Question

The set of exhaustive values of x satisfying $\frac{[x]}{sgn(x)} < \frac{|x|}{x}$ is (a, b). Then the value of a + b is (where [] and sgn denote the greatest integer function and the signum function respectively)

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

Answer: (A)

1

Answer 2

The given inequality is: $\frac{[x]}{sgn(x)} < \frac{|x|}{x}$ For any non-zero real number $x$, we know that $\frac{|x|}{x} = sgn(x)$. Thus, the inequality can be rewritten as: $\frac{[x]}{sgn(x)} < sgn(x)$ We must have $x \neq 0$ for the expressions to be defined. We analyze the inequality in two cases based on the sign of $x$.

Case 1: $x > 0$ If $x > 0$, then $sgn(x) = 1$. The inequality becomes: $\frac{[x]}{1} < 1$ $[x] < 1$ For $x > 0$, the possible values of $[x]$ are $0, 1, 2, \dots$. The condition $[x] < 1$ implies that $[x]$ must be $0$. If $[x] = 0$, then $0 \le x < 1$. Considering that we are in the case $x > 0$, the solution for this case is $0 < x < 1$.

Case 2: $x < 0$ If $x < 0$, then $sgn(x) = -1$. The inequality becomes: $\frac{[x]}{-1} < -1$ Multiplying both sides by $-1$ and reversing the inequality sign, we get: $[x] > 1$ For $x < 0$, the possible values of $[x]$ are $-1, -2, -3, \dots$. The condition $[x] > 1$ means $[x]$ must be an integer greater than 1 (i.e., $2, 3, 4, \dots$). However, for $x < 0$, $[x]$ is always less than or equal to $-1$. Therefore, there are no values of $x < 0$ that satisfy $[x] > 1$.

Conclusion The set of all values of $x$ satisfying the inequality is the union of the solutions from both cases. The only solutions are from Case 1, which is the interval $(0, 1)$. The problem states that the set of exhaustive values of $x$ is $(a, b)$. Comparing $(0, 1)$ with $(a, b)$, we find $a = 0$ and $b = 1$. The value of $a + b$ is $0 + 1 = 1$.