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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2024 · Mathematics · Continuity and Differentiability

Paragraph: Let f(x)=[x]^2- [x^2] . Question: Consider the following...

Paragraph: Let f(x)=[x]2−[x2]f(x)=[x]^2- \left[x^2\right].
Question: Consider the following statements: A. f(x)f(x) is continuous at x=0x=0. B. f(x)f(x) is continuous at x=1Whichofthestatementsgivenaboveis/arecorrect?x=1 Which of the statements given above is/are correct?

Options

  1. AA only
  2. BB only
  3. CBoth A and B
  4. DNeither A nor B

Correct answer

B. B only

Step-by-step solution

Given f(x)=[x]2−[x2]Given,f(x)=[x]^2- \left[x^2\right] Given,f(x)=[x]^2- \left[x^2\right]LetuscheckcontinuityofLet us check continuity off(x)atatx=0LHL=lim⁡x→000−=[0−h]2− lleft[(0−h)2]=(−1)2−0=1RHL=lim⁡x→111+=[0+h]2− lleft[(0+h)2]=0−0=0\begin{aligned} & \mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 000^{-}}=[0-\mathrm{h}]^2-\ lleft[(0-\mathrm{h})^2]=(-1)^2-0=1 \\ & \mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 111^{+}}=[0+\mathrm{h}]^2-\ lleft[(0+\mathrm{h})^2]=0-0=0 \end{aligned}LHL≠RHL,soLHL ≠ RHL , sof(x)isnotcontinousatis not continous atx=0LetuscheckcontinuityofLet us check continuity off(x)atatx=1LHL=lim⁡x→111−=[1−h]2− lleft[(1−h)2]=0−0=0RHL=lim⁡x→111+=[1+h]2− lleft[(1+h)2]=1−1=0\begin{aligned} & \mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 111^{-}}=[1-\mathrm{h}]^2-\ lleft[(1-\mathrm{h})^2]=0-0=0 \\ & \mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 111^{+}}=[1+\mathrm{h}]^2-\ lleft[(1+\mathrm{h})^2]=1-1=0 \end{aligned}\mathrm{f}(\mathrm{x})iscontinuousatis continuous at\mathrm{x}=1 or LHL=RHL\mathrm{LHL}=\mathrm{RHL}
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