Let fx- sin-1x and g x-x2-x-2-2x2-x-6- If g 2 - x- 2lim g x- then then the domain of the function f- gis-

Explanation for Correct AnswerStep 1: Value of g(2) using limit: f(x)= sin^ 1x g (x)=x^2 x 2/2x^2 x 6Finding the value of g(2)g (2) = x→ 2lim g (x)0ex0ex ...

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Byju's Answer

Standard XII

Mathematics

Continuity of a Function

Let fx= sin-1...

Question

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 then the domain of the function $f \circ g$ is:

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

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

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

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

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Solution

The correct option is A
$(- \infty, - 2] \cup [\frac{- 4}{3}, \infty)$

Explanation for Correct Answer
Step-1: Value of $g (2)$ using limit:
$f (x) = \sin^{- 1} x$
$g (x) = \frac{x^{2} - x - 2}{2 x^{2} - x - 6}$
Finding the value of $g (2)$
$g (2) = \lim_{x \to 2} g (x) = \lim_{x \to 2} \frac{x^{2} - x - 2}{2 x^{2} - x - 6} = \lim_{x \to 2} \frac{(x - 2) (x + 1)}{(2 x + 3) (x - 2)} = \lim_{x \to 2} \frac{(x + 1)}{(2 x + 3)} = \frac{(2 + 1)}{(2 \times 2 + 3)} = \frac{3}{7}$
Therefore $g (2) = \frac{3}{7}$
Step-2 :Domain of composite function:
Calculating $f \circ g$
$f \circ g = f (g (x)) = \sin^{- 1} (g (x))$
Therefore,
$\Rightarrow - 1 \leq g (x) \leq 1 \Rightarrow - 1 \leq \frac{(x + 1)}{(2 x + 3)} \leq 1$
Solving the first inequalities :
$\Rightarrow \frac{(x + 1)}{(2 x + 3)} \leq 1 \Rightarrow \frac{(x + 1)}{(2 x + 3)} - 1 \leq 0 \Rightarrow \frac{(x + 1) - (2 x + 3)}{(2 x + 3)} \leq 0 \Rightarrow \frac{- x - 2}{2 x + 3} \leq 0 \Rightarrow \frac{x + 2}{2 x + 3} \geq 0$
Using the
number line we get,

Similarly for the second inequality :
$\Rightarrow - 1 \leq \frac{(x + 1)}{(2 x + 3)} \Rightarrow 0 \leq \frac{(x + 1)}{(2 x + 3)} + 1 \Rightarrow 0 \leq \frac{(x + 1) + (2 x + 3)}{(2 x + 3)} \Rightarrow 0 \leq \frac{3 x + 4}{2 x + 3}$
Using the number line we get,
Therefore $x$ is belongs to $(- \infty, - 2] \cup [\frac{- 4}{3}, \infty)$

Hence, option (A) is the correct answer.

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

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 then the domain of the function $f \circ g$ is: