MHT CET · Maths · Continuity and Differentiability
If the function
\(f(x)= \begin{cases}3 a x+b, \text { for } x < 1 \ 11, \text { for } x=1 \end{cases}\) \(\text { is continuous at } \ 5 a x\) \(-2 b,\text { for } x>1\) \(x=1\). Then, the values of \(a\) and \(b\) are
- A \(a=2, b=3\)
- B \(a=3, b=3\)
- C \(a=2, b=2\)
- D \(a=3, b=2\)
Answer & Solution
Correct Answer
(D) \(a=3, b=2\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3 a x+b=3 a+b \\
& \lim _{x \rightarrow 1^{+}} f(x) \lim _{x \rightarrow 1^{+}} 5 a x-2 b=5 a-2 b
\end{aligned}
\)
\(f(1)=11\) and function is continuous at \(x=1\), so we write
\(
3 \mathrm{a}+\mathrm{b}=11=5 \mathrm{a}-2 \mathrm{~b} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}
\)
This condition is satisfied when \(\mathrm{a}=3, \mathrm{~b}=2\)
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3 a x+b=3 a+b \\
& \lim _{x \rightarrow 1^{+}} f(x) \lim _{x \rightarrow 1^{+}} 5 a x-2 b=5 a-2 b
\end{aligned}
\)
\(f(1)=11\) and function is continuous at \(x=1\), so we write
\(
3 \mathrm{a}+\mathrm{b}=11=5 \mathrm{a}-2 \mathrm{~b} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}
\)
This condition is satisfied when \(\mathrm{a}=3, \mathrm{~b}=2\)
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