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

Concept: This problem combines inverse trigonometric identities with equation solving. First, simplify the constant \( k \) using standard half–angle identities. Then reduce the equation using identities involving inverse sine and cosine, carefully applying domain restrictions.
Step 1: Evaluate the value of \( k \) Let \[ \alpha=\cos^{-1}\!\left(\frac{2}{3}\right) \Rightarrow \sin\alpha=\frac{\sqrt5}{3} \] Using the identity: \[ \tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha} \] \[ \tan\frac{\alpha}{2} =\frac{1-\frac{2}{3}}{\frac{\sqrt5}{3}} =\frac{1}{\sqrt5} \] Now, \[ \tan\!\left(\frac{\pi}{4}+\frac{\alpha}{2}\right) =\frac{1+\tan(\alpha/2)}{1-\tan(\alpha/2)} =\frac{1+\frac{1}{\sqrt5}}{1-\frac{1}{\sqrt5}} =\frac{3+\sqrt5}{2} \] Next, let \[ \beta=\sin^{-1}\!\left(\frac{2}{3}\right) \Rightarrow \cos\beta=\frac{\sqrt5}{3} \] \[ \tan\frac{\beta}{2} =\frac{1-\cos\beta}{\sin\beta} =\frac{3-\sqrt5}{2} \] Hence, \[ k=\frac{3+\sqrt5}{2}+\frac{3-\sqrt5}{2}=3 \]
Step 2: Simplify the given equation Substitute \( k=3 \): \[ \sin^{-1}(3x-1)=\sin^{-1}x-\cos^{-1}x \] Using the identity: \[ \sin^{-1}x-\cos^{-1}x =\sin^{-1}x-\left(\frac{\pi}{2}-\sin^{-1}x\right) =2\sin^{-1}x-\frac{\pi}{2} \] So the equation becomes: \[ \sin^{-1}(3x-1)=2\sin^{-1}x-\frac{\pi}{2} \]
Step 3: Determine the domain For \( \sin^{-1}(3x-1) \) to be defined: \[ -1\le 3x-1\le 1 \Rightarrow 0\le x\le \frac{2}{3} \] For the right-hand side to lie in \( [-\tfrac{\pi}{2},\tfrac{\pi}{2}] \): \[ 0\le \sin^{-1}x\le \frac{\pi}{2} \Rightarrow 0\le x\le 1 \] Hence, the effective domain is: \[ 0\le x\le \frac{2}{3} \]
Step 4: Solve the equation Let \( t=\sin^{-1}x \). Then: \[ \sin(2t-\tfrac{\pi}{2})=3x-1 \] Since \( \sin(2t-\tfrac{\pi}{2})=-\cos 2t \), \[ -\cos 2t=3x-1 \] But \[ \cos 2t=1-2\sin^2 t=1-2x^2 \] So, \[ -(1-2x^2)=3x-1 \Rightarrow 2x^2=3x \Rightarrow x(2x-3)=0 \]
Step 5: Check admissible solutions \[ x=0 \quad \text{or} \quad x=\frac{3}{2} \] Only \( x=0 \) lies in \( [0,\tfrac{2}{3}] \).
Step 6: Verify At \( x=0 \): \[ \sin^{-1}(-1)=-\frac{\pi}{2},\quad \sin^{-1}(0)-\cos^{-1}(0)=0-\frac{\pi}{2}=-\frac{\pi}{2} \] Hence, it satisfies the equation. Final Answer: \[ \boxed{1} \]