Question

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)​

Answer 1

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.