Q84 / 120 · NDA 2024 (Phase 2) · Mathematics

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

  • A
    A only
  • Correct answer
    B
    B only
  • C
    Both A and B
  • D
    Neither A nor B

Solution

Given $f(x)=[x]^2-\left[x^2\right]$ Given, $f(x)=[x]^2-\left[x^2\right]$ Let us check continuity of $f(x)$ at $x=0$ $\begin{aligned} & \mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 0^{-}}=[0-\mathrm{h}]^2-\left[(0-\mathrm{h})^2\right]=(-1)^2-0=1 \\ & \mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 1^{+}}=[0+\mathrm{h}]^2-\left[(0+\mathrm{h})^2\right]=0-0=0 \end{aligned}$ LHL $\neq$ RHL , so $f(x)$ is not continous at $x=0$ Let us check continuity of $f(x)$ at $x=1$ $\begin{aligned} & \mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 1^{-}}=[1-\mathrm{h}]^2-\left[(1-\mathrm{h})^2\right]=0-0=0 \\ & \mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 1^{+}}=[1+\mathrm{h}]^2-\left[(1+\mathrm{h})^2\right]=1-1=0 \end{aligned}$ $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$ or $\mathrm{LHL}=\mathrm{RHL}$

Hence, the correct answer is option: 'B'.

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