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:
A.4$4$
B.2$2$
C.8$8$
D.6$6$
Solution & Explanation
Related Formula
Logarithmic addition property: (a) + (b) = (ab)$\log(a) + \log(b) = \log(ab)$.
Double angle sine formula: 2 x x = 2x$2\sin x \cos x = \sin 2x$.
Since π² ≈ 9.87$\pi^2 \approx 9.87$, (8)/(π²) ≈ 0.81$\frac{8}{\pi^2} \approx 0.81$, which is less than 1. Plotting | 2x| = (8)/(π²)$|\sin 2x| = \frac{8}{\pi^2}$ over the specified range x in (0, π)$x \in (0, \pi)$ yields exactly 4 real intersection points.
Trigonometric Equations graph for Q55 - JEE Main 2025 Evening
Hence, n(A) = 4$n(A) = 4$.
Step 2: Simplify Set B
Let √(x) = t$\sqrt{x} = t$ where t ≥ 0$t \ge 0$. The equation becomes:
Hence, set B = 0, 1, 9, 16$B = \{0, 1, 9, 16\}$, giving n(B) = 4$n(B) = 4$.
Step 3: Calculate Union
Since all elements of set A are non-integral angles in (0, π)$(0, \pi)$ and elements of set B are pure integers, the sets are completely disjoint (A B =$A \cap B = \emptyset$).
Always separate functions into disjoint numeric domains (e.g., angles vs. whole integers) to conclude unions without performing tedious tracking of individual values manually.
Chapter Mix
Class 11 Physics: Trigonometric Functions
Class 11 Mathematics: Sets
Since t > 0$t > 0$, we discard 1 - √(3)$1 - \sqrt{3}$.
Thus, t = 1 + √(3) ≈ 2.732$t = 1 + \sqrt{3} \approx 2.732$.
Now, equate back:
ex = 1 + √(3) x = ln(1 + √(3))$$e^{\sin x} = 1 + \sqrt{3} \implies \sin x = \ln(1 + \sqrt{3})$$
We know e ≈ 2.718$e \approx 2.718$. Since 1 + √(3) > e$1 + \sqrt{3} > e$, it follows that ln(1 + √(3)) > 1$\ln(1 + \sqrt{3}) > 1$.
But the range of x$\sin x$ is [-1, 1]$[-1, 1]$. Therefore, x$\sin x$ cannot equal a value strictly greater than 1.
No real solution exists.
Chapter Mix
Class 11 Maths: Trigonometric Functions
Class 12 Maths: Continuity and Differentiability
Q16jee_main_2024_31_jan_eveningProperties of ITFs
If a = ⁻¹( (5))$a = \sin^{-1}(\sin(5))$ and b = ⁻¹( (5))$b = \cos^{-1}(\cos(5))$, then a² + b²$a^2 + b^2$ is equal to
A.4π² + 25$4\pi^2 + 25$
B.8π² - 40π + 50$8\pi^2 - 40\pi + 50$
C.4π² - 20π + 50$4\pi^2 - 20\pi + 50$
D.25$25$
Solution
Related Formula
⁻¹( x) = x - 2π for x in [3π/2, 5π/2]$$\sin^{-1}(\sin x) = x - 2\pi \text{ for } x \in [3\pi/2, 5\pi/2]$$⁻¹( x) = 2π - x for x in [π, 2π]$$\cos^{-1}(\cos x) = 2\pi - x \text{ for } x \in [\pi, 2\pi]$$
Core Logic
Evaluate a = ⁻¹( 5)$a = \sin^{-1}(\sin 5)$:
The principal branch of ⁻¹ x$\sin^{-1} x$ is [-π/2, π/2]$[-\pi/2, \pi/2]$.
5$5$ radians is approximately 5 × 57.3^° ≈ 286.5^°$5 \times 57.3^\circ \approx 286.5^\circ$ (in 4th quadrant).
The equivalent angle in the principal domain is 5 - 2π$5 - 2\pi$.
Thus, a = 5 - 2π$a = 5 - 2\pi$.
Evaluate b = ⁻¹( 5)$b = \cos^{-1}(\cos 5)$:
The principal branch of ⁻¹ x$\cos^{-1} x$ is [0, π]$[0, \pi]$.
5$5$ radians is in [π, 2π]$[\pi, 2\pi]$. The equivalent angle is 2π - 5$2\pi - 5$.
Thus, b = 2π - 5$b = 2\pi - 5$.
Let ⁻¹α = A, ⁻¹β = B, ⁻¹γ = C$\sin^{-1}\alpha = A, \sin^{-1}\beta = B, \sin^{-1}\gamma = C$.
Given A + B + C = π$A + B + C = \pi$.
Since A = α, B = β, C = γ$\sin A = \alpha, \sin B = \beta, \sin C = \gamma$, α, β, γ$\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by 2R$2R$ by Sine rule. However, directly dealing with the relation:
By Cosine Rule, C = (1)/(2)$\cos C = \frac{1}{2}$.
Since C = ⁻¹γ$C = \sin^{-1}\gamma$, we know C = γ$\sin C = \gamma$.
C = √(1 - γ²) = (1)/(2)$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
