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Question: If the solution set of \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \righ...

If the solution set of sin⁡−1(sin⁡(2x2+4x2+1))<π−3{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)} \right) < \pi - 3 is (a, b), where a,b∈Ia,b \in I , then find (b-a+5)?

Explanation

Solution

Hint : We need to know the range and domain of sin⁡−1x{\sin ^{ - 1}}x . We also know that the supplementary angle sin⁡(π−θ)=sin⁡θ\sin (\pi - \theta ) = \sin \theta , positive sign because π−θ\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⁡−1x{\sin ^{ - 1}}x is x∈[−π2,π2]x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and range is x∈[−1,1]x \in \left[ { - 1,1} \right] .
Given, sin⁡−1(sin⁡(2x2+4x2+1))<π−3{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)} \right) < \pi - 3 . ----- (1)
Take, (2x2+4x2+1)\left( {\dfrac{{2{x^2} + 4}}{{{x^2} + 1}}} \right)
4 can be written as 2+2, that is
=2x2+2+2x2+1= \dfrac{{2{x^2} + 2 + 2}}{{{x^2} + 1}}
Separating we get,
=2x2+2x2+1+2x2+1= \dfrac{{2{x^2} + 2}}{{{x^2} + 1}} + \dfrac{2}{{{x^2} + 1}}
Taking 2 as common, we get:
=2(x2+1)x2+1+2x2+1= \dfrac{{2({x^2} + 1)}}{{{x^2} + 1}} + \dfrac{2}{{{x^2} + 1}}
Cancelling, (x2+1)({x^2} + 1) term, we get:
=2+2x2+1∈(2,4]= 2 + \dfrac{2}{{{x^2} + 1}} \in (2,4] .
Because, it is obvious that for the value of xx it lies between open interval 2 and closed interval 4.
We know, sin⁡(π−θ)=sin⁡θ\sin (\pi - \theta ) = \sin \theta , using this in (1).
⇒sin⁡−1(sin⁡(π−2x2+4x2+1))<π−3\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{\sin ^{ - 1}}(\sin (x)) = x using this in above we get,
⇒π−2x2+4x2+1<π−3\Rightarrow \pi - \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} < \pi - 3
Cancelling π\pi on both side we get,
⇒−2x2+4x2+1<\-3\Rightarrow - \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} < \- 3
Multiply by -1 and also the less than changes to greater than, that is:
⇒2x2+4x2+1>3\Rightarrow \dfrac{{2{x^2} + 4}}{{{x^2} + 1}} > 3
⇒2x2+4>3(x2+1)\Rightarrow 2{x^2} + 4 > 3({x^2} + 1)
⇒2x2+4>3x2+3\Rightarrow 2{x^2} + 4 > 3{x^2} + 3
Subtract -4 on both side, we get:
⇒2x2+4−4>3x2+3−4\Rightarrow 2{x^2} + 4 - 4 > 3{x^2} + 3 - 4
⇒2x2>3x2−1\Rightarrow 2{x^2} > 3{x^2} - 1
Subtract 2x22{x^2} on both side, we get:
⇒2x2−2x2>3x2−2x2−1\Rightarrow 2{x^2} - 2{x^2} > 3{x^2} - 2{x^2} - 1
⇒0>x2−1\Rightarrow 0 > {x^2} - 1
Rearranging this, we get:
⇒x2−1<0\Rightarrow {x^2} - 1 < 0
⇒x∈(−1,1)\Rightarrow x \in ( - 1,1)
Thus, we obtained the solution. That is (a,b)=(−1,1)(a,b) = ( - 1,1)
Now, we need to find
(b−a+5)=(1−(−1)+5)(b - a + 5) = (1 - ( - 1) + 5)
=1+1+5= 1 + 1 + 5
=7= 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>12 > 1 we multiply -1 we get, −1<\-2 - 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∈(−1,1)x \in ( - 1,1) . Careful in the calculation part.