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

(A) $f(x)$ has three real roots if $a>4$

(B) $f(x)$ has only one real root if $a<-4$

(C) $f(x)$ has three real roots if $a<-4$

(D) $f(x)$ has three real roots if $-4<a<4$

In this question, I first differentiated $f(x)$ and named it another function $g(x)$.

$f'(x) = g(x) = 5x^4 - 5$

Also, $g(0) = -5$, which means it has two roots.

$\implies g(x)$ is strictly decreasing in $(-\infty, -5]$ and strictly increasing in $(5, \infty)$.

I am stuck now. I don't understand how to proceed from here. I request to whoever answering this question, to tell me how to solve this using the method I am following. I am well aware that this question can be solved in other easier ways too. However, I want to solve this using ONLY this method. Thank you.

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You know that $f$ is strictly increasing on $(-\infty,-1]$ and on $[1,\infty)$ and strictly decreasing on $[-1,1]$. So

  1. If $a>4$, then, since $f(1)=a-4>0$, $f$ has only one real root.
  2. If $a<-4$, then, since $f(-1)=a+4<0$, $f$ has only one real root.
  3. If $-4<a<4$, then, since $f(1)<0$ and $f(-1)>0$, $f$ has three real roots.

So, assertions B. and D. are true, whereas assertions A. and C. are false.

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  • $\begingroup$ Understood it finally. Thank you! $\endgroup$ Jul 17 at 14:23
  • $\begingroup$ I'm glad I could help. $\endgroup$ Jul 17 at 14:25

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