Let limlimits_x → 2 (tan(x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p ∈ mathbbR. If the set of all possible values of q, such that the roots…

Given limit is $\displaystyle \lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} \cdot \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5$x→2lim​x−2tan(x−2)​⋅x−2rx2+(p−2)x−2p​=5

Since $\displaystyle \lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} = 1$x→2lim​x−2tan(x−2)​=1, we have:

$$\lim\limits_{x \to 2} \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5$$x→2lim​x−2rx2+(p−2)x−2p​=5

For the limit to exist, the numerator must be zero at $x = 2$x=2:

$$r(2)^2 + (p-2)(2) - 2p = 0$$r(2)2+(p−2)(2)−2p=0

$$4r + 2p - 4 - 2p = 0 \Rightarrow 4r = 4 \Rightarrow r = 1$$4r+2p−4−2p=0⇒4r=4⇒r=1

Substituting $r = 1$r=1 into the limit expression:

$$\lim\limits_{x \to 2} \dfrac{x^2 + (p-2)x - 2p}{x-2} = 5$$x→2lim​x−2x2+(p−2)x−2p​=5

$$\lim\limits_{x \to 2} \dfrac{(x-2)(x+p)}{x-2} = 5$$x→2lim​x−2(x−2)(x+p)​=5

$$\lim\limits_{x \to 2} (x+p) = 5 \Rightarrow 2 + p = 5 \Rightarrow p = 3$$x→2lim​(x+p)=5⇒2+p=5⇒p=3

The quadratic equation is $rx^2 - px + q = 0$rx2−px+q=0, which becomes $x^2 - 3x + q = 0$x2−3x+q=0.

For both roots of $f(x) = x^2 - 3x + q = 0$f(x)=x2−3x+q=0 to lie in the interval $(0, 2)$(0,2), the following conditions must hold:

1. Discriminant $D \ge 0 \Rightarrow (-3)^2 - 4(1)(q) \ge 0 \Rightarrow 9 - 4q \ge 0 \Rightarrow q \le \dfrac{9}{4}$D≥0⇒(−3)2−4(1)(q)≥0⇒9−4q≥0⇒q≤49​

2. $f(0) > 0 \Rightarrow q > 0$f(0)>0⇒q>0

3. $f(2) > 0 \Rightarrow 2^2 - 3(2) + q > 0 \Rightarrow q - 2 > 0 \Rightarrow q > 2$f(2)>0⇒22−3(2)+q>0⇒q−2>0⇒q>2

4. The $x$x-coordinate of the vertex must lie in $(0, 2) \Rightarrow 0 < \dfrac{-(-3)}{2(1)} < 2 \Rightarrow 0 < 1.5 < 2$(0,2)⇒0<2(1)−(−3)​<2⇒0<1.5<2, which is true.

Taking the intersection of all conditions, we get $q \in (2, \dfrac{9}{4}]$q∈(2,49​].

Comparing this with $(\alpha, \beta]$(α,β], we have $\alpha = 2$α=2 and $\beta = \dfrac{9}{4}$β=49​.

Therefore, $4(\alpha + \beta) = 4(2 + \dfrac{9}{4}) = 8 + 9 = 17$4(α+β)=4(2+49​)=8+9=17.

Answer: $17$17