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Question: The domain of $f(x) = log_{[\frac{x+1}{2}]}(x^2+x-2)$ is [Note: [k] denotes greatest integer functi...

The domain of f(x)=log[x+12](x2+x−2)f(x) = log_{[\frac{x+1}{2}]}(x^2+x-2) is

[Note: [k] denotes greatest integer function less than or equal to k.]

A

[32,∞)[\frac{3}{2},\infty)

B

[32,∞)−{2}[\frac{3}{2},\infty)-\{2\}

C

(2,∞)(2, \infty)

D

[12,∞)−{2}[\frac{1}{2},\infty)-\{2\}

Answer

(2, \infty)

Explanation

Solution

For the function f(x)=log[x+12](x2+x−2)f(x) = log_{[\frac{x+1}{2}]}(x^2+x-2) to be defined, the following conditions must be met:

  1. The base must be positive: [x+12]>0[\frac{x+1}{2}] > 0. The greatest integer function [k][k] is positive if and only if k≥1k \ge 1. So, x+12≥1  ⟹  x+1≥2  ⟹  x≥1\frac{x+1}{2} \ge 1 \implies x+1 \ge 2 \implies x \ge 1.

  2. The base must not be equal to 1: [x+12]≠1[\frac{x+1}{2}] \neq 1. The greatest integer function [k][k] is equal to 1 if and only if 1≤k<21 \le k < 2. So, [x+12]=1[\frac{x+1}{2}] = 1 if 1≤x+12<2  ⟹  2≤x+1<4  ⟹  1≤x<31 \le \frac{x+1}{2} < 2 \implies 2 \le x+1 < 4 \implies 1 \le x < 3. Therefore, for [x+12]≠1[\frac{x+1}{2}] \neq 1, we must have x<1x < 1 or x≥3x \ge 3.

Combining conditions 1 and 2: We need x≥1x \ge 1 and (x<1x < 1 or x≥3x \ge 3). The intersection is x≥3x \ge 3. So, from the base conditions, x∈[3,∞)x \in [3, \infty).

  1. The argument must be positive: x2+x−2>0x^2+x-2 > 0. Factor the quadratic expression: (x+2)(x−1)>0(x+2)(x-1) > 0. This inequality holds when xx is less than the smaller root or greater than the larger root. The roots are x=−2x=-2 and x=1x=1. So, x<−2x < -2 or x>1x > 1.

Now, we need to find the intersection of the conditions from the base and the argument. We need x∈[3,∞)x \in [3, \infty) AND (x<−2x < -2 or x>1x > 1). The intersection of [3,∞)[3, \infty) and (−∞,−2)(-\infty, -2) is empty. The intersection of [3,∞)[3, \infty) and (1,∞)(1, \infty) is [3,∞)[3, \infty). So, the domain is [3,∞)[3, \infty).

Let's examine the given options: (1) [32,∞)=[1.5,∞)[\frac{3}{2},\infty) = [1.5, \infty) (2) [32,∞)−{2}=[1.5,∞)−{2}[\frac{3}{2},\infty)-\{2\} = [1.5, \infty) - \{2\} (3) (2,∞)(2, \infty) (4) [12,∞)−{2}=[0.5,∞)−{2}[\frac{1}{2},\infty)-\{2\} = [0.5, \infty) - \{2\}

Our derived domain is [3,∞)[3, \infty). Let's check if any option is equivalent to or contains [3,∞)[3, \infty) while satisfying the conditions.

Consider option (3) (2,∞)(2, \infty). If x∈(2,3)x \in (2, 3), say x=2.5x=2.5. Base: [2.5+12]=[3.52]=[1.75]=1[\frac{2.5+1}{2}] = [\frac{3.5}{2}] = [1.75] = 1. Argument: 2.52+2.5−2=6.25+2.5−2=6.75>02.5^2+2.5-2 = 6.25+2.5-2 = 6.75 > 0. The base is 1, so the logarithm is undefined. Thus, (2,3)(2, 3) is not in the domain. So, (2,∞)(2, \infty) is not the correct domain.