COMEDK · Maths · 23. Inverse Trigonometric Functions
\(\sin ^{-1}(x-1)+\cos ^{-1}(x-3)+\tan ^{-1}\left(\dfrac{x}{2-x^2}\right)=\cos ^{-1} k+\pi\), then the value of ' \(k\) ' is
- A \(\dfrac{1}{\sqrt{2}}\)
- B \(1\)
- C \(0\)
- D \(-\dfrac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
The domain of \(\sin^{-1}(x-1)\) is \(-1 \le x-1 \le 1\), which implies \(0 \le x \le 2\). The domain of \(\cos^{-1}(x-3)\) is \(-1 \le x-3 \le 1\), which implies \(2 \le x \le 4\). For the expression to be defined, \(x\) must satisfy both conditions, so \(x = 2\). Substituting…
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