JEE Mathematics 2025 Question with Solution

Given:

• $A = \left\{ x \in (0, \pi) \mid - \log\left(\frac{2}{\pi}\right)\sin x + \log\left(\frac{2}{\pi}\right)\cos x = 2 \right\}$
• $B = \left\{ x \geq 0 : \sqrt{x}\sqrt{x - 4} - 3\sqrt{x - 2} + 6 = 0 \right\}$

Find: $n(A \cup B)$

For set $A$, factor the given equation:

$$\log\left(\frac{2}{\pi}\right)(-\sin x + \cos x) = 2$$

Since $\log\left(\frac{2}{\pi}\right) < 0$, we get

$$-\sin x + \cos x = \frac{2}{\log\left(\frac{2}{\pi}\right)}$$

The solution states that the right-hand side is negative and lies outside the range $[-\sqrt{2}, \sqrt{2}]$ of $-\sin x + \cos x$. Therefore, there is no solution in $x \in (0, \pi)$. Hence,

$$n(A) = 0$$

For set $B$, the domain requires

$$x \geq 4$$

Let

$$t = \sqrt{x - 2}$$

Then the equation becomes

$$\sqrt{t^2 + 2} \, \sqrt{t^2 - 2} - 3t + 6 = 0$$

So,

$$\sqrt{t^4 - 4} = 3t - 6$$

Squaring both sides,

$$t^4 - 4 = 9t^2 - 36t + 36$$

Rearranging,

$$t^4 - 9t^2 + 36t - 40 = 0$$

the solution states that this quartic gives four distinct solutions for $x$ in set $B$. Thus,

$$n(B) = 4$$

Therefore,

$$n(A \cup B) = 0 + 4 = 4$$

So, the correct option is D.

Given: the two sets $A$ and $B$ as defined in the question.

Find: the number of distinct elements in $A \cup B$.

The hint suggests solving each set separately and then counting the unique solutions in their union.

For $A$, first simplify the trigonometric equation by taking out the common factor:

$$\log\left(\frac{2}{\pi}\right)(-\sin x + \cos x) = 2$$

This gives

$$-\sin x + \cos x = \frac{2}{\log\left(\frac{2}{\pi}\right)}$$

The solution notes that the obtained value is not attainable by $-\sin x + \cos x$, because that expression can vary only within

$$[-\sqrt{2}, \sqrt{2}]$$

Therefore $A$ has no element.

For $B$, use the substitution

$$t = \sqrt{x - 2}$$

with the domain coming from the radicals. Then rewrite the equation as

$$\sqrt{t^2 + 2} \, \sqrt{t^2 - 2} - 3t + 6 = 0$$

which becomes

$$\sqrt{t^4 - 4} = 3t - 6$$

Now square:

$$t^4 - 4 = 9t^2 - 36t + 36$$

Hence,

$$t^4 - 9t^2 + 36t - 40 = 0$$

According to the provided solution, this leads to four distinct admissible values of $x$ in set $B$.

Thus,

$$n(A \cup B) = n(A) + n(B) = 0 + 4 = 4$$

Therefore, the correct option is D.