Let for some . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval (], then 4() equals: [2026]

Question

Let $\lim_{x \to 2} \frac{(\tan (x - 2)) (r x^{2} + (p - 2) x - 2 p)}{{(x - 2)}^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval ( $α, β$ ], then 4( $α + β$ ) equals: [2026]

Smart Achievers · Accepted Answer

(3)

$\lim_{x \to 2} [\frac{\tan (x - 2)}{(x - 2)}] [\frac{r x^{2} + (p - 2) x - 2 p}{x - 2}] = 5$

$\Rightarrow \lim_{x \to 2} \frac{r x^{2} + (p - 2) x - 2 p}{x - 2} = 5$ $[∵ \lim_{x \to 2} [\frac{\tan (x - 2)}{(x - 2)} = 1]$

Since $D^{r} \to 0, N^{r} \to 0$

$\Rightarrow r = 1$

$\lim_{x \to 2} \frac{x^{2} + (p - 2) x - 2 p}{x - 2} = 5$

$\Rightarrow \lim_{x \to 2} \frac{x (x - 2) + p (x - 2)}{(x - 2)} = 5$

$\Rightarrow \lim_{x \to 2} (x + p) = 5$

$\Rightarrow 2 + p = 5 \Rightarrow p = 3$

Now given equation : $r x^{2} - p x + q = 0$

$\Rightarrow x^{2} - 3 x + q = 0$

$f (0) > 0 \Rightarrow q > 0, f (2) > 0 \Rightarrow q - 2 > 0 \Rightarrow q > 2$

And $D \geq 0 \Rightarrow 9 - 4 q \geq 0 \Rightarrow \frac{9}{4} \geq q$

$\Rightarrow q \in (2, \frac{9}{4}]$

Thus, $α = 2, β = \frac{9}{4}$

$∴ 4 (α + β) = 4 (2 + \frac{9}{4}) = 17$

Solution

Q.
Let $\lim_{x \to 2} \frac{(\tan (x - 2)) (r x^{2} + (p - 2) x - 2 p)}{{(x - 2)}^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval ( $α, β$ ], then 4( $α + β$ ) equals: [2026]

1 11

2 13

3 17

4 21

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