Let A=xin(0,π)-(π)/(2): (2/π)| x|+ (2/π)| x|=2$A=\left\{x\in(0,\pi)-\left\{\frac{\pi}{2}\right\}:\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\}$ and B=x≥0:√(x)(√(x)-4)-3|√(x)-2|+6=0.$B=\left\{x\ge0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\right\}.$ Then n(A B)$n(A\cup B)$ is equal to:

Related Formula

(2θ) = 1 - 2 ²θ$\cos(2\theta) = 1 - 2\sin^2\theta$

Core Logic

Let ⁻¹x = θ θ = x$\sin^{-1}x = \theta \implies \sin\theta = x$. The equation matches (2θ) = (1)/(9)$\cos(2\theta) = \frac{1}{9}$:

Since m$m$ and n$n$ are natural numbers, we pick x = (2)/(3) = (m)/(n)$x = \frac{2}{3} = \frac{m}{n}$. Because 2$2$ and 3$3$ are co-prime, we choose m = 2$m = 2$ and n = 3$n = 3$.

Step 1: Formulating Quadratic Equations

Substituting values into mx² - nx - m + n = 0$mx^2 - nx - m + n = 0$:

Factoring the equations:

Given α > β$\alpha > \beta$, we have α = 1$\alpha = 1$ and β = (1)/(2)$\beta = \frac{1}{2}$.

Step 2: Checking Options

Let us check the coordinates (1, (1)/(2))$\left(1, \frac{1}{2}\right)$ against option line configurations:

This exactly matches option (4).

Pattern Recognition

Co-prime conditions uniquely lock fractional values down to absolute integers. This bridges variables directly into standard algebraic calculations.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 10 Mathematics: Quadratic Equations

Related Formula

Core Logic

Transform the trigonometric equation into a quadratic in terms of x$\sin x$:

Factorizing the quadratic:

Step 1: Finding bounds for a

Since x = 2$\sin x = 2$ has no real solution, we must have:

For this to have a solution, the root must lie in the standard domain of sine:

2 ≤ a ≤ 6$2 \le a \le 6$ Thus, the solution set is [p, q] = [2, 6]$[p, q] = [2, 6]$, meaning p = 2$p = 2$ and q = 6$q = 6$.

Step 2: Evaluating r

Evaluate r = 9° - 27° - 1 63° + 81°$r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$. Using complementary angles ((90 - θ) = θ$\tan(90 - \theta) = \cot \theta$): 81° = 9°$\tan 81^{\circ} = \cot 9^{\circ}$ 1 63° = 63° = 27°$\frac{1}{\cot 63^{\circ}} = \tan 63^{\circ} = \cot 27^{\circ}$ Substitute these in:

Apply the formula θ + θ = (2)/( 2θ)$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$:

We know 18° = √(5)-14$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and 54° = 36° = √(5)+14$\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.

Step 3: Final Output Calculation

We need the value of pqr$pqr$:

Pattern Recognition

Converting mixed trig degrees like 9, 27, 63, 81 entirely into cot/tan pairs ALWAYS drops them into the (2)/( 2θ)$\frac{2}{\sin 2\theta}$ double-angle trap, bringing them natively to 18 and 54 degrees.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

If a θ + b θ = c$a\cos\theta + b\sin\theta = c$, dividing by θ$\cos\theta$ transforms the equation into a quadratic in terms of θ$\tan\theta$ and θ$\sec\theta$.

Core Logic

Given the equation:

Divide the entire equation by θ$\cos\theta$:

Square both sides to convert the θ$\sec\theta$ into a θ$\tan\theta$ expression:

Rearranging into a standard quadratic equation in terms of θ$\tan\theta$:

Step 1: Apply Quadratic Formula

Solve for θ$\tan\theta$ using the quadratic formula:

This gives two possible values:

Step 2: Check Extraneous Roots

When we squared the equation 4 + 5 θ = θ$4 + 5\tan\theta = \sec\theta$, we introduced the possibility of extraneous roots where θ$\sec\theta$ might be strictly negative while 4 + 5 θ$4 + 5\tan\theta$ is negative, but α in (-π/2, π/2)$\alpha \in (-\pi/2, \pi/2)$ restricts α gt 0$\cos\alpha \gt 0$, hence α gt 0$\sec\alpha \gt 0$. For α$\sec\alpha$ to be positive, 4 + 5 α gt 0$4 + 5\tan\alpha \gt 0$. If α = - 10 + √(10)12$\tan\alpha = -\frac{10 + \sqrt{10}}{12}$ (approx -1.09$-1.09$):

This contradicts α gt 0$\sec\alpha \gt 0$. Hence, this root is rejected.

Therefore, the only valid solution is:

Pattern Recognition

Whenever you square a trigonometric equation (like converting $\sec$ to $\tan$), always map the proposed roots back to the domain limits to prune out extraneous negative parity roots.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Core Logic

Given equation:

Expanding both sides:

Step 1: Rearranging Terms

Grouping like terms together:

Dividing both sides by α β$\cos\alpha\cos\beta$:

Step 2: Conclusion

Comparing with the given equation α = k β$\tan\alpha = k\tan\beta$, we get k = -5$k = -5$.

Note by our answer (Bonus): Since α, β in (0, (π)/(2))$\alpha, \beta \in (0, \frac{\pi}{2})$, both α$\tan\alpha$ and β$\tan\beta$ must be positive. Hence, α = -5 β$\tan\alpha = -5\tan\beta$ is not possible. The data is inconsistent, but the NTA key marks option (2) as correct.

Pattern Recognition

Standard expansion of (A± B)$\sin(A\pm B)$ and grouping identical products to isolate (A)$\tan(A)$ and (B)$\tan(B)$.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Given equation: 2 ³ x + 2x x + 4 x - 4 = 0$2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$ Expand 2x$\sin 2x$:

Factor out 2 x$2 \sin x$ from the first two terms:

Since ² x + ² x = 1$\sin^2 x + \cos^2 x = 1$:

Step 1: Finding appropriate interval for exactly 3 roots

We need exactly 3 solutions in [0, (nπ)/(2)]$\left[0, \frac{n\pi}{2}\right]$. The line y = 2/3$y = 2/3$ intersects the sine wave y = x$y = \sin x$ twice in every 2π$2\pi$ interval. In [0, π]$[0, \pi]$, there are 2 solutions. In [π, 2π]$[\pi, 2\pi]$, there are 0 solutions. In [2π, 3π]$[2\pi, 3\pi]$, there are 2 solutions (total 4 solutions). To get exactly 3 solutions, the interval must stretch past the first root in [2π, 3π]$[2\pi, 3\pi]$, but not reach the second root in that interval. However, the interval is defined as (nπ)/(2)$\frac{n\pi}{2}$. Let's check endpoints (nπ)/(2)$\frac{n\pi}{2}$: For n=4$n=4$: [0, 2π]$[0, 2\pi]$ has 2 solutions. For n=5$n=5$: [0, (5π)/(2)]$[0, \frac{5\pi}{2}]$ includes [2π, 2π + (π)/(2)]$[2\pi, 2\pi + \frac{\pi}{2}]$. Since x = 2/3$\sin x = 2/3$ happens in (0, π/2)$(0, \pi/2)$, there is exactly 1 solution in [2π, 5π/2]$[2\pi, 5\pi/2]$. Thus, total solutions = 3 for n=5$n=5$.

Step 2: Solving quadratic equation

Given n = 5$n = 5$, the quadratic equation is:

Using quadratic formula:

The roots are approximately (-5 ± 4.12)/(2)$\frac{-5 \pm 4.12}{2}$, which evaluates to roughly -0.44$-0.44$ and -4.56$-4.56$. Both roots are strictly negative.

Step 3: Determining interval membership

Since both roots are negative, they belong to the interval (-∞, 0)$(-\infty, 0)$.

Pattern Recognition

Collapsing complex trigonometric expressions often yields c₁ x = c₂$c_1\sin x = c_2$. Overlaying horizontal line intersections on the sine graph bounds n$n$ rapidly by counting nodes.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 11 Maths: Complex Numbers and Quadratic Equations