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If $f(x)$ defined as given below, is continuous on $R$, then the value of $a+b$ is equal to $…
If $f(x)$ defined as given below, is continuous on $R$, then the value of $a+b$ is equal to
$
f(x)=\left\{\begin{array}{cc}
\sin x, & x \leq 0 \\
x^2+a, & 0 < x < 1 \\
b x+3, & 1 \leq x \leq 3 \\
-3, & x>3
\end{array}\right.
$
0 2 -2 3 Solution
$f(x)=\left\{\begin{array}{cc}\sin x, & x \leq 0 \\ x^2+a, & 0 < x < 1 \\ b x+3, & 1 \leq x \leq 3 \\ -3, & x>3\end{array}\right.$
Given that, $f(x)$ is continuous on $R$.
Then, $f(x)$ is continuous at $n=0,3$
$
\Rightarrow f(0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)
$
Now, $f(0)=\sin 0=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^2+a\right)=\lim _{h \rightarrow 0}\left(h^2+a\right)=a \\
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(\sin x)=\lim _{h \rightarrow 0} \sin (0-h)=0 \\
& \Rightarrow a=0
\end{aligned}
$
Again, we must have
$
f(3)=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)
$
Now, $f(3)=b(3)+3=3 b+3$
$
\lim _{x \rightarrow 3^{+}} f(x)=-3
$
$
\Rightarrow 3 b+3=-3 \Rightarrow b=-2
$
Then, $a+b=0-2=-2$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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