Define $f(x)=\left\{\begin{array}{cc}x^2+b x+c, & x < 1 \\ x, & x \geq 1\end{array}\right.$. If $f(x)$ is…
Define $f(x)=\left\{\begin{array}{cc}x^2+b x+c, & x < 1 \\ x, & x \geq 1\end{array}\right.$. If $f(x)$ is differentiable at $x=1$, then $(b-c)$ is equal to
$-2$
$0$
$1$
$2$
Solution
Given function,
$\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x^2+b x+c, & x < 1 \\
x, & x \geq 1
\end{array}\right. \\
& f^{\prime}(x)=\left\{\begin{array}{cc}
2 x+b, & x < 1 \\
1, & x \geq 1
\end{array}\right.
\end{aligned}$
Since, $f(x)$ is differentiable at $x=1$.
$\begin{aligned}
& \lim _{x \rightarrow 1^{+}} f^{\prime}(x)=\lim _{x \rightarrow 1^{+}} f^{\prime}(x) \\
& \lim _{x \rightarrow 1^{+}}(2 x+b)=\lim _{x \rightarrow 1^{+}} 1 \\
& 2+b=1 \Rightarrow b=-1
\end{aligned}$
As form is differentiable at $n=1$ So, it will be continuous at $x=1$ also.
$\begin{aligned}
& \Rightarrow \quad \lim _{x \rightarrow 1^{+}} f(x)=\lim _{\substack{x \rightarrow 1^{+}}} f^{\prime}(x)=f(1) \\
& \Rightarrow \quad \lim _{x \rightarrow 1^{+}} x^2+b x+c=\lim _{x \rightarrow 1^{+}} x=1 \\
& \Rightarrow \quad 1+b+c=1 \\
& \Rightarrow \quad 1-1+c=1 \Rightarrow c=1
\end{aligned}$
Hence, $b-c=-1-1=-2$