Q100 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

Let $f(x)$ be a real valued function. If $f^{\prime}(x)$ is a constant for all $x \in \mathbb{R}, f(0)=2$ and $f^{\prime}(0)=1$, then

  • A
    $\mathrm{f}(\mathrm{x})$ is not continuous on $\mathbb{R}$
  • B
    $f(x)$ is continuous at $x=0,1,2$ and 3 only
  • C
    $f(x)$ is continuous only on $[0, \infty)$
  • Correct answer
    D
    $\mathrm{f}(\mathrm{x})$ is continuous on $\mathbb{R}$

Solution

Since $f^{\prime}(x)$ is a constant $$ \therefore f^{\prime}(x)=a \text { (say) }...(1) $$ $\Rightarrow f(x)=a x+b$ where $b$ is arbitrary constant. ...(2) Since $f(0)=2 \Rightarrow b=2$ Since $f^{\prime}(0)=1 \Rightarrow a=1$ $$ \therefore f(x)=(x+2) $$ Which is continuous on $(-\infty, \infty)$ i.e $\mathbb{R}$

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

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