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QJEEM21Ma0262: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0262 Solution: Let and . If , then the domain of the function fog is :

Let f(x)=sin⁡−1xf(x) = {\sin ^{ - 1}}x and g(x)=x2−x−22x2−x−6g(x) = {{{x^2} - x - 2} \over {2{x^2} - x - 6}} . If g(2)=lim⁡x→2g(x)g(2) = \mathop {\lim }\limits_{x \to 2} g(x) , then the domain of the function fog is :

Multiple Choice Options

  • A (−∞,−2]∪[−43,∞)( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)
  • B (−∞,−2]∪[−1,∞)( - \infty , - 2] \cup [ - 1,\infty )
  • C (−∞,−2]∪[−32,∞)( - \infty , - 2] \cup \left[ { - {3 \over 2},\infty } \right)
  • D (−∞,−1]∪[2,∞)( - \infty , - 1] \cup [2,\infty )

Step-by-Step Verified Solution

g(2)=lim⁡x→2(x−2)(x+1)(2x+3)(x−2)=37g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}

Domain of fog(x)=sin⁡−1(g(x))fog(x) = {\sin ^{ - 1}}(g(x))

⇒∣g(x)∣ ≤1\Rightarrow |g(x)|\, \le 1

∣x2−x−22x2−x−6∣≤1\left| {{{{x^2} - x - 2} \over {2{x^2} - x - 6}}} \right| \le 1

∣(x+1)(x−2)(2x+3)(x−2)∣≤1\left| {{{(x + 1)(x - 2)} \over {(2x + 3)(x - 2)}}} \right| \le 1

x+12x+3≤1{{x + 1} \over {2x + 3}} \le 1 and x+12x+3≥−1{{x + 1} \over {2x + 3}} \ge - 1

x+1−2x−32x+3≤0{{x + 1 - 2x - 3} \over {2x + 3}} \le 0 and x+1+2x+32x+3≥0{{x + 1 + 2x + 3} \over {2x + 3}} \ge 0

x+22x+3≥0{{x + 2} \over {2x + 3}} \ge 0 and 3x+42x+3≥0{{3x + 4} \over {2x + 3}} \ge 0

x∈(−∞,−2]∪[−43,∞)x \in ( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)

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