11sir, in the first solution that u gave ,after solving
f(t)=g(t)
we are getting (a2 +a-1)t-2a2 -a+1=0
u are getting another one........
f(x)=x2 -2x-a2 +1
g(x)=x2 -(a2 +a+1)x+a2 +a
Q1. The number of values of a for which f(x)=0 and g(x)=0 have exactly one common root.
Q2. The range of a for which the bigger root of g(x)=0 is greater than the bigger root of
f(x)=0.
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8 Answers
11
62sorry my wrong..
i will fix this one..
right now am doing the other question you just gave .. :)
62first one has roots
a-1 and -a-1
2nd equation g()
x2 -(a2 +a+1)x+a2 +a
=x2 -2x +1 -(a2 +a-1)x+a2 +a - 1
= (x-1)2 + (a2 +a - 1) (1-x)
take both the cases, x=a-1 and x=-1-a
62first one has roots
a-1 and -a-1 (This is from observation...)
2nd equation g()
x2 -(a2 +a+1)x+a2 +a
=x2 -2x +1 -(a2 +a-1)x+a2 +a - 1
= (x-1)2 + (a2 +a - 1) (1-x)
since we want both to have common roots,
therefore, either a-1 or -a-1 should be the root of g..
so we substitute x=a-1 in the 2nd equation. it should give zero..
is it clear now? Could you tell me which is the first step you were not able to understand.
341The roots of f(x) are 1-a and 1+a
The roots of g(x) are 1 and a+a2
Now if 1 is the common root, 1 equals either 1+a or 1-a, and so a = 0
If a+a2 equals 1+a, then a = ±1. We can verify that exactly one root is common satisfies the condition
If a+a2 equals 1-a, then similarly we have a = 1 \pm \sqrt 2