JEE Main 2026 Apr 6 Shift 2, Mathematics Q16: Methods to Evaluate Limits

JEE Main2026Apr 6, Shift 2Mathematics

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

A
11
B
13
C
17
D
21

Solution

Since $tan (x - 2) / (x - 2) \to 1$ , the limit reduces to $x \to 2 lim \frac{r x ^{2} + ( p - 2 ) x - 2 p}{x - 2} = 5$ .

For a finite limit the numerator must vanish at $x = 2$ : $4 r + 2 (p - 2) - 2 p = 0$ $\Rightarrow$ $4 r - 4 = 0$ $\Rightarrow$ $r = 1$ .

Factor: $x^{2} - 2 x + p (x - 2) = (x - 2) (x + p)$ , so the limit equals $lim_{x \to 2} (x + p) = 2 + p = 5$ $\Rightarrow$ $p = 3$ .

Now $x^{2} - 3 x + q = 0$ must have both roots in $(0, 2)$ . Conditions:

(i) Discriminant $\geq 0$ : $9 - 4 q \geq 0$ $\Rightarrow$ $q \leq \frac{9}{4}$ .

(ii) Vertex inside: $0 < \frac{3}{2} < 2$ ✓.

(iii) $f (0) > 0$ : $q > 0$ . (iv) $f (2) > 0$ : $4 - 6 + q > 0$ $\Rightarrow$ $q > 2$ .

Combining: $q \in (2, \frac{9}{4}]$ . So $α = 2$ , $β = \frac{9}{4}$ , and $4 (α + β) = 4 (\frac{17}{4}) = 17$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q16: Methods to Evaluate Limits

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