Let ϕ (x) =\(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\) , where x �

1.	Let ϕ (x) =\(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\) , where x ∈ R and a and b are fixed numbers with a ≠ b. Then ϕ (a + b) is equal to(a) ϕ (0) (b) ϕ (a – b) (c) ϕ (ab) (d) ϕ (a) + ϕ (b)
Answer» Answer : (d) = ϕ (a) + ϕ (b) ϕ (x) = \(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\) Then, ϕ (a + b) = \(\frac{b(a+b-a)}{b-a} + \frac{a(a+b-b)}{a-b}\) = \(\frac{b^2}{b-a} +\frac{a^2}{a-b} = \frac{b^2-a^2}{b-a} =b+a\) ...(i) Now ϕ (a) = \(\frac{b\times 0}{b-a}\) + \(\frac{a(a-b)}{a-b}\) = a ...(ii) ϕ (b) = \(\frac{b(b-a)}{b-a}\) + \(\frac{a\times 0}{a-b}\) ...(iii) (i), (ii) and (iii) ⇒ ϕ a + b) = ϕ(a) + ϕ(b).

Let ϕ (x) =\(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\) , where x ∈ R and a and b are fixed numbers with a ≠ b. Then ϕ (a + b) is equal to(a) ϕ (0) (b) ϕ (a – b) (c) ϕ (ab) (d) ϕ (a) + ϕ (b)

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Answer : (d) = ϕ (a) + ϕ (b)

ϕ (x) = \(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\)

Then, ϕ (a + b) = \(\frac{b(a+b-a)}{b-a} + \frac{a(a+b-b)}{a-b}\)

= \(\frac{b^2}{b-a} +\frac{a^2}{a-b} = \frac{b^2-a^2}{b-a} =b+a\) ...(i)

Now ϕ (a) = \(\frac{b\times 0}{b-a}\) + \(\frac{a(a-b)}{a-b}\) = a ...(ii)

ϕ (b) = \(\frac{b(b-a)}{b-a}\) + \(\frac{a\times 0}{a-b}\) ...(iii)

(i), (ii) and (iii) ⇒ ϕ a + b) = ϕ(a) + ϕ(b).