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

Let $f\left(x\right)=x^5-5x+a$
$f^{\prime }\left(x\right)=5x^4-5$
$f^{\prime }\left(x\right)=0$ $\Rightarrow x=\pm 1$
$f^{\prime \prime }\left(x\right)=20x^3$
Hence,$x=1$ will be a point of local minimum and $x=-1$ will be a point of local maximum.
$f\left(1\right)=-4+a$, $f(-1)=4+a$.
As $x\to \infty$, $f\left(x\right)\to \infty$ and as $x\to -\infty$ , $f\left(x\right)\to -\infty$. Hence, $f\left(x\right)$ has atleast one root.

Option A : $a>4$
For $a>4$,
$f(-1)>0$ and $f\left(1\right)>0$. Hence, there will be only one real root.
Option B : $a>4$
$f(-1)>0$ and $f\left(1\right)>0$.
Hence, it will have one real root.
Option C : $a<-4$
$f(-1)<0$ and $f\left(1\right)>0$. Hence, there will be one real root.
Option D : $-4<a<4$
$f(-1)>0$ and $f\left(1\right)<0$. Hence, there will be three real roots.