question_answer

If the equation \[{{a}_{n}}{{X}^{n}}+{{a}_{n-1}}{{X}^{n-1}}+....+{{a}_{1}}x=0,\] \[{{a}_{1}}\ne 0,n\ge 2,\]has a positive root \[x=\alpha ,\] then the equation \[n{{a}_{n}}{{x}^{n-1}}+(n-1){{a}_{n-1}}{{X}^{n-2}}+....+{{a}_{1}}=0\] has a positive root, which is     AIEEE  Solved  Paper-2005

A) equal to\[\alpha \]

B) greater than or equal to\[\alpha \]

C) smaller than\[\alpha \]

D) greater than \[\alpha \]

Correct Answer: C

Solution :

Let\[f(x)={{a}_{n}}\,{{x}^{n}}+{{a}_{n-1}}\,{{x}^{n-1}}+...+{{a}_{1}}\,x=0,\] \[{{a}_{1}}\ne 0\] It have rootsand So, definitely its derivative is zero between 0 and. According to the Rolle's theorem, has a positive root smaller than.