Let $a \in R$ and let $f : R \to R$ be given by $f(x) = x^{5} - 5x + a$. Then

Let a ∈ R and let f : R → R be given by f(x) = x⁵ - 5x + a. Then $\text{Given: } f(x) = x^5 - 5x + a,\quad a \in \mathbb{R}$ $\text{To find: number of real roots of } f(x) \text{ as } a \text{ varies}$ $1.\ \text{differentiate } f$ $2.\ \text{local max/min from } f'$ $3.\ \text{count } x\text{-axis crossings as } a \text{ varies}$ $y = x^5 - 5x \quad (a=0),\quad \text{changing } a \text{ shifts the curve vertically}$ $f'(x) = 5x^4 - 5 = 5(x^2-1)(x^2+1)$ $f'(x) = 0 \Rightarrow x^2 = 1 \Rightarrow x = -1,\ 1$ $f' > 0\ (x<-1),\quad f' < 0\ (-1<x<1),\quad f' > 0\ (x>1)$ $\text{local max at } x=-1,\quad \text{local min at } x=1$ $f(-1) = a+4 \ \text{(local max)}$ $f(1) = a-4 \ \text{(local min)}$ $\text{real roots of } f \ =\ x\text{-axis crossings of the curve}$ $a > 4:\ a-4>0 \Rightarrow f(x) \text{ has only one real root}$ $\Rightarrow \text{(2) TRUE},\quad \text{(1) FALSE}$ $a < -4:\ a+4<0 \Rightarrow f(x) \text{ also has only one real root}$ $\Rightarrow \text{(3) FALSE}$ $-4<a<4:\ a+4>0,\ a-4<0 \Rightarrow \text{the curve crosses } 3 \text{ times}$ $\Rightarrow \text{(4) TRUE}$ $\boxed{\text{Answer: (2) and (4)}}$