MHT CET · Maths · Continuity and Differentiability
\(
=1+\sin \frac{\pi}{2}, \quad-\infty < x \leq 1
\)
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.
- A 1,1
- B 2,1
- C 0,2
- D 2,0
Answer & Solution
Correct Answer
(D) 2,0
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1+\sin \frac{\pi}{2}=1+1=2 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} a+b=a+b
\end{aligned}
\)
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 \(=2, b=0\)
\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 \(=2, b=0\)
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