JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

Let $I (x) = \int \frac{3 d x}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where $a, b, c \in N$ , $g cd (a, b) = 1$ , then $a + b + c$ is equal to :

A
29
B
28
C
31
D
30

Solution

Note $4 x^{2} + 8 x + 3 = (2 x + 1) (2 x + 3) = 4 (x + 1)^{2} - 1$ . Substitute $t = \frac{1}{4 x + 6}$ , so $4 x + 6 = 1/ t$ , $x + 1 = \frac{1 - 2 t}{4 t}$ , and $4 d x = - \frac{d t}{t ^{2}}$ .

After substitution the integrand reduces to $- \frac{3}{2} \cdot \frac{d t}{1 - 4 t}$ .

Integrate: $- \frac{3}{2} \cdot \frac{1 - 4 t}{- 2} + c = \frac{3}{4} 1 - 4 t + c$ .

Back-substitute $t = \frac{1}{4 x + 6}$ :

$I (x) = \frac{3}{4} \frac{4 x + 2}{4 x + 6} + c$ .

Use $I (0) = \frac{3}{4} \frac{2}{6} + c = \frac{3}{4} + 20$ , giving $c = 20$ .

At $x = 1/2$ : $I (1/2) = \frac{3}{4} \frac{4}{8} + 20 = \frac{3}{4 2} + 20 = \frac{3 2}{8} + 20$ .

So $a = 3$ , $b = 8$ , $c = 20$ and $a + b + c = 31$ .

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