Let f


Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

$f(x)=\left\{\begin{array}{rr}x^{5} \sin \left(\frac{1}{x}\right)+5 x^{2}, & x<0 \\ 0, & x=0 \\ x^{5} \cos \left(\frac{1}{x}\right)+\lambda x^{2}, & x>0\end{array}\right.$

The value of $\lambda$ for which $f^{\prime \prime}(0)$ exists, is_________.


$f^{\prime}(x)=\left\{\begin{array}{cc}5 x^{4} \cdot \sin \left(\frac{1}{x}\right)-x^{3} \cos \left(\frac{1}{x}\right)+10 x, & x<0 \\ 0, & x=0 \\ 5 x^{4} \cos \left(\frac{1}{x}\right)+x^{3} \sin \left(\frac{1}{x}\right)+2 \lambda x, & x>0\end{array}\right.$

$f^{\prime \prime}(x)=\left\{\begin{array}{cc}\left(20 x^{3}-x\right) \sin \left(\frac{1}{x}\right)-8 x^{2} \cos \left(\frac{1}{x}\right)+10, & x<0 \\ 0, & x=0 \\ \left(20 x^{3}-x\right) \cos \left(\frac{1}{x}\right)+8 x^{2} \sin \left(\frac{1}{x}\right)+2 \lambda, & x>0\end{array}\right.$

Now, $f^{\prime \prime}\left(0^{+}\right)=f^{\prime \prime}\left(0^{-}\right) \Rightarrow 2 \lambda=10 \Rightarrow \lambda=5$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now