simplify the second term

Let x=n+r,n=⌊x⌋,0≤r<1.x=n+r, \quad n=\lfloor x\rfloor, \quad 0\le r<1.x=n+r,n=⌊x⌋,0≤r<1. Then −x+⌊x⌋=−(n+r)+n=−r.-x+\lfloor x\rfloor = -(n+r)+n=-r.−x+⌊x⌋=−(n+r)+n=−r. So ∣−x+⌊x⌋∣=∣−r∣=r=x−⌊x⌋,\left|-x+\lfloor x\rfloor\right|=|-r|=r=x-\lfloor x\rfloor,∣−x+⌊x⌋∣=∣−r∣=r=x−⌊x⌋, which is the fractional part of xxx.

Thus,

f(x)=⌊x2−x⌋+(x−⌊x⌋).f(x)=\lfloor x^2-x\rfloor + (x-\lfloor x\rfloor).f(x)=⌊x2−x⌋+(x−⌊x⌋).

Now check continuity at the required points.

  1. Continuity at x=0x=0x=0

(i) Value at x=0x=0x=0

f(0)=⌊02−0⌋+∣−0+⌊0⌋∣=⌊0⌋+0=0.f(0)=\lfloor 0^2-0\rfloor + |-0+\lfloor 0\rfloor|=\lfloor 0\rfloor+0=0.f(0)=⌊02−0⌋+∣−0+⌊0⌋∣=⌊0⌋+0=0.

(ii) Left-hand limit as x→0−x\to 0^-x→0−

Take x∈(−1,0)x\in(-1,0)x∈(−1,0). Then ⌊x⌋=−1,\lfloor x\rfloor=-1,⌊x⌋=−1, and hence x−⌊x⌋=x+1.x-\lfloor x\rfloor=x+1.x−⌊x⌋=x+1. Also, x2−x=x(x−1).x^2-x=x(x-1).x2−x=x(x−1). For x∈(−1,0)x\in(-1,0)x∈(−1,0), we have x2−x>0x^2-x>0x2−x>0, and near 0−0^-0− it lies in (0,1)(0,1)(0,1), so ⌊x2−x⌋=0.\lfloor x^2-x\rfloor=0.⌊x2−x⌋=0. Therefore, f(x)=0+(x+1)=x+1.f(x)=0+(x+1)=x+1.f(x)=0+(x+1)=x+1. So, lim⁡x→0−f(x)=1.\lim_{x\to 0^-}f(x)=1.limx→0−​f(x)=1.

(iii) Right-hand limit as x→0+x\to 0^+x→0+

Take x∈[0,1)x\in[0,1)x∈[0,1). Then ⌊x⌋=0,\lfloor x\rfloor=0,⌊x⌋=0, so x−⌊x⌋=x.x-\lfloor x\rfloor=x.x−⌊x⌋=x. Also for x∈(0,1)x\in(0,1)x∈(0,1), x2−x<0,x^2-x<0,x2−x<0, and near 0+0^+0+ it lies in (−1,0)(-1,0)(−1,0), hence ⌊x2−x⌋=−1.\lfloor x^2-x\rfloor=-1.⌊x2−x⌋=−1. Therefore, f(x)=−1+x=x−1.f(x)=-1+x=x-1.f(x)=−1+x=x−1. So, lim⁡x→0+f(x)=−1.\lim_{x\to 0^+}f(x)=-1.limx→0+​f(x)=−1.

Since lim⁡x→0−f(x)=1≠−1=lim⁡x→0+f(x),\lim_{x\to 0^-}f(x)=1 \ne -1=\lim_{x\to 0^+}f(x),limx→0−​f(x)=1=−1=limx→0+​f(x), fff is not continuous at x=0x=0x=0.

  1. Continuity at x=1x=1x=1

(i) Value at x=1x=1x=1

f(1)=⌊12−1⌋+∣−1+⌊1⌋∣=⌊0⌋+0=0.f(1)=\lfloor 1^2-1\rfloor+|-1+\lfloor 1\rfloor|=\lfloor 0\rfloor+0=0.f(1)=⌊12−1⌋+∣−1+⌊1⌋∣=⌊0⌋+0=0.

(ii) Left-hand limit as x→1−x\to 1^-x→1−

Take x∈(0,1)x\in(0,1)x∈(0,1). Then ⌊x⌋=0,\lfloor x\rfloor=0,⌊x⌋=0, so x−⌊x⌋=x.x-\lfloor x\rfloor=x.x−⌊x⌋=x. Also,

for x∈(0,1)x\in(0,1)x∈(0,1) lies in (−1,0)(-1,0)(−1,0), hence ⌊x2−x⌋=−1.\lfloor x^2-x\rfloor=-1.⌊x2−x⌋=−1. Therefore, f(x)=−1+x=x−1.f(x)=-1+x=x-1.f(x)=−1+x=x−1. So, lim⁡x→1−f(x)=0.\lim_{x\to 1^-}f(x)=0.limx→1−​f(x)=0.

(iii) Right-hand limit as x→1+x\to 1^+x→1+

Take x∈[1,2)x\in[1,2)x∈[1,2). Then ⌊x⌋=1,\lfloor x\rfloor=1,⌊x⌋=1, so x−⌊x⌋=x−1.x-\lfloor x\rfloor=x-1.x−⌊x⌋=x−1. Also, for xxx just greater than 111,

is positive and less than 111, so ⌊x2−x⌋=0.\lfloor x^2-x\rfloor=0.⌊x2−x⌋=0. Therefore, f(x)=0+(x−1)=x−1.f(x)=0+(x-1)=x-1.f(x)=0+(x−1)=x−1. So, lim⁡x→1+f(x)=0.\lim_{x\to 1^+}f(x)=0.limx→1+​f(x)=0.

Thus, lim⁡x→1−f(x)=lim⁡x→1+f(x)=f(1)=0,\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)=f(1)=0,limx→1−​f(x)=limx→1+​f(x)=f(1)=0, so fff is continuous at x=1x=1x=1.

  1. Conclusion
  • Not continuous at x=0x=0x=0
  • Continuous at x=1x=1x=1

Hence the correct option is: C\boxed{\text{C}}C​

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.

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