Let f(x) = sin ⁻¹x and g(x) = (x² - x - 2)/(2x² - x - 6). If g(2) = lim _(x → 2) g(x), then the domain of the function fog is :

Solution

Maths sir · 4 steps

Cancel the common factor (x-2) first, then make g(x) sit between -1 and 1 for sin⁻¹.

1. Simplify g. Factorise: x²-x-2=(x-2)(x+1) and 2x²-x-6=(x-2)(2x+3). So g(x)=(x+1)/(2x+3) for x≠ 2, and defining g(2) as the limit gives g(2)=3/7. Now g is defined for every x≠ -3/2.
2. Condition for fog. sin⁻¹ accepts only inputs in [-1,1], so we need -1≤ (x+1)/(2x+3)≤ 1. Solve the two halves separately.

   Inequality	Rearranged	Solution
   (x+1)/(2x+3)≤ 1	(x+2)/(2x+3)≥ 0	x≤ -2 or x>-3/2
   (x+1)/(2x+3)≥ -1	(3x+4)/(2x+3)≥ 0	x<-3/2 or x≥ -4/3

3. Intersect. Both conditions hold for x≤ -2 and for x≥ -4/3. The gap (-2,-4/3) fails. For example, x=-1.4 gives g=-2. The point x=2 is allowed because g(2)=3/7.
4. So. The domain of fog is (-∞,-2]∪[-4/3,∞), which is option (a).

The idea to keep. For sin⁻¹(g(x)), solve -1≤ g(x)≤ 1 as two separate rational inequalities. Bring everything to one side over a common denominator, then intersect the two solution sets.

Answer from the NTA final answer key. Solution by PrepWiser.