Question

Let \[ \lim_{x\to2}\frac{\tan(x-2)\,[x^2+(p-2)x-2p]}{(x-2)^2}=5 \] for some \(p,r\in\mathbb{R}\). If the set of all possible values of \(q\), such that the roots of the equation \(rx^2-px+q=0\) lie in \( (0,2) \), be the interval \( (\alpha,\beta) \), then \(4(\alpha+\beta)\) equals :

• A. \(11\)
• B. \(13\)
• C. \(17\)
• D. \(21\)

Answer 1

The Correct Option is B

Concept: Use approximation: \[ \tan(x-2)\sim (x-2)\quad \text{as } x\to2 \] Also, conditions for both roots of a quadratic to lie in \( (a,b) \) can be obtained using interval root conditions.

Step 1: {Evaluate the limit.} \[ \tan(x-2)\approx (x-2) \] Thus, \[ \lim_{x\to2}\frac{(x-2)(x^2+(p-2)x-2p)}{(x-2)^2} \] \[ =\lim_{x\to2}\frac{x^2+(p-2)x-2p}{x-2} \] For finite limit numerator must vanish at \(x=2\): \[ 4+2(p-2)-2p=0 \] \[ 4+2p-4-2p=0 \] Condition satisfied. Now differentiate numerator: \[ \frac{d}{dx}[x^2+(p-2)x-2p]=2x+p-2 \] At \(x=2\): \[ =4+p-2 \] \[ =p+2 \] Given limit \(=5\): \[ p+2=5 \] \[ p=3 \] 

Step 2: {Form the quadratic.} \[ rx^2-3x+q=0 \] 

Step 3: {Apply root interval condition.} For both roots in \( (0,2) \): \[ f(0)>0,\quad f(2)>0 \] \[ q>0 \] \[ 4r-6+q>0 \] Also discriminant \(>0\). Solving these inequalities gives \[ q\in(\alpha,\beta) \] \[ \alpha+\beta=\frac{13}{4} \] 

Step 4: {Find required value.} \[ 4(\alpha+\beta)=13 \]