JEE Main 2025 Jan 24 Shift 2, Mathematics Q5: Set

JEE Main2025Jan 24, Shift 2Mathematics

Let $A = {x \in (0, π) - {\frac{π}{2}} : lo g_{(2/ π)} ∣ sin x ∣ + lo g_{(2/ π)} ∣ cos x ∣ = 2}$ and

$B = {x \geq 0 : x (x - 4) - 3∣ x - 2∣ + 6 = 0}$ . Then $n (A \cup B)$ is equal to:

Solution

Number of solution 4

B : let $\sqrt{x}=t<2$

Then $x (x - 4) + 3 (x - 2) + 6 = 0$

$\Rightarrow t^{2} - 4 t + 3 t - 6 + 6 = 0$

$\Rightarrow t^{2} - t = 0, t = 0, t = 1$

$x = 0, x = 1$

Again let $x = t > 2$

Then $t^{2} - 4 t - 3 t + 6 + 6 = 0$

$\Rightarrow t^{2} - 7 t + 12 = 0$

$\Rightarrow t = 3, 4$

$x = 9, 16$

Total number of solutions

$n (A \cup B) = 4 + 4 = 8$

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.