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Let \[f(x)={{x}^{13}}+{{x}^{11}}+{{x}^{9}}+{{x}^{7}}+{{x}^{5}}+{{x}^{3}}+x+19\] . Then, $f(x)=0$ has
A. $13$ real roots
B. Only one positive and only two negative real roots
C. Not more than one real root
D. Has two positive and one real root

Answer
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Hint: The given problem is related to the application of derivatives. If the derivative of a function is greater than or equal to zero, then the function is increasing and hence, it will intersect the $x$ axis at most one point.

Complete step-by-step answer:
Let $f\left( x \right)$ be any function. We know the derivative of the function, with respect to $x$ gives the slope of the function. If the derivative of the function is greater than or equal to zero at all points, then the slope of the function at any point will be non-negative. In such a case, the value of the function will not decrease with an increase in the value of $x$ . Such a function is called an increasing function. As the value of the function does not decrease with an increase in $x$ , the curve of the function will intersect the $x$ axis at most one point. So, such functions have at most one real root.
Now, the given function is \[f(x)={{x}^{13}}+{{x}^{11}}+{{x}^{9}}+{{x}^{7}}+{{x}^{5}}+{{x}^{3}}+x+19\] . On differentiating the function with respect to $x$ , we get: \[f'(x)=13{{x}^{12}}+11{{x}^{10}}+9{{x}^{8}}+7{{x}^{6}}+5{{x}^{4}}+3{{x}^{2}}+1\] . We can see that all the terms in $f'(x)$ have even powers of $x$ . So, the value of $f'(x)$ will always be positive. Hence, \[f(x)={{x}^{13}}+{{x}^{11}}+{{x}^{9}}+{{x}^{7}}+{{x}^{5}}+{{x}^{3}}+x+19\] is an increasing function. So, $f(x)$ can intersect the $x$ axis in at most one point. So, $f(x)=0$ can have no more than one real root. Hence, option C. is the correct option.

Note: The number of roots of an equation represents the number of points in which the curve of the equation intersects the $x$ axis.
Last updated date: 30th Sep 2023
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