Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\small f(x)= \begin{cases}x+1, & x \leq 1 \\ 3-a x^2, & x>1\end{cases}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20f%28x%29%3D%20%5Cbegin%7Bcases%7Dx&p

\small f(x)= \begin{cases}x+1, & x \leq 1 \\ 3-a x^2, & x>1\end{cases}   

Value of ‘a’ for which f(x) is continuous, is 

Option: 1

1


Option: 2

2


Option: 3

-1


Option: 4

-2


Answers (1)

best_answer

\begin{aligned} & f(x)= \begin{cases}x+1 & x \leq 1 \\ 3-a x^2 & x>1\end{cases} \\ & \lim _{x \rightarrow 1^{-}}(x+1)=2=\lim _{x \rightarrow 1^{+}}\left(3-a x^2\right)=3-a \\ & 3-a=2 \quad \Rightarrow a=3-2 \Rightarrow a=1 \end{aligned}

Hence (A) is the correct answer.

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question