IIT JEE , AIEEE Forum - Mathematics , Algebra

awesomegupta

f(x)=

. The range of this function is[-5,4], a, b N, then find the value of ()

.............................. DOUBT: Here , I could not understand wether the value of x will be fixed or will it b changing. with the values of result.... like different value of x when the result is -5 and a different value when the result is 4. If it is fixed then the values of a and b will change then how can we have a fixed result for ..........

ankurgupta91

(x^2 + ax+b) / (x^2+2x+3) = y

x^2(1-y) + x(a-2y) + b -3y =0
as x is real
D>=0
(a-2y)^2 - 4(1-y)(b-3y) >=0
-8y^2 + y(4b-4a+12) +a^2 - 4b >=0
8y^2 + y (4a-4b-12) + 4b-a^2<=0 ----------(1)

nw as range is [-5,4]
so , (y+5)(y-4) <=0
8y^2 +8y -160 <=0
comparing it with (1)

4a-4b-12 = 8 ------(2) and 4b-a^2 = -160
add these two
4a - a^2 = -140
a^2 - 4a -140=0
(a-14)(a+10) = 0
a= 14 or a= -10

nw by putting it in (2)
b=9 or b = -15

so a^2+b^2 = 277 or a^2+b^2 = 325
thats the answer..........

puneet

hii

i think it has been solved well .. this is the standard way of solving these things .. always take it as y and thn do the way he has done for u ..

regarding ur doubt .. x ofcourse changes ..

cheers

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