Question

If \( 4\cot^2 x - 16\cot x + 15 < 0 \) and x is real then cot x lies in the interval:

• A. \( (0, -1/2) \)
• B. \( (3/2, 5/2) \)
• C. \( (-1, 5/2) \)
• D. None of the above

Hint:

To solve a quadratic inequality \( ay^2 + by + c < 0 \) (or \( > 0 \)): 1. Find the roots (\( y_1, y_2 \)) of the equation \( ay^2 + by + c = 0 \). 2. If \( a > 0 \), the parabola opens upwards. The inequality \( ay^2 + by + c < 0 \) holds for \( y \) between the roots (\( y_1 < y < y_2 \)), and \( ay^2 + by + c > 0 \) holds for \( y \) outside the roots (\( y < y_1 \) or \( y > y_2 \)). 3. If \( a < 0 \), the parabola opens downwards, and the conditions are reversed.

Answer 1

The Correct Option is B

Step 1: Let \( y = \cot x \). Substitute this into the given inequality. The inequality becomes: \[ 4y^2 - 16y + 15 < 0 \] Step 2: Find the roots of the corresponding quadratic equation \( 4y^2 - 16y + 15 = 0 \). We can solve this by factoring or using the quadratic formula. Let's factor the quadratic expression: We look for two numbers that multiply to \( (4)(15) = 60 \) and add up to \( -16 \). These numbers are \( -10 \) and \( -6 \). \[ 4y^2 - 10y - 6y + 15 = 0 \] \[ 2y(2y - 5) - 3(2y - 5) = 0 \] \[ (2y - 3)(2y - 5) = 0 \] The roots are \( y = \frac{3}{2} \) and \( y = \frac{5}{2} \). Step 3: Determine the interval where the quadratic expression \( 4y^2 - 16y + 15 \) is less than 0. Since the coefficient of \( y^2 \) (which is 4) is positive, the parabola \( f(y) = 4y^2 - 16y + 15 \) opens upwards. Therefore, the expression is negative between its roots. \[ \frac{3}{2} < y < \frac{5}{2} \] Step 4: Substitute back \( y = \cot x \). \[ \frac{3}{2} < \cot x < \frac{5}{2} \] This means that \( \cot x \) lies in the open interval \( \left(\frac{3}{2}, \frac{5}{2}\right) \). Step 5: Compare the result with the given options. The interval \( \left(\frac{3}{2}, \frac{5}{2}\right) \) matches option (B). Option (A) is invalid as \( 0 > -1/2 \). Option (C) contains the correct interval but is wider.