Maths Q16 – JEE-MAIN 2026

Home Maths Limits and DerivativesQuestion #266

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

r x^{2} - p x + q = 0

lie in

(0, 2)

, be the interval

(α, β)

, then

4 (α + β)

equals :

Clue 1 of 3: Limit Evaluation

First, evaluate the given limit to find the values of $r$ and $p$ . Use L'Hopital's rule or series expansion for indeterminate forms.

🧠 Full Solution Path

Step 1: Evaluate the Limit to Find r and p✦ Active

Let $h = x - 2$ . As $x \to 2$ , $h \to 0$ . The limit becomes:

lim_{h \to 0} \frac{(\tan h) (r (h + 2)^{2} + (p - 2) (h + 2) - 2 p)}{h^{2}}

Using $lim_{h \to 0} \frac{\tan h}{h} = 1$ , the limit simplifies to:

lim_{h \to 0} \frac{r (h + 2)^{2} + (p - 2) (h + 2) - 2 p}{h}

For this limit to be finite, the numerator must be zero when $h = 0$ . Let $N (h) = r (h + 2)^{2} + (p - 2) (h + 2) - 2 p$ . Setting $h = 0$ :

N (0) = r (2)^{2} + (p - 2) (2) - 2 p = 4 r + 2 p - 4 - 2 p = 4 r - 4

For the limit to exist, $N (0) = 0 ⟹ 4 r - 4 = 0 ⟹ r = 1$ . Substitute $r = 1$ into $N (h)$ :

N (h) = (h + 2)^{2} + (p - 2) (h + 2) - 2 p = (h^{2} + 4 h + 4) + (p - 2) h + 2 (p - 2) - 2 p

Simplifying $N (h)$ :

N (h) = h^{2} + 4 h + 4 + p h - 2 h + 2 p - 4 - 2 p = h^{2} + (2 + p) h = h (h + 2 + p)

Now, the limit is:

lim_{h \to 0} \frac{h (h + 2 + p)}{h} = lim_{h \to 0} (h + 2 + p) = 2 + p

Given that the limit is $5$ , we have $2 + p = 5 ⟹ p = 3$ . So, $r = 1$ and $p = 3$ .

Step 2: Determine the Range of q for Roots in (0, 2)○ Expand

The quadratic equation is $r x^{2} - p x + q = 0$ . Substituting $r = 1$ and $p = 3$ , we get $x^{2} - 3 x + q = 0$ . Let $f (x) = x^{2} - 3 x + q$ . For the roots of $f (x) = 0$ to lie in the interval $(0, 2)$ , we apply the following conditions:

1. **Discriminant ( $D \geq 0$ ):** $D = (- 3)^{2} - 4 (1) (q) = 9 - 4 q$ . For real roots, $9 - 4 q \geq 0 ⟹ 4 q \leq 9 ⟹ q \leq \frac{9}{4}$ .

2. **Values at endpoints ( $a f (k_{1}) > 0$ and $a f (k_{2}) > 0$ ):** Here $a = 1 > 0$ . So, $f (0) > 0$ and $f (2) > 0$ .

$f (0) = 0^{2} - 3 (0) + q = q > 0$ .

$f (2) = 2^{2} - 3 (2) + q = 4 - 6 + q = q - 2 > 0 ⟹ q > 2$ .

3. **Axis of symmetry ( $k_{1} < - b / (2 a) < k_{2}$ ):** The axis of symmetry is $x = - (- 3) / (2 (1)) = 3 / 2$ . We need $0 < 3 / 2 < 2$ , which is true.

Combining all conditions: $q \leq \frac{9}{4}$ , $q > 0$ , and $q > 2$ . The intersection of these conditions is $2 < q \leq \frac{9}{4}$ .

💡 Teacher's Secret Hint

Remember that 'roots lie in (0,2)' implies $f (0) > 0$ and $f (2) > 0$ , not $f (0) \geq 0$ or $f (2) \geq 0$ .

Step 3: Calculate

4 (α + β)

○ Expand

The set of all possible values of $q$ is $(2, \frac{9}{4}]$ . The problem states this set is the interval $(α, β)$ . This implies that $α = 2$ and $β = \frac{9}{4}$ (interpreting the given notation $(α, β)$ as the open interval corresponding to the bounds, even if one bound is inclusive in the derived set).

Now, we calculate $4 (α + β)$ :

4 (α + β) = 4 (2 + \frac{9}{4})

= 4 (\frac{8}{4} + \frac{9}{4})

= 4 (\frac{17}{4})

= 17

💡 Teacher's Secret Hint

Carefully combine the inequalities for $q$ and identify the correct $α$ and $β$ based on the interval notation provided.

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Limit Evaluation

🧠 Full Solution Path