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```        find the no. of solution of
-x^2+x-1 = x^2008 + |x|```
8 years ago

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.
```
8 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
```
7 years ago
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