Mathematics - Functions Question with Solution | TestHub

The given functional equation is $f(x+y)=f(x)+f(y)+f(x)f(y)$. Adding 1 to both sides, we get $1+f(x+y)=1+f(x)+f(y)+f(x)f(y)=(1+f(x))(1+f(y))$. Let $g(x)=1+f(x)$. Then the equation becomes $g(x+y)=g(x)g(y)$.

Since $f(x)$ is differentiable, $g(x)$ is also differentiable. The only differentiable solutions to $g(x+y)=g(x)g(y)$ are of the form $g(x)=e^{kx}$ for some constant $k$. So $1+f(x)=e^{kx}$, which means $f(x)=e^{kx}-1$. Given $f(1)=2$, we have $e^k-1=2⟹e^k=3$. Thus, $k=\ln 3$. Therefore, the function is $f(x)=3^x-1$. Option A: $f^{\prime }(x)=3^x\ln 3$. Since $3^x>0$ and $\ln 3>0$, $f^{\prime }(x)>0$ for all $x\in \mathbb{R}$. Hence, $f(x)$ is strictly increasing and therefore injective. So, A is correct.

Option B: The range of $f(x)=3^x-1$. As $x\to -\infty$, $3^x\to 0$, so $f(x)\to -1$. As $x\to \infty$, $3^x\to \infty$, so $f(x)\to \infty$. The range is $(-1,\infty )$. Since the codomain is $\mathbb{R}$, $f(x)$ is not surjective. So, B is incorrect. Option C: From $f(x)=3^x-1$, we have $f(0)=3^0-1=1-1=0$. So, C is correct.

Option D: Since $f(x)$ is injective (from A), its inverse $f^{-1}(x)$ exists. The domain of $f^{-1}(x)$ is the range of $f(x)$. From B, the range of $f(x)$ is $(-1,\infty )$. Therefore, the domain of $f^{-1}(x)$ is $(-1,\infty )$, not $\mathbb{R}$. So, D is incorrect.