Let f(x) = sin^-1x and g(x) = $\frac{x^{2} - x - 2}{2 x^{2} - x - 6}$ . If g(2) = $\underset{x \to 2}{l i m} g (x),$ then the domain of the function fog is:

Option 1 -

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

Option 2 -

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

Option 3 -

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

Option 4 -

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

2 Views | Posted 2 weeks ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 3

Detailed Solution:

$g (2) = \underset{x \to 2}{l i m} g (x) = \underset{x \to 2}{l i m} \frac{x^{2} - x - 2}{2 x^{2} - x - 6} = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 1)}{2 x (x - 2) + 3 (x - 2)}$

$N o w f o g = f (g (x)) = s i n^{- 1} g (x) = s i n^{- 1} (\frac{x^{2} - x - 2}{2 x^{2} - x - 6})$

$\frac{3 x^{2} - 2 x - 8}{2 x^{2} - x - 6} \geq 0 & \frac{- x^{2} + 4}{2 x^{2} - x - 6} \leq 0$

On solving we get $x \in (- \infty, - 2) \cup [\frac{- 4}{3}, \infty)$

As x = 2 also lies in domain since g(2) $= \underset{x \to 2}{l i m} g (x)$

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Let f(x) = sin^-1x and g(x) = $\frac{x^{2} - x - 2}{2 x^{2} - x - 6}$ . If g(2) = $\underset{x \to 2}{l i m} g (x),$ then the domain of the function fog is:

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Let f(x) = sin-1x and g(x) = x 2 − x − 2 2 x 2 − x − 6 . If g(2) = l i m x → 2 g ( x ) , then the domain of the function fog is:

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Let f(x) = sin^-1x and g(x) = $\frac{x^{2} - x - 2}{2 x^{2} - x - 6}$ . If g(2) = $\underset{x \to 2}{l i m} g (x),$ then the domain of the function fog is: