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Question:
Answered & Verified by Expert
If $f: R \rightarrow R$ defined by
$$
f(x)=\left\{\begin{array}{cc}
a^2 \cos ^2 x+b^2 \sin ^2 x, & x \leq 0 \\
e^{a x+b}, & x>0
\end{array}\right.
$$
is a continuous function, then
Options:
$$
f(x)=\left\{\begin{array}{cc}
a^2 \cos ^2 x+b^2 \sin ^2 x, & x \leq 0 \\
e^{a x+b}, & x>0
\end{array}\right.
$$
is a continuous function, then
Solution:
1316 Upvotes
Verified Answer
The correct answer is:
$b=2 \log |a|$
We have,
$$
\begin{aligned}
& \lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} e^{a h+b}=e^b \\
& \text { and } \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} a^2 \cos ^2 h \\
&+b^2 \sin ^2 h=a^2 0 \\
& \text { Thus, } \quad e^b=a^2 \\
& b=2 \log |a|
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} e^{a h+b}=e^b \\
& \text { and } \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} a^2 \cos ^2 h \\
&+b^2 \sin ^2 h=a^2 0 \\
& \text { Thus, } \quad e^b=a^2 \\
& b=2 \log |a|
\end{aligned}
$$
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