Question

If both the roots of the equation \[ x^2-2ax+a^2-1=0 \quad (a\in\mathbb{R}) \] lie in the interval \((-2,2)\), then the equation \[ x^2-(a^2+1)x-(a^2+2)=0 \] has:

• A. both roots in \((-3,0)\)
• B. one root in \((0,2)\) and another root in \((-2,0)\)
• C. one root in \((2,3)\) and another root in \((-2,0)\)
• D. one root in \((-3,-2)\) and another root in \((0,2)\)

Hint:

When roots are restricted to an interval, first determine the parameter range, then use sign analysis of the polynomial to locate roots in specific subintervals.

Answer 1

The Correct Option is C

Step 1: Analyze the first equation \[ x^2-2ax+a^2-1=0 \] Rewrite: \[ (x-a)^2=1 \] So the roots are: \[ x=a\pm 1 \] Given that both roots lie in \((-2,2)\): \[ -2<a-1<2 \quad \text{and} \quad -2<a+1<2 \] From these: \[ -1<a<3 \quad \text{and} \quad -3<a<1 \] Combining: \[ -1<a<1 \] Step 2: Analyze the second equation \[ x^2-(a^2+1)x-(a^2+2)=0 \] Sum of roots: \[ \alpha+\beta=a^2+1>0 \] Product of roots: \[ \alpha\beta=-(a^2+2)<0 \] Hence, one root is positive and the other is negative. Step 3: Locate the roots Evaluate the polynomial at key points (using \(|a|<1\)): \[ f(0)=-(a^2+2)<0 \] \[ f(2)=4-2(a^2+1)-(a^2+2)=-3a^2<0 \] \[ f(3)=9-3(a^2+1)-(a^2+2)=4-4a^2>0 \] Thus, the positive root lies in \((2,3)\). Now check negative side: \[ f(-2)=4+2(a^2+1)-(a^2+2)=4+a^2>0 \] Since \(f(-2)>0\) and \(f(0)<0\), the negative root lies in \((-2,0)\). Final Conclusion: One root lies in \((2,3)\) and the other lies in \((-2,0)\). \[ \boxed{\text{Option (3)}} \]