JEE Main 2020 (Online) 6th September Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be defined as
$f(x)=\begin{cases}x^{5}\sin\left(\dfrac{1}{x}\right)+5x^{2}, & x<0\\ 0, & x=0\\ x^{5}\cos\left(\dfrac{1}{x}\right)+\lambda x^{2}, & x>0\end{cases}$
The value of $\lambda$ for which $f''(0)$ exists, is $\_\_\_\_\_\_\_\_$ .
Limits & Continuity
Answer$5$
Solution
$f'(0)=0$ from both sides. $f''(0)$ from the left is $\lim\frac{f'(x)}x=10$ (the $x^5$ terms vanish) and from the right $2\lambda$. So $\lambda=5$.