Let p be the number of real solutions of the equation 3x+x-3=0 in the interval [0; 1] , and let q be the number of real solutions that are not in [0; 1].

Question

Let p be the number of real solutions of the equation 3^x+x-3=0 in the interval [0; 1] , and let q be the number of real solutions that are not in [0; 1]. Which of the following is true

a) p =0 and q =1;

b) p = 1 and q =1;

c) p =1 and q = 0 ;

d) p = 1 and q >1

Answer 1

Hello Irakli,

Let f(x) = 3^x + x - 3. This function is continuous on the given interval [0,1]. Also note that f(0) = -2 and f(1)= 1. Since f(x) is negative at x=0, and positive at x = 1, there must exist a real number (let us call it c) in the open interval (0,1) such that f(c) = 0. This follows from the Intermediate Value Theorem. Note that saying f(c)=0 (i.e, c is a zero of f(x)) is the same thing as saying that c is a solution of the equation

3^x + x - 3 = 0.

So this equation has at least one real solution on the interval [0,1]. To determine if the equation has any more real solutions (that is, if f(x) has any more real zeros), we need to study the behavior of this function further. We will use the derivative to do this. The derivative of f is given by

f ' (x) = (Ln(3))3^x + 1.

It is easy to see that f ' (x) exists everywhere, and also, that f '(x) = (Ln(3)3^x + 1 = 0 has no real solutions. Thus, f(x) has no critical points. It is easy to check that f '(x) > 0 for all x, so, in fact, f(x) is increasing on the entire real line (-∞,∞). This implies that f(x) can have at most one zero, and we have already proven that f(x) has the zero in the interval [0,1]. Therefore the equation 3^x + x - 3 = 0 has exactly one real solution, and that solution is in the interval [0,1]. Therefore, choice C (p=1 and q = 0) is correct. Let me know if you need any clarification.

William