Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let be a function defined by <img alt="f(x)= \begin{cases}\frac{\tan x}{x} & , x \neq 0 \\ 1 & , x=0\end{cases

Let f be a function defined by
f(x)= \begin{cases}\frac{\tan x}{x} & , x \neq 0 \\ 1 & , x=0\end{cases}
Statement I: x=0 is a point of minima of \int.
Statement II : f(0)=0

Option: 1

Statement I is false, statement II is true. 


Option: 2

Statement I is true, statement II is true. statement II is a correct explanation of statement I.


Option: 3

Statement I is true, statement II is true. Statement II is not a correct explanation of the statement I. 


Option: 4

Statement I is true, statement II is false. 


Answers (1)

best_answer

f(x)= \begin{cases}\frac{\tan x}{x}, & x \neq 0 \\ 1 & , x=0\end{cases} \\ \begin{aligned} & \text { As, } \frac{\tan x}{x}>1 \forall x \neq 0 \\ & \therefore \quad f(0+h)>f(0) \text { and } \quad f(0-h)>f(0) \\ & \Rightarrow \text { At } x=0, f(x) \text { attains minima. } \end{aligned} \\ \\ \begin{aligned} f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\frac{\tan h}{h}-1}{h} \\ & =\lim _{h \rightarrow 0} \frac{\tan h-h}{h^2}=\lim _{h \rightarrow 0} \frac{h+\frac{h^3}{3}+\ldots-h}{h^2}=0 \end{aligned}

Posted by

manish painkra

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 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question