Q237 / 270 · NDA 2024 (Phase 1) · Mathematics

Let $f(x)=|x|+1$ and $g(x)=[x]-1$, where [.] is the greatest integer function. Let $h(x)=\frac{f(x)}{g(x)}$ Consider the following statements: 1. $f(x)$ is differentiable for all $x \lt 0$ 2. $g(x)$ is continuous at $x=0.0001$ 3. The derivative of $g(x)$ at $x=2.5$ is 1 Which of the statements given above are correct?

  • Correct answer
    A
    1 and 2 only
  • B
    2 and 3 only
  • C
    1 and 3 only
  • D
    1,2 and 3

Solution

and $\begin{aligned} f(x) & =|x|+1 \\ g(x) & =[x]-1 \\ h(x) & =\frac{f(x)}{g(x)}=\frac{|x|+1}{[x]-1}\end{aligned}$ (1) For, $x \lt 0$ $\begin{aligned} & f(x)=-x+1 \\ & \therefore \quad f(x)=-1 \end{aligned}$ $f(x)$ is differentiable $\forall x \lt 0$ $\Rightarrow$ Statement (1) is correct. (2) At $x=0.0001$ $g(x)$ is continuous everywhere except for integer value as [ $x$ ] is continuous $\forall x \in \mathrm{R}$ except integers. So, $g(x)$ is continuous at $x=0.0001$. Statement (2) is correct. $\begin{aligned} &\text{(3)} \quad g(x)=[x]-1 \\ & {[x] \Rightarrow \text { Always } a b \text { integer }} \\ & \qquad \begin{aligned} g^{\prime}(x) & =\frac{d}{d x}[x]-0 \\ g^{\prime}(x) & =0-0=0 \end{aligned} \end{aligned}$ $\Rightarrow$ Statement (3) is incorrect. Hence, (1) and (2) only correct.

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

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