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no. of real roots of equation is equal to x 8 -x 5 +x 2 -x+1=0 a) 0 b) 2 c) 4 d) 6
Let f(x) = x8-x5+x2-x+1 f(0) = 1 = f(1) Its obvious that for x<0, f(x)>0 For x>1 we have x8>x5 and x2>x so that f(x)>1 Again for 0<x<1, we write f(x) = x8 +(x2-x5) + (1-x) and we know that in this domain we have (x2>x5) and (1>x). So again f(x)>0 Thus f(x)>0 for all real x and hence this polynomial has no real roots
Let f(x) = x8-x5+x2-x+1
f(0) = 1 = f(1)
Its obvious that for x<0, f(x)>0
For x>1 we have x8>x5 and x2>x so that f(x)>1
Again for 0<x<1, we write f(x) = x8 +(x2-x5) + (1-x) and we know that in this domain we have (x2>x5) and (1>x).
So again f(x)>0
Thus f(x)>0 for all real x and hence this polynomial has no real roots
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