Question

Let \(1\) lie between the roots of the equation \(y^{2}-my+1=0\) and \([x]\) denotes the greatest integer function. Then the value of \(\left[\left(\frac{4|x|}{x^{2}+16}\right)^{m}\right]\) is:

• A. 5
• B. 4
• C. 0
• D. 1

Hint:

Whenever you see a floor function question where a fraction bounded below 1 is raised to a large positive exponent, the output inside the brackets will almost always shrink down close to zero. This allows you to evaluate the floor value as 0 by inspection.

Answer 1

The Correct Option is C

Concept: For a given target number $k$ to lie strictly between the two roots of a quadratic function $P(y) = Ay^2 + By + C$ (where $A > 0$), the value of the function evaluated at that target point must be strictly negative ($P(k) < 0$). We can use this condition to establish an algebraic inequality constraint for the parameter $m$. Step 1: Apply the root separation condition.
Let our quadratic function expression be defined as $P(y) = y^2 - my + 1$. We are given that the number $1$ lies between its roots. Since the leading coefficient is positive ($1 > 0$), setting $y = 1$ forces the function output to be negative: \[ P(1) 2 \quad \cdots (1) \]

Step 2: Find the maximum range bound of the internal fraction.
Let us analyze the internal fraction function $\phi(x) = \frac{4|x|}{x^2 + 16}$. Since both the numerator and denominator parameters are non-negative, $\phi(x) \ge 0$. Let us divide the top and bottom by $|x|$ (assuming $x \neq 0$): \[ \phi(x) = \frac{4}{|x| + \frac{16}{|x|}} \] Apply the AM-GM inequality to the denominator expression: $|x| + \frac{16}{|x|} \ge 2\sqrt{|x| \cdot \frac{16}{|x|}} = 2\sqrt{16} = 8$. Since the minimum value of the denominator is 8, the maximum value of the fraction occurs at this boundary point: \[ \phi(x) \le \frac{4}{8} = \frac{1}{2} \quad \Rightarrow \quad 0 \le \frac{4|x|}{x^2 + 16} \le \frac{1}{2} \quad \cdots (2) \]

Step 3: Combine constraints to evaluate the high power exponent.
Now let us raise our fraction inequality to the power of the exponent $m$. Since $m > 2$, raising a positive fractional base strictly less than 1 to a power greater than 2 causes it to shrink further: \[ \left(\frac{4|x|}{x^2 + 16}\right)^m \le \left(\frac{1}{2}\right)^m < \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] This shows that the entire internal expression is strictly bounded within the following fractional range: \[ 0 \le \left(\frac{4|x|}{x^2 + 16}\right)^m < \frac{1}{4} \]

Step 4: Evaluate the greatest integer floor operation.
By definition, the greatest integer function $[y]$ returns the largest integer less than or equal to $y$. For any real value sitting safely inside the fractional interval $\left[0, \frac{1}{4}\right)$, the lower floor integer step is exactly: \[ \left[\left(\frac{4|x|}{x^{2}+16}\right)^{m}\right] = 0 \] This matches option (C) perfectly.