Let $f: R \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{cl}x^{5} \sin \left(\frac{1}{x}\ri

Question

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

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