Question

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

• A. 5.0
• B. 5
• C. 5.00

Answer 1

Answer: (A), (B), (C)

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

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

$f^{\prime \prime }\left(x\right)=\{\begin{array}{ccc}20x^3\sin \left(\frac{1}{x}\right)-5x^2\cos \left(\frac{1}{x}\right)-3x^2\cos \left(\frac{1}{x}\right)-x\sin \left(\frac{1}{x}\right)+10,& & x<0\\ 0& & x=0\\ 20x^3\cos \left(\frac{1}{x}\right)+5x^2\sin \left(\frac{1}{x}\right)+3x^2\sin \left(\frac{1}{x}\right)-x\cos \left(\frac{1}{x}\right)+2\lambda ,& & x>0\end{array}$
if $f^{\prime \prime }\left(0\right)$ exists then
$f^{\prime \prime }\left(0^{+}\right)=f^{\prime \prime }\left(0^{-}\right)$
$2\lambda =10\Rightarrow \lambda =5$