Question

If the arithmetic mean of \(\dfrac{1}{a}\) and \(\dfrac{1}{b}\) is \(\dfrac{5}{16}\) and \(a,\,4,\,\alpha,\,b\) are in increasing A.P., then both the roots of the equation \[ \alpha x^2-ax+2(\alpha-2b)=0 \] lie between:

• A. \((-3,0)\)
• B. \((-2,3)\)
• C. \((0,3)\)
• D. \((-1,1)\)

Hint:

When roots are asked to lie in an interval, explicitly compute them after simplifying parameters using A.P. relations.

Answer 1

The Correct Option is B

Concept:
The signum function is defined as: \[ \operatorname{sgn}(p) = \begin{cases} +1, & p>0\\ 0, & p=0\\ -1, & p<0 \end{cases} \] For this problem, we only consider intervals where the trigonometric functions are defined and non-zero.
Step 1: Analyze by Quadrants 
Consider one full cycle \(0<x<2\pi\). First Quadrant \((0, \frac{\pi}{2})\):
\[ \sin x>0,\ \cos x>0,\ \tan x>0,\ \cot x>0 \] \[ y = 1+1+1+1 = 4 \] Second Quadrant \((\frac{\pi}{2}, \pi)\):
\[ \sin x>0,\ \cos x<0,\ \tan x<0,\ \cot x<0 \] \[ y = 1-1-1-1 = -2 \] Third Quadrant \((\pi, \frac{3\pi}{2})\):
\[ \sin x<0,\ \cos x<0,\ \tan x>0,\ \cot x>0 \] \[ y = -1-1+1+1 = 0 \] Fourth Quadrant \((\frac{3\pi}{2}, 2\pi)\):
\[ \sin x<0,\ \cos x>0,\ \tan x<0,\ \cot x<0 \] \[ y = -1+1-1-1 = -2 \] 
Step 2: Determine the Range of \(y\)
Possible values of \(y\): \[ \{4,\ -2,\ 0\} \] 
Step 3: Sum of Elements in the Range
\[ 4 + (-2) + 0 = 2 \] \[ \boxed{2} \]