Calculus - solv.this one on differentiation

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k=1.
let the roots be p, p+1.
b^2-4ac=a^2

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For the function to be continuous, k=1.

So, b² -4ac = a²

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Asish Mahapatra ·2009-03-19 00:07:26

as 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

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thanxxxxxxxxx.......

solv.this one on differentiation