Step 1: Analyze Question 1. The question asks to identify the graph of a polynomial whose both zeros are positive. Zeros of a polynomial are the x-intercepts of its graph. For both zeros to be positive, the graph must intersect the positive x-axis at two distinct points. Graph (a) shows one zero at $x=0$ and another positive zero. This does not represent two positive zeros. Graph (b) is a circle, not a polynomial graph. Graph (c) shows a parabola intersecting the x-axis at two points, both of which are on the positive side of the x-axis. This means both zeros are positive. Graph (d) shows one negative zero and one positive zero. Therefore, graph (c) represents a polynomial with two positive zeros. The final answer is $\boxed{\text{c}}$. Step 2: Analyze Question 2. The question asks for the nature of the solution to the system of equations $x=2$ and $x=3$. For a system of equations to have a solution, there must be a value (or set of values) that satisfies all equations simultaneously. Here, we have two conditions for $x$: 1. $x=2$ 2. $x=3$ It is impossible for $x$ to be equal to 2 and 3 at the same time. Therefore, there is no value of $x$ that can satisfy both equations. The final answer is $\boxed{\text{c}}$. Step 3: Analyze Question 3. The question states that the numbers $1-a$, $3-a$, and $5-a$ are in an arithmetic progression (AP) and asks for the common difference. In an arithmetic progression, the common difference ($d$) is the difference between any term and its preceding term. Let the terms be $t_1 = 1-a$, $t_2 = 3-a$, and $t_3 = 5-a$. Calculate the common difference using the first two terms: $$d = t_2 - t_1$$ $$d = (3-a) - (1-a)$$ $$d = 3 - a - 1 + a$$ $$d = 2$$ Verify the common difference using the second and third terms: $$d = t_3 - t_2$$ $$d = (5-a) - (3-a)$$ $$d = 5 - a - 3 + a$$ $$d = 2$$ Both calculations yield the same common difference, which is 2. The final answer is $\boxed{\text{c}}$.