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