JEE Main 2026 Apr 8 Shift 2, Mathematics Q21: Inverse Trigonometric Functions

JEE Main2026Apr 8, Shift 2Mathematics

The sum of squares of all real solution of the equation

$lo g_{(x + 1)} (2 x^{2} + 5 x + 3) = 4 - lo g_{(2 x + 3)} (x^{2} + 2 x + 1)$

is equal to _______.

Your answer

Solution

Factor: $2 x^{2} + 5 x + 3 = (2 x + 3) (x + 1)$ and $x^{2} + 2 x + 1 = (x + 1)^{2}$ .

$lo g_{x + 1} [(2 x + 3) (x + 1)] + lo g_{2 x + 3} (x + 1)^{2} = 4$

$1 + lo g_{x + 1} (2 x + 3) + 2 lo g_{2 x + 3} (x + 1) = 4$

Let $t = lo g_{x + 1} (2 x + 3)$ . Then $lo g_{2 x + 3} (x + 1) = \frac{1}{t}$ and the equation becomes:

$t + \frac{2}{t} = 3 ⟹ t^{2} - 3 t + 2 = 0 ⟹ t = 1 or 2$ .

Case $t = 1$ : $2 x + 3 = x + 1 ⟹ x = - 2$ . Rejected: base $x + 1 = - 1$ invalid.

Case $t = 2$ : $2 x + 3 = (x + 1)^{2} = x^{2} + 2 x + 1 ⟹ x^{2} = 2 ⟹ x = \pm 2$ .

Check domains (need $x + 1 > 0, x + 1 \neq = 1, 2 x + 3 > 0, 2 x + 3 \neq = 1$ ): $x = 2$ satisfies all; $x = - 2 \approx - 1.414$ gives $x + 1 < 0$ , rejected.

Only valid root is $x = 2$ , so sum of squares $= 2$ .

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