If $f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x < 0 \\ b\left(\frac{5 x^2+a,}{x^2-3…
If $f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x < 0 \\ b\left(\frac{5 x^2+a,}{x^2-3 x+2}\right), & \text { when } 0 \leq x \leq 1 \\ -14, & \text { when } x \geq 3\end{array}\right.$ is a continuous function on $R$, then $(a, b)=$
$\left(2,-\frac{7}{2}\right)$
$(2,-14)$
$\left(-\frac{7}{2},-14\right)$
$(2,7)$
Solution
Given, $f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x} ; & \text { when } x < 0 \\ b\left(\frac{x^2+a ;}{x^2-3 x+2}\right) ; & \text { when } 0 \leq x \leq 1 \\ -14 ; & \text { when } x \geq 3\end{array}\right.$
Since, given $f(x)$ is continuous at $x=0$ and $x=3$ $f(x)$ is continuous at $x=0$.
$
\begin{aligned}
& \Rightarrow \text { LHL at } x=0=\text { RHL at } x=0 \\
& =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)
\end{aligned}
$
$
\begin{aligned}
& =\lim _{x \rightarrow 0^{-}} \frac{x-|x|}{x}=\lim _{x \rightarrow 0^{+}} 5 x^2+a \\
& =\lim _{x \rightarrow 0^{-}} \frac{x-(-x)}{x}=\lim _{x \rightarrow 0} 5 x^2+a \\
& =\lim _{x \rightarrow 0^{-}} \frac{2 x}{x}=5(0)^2+a \\
\therefore \quad & \quad[\text { as } x < 0, \text { then }|x|=-x] \\
\therefore \quad & a \\
\therefore \quad & 2
\end{aligned}
$
$f(x)$ is also continuous at $x=3$
$\therefore$ LHL at $(x=3)=$ RHL at $(x=3)$
$
\begin{aligned}
& =\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x) \\
& =\lim _{x \rightarrow 3^{-}} b\left(\frac{x^2-1}{x^2-3 x+2}\right)=\lim _{x \rightarrow 3^{+}}-14 \\
& =\lim _{x \rightarrow 3^{-}} b\left(\frac{(x+1)(x-1)}{(x-1)(x-2)}\right)=-14 \\
& =\lim _{x \rightarrow 3^{-}} b\left(\frac{x+1}{x-2}\right)=-14
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \quad b\left(\frac{3+1}{3-2}\right)=-14 \\
& \Rightarrow \quad 4 b=-14 \\
& \Rightarrow \quad b=-\frac{14}{4} \\
& \Rightarrow \quad b=-\frac{7}{2} \\
& \therefore \quad(a, b)=\left(2, \frac{-7}{2}\right) \\
&
\end{aligned}
$
$\therefore$ Hence, option (a) is correct