JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $f (x) = \int \frac{d x}{x ^{(\frac{2}{3})} + 2 x ^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 lo g_{e} (2)$ . If $f (1) = a + b lo g_{e} (3)$ , where $a, b \in Z$ , then $a + b$ is equal to :

A
$- 18$
B
$- 5$
C
$- 11$
D
$- 26$

Solution

Substitute $x = t^{6}$ , so $d x = 6 t^{5} d t$ , $x^{2/3} = t^{4}$ , $x^{1/2} = t^{3}$ .

$f = \int \frac{6 t ^{5} d t}{t ^{4} + 2 t ^{3}} = \int \frac{6 t ^{2} d t}{t + 2} = 6 \int \frac{( t ^{2} - 4 ) + 4}{t + 2} d t = 6 \int (t - 2) d t + 24 \int \frac{d t}{t + 2}$ .

$= 6 [\frac{t ^{2}}{2} - 2 t] + 24 ln (t + 2) + C = 3 t^{2} - 12 t + 24 ln (t + 2) + C$ .

Back to $x$ via $t = x^{1/6}$ : $f (x) = 3 x^{1/3} - 12 x^{1/6} + 24 ln (x^{1/6} + 2) + C$ .

$f (0) = 0 - 0 + 24 ln 2 + C = - 26 + 24 ln 2 \Rightarrow C = - 26$ .

$f (1) = 3 - 12 + 24 ln 3 - 26 = - 35 + 24 ln 3$ . So $a = - 35, b = 24 \Rightarrow a + b = - 11$ .

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