Question

How many real solutions does the equation ${x^7} + 14{x^5} + 16{x^3} + 30x - 560$ has?
A) 3
B) 5
C) 7
D) 1

Answer 1

Here to solve this problem, we will first differentiate the equation and we will check whether this function is increasing or decreasing. If the differentiation of the equation is greater than zero then it will be increasing function otherwise decreasing function.

Complete step by step solution:
Let’s first consider the given equation as-
$f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560$
Now, we will differentiate the given equation with respect to $x$ on both sides.
$\dfrac{{df\left( x \right)}}{{dx}} = \dfrac{{d\left( {{x^7} + 14{x^5} + 16{x^3} + 30x - 560} \right)}}{{dx}}$
Differentiating both the side, we get
$f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30$
As the exponents are even so even if we put a negative number, the result will be always greater than zero. So we can write,
$f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0 \, \forall x \in R$
Therefore, we can say that the function is a strictly increasing function. So, it will cut the axis at only one point and has only one real solution.

Hence, the number of real solutions of the equation ${x^7} + 14{x^5} + 16{x^3} + 30x - 560$ is one.

Note:
We need to know the important properties of functions that we have used here.
Functions are said to be strictly increasing functions if the value of $f'\left( x \right)$ is greater than zero.
Functions are said to be strictly decreasing functions if the value of $f'\left( x \right)$ is less than zero.
Here we have also obtained the derivative of a function, which measures the rate of change of one variable with respect to the change of another variable.