prove that f(x)=\frac{ax^{2}+x-2}{a+x-2x^{2}} has the range R when x belongs to R if a\epsilon [1,3]
-
UP 0 DOWN 0 0 1
1 Answers
11letf(x) = ax2 + x -2 a + x - 2 x2 = y.
thus, ax2+ x - 2 =ay +xy -2x2y
or, (a+2y)x2 +x(1-y) - (2+ay) = 0.....................1
as x is real, discriminant should be non negetive..
b2 - 4ac ≥ 0.
(1-y)2 + 4 (a+2y)(2+ay) ≥ 0
thus,we get,
y2(1+8a) +y(14+4a2) + (1+8a) ≥ 0........ 2
as expression 2 is positive..
its discriminant should be -ve
(14-4a2)2 -4(1+8a)2 ≤ 0
(7-2a2)2 - (1+8a)2 ≤0
(a2-4a+3)(a2+4a+a) ≤ 0
(a-1)(a-3) (a+2)2 ≤ 0
(a-1)(a-3) ≤ 0
thus, for the range of f(x) to b R a belongs to [1,3]
hence proved.. :)