Question

Let the arithmetic mean of \(\frac{1}{a}\) and \(\frac{1}{b}\) be \(\frac{5}{16}\), where \(a>2\). If \(a,4,b\) are in A.P., then the equation \[ ax^2-ax+2(a-2b)=0 \] has:

• A. one root in \((1,4)\) and another in \((-2,0)\)
• B. complex roots of magnitude less than \(2\)
• C. both roots in the interval \((-2,0)\)
• D. one root in \((0,2)\) and another in \((-4,-2)\)

Hint:

To locate roots, always evaluate the polynomial at strategic test points.

Answer 1

The Correct Option is A

Step 1: Use the arithmetic mean condition \[ \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{5}{16} \Rightarrow \frac{a+b}{ab}=\frac{5}{8} \]
Step 2: Use A.P. condition Since \(a,4,b\) are in A.P.: \[ 4=\frac{a+b}{2}\Rightarrow a+b=8 \] Substitute into the previous relation: \[ \frac{8}{ab}=\frac{5}{8} \Rightarrow ab=\frac{64}{5} \]
Step 3: Form the quadratic Given: \[ ax^2-ax+2(a-2b)=0 \] Using \(b=8-a\): \[ a-2b=a-2(8-a)=3a-16 \] Thus equation becomes: \[ ax^2-ax+2(3a-16)=0 \]
Step 4: Nature and location of roots Evaluate \(f(x)=ax^2-ax+2(3a-16)\): \[ f(1)=a-a+6a-32=6a-32>0 \] \[ f(4)=16a-4a+6a-32=18a-32>0 \] \[ f(0)=6a-32<0 \] \[ f(-2)=4a+2a+6a-32=12a-32>0 \] Thus:
One root lies between \((1,4)\)
Another root lies between \((-2,0)\)
Final Answer: \[ \boxed{\text{Option (A)}} \]