If `alpha, beta` are roots of the equation`x^(2)-p(x+1)-c=0` show that `(alpha+1)(beta+1)=1-c` Hence prove that `(alpha^(2)+2alpha+1)/(alpha^(2)+2alpha+c)+(beta^(2)+2beta+1)/(beta^(2)+2beta+c)=1`

Options

A. 1
B. 2
C. 3
D. 0

Answer

A

Explanation

The given equation is `x^(2) - px - (p+q) = 0`
`therefore" "alpha+beta = p and alpha beta = -(p+q)`
`rArr" "alpha beta + alpha + beta = - q`
`rArr" "alpha beta + alpha + beta + 1 = - q+1`
`rArr" "(alpha + 1)(beta + 1) = 1 - q" "....(i)`
Now,
`(alpha^(2)+2 alpha + 1)/(alpha^(2)+2 alpha + q)+(beta^(2)+2 beta+1)/(beta^(2)+2 beta + q)`
`=((alpha+1)^(2))/((alpha+1)^(2)+q-1)+((beta+1)^(2))/((beta+1)^(2)+(q-1))`
`=((alpha+1)^(2))/((alpha+1)^(2)-(alpha+1)(beta+1))+((beta+1)^(2))/((beta+1)^(2)-(alpha+1)(beta+1))" "["using (i)"]`
`(alpha+1)/(alpha+beta)+(beta+1)/(beta-alpha)=(alph-beta)/(alpha-beta) = 1`.