111) It is given that a<b<c<d,
Let f(x) = (x-a)(x-c) + 2(x-b)(x-d)
Then f(a) = 0 + 2 (a-b) ( a-d) > 0 {bcoz a-b < 0, a-d < 0}
and f(b) = (b-a)(b-c) + 0 < 0 {bcoz b-a>0, b-c<0}
Thus,, if f(x) = be a polynomial eqn and a and b are two real numbers, then f(x) = 0 will have atleast one real rootb/w and b if f(a) and f(b) are of opp signs.
And, f(c) = 0 +2(c-b) (c-d) < 0 { bocz c-b > 0 , c-d < 0}
and f(d) = (d-a) (d-c) + 0 > 0 { bcoz d-a > 0, d-c > 0}
Thus one root will lie b/w c and d. Hence the roots of eqn are real and disinct.
13) f(x)>0=>that parabola is above x-axis n openig up....
so a>0
for g(x)>0....its discriminant<0(vertex of parabola shud b +ve)
discriminant is 5b2-12ac<0......threfor c has to be +ve
ans a.
11
Tush Watts
·2009-08-29 01:47:37
2) Ans is d
Since α is a root of a2x2 + bx +c =0
then a2 α2 + b α +c = 0 ...........................(1)
and β is a root of a2x2 - bx-c=0
then a2β 2 - bβ -c =0 ..............................................(2)
Let f(x) = a2x2+2bx+2c
f( α) = a2 α2 + 2b α +2c
= a2 α2 - 2a2 α2 [from(1)]
= - a2 α2
That implies, f( α) < 0
and f(β) = a2β2 + 2bβ + 2c
= a2β2 + 2a2β2 [ from (2)]
= 3a2β2
That implies f(β) > 0
Since f(α) and f(β) are of opp signs, then it is clear that a root γ of the eqn f(x) = 0 lies b/w α and β.
4@Archana
Thanks for the solution but ans for the prob u've amswered is c i.e. b<0
1
archana anand
·2009-08-29 10:21:36
ans should be (a) c>0.....coz dats the necessary condition
plz check the discriminant of equation......donno wer m i makin mistake.[7]
1
archana anand
·2009-08-29 12:43:25
okay lets see
g(x)=a(ax2+bx+c)+b(2xa+b)+c(2a)
g(x)=a2x2+3abx+3ac+b2
discriminant=(5a2b2-12a3c)=a2(5b2-12ac)wich is <0
obviously a2cannot be<0 and for 5b2-12ac to be -ve c
has to be >0..........now u temme wer m i makin mistake??
4
UTTARA
·2009-08-30 22:42:04
Ur working is right Archana May be there is some mistake in the ans
