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If $a$ and $b(a>b)$ are points of discontinuity of the function
$$
f(x)=\left\{\begin{array}{ccc}
3-2 x^2, & \text { for } & x \leq 0 \\
2 x+3, & \text { for } & 0 < x \leq 1 \\
2 x^2-3 x, & \text { for } & 1 < x < 2 \\
2 x-3, & \text { for } & 2 \leq x < 3 \\
|x|, & \text { for } & x \geq 3
\end{array}\right.
$$
then $3 a-b=$
Options:
$$
f(x)=\left\{\begin{array}{ccc}
3-2 x^2, & \text { for } & x \leq 0 \\
2 x+3, & \text { for } & 0 < x \leq 1 \\
2 x^2-3 x, & \text { for } & 1 < x < 2 \\
2 x-3, & \text { for } & 2 \leq x < 3 \\
|x|, & \text { for } & x \geq 3
\end{array}\right.
$$
then $3 a-b=$
Solution:
1169 Upvotes
Verified Answer
The correct answer is:
5
Since, LHL (at $x=1)=5$
and RHL (at $x=1$ ) $=-1$, so function is discontinuous at $x=1$,
And LHL (at $x=2)=2$
and RHL (at $x=2$ ) $=1$, so function is discontinuous at $x=2$.
So, $a=2$ and $b=1(\because a>b)$, $\Rightarrow \quad 3 a-b=5$
and RHL (at $x=1$ ) $=-1$, so function is discontinuous at $x=1$,
And LHL (at $x=2)=2$
and RHL (at $x=2$ ) $=1$, so function is discontinuous at $x=2$.
So, $a=2$ and $b=1(\because a>b)$, $\Rightarrow \quad 3 a-b=5$
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