1k=1.
let the roots be p, p+1.
b^2-4ac=a^2
f(x)=[x] ,x≥0
f(x)=K+[x] ,x<0,
{[.] is the greatest integer function......}
is continuous at x=0, and one root of ax^{2}+bx+c=0 exceeds the other by K,....then b^2-4ac =?
a) a b)b c)a^2 d)b^2
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4 Answers
106as f(x) is continous at x=0
at x=0- .. f(x) = K + [0-] = K-1
at x=0+... f(x) = [0+] = 0
as continous... K-1=0 ==> K=1
let one root be m and other be m+1
now, x = (-b+√b2-4ac)/2a and (-b-√b2-4ac)/2a = p and p+1
subtracting both..
√b2-4ac/a = (p+1) - p = 1
So, √b2-4ac = a
