Question

The number of integral values of x satisfying the equation $sgn \left( \left[ \frac{15}{1 + x^2} \right] \right) = [1 + \{2x\}]$ is:

[Note: sgn(y), [y] and {y} denote signum function, greatest integer function and fractional part function respectively]

• A. 5
• B. 7
• C. 15
• D. 16

Answer 1

Answer: (B)

7

Answer 2

The given equation is $sgn \left( \left[ \frac{15}{1 + x^2} \right] \right) = [1 + \{2x\}]$.

Analysis of the Right Hand Side (RHS):

The term $\{2x\}$ represents the fractional part of $2x$. By definition, $0 \le \{2x\} < 1$ for any real number $x$. Adding 1 to this inequality, we get $1 \le 1 + \{2x\} < 1 + 1$, which simplifies to $1 \le 1 + \{2x\} < 2$. The term $[1 + \{2x\}]$ represents the greatest integer less than or equal to $1 + \{2x\}$. Since $1 \le 1 + \{2x\} < 2$, the greatest integer less than or equal to $1 + \{2x\}$ is 1. So, $[1 + \{2x\}] = 1$ for all real values of $x$.

Analysis of the Left Hand Side (LHS):

Now, the given equation becomes $sgn \left( \left[ \frac{15}{1 + x^2} \right] \right) = 1$.

The term $sgn(y)$ is the signum function, defined as:

$sgn(y) = \begin{cases} 1 & \text{if } y > 0 \\ 0 & \text{if } y = 0 \\ -1 & \text{if } y < 0 \end{cases}$

For the equation $sgn \left( \left[ \frac{15}{1 + x^2} \right] \right) = 1$ to hold, the argument of the signum function must be positive. The argument is $\left[ \frac{15}{1 + x^2} \right]$. So, we must have $\left[ \frac{15}{1 + x^2} \right] > 0$.

The term $\left[ \frac{15}{1 + x^2} \right]$ represents the greatest integer less than or equal to $\frac{15}{1 + x^2}$. For the greatest integer of a number to be positive, the number itself must be greater than or equal to 1. So, we must have $\frac{15}{1 + x^2} \ge 1$.

Solving the Inequality:

We need to solve the inequality $\frac{15}{1 + x^2} \ge 1$ for integral values of $x$. Since $x$ is a real number, $x^2 \ge 0$. Therefore, $1 + x^2 \ge 1$. This means $1 + x^2$ is always positive. We can multiply both sides of the inequality by $1 + x^2$ without changing the direction of the inequality: $15 \ge 1 \cdot (1 + x^2)$ $15 \ge 1 + x^2$ Subtracting 1 from both sides, we get: $15 - 1 \ge x^2$ $14 \ge x^2$ $x^2 \le 14$.

We are looking for integral values of $x$ that satisfy $x^2 \le 14$. This inequality is equivalent to $-\sqrt{14} \le x \le \sqrt{14}$. We need to find the integers in the interval $[-\sqrt{14}, \sqrt{14}]$. We know that $3^2 = 9$ and $4^2 = 16$. So, $3 < \sqrt{14} < 4$. Approximating $\sqrt{14}$, we have $\sqrt{14} \approx 3.74$. So the inequality is approximately $-3.74 \le x \le 3.74$.

The integral values of $x$ that lie in the interval $[-3.74, 3.74]$ are $-3, -2, -1, 0, 1, 2, 3$. Let's check these values: If $x = -3$, $x^2 = 9 \le 14$. If $x = -2$, $x^2 = 4 \le 14$. If $x = -1$, $x^2 = 1 \le 14$. If $x = 0$, $x^2 = 0 \le 14$. If $x = 1$, $x^2 = 1 \le 14$. If $x = 2$, $x^2 = 4 \le 14$. If $x = 3$, $x^2 = 9 \le 14$. If $x = -4$, $x^2 = 16 \not\le 14$. If $x = 4$, $x^2 = 16 \not\le 14$.

The integral values of $x$ satisfying $x^2 \le 14$ are $-3, -2, -1, 0, 1, 2, 3$. The number of these integral values is 7.

Therefore, the number of integral values of $x$ satisfying the equation is 7.