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If $\mathrm{A}$ is the domain and $\mathrm{B}$ is the range of the function
$f(x)=\left\{\begin{array}{ll}3 x-1, & x>1 \\ x^2+1, & x \leq 1\end{array}\right.$ then $A-B=$
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$f(x)=\left\{\begin{array}{ll}3 x-1, & x>1 \\ x^2+1, & x \leq 1\end{array}\right.$ then $A-B=$
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Verified Answer
The correct answer is:
$(-\infty, 1)$
$\because f(x)= \begin{cases}3 x-1 & x>1 \\ x^2+1 & x \leq 1\end{cases}$
$\therefore$ Domain is $(-\infty, 1] \cup(1, \infty)=(-\infty, \infty)=\mathrm{A}$
When $x>1$ : $f(x)=3 x-1 \Rightarrow 2 \leq f(x) < \infty$ and $\mathrm{x} \leq 1: \mathrm{f}(\mathrm{x})=\mathrm{x}^2+1 \Rightarrow 1 \leq \mathrm{f}(\mathrm{x}) < \infty$.
$\therefore \quad$ Range $B=(1, \infty) \cup(2, \infty)=[1, \infty)$
Now, $A-B=(-\infty, \infty)-[1, \infty)=(-\infty, 1)$
$\therefore$ Domain is $(-\infty, 1] \cup(1, \infty)=(-\infty, \infty)=\mathrm{A}$
When $x>1$ : $f(x)=3 x-1 \Rightarrow 2 \leq f(x) < \infty$ and $\mathrm{x} \leq 1: \mathrm{f}(\mathrm{x})=\mathrm{x}^2+1 \Rightarrow 1 \leq \mathrm{f}(\mathrm{x}) < \infty$.
$\therefore \quad$ Range $B=(1, \infty) \cup(2, \infty)=[1, \infty)$
Now, $A-B=(-\infty, \infty)-[1, \infty)=(-\infty, 1)$
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