MHT CET · Maths · Continuity and Differentiability
The function \(\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}\) where \(\mathrm{t}=\frac{1}{x-1}\) is discontinuous at
- A \(-2,1\)
- B \(2, \frac{1}{2}\)
- C \(\frac{1}{2}, 1\)
- D 2,1
Answer & Solution
Correct Answer
(B) \(2, \frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(
f(t)=\frac{1}{t^2+t-2}=\frac{1}{(t+2)(t-1)}
\)
\(\mathrm{f}(\mathrm{t})\) is not defined at \(\mathrm{t}=-2\) and \(\mathrm{t}=1\).
\(
\begin{aligned}
& \mathrm{t}=-2 \\
& \Rightarrow \frac{1}{x-1}=-2 \\
& \Rightarrow x=\frac{1}{2} \\
& \mathrm{t}=1 \\
& \Rightarrow \frac{1}{x-1}=1 \\
& \Rightarrow x=2
\end{aligned}
\)
\(\therefore \quad\) The function \(\mathrm{f}(\mathrm{t})\) is discontinuous at \(x=\frac{1}{2}\) and \(x=2\).
f(t)=\frac{1}{t^2+t-2}=\frac{1}{(t+2)(t-1)}
\)
\(\mathrm{f}(\mathrm{t})\) is not defined at \(\mathrm{t}=-2\) and \(\mathrm{t}=1\).
\(
\begin{aligned}
& \mathrm{t}=-2 \\
& \Rightarrow \frac{1}{x-1}=-2 \\
& \Rightarrow x=\frac{1}{2} \\
& \mathrm{t}=1 \\
& \Rightarrow \frac{1}{x-1}=1 \\
& \Rightarrow x=2
\end{aligned}
\)
\(\therefore \quad\) The function \(\mathrm{f}(\mathrm{t})\) is discontinuous at \(x=\frac{1}{2}\) and \(x=2\).
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