Solution:
f(x) being a polynomial is continuous and differentiable for all real values of x.
We also have f(a) = f(b) = f(c).
If we apply Rolle’s theorem to f(x) in [a, b] and [b, c] we will observe that f(x) = 0
will have at least one root in (a, b) and at least one root in (b, c).
But f(x) is a polynomial of degree two, so that f(x) = 0
can’t have more than two roots. It implies that exactly one root of f(x) = 0
will lie in (a, b) and exactly one root of f(x) = 0 will lie in (b, c).
Let y = f(x) be a polynomial function of degree n. If f(x) = 0 has real roots only,
then f(x) = 0, f(x) = 0, … , fn-1(x) = 0 will have real roots.
It is in fact the general version of above mentioned application,
because if f(x) = 0 have all real roots, then between two consecutive roots of f(x) = 0,
exactly one root of f(x) = 0 will lie.
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Equation of horizontal line below x-axis at 5 units from x-axis is:
The negation of the statement.
“A circle is an ellipse” is.
