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Х- а
x-b
If f(x) = b.x-a÷b-a + a.x-b÷a - b
Prove that: f (a) + f(b) = f (a + b)​

User Mity
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Given:

Consider the given function:


f(x)=(b\cdot(x-a))/(b-a)+(a\cdot(x-b))/(a-b)

To prove:


f(a)+f(b)=f(a+b)

Solution:

We have,


f(x)=(b\cdot(x-a))/(b-a)+(a\cdot (x-b))/(a-b)

Substituting
x=a, we get


f(a)=(b\cdot(a-a))/(b-a)+(a\cdot (a-b))/(a-b)


f(a)=(b\cdot 0)/(b-a)+(a)/(1)


f(a)=0+a


f(a)=a

Substituting
x=b, we get


f(b)=(b\cdot(b-a))/(b-a)+(a\cdot (b-b))/(a-b)


f(b)=(b)/(1)+(a\cdot 0)/(a-b)


f(b)=b+0


f(b)=b

Substituting
x=a+b, we get


f(a+b)=(b\cdot(a+b-a))/(b-a)+(a\cdot (a+b-b))/(a-b)


f(a+b)=(b\cdot (b))/(b-a)+(a\cdot (a))/(-(b-a))


f(a+b)=(b^2)/(b-a)-(a^2)/(b-a)


f(a+b)=(b^2-a^2)/(b-a)

Using the algebraic formula, we get


f(a+b)=((b-a)(b+a))/(b-a)
[\because b^2-a^2=(b-a)(b+a)]


f(a+b)=b+a


f(a+b)=a+b [Commutative property of addition]

Now,


LHS=f(a)+f(b)


LHS=a+b


LHS=f(a+b)


LHS=RHS

Hence proved.

User Hagne
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