question_answer

How many real solutions does the equation \[{{x}^{7}}+14{{x}^{5}}+16{{x}^{3}}+30x-560=0\] have?       AIEEE  Solved  Paper-2007

A) 3                             

B)        5                             

C)        7                             

D)        1

Correct Answer: D

Solution :

                \[f\left( x \right)={{x}^{7}}+14{{x}^{5}}+16{{x}^{3}}+30x-560\] \[f''\left( x \right)=7{{x}^{6}}+70{{x}^{4}}+48{{x}^{2}}+30>0\] \[\therefore \] \[f\] is increasing also \[\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\infty \,;\,\underset{x\to -\infty }{\mathop{\lim }}\,f\left( x \right)=-\infty \]                                 Clearly \[f\left( x \right)=0\] have exactly one real root.