Discussion Forum



1)

Using  Rolle's theorm , the equation 

$a_{0}x^{n}+a_{1}x^{n-1}+.........+a_{n}=0$

 has atleast one root betwwen 0 and 1, if


A) $\frac{a_{0}}{n}+ \frac{a_{1}}{n-1}+.....+a_{n-1}=0$

B) $\frac{a_{0}}{n-1}+ \frac{a_{1}}{n-2}+.....+a_{n-2}=0$

C) $na_{0}+(n-1) a_{1}+.......+a_{n-1}=0$

D) $\frac{a_{0}}{n+1}+ \frac{a_{1}}{n}+.....+a_{n}=0$

Answer:

Option D

Explanation:

 Consider the function  f defined  by 

$f(x)=a_{0}\frac{x^{n+1}}{n+1}+a_{n}\frac{x^{n}}{n}+....+a_{n-1}\frac{x^{2}}{2}+a_{n} x$

 since, f(x) ia a polynomial , so it is continuous and differentiable  for all x. f(x)  is continuous in the closed interval [0,1] and differentiable in the open interval (0,1).

 Also, f(0)=0

 and

$f(1)=\frac{a_{0}}{n+1}+\frac{a_{1}}{n}+....+\frac{a_{n-1}}{2}+a_{n} =0$ [say]

i.e, f(0)= f(1)

Thus , all the three conditions of Rolle's theorm are satisfied . Hence , there is atleast one value of x in the open interval  (0,1)

 where f '(x)=0

 i.e,  $a_{0}x^{n}+a_{1}x^{n-1}+a_{n}=0$