1. The roots of the quad. eqn. :       

    (a+b-2c)- (2a-b-c)x +(a-2b+c)=0

a)a+b+c and a-b+c                  b)1/2 and a-2b+c                c)a-2b+c and             d)none      

let A= a+b-2c , -B= 2a-b-c , C= a-2b+c

we have A+B +C= = 0

So -B = A +C

so discriminant = (A+C)^2-4AC =(A-C)^2 = (3(b-c))^2

so roots are x= (2a-b-c)+-(3(b-c))/(2(a+b-2c))

on calculation it becomes 1 and (a-2b+c)/(a+b-2c)

so I think it is (d)

But plz check the calculation the trick remaining the same .

Most straighforward method:

user Pawan JE has given a much more straighforward solution in another thread. worth looking up

Q1- If    are the roots of the eqn.,   then the range of values of m for which is:

a)  (-1,3)        b)   (1,3)           c)           d)  None

Q2- If a,b,p,q are non zero real nos.,the two eqns. ,   and have:

a) no common root                                      b) One common root if                  

c) two common roots if  3pq=2ab             d) two common roots if 3qb=2ap

Q3- If coefficients of the eqn.   ae real and roots of the eqn. are non-real complex and a+c<b then,

a)4a+c>2b       b)4a+c<2b          c)4a+c=2b              d)none

Q4- If a,b are non-zero real nos. and the roots of  then prove    are the roots of

for d first one guess ur q is wrng i mean u hav mistyped it it shud b ......

x^2-2mx+m^2-1=0 ................rite ?? aapne +m^2 ki jagah -m^2 likha hai

use .......................a*f(k)<0 ......for roots to lie in bewtn (-2,4)

thus a=1 .............f(-2)=m^2-4m+m^2-1

solvin for m u get d anser as a )

hope i m rite ..............

 (a+b-2c)- (2a-b-c)x +(a-2b+c)=0

by inspection we see that x=1 satisfies the eqn

so one root=1

so the other root =(2a-b-c)/(a+b-2c)  -1 ........[since the sum of the roots =-B/A

or this is same as (a-2b+c)/(a+b-2c).....[ sinc ethe prod. of the roots = C/A]