Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the function<img alt="f(x)=\left\{\begin{matrix} 3+\left | x-k \right | ,& c\leq k\\ a^{2}-2+\frac{sin(x-k)}{x-k}& x>k \end{matrix}\right." src="https://entrancecorner.oncod

If the functionf(x)=\left\{\begin{matrix} 3+\left | x-k \right | ,& c\leq k\\ a^{2}-2+\frac{sin(x-k)}{x-k}& x>k \end{matrix}\right.. has local minimum at the point x=k(\forall k \in R), then a can be equal to


 

Option: 1

-1


Option: 2

3


Option: 3

1


Option: 4

5


Answers (1)

best_answer

\begin{aligned} & \text { (a,c) } \text { LHL } =\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{-}}(3+|x-k|) \\ & =\lim _{\rightarrow \rightarrow 0^{+}}(3+h)=3 \\ f(k) & =3 \\ \text { RHL } & =f\left(k^{+}\right)=\lim _{x \rightarrow k^{+}} f(x) \\ & =\lim _{x \rightarrow k^{+}}\left(a^2-2+\frac{\sin (x-k)}{x-k}\right)=a^2-2+1=a^2-1 \end{aligned}

For local minimum to exist at the point x=k(\forall k \in R),  we have

\Rightarrow \quad \begin{aligned} & a^2-1 \geq 3 \\ & a^2>4 \Rightarrow|a|>2 \end{aligned}

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question