Question

Let $f:R\to R$ be defined by $f\left(x\right)=\left\{\begin{array}{l}{x}^5\sin \left(\frac{1}{x}\right)+5x^2,x<0\\ 0,x=0\\ x^5\cos \left(\frac{1}{x}\right)+\lambda x^2,x>0\end{array}\right.$ The value of $\lambda$ for which $f\text{'}\text{'}\left(0\right)$ exists, is:

Answer 1

Calculating $\lambda$ for which $f\text{'}\text{'}\left(0\right)$ exists:

Differentiating $f\left(x\right)$, we have,

$f\text{'}\left(x\right)=\left\{\begin{array}{l}5x^4\sin \left(\frac{1}{x}\right)-x^5\cos \left(\frac{1}{x}\right)\times \frac{1}{x^2}+10x,x<0\\ 0,x=0\\ 5x^4\cos \left(\frac{1}{x}\right)+x^5.\sin \left(\frac{1}{x}\right)\times \frac{1}{x^2}+2\lambda x,x>0\end{array}\right.$

$f\text{'}\text{'}\left(x\right)=\left\{\begin{array}{l}20x^3\sin \left(\frac{1}{x}\right)-5x^2\cos \left(\frac{1}{x}\right)-3x^2\cos \left(\frac{1}{x}\right)-x\sin \left(\frac{1}{x}\right)+10,x<0\\ 0,x=0\\ 20x^3\cos \left(\frac{1}{x}\right)+5x^2\sin \left(\frac{1}{x}\right)+3x^2.\sin \left(\frac{1}{x}\right)-x\cos \left(\frac{1}{x}\right)+2\lambda ,x>0\end{array}\right.$

Now, if $f\text{'}\text{'}\left(x\right)$ exists then,

$$\begin{gathered} f\text{'}\text{'}\left(0^{+}\right)=f\text{'}\text{'}\left(0^{-1}\right) \\ \Rightarrow 2\lambda =10 \\ \Rightarrow \lambda =5 \end{gathered}$$

Hence, the correct value of $\lambda$ is $5.$