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Value of Combined Expression of Additive and Multiplicative Functional Equations

Let f:RRf : \mathbb{R} \rightarrow \mathbb{R} be a function such that f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) for all x,yRx, y \in \mathbb{R}, and g:R(0,)g : \mathbb{R} \rightarrow (0, \infty) be a function such that g(x+y)=g(x)g(y)g(x + y) = g(x) g(y) for all x,yRx, y \in \mathbb{R}. If f(35)=12f\left(\frac{-3}{5}\right) = 12 and g(13)=2g\left(\frac{-1}{3}\right) = 2, then the value of (f(14)+g(2)8)g(0)\left( f\left(\frac{1}{4}\right) + g(-2) - 8 \right) g(0) is ________.

Official Numerical Answer51

Step-by-Step Solution

To find the value of the given expression, we analyze the properties of the functional equations for f(x)f(x) and g(x)g(x).

Step 1: Evaluation of f(14)f\left(\frac{1}{4}\right)

The function f:RRf: \mathbb{R} \rightarrow \mathbb{R} satisfies Cauchy's additive functional equation: f(x+y)=f(x)+f(y)for all x,yRf(x + y) = f(x) + f(y) \quad \text{for all } x, y \in \mathbb{R}

By mathematical induction, for any rational number rQr \in \mathbb{Q} and any xRx \in \mathbb{R}, we have: f(rx)=rf(x)f(r x) = r f(x)

We are given f(35)=12f\left(-\frac{3}{5}\right) = 12. Expressing 14\frac{1}{4} in terms of 35-\frac{3}{5}, we have: 14=(512)×(35)\frac{1}{4} = \left(-\frac{5}{12}\right) \times \left(-\frac{3}{5}\right)

Since 512-\frac{5}{12} is a rational number, we apply the property: f(14)=f(512(35))=512f(35)f\left(\frac{1}{4}\right) = f\left(-\frac{5}{12} \cdot \left(-\frac{3}{5}\right)\right) = -\frac{5}{12} f\left(-\frac{3}{5}\right)

Substitute f(35)=12f\left(-\frac{3}{5}\right) = 12: f(14)=512×12=5f\left(\frac{1}{4}\right) = -\frac{5}{12} \times 12 = -5


Step 2: Evaluation of g(0)g(0) and g(2)g(-2)

The function g:R(0,)g: \mathbb{R} \rightarrow (0, \infty) satisfies the exponential functional equation: g(x+y)=g(x)g(y)for all x,yRg(x + y) = g(x) g(y) \quad \text{for all } x, y \in \mathbb{R}

  1. Finding g(0)g(0): Setting x=y=0x = y = 0: g(0)=g(0+0)=g(0)g(0)=(g(0))2g(0) = g(0 + 0) = g(0) g(0) = (g(0))^2 Since the codomain of gg is (0,)(0, \infty), g(0)0g(0) \neq 0. Dividing both sides by g(0)g(0), we get: g(0)=1g(0) = 1

  2. Finding g(2)g(-2): For any rational number rQr \in \mathbb{Q} and any xRx \in \mathbb{R}, the relation g(rx)=(g(x))rg(r x) = (g(x))^r holds.

    Expressing 2-2 in terms of 13-\frac{1}{3}, we have: 2=6×(13)-2 = 6 \times \left(-\frac{1}{3}\right)

    Therefore: g(2)=g(6(13))=(g(13))6g(-2) = g\left(6 \cdot \left(-\frac{1}{3}\right)\right) = \left(g\left(-\frac{1}{3}\right)\right)^6

    Given g(13)=2g\left(-\frac{1}{3}\right) = 2: g(2)=26=64g(-2) = 2^6 = 64


Step 3: Calculating the Final Expression

We substitute f(14)=5f\left(\frac{1}{4}\right) = -5, g(2)=64g(-2) = 64, and g(0)=1g(0) = 1 into the target expression: (f(14)+g(2)8)g(0)=(5+648)×1\left( f\left(\frac{1}{4}\right) + g(-2) - 8 \right) g(0) = (-5 + 64 - 8) \times 1

=(598)×1=51= (59 - 8) \times 1 = 51

Thus, the value of the expression is 51.

Value of Combined Expression of Additive and Multiplicative Functional Equations | Mathematics PYQ Solution - JEE Challenger