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

Step 1: Understanding the Concept:
For a quadratic equation \( f(y) = y^2 - my + 1 = 0 \), if a number \( k \) lies between the roots, then \( a \cdot f(k)<0 \). Here \( a = 1 \). Also, the greatest integer function \( [x] \) returns 0 for any value in the interval \( [0, 1) \).
Step 2: Key Formula or Approach:
1. Use condition for roots: \( f(1)<0 \).
2. Find the range of the expression \( \frac{4|x|}{x^2+16} \).
3. Apply the exponent \( m \).
Step 3: Detailed Explanation:
Condition for 1 lying between roots:
\( f(1) = 1^2 - m(1) + 1<0 \)
\[ 2 - m<0 \implies m>2 \]
Now, let \( g(x) = \frac{4|x|}{x^2+16} \).
Since \( x^2+16 \ge 8|x| \) (by AM-GM inequality \( x^2 + 16 \ge 2\sqrt{16x^2} = 8|x| \)):
\[ 0 \le g(x) = \frac{4|x|}{x^2+16} \le \frac{4|x|}{8|x|} = \frac{1}{2} \]
The expression is always in the range \( [0, 1/2] \).
Since \( m>2 \), let's consider \( (g(x))^m \).
Since \( 0 \le g(x) \le 0.5 \), raising it to a power \( m>2 \) makes it even smaller.
Specifically, \( 0 \le (g(x))^m \le (1/2)^m \).
Since \( m>2 \), \( (1/2)^m<1/4 \).
Thus, the value \( \left( \frac{4|x|}{x^2+16} \right)^m \) is strictly in the interval \( [0, 0.25) \).
The greatest integer part of any number in \( [0, 1) \) is 0.
\[ \left[ \left( \frac{4|x|}{x^2+16} \right)^m \right] = 0 \]
Step 4: Final Answer:
The value is 0, which corresponds to option (C).