JEE MAIN - Mathematics (2008 - No. 11)
How many real solutions does the equation
$${x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0$$ have?
$${x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0$$ have?
$$7$$
$$1$$
$$3$$
$$5$$
Explanation
Let $$f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560$$
$$ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R$$
$$ \Rightarrow f$$ is an increasing function on $$R$$
Also $$\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty $$ and $$\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty $$
$$ \Rightarrow $$ The curve $$y = f\left( x \right)$$ crosses $$x$$-axis only once.
$$\therefore$$ $$f\left( x \right) = 0$$ has exactly one real root.
$$ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R$$
$$ \Rightarrow f$$ is an increasing function on $$R$$
Also $$\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty $$ and $$\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty $$
$$ \Rightarrow $$ The curve $$y = f\left( x \right)$$ crosses $$x$$-axis only once.
$$\therefore$$ $$f\left( x \right) = 0$$ has exactly one real root.
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