Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If the function $f(x)= \begin{cases}3 a x+b, & \text { for } x < 1 \\ 11, & \text { for } x=1 \text { is continuous at } \\ 5 a x-2 b, & \text { for } x>1\end{cases}$ $x=1$. Then, the values of $a$ and $b$ are
MathematicsContinuity and DifferentiabilityMHT CETMHT CET 2021 (23 Sep Shift 1)
Options:
  • A $a=2, b=3$
  • B $a=3, b=3$
  • C $a=2, b=2$
  • D $a=3, b=2$
Solution:
2678 Upvotes Verified Answer
The correct answer is: $a=3, b=2$
$$
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3 a x+b=3 a+b \\
& \lim _{x \rightarrow 1^{+}} f(x) \lim _{x \rightarrow 1^{+}} 5 a x-2 b=5 a-2 b
\end{aligned}
$$
$f(1)=11$ 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 $\mathrm{a}=3, \mathrm{~b}=2$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.