Question

If the solution set of \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)} \right) < \pi - 3\] is (a, b), where \[a,b \in I\] , then find (b-a+5)?

Answer 1

Hint: We need to know the range and domain of \[{\sin ^{ - 1}}x\] . We also know that the supplementary angle \[\sin (\pi - \theta ) = \sin \theta \] , positive sign because \[\pi - \theta \] lies in the second quadrant and sine is positive in the second quadrant. Using this, we get the solution (a, b) and substituting in (b-a+5) we get the required answer.

Complete step-by-step answer:
We know that the domain of \[{\sin ^{ - 1}}x\] is \[x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \] and range is \[x \in \left[ { - 1,1} \right] \] .
Given, \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)} \right) < \pi - 3\] . ----- (1)
Take, \[\left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)\]
4 can be written as 2+2, that is
 \[ = \dfrac{{2{x^2} + 2 + 2}}{{{x^2} + 1}}\]
Separating we get,
 \[ = \dfrac{{2{x^2} + 2}}{{{x^2} + 1}} + \dfrac{2}{{{x^2} + 1}}\]
Taking 2 as common, we get:
 \[ = \dfrac{{2({x^2} + 1)}}{{{x^2} + 1}} + \dfrac{2}{{{x^2} + 1}}\]
Cancelling, \[({x^2} + 1)\] term, we get:
 \[ = 2 + \dfrac{2}{{{x^2} + 1}} \in (2,4] \] .
Because, it is obvious that for the value of \[x\] it lies between open interval 2 and closed interval 4.
We know, \[\sin (\pi - \theta ) = \sin \theta \] , using this in (1).
 \[ \Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)} \right) < \pi - 3\]
We also know, \[{\sin ^{ - 1}}(\sin (x)) = x\] using this in above we get,
 \[ \Rightarrow \pi - \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} < \pi - 3\]
Cancelling \[\pi \] on both side we get,
 \[ \Rightarrow - \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} < - 3\]
Multiply by -1 and also the less than changes to greater than, that is:
 \[ \Rightarrow \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} > 3\]
 \[ \Rightarrow 2{x^2} + 4 > 3({x^2} + 1)\]
 \[ \Rightarrow 2{x^2} + 4 > 3{x^2} + 3\]
Subtract -4 on both side, we get:
 \[ \Rightarrow 2{x^2} + 4 - 4 > 3{x^2} + 3 - 4\]
 \[ \Rightarrow 2{x^2} > 3{x^2} - 1\]
Subtract \[2{x^2}\] on both side, we get:
 \[ \Rightarrow 2{x^2} - 2{x^2} > 3{x^2} - 2{x^2} - 1\]
 \[ \Rightarrow 0 > {x^2} - 1\]
Rearranging this, we get:
 \[ \Rightarrow {x^2} - 1 < 0\]
 \[ \Rightarrow x \in ( - 1,1)\]
Thus, we obtained the solution. That is \[(a,b) = ( - 1,1)\]
Now, we need to find
 \[(b - a + 5) = (1 - ( - 1) + 5)\]
 \[ = 1 + 1 + 5\]
 \[ = 7\]
Thus, we get (b-a+5) = 7.
So, the correct answer is “7”.

Note: Remember the supplementary angles and complementary angles. We know that if \[2 > 1\] we multiply -1 we get, \[ - 1 < - 2\] . The same thing we applied in the above calculation. They can also ask the same problem as solve for x, or the solution of x then we need to stop here \[x \in ( - 1,1)\] . Careful in the calculation part.