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  • If <img alt="\frac{a_{0}}{n+1}+\frac{a_{1}}{n}+\frac{a_{2}}{n-1}+....+\frac{a_{n-1}}{2}+a_{n}=0" src="https://learn.careers360.com/latex-image/?%5Cfrac%7Ba_%7B0%7D%7D%7Bn&plus;1%7D&plus;%5C

If \frac{a_{0}}{n+1}+\frac{a_{1}}{n}+\frac{a_{2}}{n-1}+....+\frac{a_{n-1}}{2}+a_{n}=0, then the equation a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+......+a_{n-1}x+a_{n}=0 has

Option: 1

exactly one root in (0,1)
 


Option: 2

at least one root in (0,1)


Option: 3

no root in (0,1)

 


Option: 4

at the most one root in (0,1)


Answers (1)

best_answer

 

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1.  f(x) is continuous function of    x:x\epsilon [a,b]

2.  f'(x) is exists for every point :  x\epsilon [a,b]

3.  f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.

-

 

 

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

Then f(0)=0 and f(1)=0

hence f'(x)=0 has at least one solution in (0,1)

Posted by

Sanket Gandhi

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