Q83 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: Let $f(x)=[x]^2-\left[x^2\right]$.
Question: What is $f(0.999)+f(1.001)$ equal to?

  • A
    $-1$
  • Correct answer
    B
    $0$
  • C
    $1$
  • D
    $2$

Solution

Given $f(x)=[x]^2-\left[x^2\right]$ According to the question $f(0.999)=[0.999]^2-\left[(0.999)^2\right]$ $=0-$ [value lies $\mathrm{b} / \mathrm{w} 0$ and 1 ] $\begin{aligned} & f(0.999)=0-0=0 \quad \ldots \text { (i) } \\ & f(1.001)=[1.001]^2-\left[(1.0001)^2\right]=1-1= ....(ii) \end{aligned}$ Adding (i) and (ii) $f(0.999)+f(1.001)=0-0=0$

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

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