If the function f(x)= c 1+ x 2 , for - < x 1 x+b, for 1 < x < 3 6… - Vidyadip

MCQ

If the function
$f(x)=\left\{\begin{array}{c}1+\sin \frac{\pi x}{2}, \text { for }-\infty < x \leq 1 \\a x+b, \text { for } 1 < x < 3 \\6 \tan \frac{\pi x}{12}, \text { for } 3 \leq x < 6\end{array}\right.$
is continuous in the interval $(-\infty, 6)$, then the values of a and b are respectively

A
0, 2
B
1, 1
✓
2, 0
D
2, 1

✓

Answer

Correct option: C.

2, 0

(C)
Since $f(x)$ is continuous in $(-\infty, 6)$.
$\therefore $ it is continuous at $x=1$ and $x=3$.
$\therefore \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}\left(1+\sin \frac{\pi x}{2}\right)=\lim _{x \rightarrow 1^{+}}(a x+b)$
$\Rightarrow 1+\sin \frac{\pi}{2}=a+b$
$\Rightarrow a+b=2$ ...(i)
Also, $\lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 3^{-}}(a x+b)=\lim _{x \rightarrow 3^{+}}\left(6 \tan \frac{\pi x}{12}\right)$
$\Rightarrow 3 a+b=6 \tan \frac{3 \pi}{12}$
$\Rightarrow 3 a+b=6$ ...(ii)
From (i) and (ii), we get $a=2, b=0$

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