Question:
Let $f(x) = x^{13} + x^{11} + x^9 + x^7 + x^5 + x^3 + x + 19$. Then $f(x) = 0$ has
Let $f(x) = x^{13} + x^{11} + x^9 + x^7 + x^5 + x^3 + x + 19$. Then $f(x) = 0$ has
Updated On: Feb 15, 2024
- 13 real roots
- only one positive and only two negative real roots
- not more than one real root
- has two positive and one negative real root
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The Correct Option is C
Solution and Explanation
We have,
$f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19 $
$\Rightarrow f^{\prime}(x)=13 x^{12}+11 x^{10}+9 x^{8}$
$+7 x^{6}+5 x^{4}+3 x^{2}+1$
$\therefore f^{\prime}(x)$ has no real root.
$\therefore f(x)=0$ has not more than one real root.
$f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19 $
$\Rightarrow f^{\prime}(x)=13 x^{12}+11 x^{10}+9 x^{8}$
$+7 x^{6}+5 x^{4}+3 x^{2}+1$
$\therefore f^{\prime}(x)$ has no real root.
$\therefore f(x)=0$ has not more than one real root.
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
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If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
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