Theory of Equations 5 Question 10
10. Let $a \in R$ and $f: R \rightarrow R$ be given by $f(x)=x^{5}-5 x+a$. 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 $-4<a<4$
Passage Based Problems
Read the following passage and answer the questions.
Passage I
Consider the polynomial $f(x)=1+2 x+3 x^{2}+4 x^{3}$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$.
(2010)
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Answer:
Correct Answer: 10. (b,d)
Solution:
- PLAN
(i) Concepts of curve tracing are used in this question.
(ii) Number of roots are taken out from the curve traced.
Let $y=x^{5}-5 x$
(i) As $x \rightarrow \infty, y \rightarrow \infty$ and as $x \rightarrow-\infty, y \rightarrow-\infty$
(ii) Also, at $x=0, y=0$, thus the curve passes through the origin.
(iii) $\frac{d y}{d x}=5 x^{4}-5=5\left(x^{4}-1\right)=5\left(x^{2}-1\right)\left(x^{2}+1\right)$
$ =5(x-1)(x+1)\left(x^{2}+1\right) $
Now, $\frac{d y}{d x}>0$ in $(-\infty,-1) \cup(1, \infty)$, thus $f(x)$ is increasing in these intervals.
Also, $\frac{d y}{d x}<0$ in $(-1,1)$, thus decreasing in $(-1,1)$.
(iv) Also, at $x=-1, d y / d x$ 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. $\quad a \in(-4,+4)$
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.
