Q.16. Let for some . If the set of all possible values of q, such that the roots of the equation l

Q: What is the correct answer for this exam Mathematics Q16?

The correct answer is C. Given limit: $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$.Let $y = x-2$. As $x \to 2$, $y \to 0$.The limit becomes $\lim_{y \to 0} \frac{(\tan y)(r(y+2)^2 + (p-2)(y+2) - 2p)}{y^2} = 5$.We know $\lim_{y \to 0} \frac{\tan y}{y} = 1$.So, $\lim_{y \to 0} \frac{r(y+2)^2 + (p-2)(y+2) - 2p}{y} = 5$.Let $f(y) = r(y+2)^2 + (p-2)(y+2) - 2p$.For the limit to be finite, $f(0)$ must be

Q: Q.16. Let for some . If the set of all possible values of q, such that the roots of the equation lie in , be the interval , then equals.

The correct answer is C. Given limit: $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$.Let $y = x-2$. As $x \to 2$, $y \to 0$.The limit becomes $\lim_{y \to 0} \frac{(\tan y)(r(y+2)^2 + (p-2)(y+2) - 2p)}{y^2} = 5$.We know $\lim_{y \to 0} \frac{\tan y}{y} = 1$.So, $\lim_{y \to 0} \frac{r(y+2)^2 + (p-2)(y+2) - 2p}{y} = 5$.Let $f(y) = r(y+2)^2 + (p-2)(y+2) - 2p$.For the limit to be finite, $f(0)$ must be

Q: What are the options for this this exam Mathematics question?

A. 11 · B. 13 · C. 17 · D. 21

Question

Q.16. Let  for some . If the set of all possible values of q, such that the roots of the equation  lie in , be the interval , then  equals.

FirstInTest · Accepted Answer

Given limit: $\lim_{x 	o 2} \frac{(	an(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$.Let $y = x-2$. As $x 	o 2$, $y 	o 0$.The limit becomes $\lim_{y 	o 0} \frac{(	an y)(r(y+2)^2 + (p-2)(y+2) - 2p)}{y^2} = 5$.We know $\lim_{y 	o 0} \frac{	an y}{y} = 1$.So, $\lim_{y 	o 0} \frac{r(y+2)^2 + (p-2)(y+2) - 2p}{y} = 5$.Let $f(y) = r(y+2)^2 + (p-2)(y+2) - 2p$.For the limit to be finite, $f(0)$ must be 0 (L'Hopital's rule condition).$f(0) = r(0+2)^2 + (p-2)(0+2) - 2p = 4r + 2p - 4 - 2p = 4r - 4$.So, $4r-4=0 \Rightarrow r=1$.Now, substitute $r=1$ into $f(y)$: $f(y) = (y+2)^2 + (p-2)(y+2) - 2p$.Expand: $f(y) = y^2+4y+4 + py+2p-2y-4 - 2p = y^2 + (4+p-2)y + (4+2p-4-2p) = y^2 + (p+2)y$.So the limit is $\lim_{y 	o 0} \frac{y^2 + (p+2)y}{y} = \lim_{y 	o 0} (y + p+2) = p+2$.Given that the limit is 5, $p+2=5 \Rightarrow p=3$.So we have $r=1$ and $p=3$.Now consider the quadratic equation $rx^2 - px + q = 0$.Substitute $r=1, p=3$: $x^2 - 3x + q = 0$.The roots of this equation lie in $(0, 2)$.Let $f(x) = x^2 - 3x + q$.This is a parabola opening upwards. The axis of symmetry is $x = -\frac{-3}{2(1)} = \frac{3}{2}$.For both roots to lie in $(0,2)$:1. Discriminant $D \ge 0$: $(-3)^2 - 4(1)(q) \ge 0 \Rightarrow 9 - 4q \ge 0 \Rightarrow 4q \le 9 \Rightarrow q \le \frac{9}{4}$.2. $f(0) > 0$: $0^2 - 3(0) + q > 0 \Rightarrow q > 0$.3. $f(2) > 0$: $2^2 - 3(2) + q > 0 \Rightarrow 4 - 6 + q > 0 \Rightarrow -2 + q > 0 \Rightarrow q > 2$.Combining these conditions: $q > 2$ and $q \le \frac{9}{4}$.So, $q \in (2, \frac{9}{4}]$.This is the interval $(\alpha, \beta]$.So $\alpha = 2$ and $\beta = \frac{9}{4}$.We need to find $4(\alpha + \beta) = 4(2 + \frac{9}{4}) = 4(\frac{8+9}{4}) = 4(\frac{17}{4}) = 17$.

Answer

11

Answer

13

Answer

17

Answer

21

Related Questionsin Mathematics

Browse Engineering Questions by Subject

Practice Other Engineering Exams

Practice More Engineering Questions