JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let $f (x) = \int (\frac{16 x + 24}{x ^{2} + 2 x - 15}) d x$ . If $f (4) = 14 lo g_{e} (3)$ and $f (7) = lo g_{e} (2^{α} \cdot 3^{β})$ , $α, β \in N$ , then $α + β$ is equal to:

A
$31$
B
$37$
C
$39$
D
$41$

Solution

Split the numerator: $16 x + 24 = 8 (2 x + 2) + 8$ , where $2 x + 2$ is the derivative of $x^{2} + 2 x - 15 = (x + 5) (x - 3)$ .

$f (x) = 8 ln ∣ x^{2} + 2 x - 15∣ + 8 \cdot \int \frac{d x}{( x - 3 ) ( x + 5 )} = 8 ln ∣ x^{2} + 2 x - 15∣ + ln \frac{x - 3}{x + 5} + C$ .

At $x = 4$ : $x^{2} + 2 x - 15 = 9$ , $\frac{x - 3}{x + 5} = \frac{1}{9}$ . So $f (4) = 8 ln 9 + ln \frac{1}{9} + C = 16 ln 3 - 2 ln 3 + C = 14 ln 3 + C$ . The condition forces $C = 0$ .

At $x = 7$ : $x^{2} + 2 x - 15 = 48$ , $\frac{x - 3}{x + 5} = \frac{4}{12} = \frac{1}{3}$ . So $f (7) = 8 ln 48 - ln 3 = ln \frac{4 8 ^{8}}{3} = ln \frac{( 2 ^{4} \cdot 3 ) ^{8}}{3} = ln (2^{32} \cdot 3^{7})$ .

Therefore $α = 32$ , $β = 7$ , and $α + β = 39$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →