The number of solutions of⁻¹4x + ⁻¹6x = (π)/(6)$\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$, where - 12√(6) < x < 12√(6)$-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to
A.3$3$
B.0$0$
C.1$1$
D.2$2$
Solution & Explanation
Related Formula
⁻¹a + ⁻¹b = ⁻¹((a + b)/(1 - ab)) for ab < 1$$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{for } ab < 1$$
Core Logic
Given x in (- 12√(6), 12√(6))$x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$, the product (4x)(6x) = 24x² < 24((1)/(24)) = 1$(4x)(6x) = 24x^2 < 24\left(\frac{1}{24}\right) = 1$. Thus, the standard identity holds without additive phase shifts.
We need to check which roots fall within (- 12√(6), 12√(6))$\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$.
The positive root is x₁ = √(396) - 10√(3)48$x_1 = \frac{\sqrt{396} - 10\sqrt{3}}{48}$.
Since √(396) > √(300) = 10√(3)$\sqrt{396} > \sqrt{300} = 10\sqrt{3}$, x₁ > 0$x_1 > 0$, making it a valid positive fraction.
The negative root is x₂ = -√(396) - 10√(3)48$x_2 = \frac{-\sqrt{396} - 10\sqrt{3}}{48}$. This is extremely negative and falls well outside the tight bound of - 12√(6)$-\frac{1}{2\sqrt{6}}$.
Only 1 solution exists in the specified domain.
Pattern Recognition
Checking the valid domain for ab < 1$ab < 1$ prevents phantom solutions. A quadratic always yields two roots, but the strict interval provided in the question stems directly from the convergence limits of the inverse tangent addition identity, aggressively discarding the far-flung negative root.
Chapter Mix
Class 12 Maths: Inverse Trigonometric Functions
Class 11 Maths: Quadratic Equations
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)$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$.
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