Question

let f and g be differentiable functions with the following properties (i) g(x)>0 for all x (ii) f(0)=1 if h(x)=f(x)g(x) and h(x)=f(x)g(x), then f(x)= : a f(x) b g(x) c e^x d 0 e 1

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. But we are given that $h^{\prime}(x)=f(x)g^{\prime}(x)$. So, $f^{\prime}(x)g(x)+f(x)g^{\prime}(x)=f(x)g^{\prime}(x)$.

Step2: Simplify the equation

Subtract $f(x)g^{\prime}(x)$ from both sides of the equation $f^{\prime}(x)g(x)+f(x)g^{\prime}(x)=f(x)g^{\prime}(x)$. We get $f^{\prime}(x)g(x)=0$. Since $g(x)>0$ for all $x$, then $f^{\prime}(x) = 0$ for all $x$.

Step3: Find the function $f(x)$

If $f^{\prime}(x)=0$ for all $x$, then $f(x)$ is a constant function. We know that $f(0)=1$. So, $f(x)=1$ for all $x$.

Answer:

E. 1