If f: R to R is a function, defined by f(x)=2^x; then f(x+y) is

Option b: f(x).f(y). Solution: Given the function: f(x) = 2^x We want to find the value of f(x+y): f(x+y) = 2^{x+y} Using the exponential algebraic identity a^{m+n} = a^m a^n, we can write: 2^{x+y} = 2^x 2^y Since f(x) = 2^x and f(y) = 2^y, we substitute these terms back in: 2^x 2^y = f(x) f(y) Therefore: f(x+y) = f(x) f(y) Hence, Option B is the correct answer.

If $f: R \to R$ is a function, defined by $f(x)=2^x$; then $f(x+y)$ is

The correct answer is Option b: $f(x).f(y)$. Given the function: $$f(x) = 2^x$$ We want to find the value of $f(x+y)$: $$f(x+y) = 2^{x+y}$$ Using the exponential algebraic identity $a^{m+n} = a^m \cdot a^n$, we can write: $$2^{x+y} = 2^x \cdot 2^y$$ Since $f(x) = 2^x$ and $f(y) = 2^y$, we substitute these terms back in: $$2^x \cdot 2^y = f(x) \cdot f(y)$$ Therefore: $$f(x+y) = f(x) \cdot f(y)$$ Hence, **Option B** is the correct answer.

Set, Relations and FunctionsMTP March 22Question 1991 of 217

All Questions

If $\displaystyle f: R \to R$ is a function, defined by $\displaystyle f(x)=2^x$ ; then $\displaystyle f(x+y)$ is

Options

\displaystyle f(x)+f(y)

\displaystyle f(x).f(y)

\displaystyle f(x) \div f(y)

Dnone

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — $\displaystyle f(x).f(y)$

All Options:

A $\displaystyle f(x)+f(y)$
B $\displaystyle f(x).f(y)$
C $\displaystyle f(x) \div f(y)$
Dnone

Detailed Solution & Explanation

Given the function:

f(x) = 2^x

We want to find the value of

\displaystyle f(x+y)

f(x+y) = 2^{x+y}

Using the exponential algebraic identity

\displaystyle a^{m+n} = a^m \cdot a^n

, we can write:

2^{x+y} = 2^x \cdot 2^y

Since

\displaystyle f(x) = 2^x

and

\displaystyle f(y) = 2^y

, we substitute these terms back in:

2^x \cdot 2^y = f(x) \cdot f(y)

Therefore:

f(x+y) = f(x) \cdot f(y)

Hence, **Option B** is the correct answer.

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If f:R→R\displaystyle f: R \to Rf:R→R is a function, defined by f(x)=2x\displaystyle f(x)=2^xf(x)=2x; then f(x+y)\displaystyle f(x+y)f(x+y) is