Inverse Trigonometric Functions · Question 2

The total number of real solutions of the equation = -1(2 )- 12 -1 ( 6 9+ 2 ) is (Here, the inverse trigonometric functions -1x and -1x assume values in [- 2, 2 ] and (- 2, 2 ), respectively.)

1. 1
2. 2
3. 3
4. 5

Answer

(3) 3

Solution

Range of RHS is within (- 3 4, 3 4 ), so lies there. Put t= ; 6t9+t2= 2(t/3)1+(t/3)2.

Case |t| 3: 12 -1( )= -1 t3, so RHS = -12t- -1 t3= -1 5t3+2t2, which has the same sign as t.

So (- 2, 2 ) and t= 5t3+2t2 t=0 or t2=1: =0, 4.

Case t>3: RHS = -12t+ -1 t3- 2 (0, 2 ), and (RHS)= 2t2-37t=t gives 5t2=-3: no solution. Case t<-3 is symmetric (odd function): none.

Total =3

Answer: (C)