Question

If \[ k=\tan\!\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\!\left(\frac{2}{3}\right)\right) +\tan\!\left(\frac{1}{2}\sin^{-1}\!\left(\frac{2}{3}\right)\right), \] then the number of solutions of the equation \[ \sin^{-1}(kx-1)=\sin^{-1}x-\cos^{-1}x \] is:

Hint:

Always check the {range} of inverse trigonometric expressions before solving equations—many algebraic solutions may be extraneous.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
First, evaluate the constant $k$ using inverse trigonometric identities. Then, solve the resulting equation within its valid domain.
Step 2: Key Formula or Approach:
Use the identity: $\tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{2}{\cos 2\theta}$.
Also, $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.
Step 3: Detailed Explanation:
Let $\cos^{-1}\left(\frac{2}{3}\right) = \alpha \implies \cos \alpha = \frac{2}{3}$.
Note that $\sin^{-1}\left(\frac{2}{3}\right) = \frac{\pi}{2} - \alpha$.
$k = \tan\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) + \tan\left(\frac{1}{2}\left(\frac{\pi}{2} - \alpha\right)\right)$
$k = \tan\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$.
Using the identity $\tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{2}{\cos 2\theta}$ with $\theta = \frac{\alpha}{2}$:
$k = \frac{2}{\cos \alpha} = \frac{2}{2/3} = 3$.
Now, solve $\sin^{-1}(3x - 1) = \sin^{-1} x - \cos^{-1} x$.
$\sin^{-1}(3x - 1) = \sin^{-1} x - \left(\frac{\pi}{2} - \sin^{-1} x\right) = 2\sin^{-1} x - \frac{\pi}{2}$.
Domain: $-1 \leq 3x - 1 \leq 1 \implies 0 \leq x \leq \frac{2}{3}$.
Taking sine on both sides:
$3x - 1 = \sin\left(2\sin^{-1} x - \frac{\pi}{2}\right) = -\cos(2\sin^{-1} x)$.
$3x - 1 = -(1 - 2x^2) = 2x^2 - 1$.
$2x^2 - 3x = 0 \implies x(2x - 3) = 0 \implies x = 0$ or $x = \frac{3}{2}$.
Checking the domain $[0, 2/3]$: $x = 0$ is the only valid solution.
Step 4: Final Answer:
The number of solutions is 1.