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?
Question: What is $f(0.999)+f(1.001)$ equal to?
- A$-1$
- Correct answerB$0$
- 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'.