Question

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

• A. $f\left(x\right)$ has three real roots if $a>4$
• B. $f\left(x\right)$ has only one real root if $a>4$
• C. $f\left(x\right)$ has three real roots if $a<-4$
• D. $f\left(x\right)$ has three real roots if $-4<a<4$

Answer 1

Answer: (B), (D)

$f\left(x\right)$ has three real roots if $-4<a<4$

$f\left(x\right)=x^5-5x+a\Rightarrow f^{\prime }\left(x\right)=5x^4-5\Rightarrow f^{\prime }\left(x\right)=5(x^4-1)=0\Rightarrow x=\pm 1$

$\Rightarrow f^{\prime \prime }\left(x\right)=20x^3$
$\Rightarrow f^{\prime \prime }\left(1\right)>0,f^{\prime \prime }(-1)<0$

$x=1$ is point of minima,
$x=-1$ is point of maxima.

$f\left(1\right)=a-4;f(-1)=a+4$

When $-4<a<4$

$f\left(x\right)$ has three real roots.

When $a>4$

$f\left(x\right)$ has only one root.

When $a<-4$

$f\left(x\right)$ has only one root.