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Rahul kumar jaiswal Grade: 12
        find the no. of solution of

-x^2+x-1 = x^2008 + |x|
7 years ago

Answers : (2)

Pratham Ashish
17 Points
										

hi rahul,


dis ques can be done by d approach of graph,,,


for LHS,


f = -x^2+x-1       ,f < 0 for all x <0,n f<0 for x>1, f >0 for 0<x>1........


 


since dy/dx < 0, n dis is a monotonus fxn so it decreases from +infinite to o at x=1,,,,,,,,,,


for rhs, f = x^2008 + |x|,   dis is a increasing fxn,,,,,n for all x f>0....


at x=1 , f=2,,at x =0,f=0,,,


so d two fxn can only intersect at one point in b/w 0<x>1...


so no. of soln = 1.

7 years ago
Rajesh Juluru
13 Points
										

for x>0,


-x^2-1=x^2008


x^2008+x^2+1=0


So no positive roots (since all are positive terms)


for x<0,


-x^2+2x-1=x^2008


x^2008+(x-1)^2=0


so no negative roots (since all are positive terms)


So i think number of solutions is zero

6 years ago
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