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Question:
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If the function $f$ defined by
$$
f(x)=\left\{\begin{array}{llc}
\cos x, & \text { if } & x \leq 0 \\
3 x+\alpha, & \text { if } & 0 < x < 2 \\
\beta x+3, & \text { if } & 2 \leq x \leq 4 \\
11, & \text { if } & x>4
\end{array}\right.
$$
where, $\alpha$ and $\beta$ are real constants, is continuous on $R$, then $\alpha^2+\beta^2=$
Options:
$$
f(x)=\left\{\begin{array}{llc}
\cos x, & \text { if } & x \leq 0 \\
3 x+\alpha, & \text { if } & 0 < x < 2 \\
\beta x+3, & \text { if } & 2 \leq x \leq 4 \\
11, & \text { if } & x>4
\end{array}\right.
$$
where, $\alpha$ and $\beta$ are real constants, is continuous on $R$, then $\alpha^2+\beta^2=$
Solution:
1527 Upvotes
Verified Answer
The correct answer is:
5
$f(x)$ is continuous on $R$, so
$$
\begin{array}{rlrl}
\text { at } x=0, & \lim _{x \rightarrow 0^{+}} f(x) & =f(0) \\
\Rightarrow & 0+\alpha=\cos 0 & =1 \\
\alpha & =1 \\
\text { at } x=2, & \lim _{x \rightarrow 2^{-}} f(x) & =f(2) \\
6+\alpha & =2 \beta+3
\end{array}
$$
$\begin{aligned} 7 & =2 \beta+3 \\ 2 \beta & =4 \Rightarrow \beta=2 \\ \text { So, } \quad \alpha^2+\beta^2 & =1+4=5 .\end{aligned}$
$$
\begin{array}{rlrl}
\text { at } x=0, & \lim _{x \rightarrow 0^{+}} f(x) & =f(0) \\
\Rightarrow & 0+\alpha=\cos 0 & =1 \\
\alpha & =1 \\
\text { at } x=2, & \lim _{x \rightarrow 2^{-}} f(x) & =f(2) \\
6+\alpha & =2 \beta+3
\end{array}
$$
$\begin{aligned} 7 & =2 \beta+3 \\ 2 \beta & =4 \Rightarrow \beta=2 \\ \text { So, } \quad \alpha^2+\beta^2 & =1+4=5 .\end{aligned}$
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