The domain of sin-1 \(\left( {\frac{{x + 1}}{3}} \right)\)

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AAI ATC Junior Executive 27 July 2022 Shift 2 Official Paper
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  1. (-4, 2)
  2. R
  3. [-1, 1]
  4. [-4, 2]

Answer (Detailed Solution Below)

Option 4 : [-4, 2]
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Explanation:

sin-1y is defined for y ∈ [-1, 1]

∴ sin-1 \(\left( {\frac{{x + 1}}{3}} \right)\) is defined for \(\frac{x+1}{3}\in [-1,1]\)

⇒ \(-1\leq \frac{x+1}{3}\leq 1\)

⇒ \(-3\leq x+1\leq 3\)

⇒ \(-3-1\leq x\leq 3-1\)

⇒ \(-4\leq x\leq 2\)

Hence, the domain of sin-1 \(\left( {\frac{{x + 1}}{3}} \right)\) is [-4, 2].

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