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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
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Verified Answer
The correct answer is:
$\mathrm{f}(\mathrm{x})$ is continuous on $\mathbb{R}$
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 $\mathbb{R}$
$$
\therefore f^{\prime}(x)=a \text { (say) }...(1)
$$
$\Rightarrow f(x)=a x+b$ 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 $\mathbb{R}$
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