Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let \mathrm{f: R \rightarrow R }  be defined as
\mathrm{f(x)=\left\{\begin{array}{cc} x^5 \sin \left(\frac{1}{x}\right)+\lambda x^2 & , x<0 \\ 0 & , x=0 \\ x^5 \cos \left(\frac{1}{x}\right)+5 x^2 & , x>0 \end{array}\right. }
The value of \lambda for which \mathrm{f^{\prime \prime}(0) } exists, is

Option: 1

10


Option: 2

5


Option: 3

20


Option: 4

None of these


Answers (1)

best_answer

 We have,
\mathrm{\begin{aligned} & f(x)=\left\{\begin{array}{c} x^5 \sin \left(\frac{1}{x}\right)+\lambda x^2, \quad x<0 \\ 0 \quad, x=0 \\ x^5 \cos \left(\frac{1}{x}\right)+5 x^2, x>0 \end{array}\right. \\ & \Rightarrow f^{\prime}(x)=\left\{\begin{array}{c} 5 x^4 \sin \left(\frac{1}{x}\right)+x^5 \cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)+2 \lambda x, x<0 \\ 0 \quad, x=0 \\ 5 x^4 \cos \left(\frac{1}{x}\right)+x^5\left(-\sin \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)\right)+10 x, x>0 \end{array}\right. \end{aligned} }
\mathrm{\begin{aligned} & \Rightarrow f^{\prime}(x)= \begin{cases}5 x^4 \sin \left(\frac{1}{x}\right)-x^3 \cos \left(\frac{1}{x}\right)+2 \lambda x & , x<0 \\ 0 & , x=0 \\ 5 x^4 \cos \left(\frac{1}{x}\right)+x^3 \sin \left(\frac{1}{x}\right)+10 x & , x>0\end{cases} \\ & \Rightarrow f^{\prime \prime}(x)= \begin{cases}{\left[20 x^3 \sin \left(\frac{1}{x}\right)-5 x^2 \cos \left(\frac{1}{x}\right)\right.} & , x<0 \\ \left.-3 x^2 \cos \left(\frac{1}{x}\right)-x \sin \left(\frac{1}{x}\right)+2 \lambda\right] & , x=0 \\ 0 & , x>0 \\ {\left[20 x^3 \cos \left(\frac{1}{x}\right)+5 x^2 \sin \left(\frac{1}{x}\right)\right.} & \\ \left.+3 x^2 \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right)+10\right] & \end{cases} \end{aligned} }
\mathrm{To exist f^{\prime \prime}(0), f^{\prime \prime}\left(0^{+}\right)=f^{\prime \prime}\left(0^{-}\right) \Rightarrow 2 \lambda=10 \Rightarrow \lambda=5 }
 

Posted by

seema garhwal

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 Tiruchirappalli cut-offs rose for major BTech programmes this year; closing ranks
anam-khan

JoSAA 2025 reconstituted; NIT Rourkela to coordinate BTech counselling for IITs, NITs, IIITs
anam-khan

JEE Main 2025: BTech Information Technology cut-offs at NITs; know opening, closing ranks

Latest Question