Let f(x)=sin-1x and g(x)=x2-x-22x2-x-6 . If g(2)=limx→2g(x) , then the domain of the function fog is

f(x)=sin-1x,g(x)=x2-x-22x2-x-6,x≠2g(2)=limx→2g(x)=37(f∘g)(x)=f(g(x))=sin-1(g,(x))⇒x2-x-22x2-x-6≤1⇒(x-2)(x+1)(x-2)(2x+3)≤1⇒x+12x+3≤1⇒-1≤x+12x+3≤1⇒3x+42x+3≥0 and -2-x2x+3≤0⇒x∈-∞,-32∪-43,∞ and x∈(-∞,-2]∪(-3/2,∞)⇒x∈(-∞,-2]∪-43,∞

Let f(x)=sin-1x and g(x)=x2-x-22x2-x-6 . If g(2)=limx→2g(x) , then the domain of the function fog is

f(x)=sin-1x,g(x)=x2-x-22x2-x-6,x≠2g(2)=limx→2g(x)=37(f∘g)(x)=f(g(x))=sin-1(g,(x))⇒x2-x-22x2-x-6≤1⇒(x-2)(x+1)(x-2)(2x+3)≤1⇒x+12x+3≤1⇒-1≤x+12x+3≤1⇒3x+42x+3≥0 and -2-x2x+3≤0⇒x∈-∞,-32∪-43,∞ and x∈(-∞,-2]∪(-3/2,∞)⇒x∈(-∞,-2]∪-43,∞

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$Let f (x) = \sin^{- 1} x and g (x) = \frac{x^{2} - x - 2}{2 x^{2} - x - 6} . If g (2) = \lim_{x \to 2} g (x), then the domain of the$ $function f o g is$

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$(- \infty, - 1] \cup [2, \infty)$

$(- \infty, - 2] \cup [- 1, \infty)$

$(- \infty, - 2] \cup [- \frac{3}{2}, \infty)$

$(- \infty, - 2] \cup [- \frac{4}{3}, \infty)$

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Detailed Solution

$\begin{array}{l} f (x) = \sin^{- 1} x, g (x) = \frac{x^{2} - x - 2}{2 x^{2} - x - 6}, x \neq 2 \\ g (2) = \lim_{x \to 2} g (x) = \frac{3}{7} \\ (f \circ g) (x) = f (g (x)) = \sin^{- 1} (g, (x)) \Rightarrow |\frac{x^{2} - x - 2}{2 x^{2} - x - 6}| \leq 1 \end{array}$

$\begin{array}{l} \Rightarrow |\frac{(x - 2) (x + 1)}{(x - 2) (2 x + 3)}| \leq 1 \Rightarrow |\frac{x + 1}{2 x + 3}| \leq 1 \Rightarrow - 1 \leq \frac{x + 1}{2 x + 3} \leq 1 \\ \Rightarrow \frac{3 x + 4}{2 x + 3} \geq 0 and \frac{- 2 - x}{2 x + 3} \leq 0 \Rightarrow x \in (- \infty, - \frac{3}{2}) \cup [\frac{- 4}{3}, \infty) \\ and x \in (- \infty, - 2] \cup (- 3 / 2, \infty) \Rightarrow x \in (- \infty, - 2] \cup [- \frac{4}{3}, \infty] \end{array}$