Q68 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
$$
=1+\sin \frac{\pi}{2}, \quad-\infty < x \leq 1
$$
If the function $\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}, \quad 1 < \mathrm{x} < 3$ is continuous in
$$
=6 \tan \frac{x \pi}{12}, \quad 3 \leq x < 6
$$
$(-\infty, 6)$, then the values of $a$ and $b$ are respectively.
- A1,1
- B2,1
- C0,2
- Correct answerD2,0
Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1+\sin \frac{\pi}{2}=1+1=2 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} a+b=a+b
\end{aligned}
$$
Since $f(x)$ is continuous at $x=1$, we get $a+b=2$
[From (1), (2)]
$$
\begin{aligned}
& \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} a x+b=3 a+b \\
& \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} 6 \tan \frac{x \pi}{12}=6 \tan \frac{3 \pi}{12}=6
\end{aligned}
$$
Since $f(x)$ is continuous at $x=3$, we get $3 a+b=6$
[From (4), (5)]
Solving (3) and (6), we get a $=2, b=0$
Hence, the correct answer is option: 'D'.