Question

\( \sin^{-1}(x - 1) + \cos^{-1}(x - 3) + \tan^{-1}\!\left(\frac{x}{2 - x^2}\right) = \cos^{-1} k + \pi \), then the value of \( k \) is __________.

• A. \( 0 \)
• B. \( -\frac{1}{\sqrt{2}} \)
• C. \( 1 \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

For inverse trigonometric questions, first check the domain restrictionsThey often give the required value of the variable directly.

Answer 1

The Correct Option is D

Step 1: Check the domain of \( \sin^{-1}(x-1) \).
For \( \sin^{-1}(x-1) \) to be defined:
\[ -1 \leq x-1 \leq 1 \]
\[ 0 \leq x \leq 2 \]

Step 2: Check the domain of \( \cos^{-1}(x-3) \).
For \( \cos^{-1}(x-3) \) to be defined:
\[ -1 \leq x-3 \leq 1 \]
\[ 2 \leq x \leq 4 \]

Step 3: Find common value of \( x \).
From both domain conditions:
\[ 0 \leq x \leq 2 \]
and
\[ 2 \leq x \leq 4 \]
Therefore:
\[ x = 2 \]

Step 4: Substitute \( x = 2 \) in the given expression.
\[ \sin^{-1}(2-1) + \cos^{-1}(2-3) + \tan^{-1}\left(\frac{2}{2-2^2}\right) \]
\[ = \sin^{-1}(1) + \cos^{-1}(-1) + \tan^{-1}\left(\frac{2}{-2}\right) \]
\[ = \sin^{-1}(1) + \cos^{-1}(-1) + \tan^{-1}(-1) \]

Step 5: Use standard inverse trigonometric values.
\[ \sin^{-1}(1) = \frac{\pi}{2} \]
\[ \cos^{-1}(-1) = \pi \]
\[ \tan^{-1}(-1) = -\frac{\pi}{4} \]

Step 6: Add the values.
\[ \frac{\pi}{2} + \pi - \frac{\pi}{4} = \frac{2\pi}{4} + \frac{4\pi}{4} - \frac{\pi}{4} \]
\[ = \frac{5\pi}{4} \]

Step 7: Compare with the given expression.
Given:
\[ \cos^{-1} k + \pi = \frac{5\pi}{4} \]
\[ \cos^{-1} k = \frac{\pi}{4} \]
Therefore:
\[ k = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
\[ \boxed{\frac{1}{\sqrt{2}}} \]