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Kisachek [45]
2 years ago
11

Х- аx-bIf f(x) = b.x-a÷b-a + a.x-b÷a - bProve that: f (a) + f(b) = f (a + b)​

Mathematics
1 answer:
GenaCL600 [577]2 years ago
5 0

Given:

Consider the given function:

f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}

To prove:

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

Solution:

We have,

f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}

Substituting x=a, we get

f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}

f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}

f(a)=0+a

f(a)=a

Substituting x=b, we get

f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}

f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}

f(b)=b+0

f(b)=b

Substituting x=a+b, we get

f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}

f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}

f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}

f(a+b)=\dfrac{b^2-a^2}{b-a}

Using the algebraic formula, we get

f(a+b)=\dfrac{(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.

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