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If the roots of the equation (a-b)x2 + (b-c) x+ (c - a)= 0 are equal. Prove that 2a=b+c.
Ans: (a-b)x2 + (b-c) x+ (c - a) = 0
T.P 2a = b + c
B2 - 4AC = 0
(b-c)2 - [4(a-b) (c - a)] = 0
b2-2bc + c2 - [4(ac-a2 - bc + ab)] = 0
⇒ b2-2bc + c2 - 4ac + 4a2 + 4bc - 4ab = 0
⇒ b2+ 2bc + c2 + 4a2 - 4ac - 4ab= 0
⇒ (b + c - 2a)2 = 0
⇒ b + c = 2a
f(x)+f(x+1/2) =1 f(x)=1-f(x+1/2) 0∫2f(x)dx=0∫21-f(x+1/2)dx 0∫2f(x)dx=2-0∫2f(x+1/2)dx take (x+1/2)=v dx=dv 0∫2f(v)dv=2-0∫2f(v)dv 2(0∫2f(v)dv)=2 0∫2f(v)dv=1 0∫2f(x)dx=1
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