Question

Let $a\in R$ and $f:R\to R$ 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

Answer 1

Answer: (B), (D)

PLAN
(i) Concepts of curve tracing are used in this question.
(ii) Number of roots are taken out from the curve traced.
Let $y=x^5-5x$
(i) As $x\to \infty ,y\to \infty$ and as $x\to -\infty ,y\to -\infty$
(ii) Also, at $x=0,y=0$, thus the curve passes through the origin.
(iii) $\frac{dy}{dx}=5x^4-5=5(x^4-1)$
$=5(x^2-1)(x^2+1)$
$=5(x-1)(x+1)(x^2+1)$

Now, $\frac{dy}{dx}>0$ in $(-\infty ,-1)\cup (1,\infty )$, thus $f(x)$ is
increasing in these intervals.
Also, $\frac{dy}{dx}<0$ in $(-1,1)$, thus decreasing in $(-1,1)$.
(iv) Also, at $x=-1,dy/dx$ changes its sign from + ve to - ve.
$\therefore x=-1$ is point of local maxima.
Similarly, $x=1$ is point of local minima
Local maximum value, $y=(-1{)}^5-5(-1)=4$
Local minimum value, $y=(1{)}^5-5(1)=-4$

Now, let $y=-a$
As evident from the graph, if $-a\in (-4,4)$
i,e., $a\in (-4,+4)$
Then, $f(x)$ has three real roots and if $-a>4$
or $-a<-4$, then $f(x)$ has one real root,
i.e. for $a<-4$ or $a>4,f(x)$ has one real root.