First slide

Theory of expressions

Question

Let $a \in R and f : R \to R$ be given by $f (x) = x^{5} - 5 x + a$ then,

Difficult

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

Solution

Let $y = x^{5} - 5 x$
$(i) As x \to \infty, y \to \infty and x \to - \infty, y \to - \infty$
(ii) Also, at x = 0, y = 0, thus the curve passes through the origin.
$(iii) \frac{dy}{dx} = 5 x^{4} - 5 = 5 (x^{4} - 1) = 5 (x^{2} - 1) (x^{2} + 1)$
$= 5 (x - 1) (x + 1) (x^{2} + 1)$
Now, $\frac{dy}{dx} > 0 in (- \infty, - 1) \cup (1, \infty)$ thus f(x) is increasing in these interval
Also, $\frac{dy}{dx} < 0 (- 1, 1)$ thus decreasing in (-1, 1).
(iv) Also , at x = -1, $\frac{dy}{dx}$ changes its sign from + ve to ve
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.

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