Let f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19$. Taking the derivative with respect to $x$: $f'(x) = 13x^{12} + 11x^{10} +

Vedclass · Accepted Answer

not more than one real root

Vedclass · Answer

(C) Given $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19$. Taking the derivative with respect to $x$: $f'(x) = 13x^{12} + 11x^{10} + 9x^{8} + 7x^{6} + 5x^{4} + 3x^{2} + 1$. Since all exponents of $x$ in $f'(x)$ are even and the coefficients are positive,$f'(x) \geq 1$ for all real $x$. Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$,which means $f(x)$ is a strictly increasing function. $A$ strictly increasing function can intersect the $x$-axis at most once. Therefore,$f(x) = 0$ has not more than one real root.

Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$. Then,$f(x)=0$ has

Similar Questions

For a real number $a$,if a real-valued function $f(x) = 4x^3 + ax^2 + 3x - 2$ is monotonic in its domain,then the range of $a$ is

The function $f(x) = x^x$ for $x > 0$ is strictly increasing at

If $f(x) = \frac{1}{x + 1} - \log(1 + x)$,where $x > 0$,then $f$ is:

Show that the function given by $f(x) = \sin x$ is strictly increasing in the interval $\left(0, \frac{\pi}{2}\right)$.

Find the interval in which the function $f(x) = \log x - \frac{2x}{x+2}$ is strictly increasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module