Statement I: The equation ( ⁻¹ x )^3+ ( ⁻¹ x )^3- a ^3=0 has a solution for all a 1 32 . Statement II: For any x R , ⁻¹ x+ ⁻¹ x= 2 and 0 ( ⁻¹ x- 4 )^2 9 ^2 16

Step-by-step solution

aligned & ⁻¹ x [- 2 , 2 ] & - 3 4 ( ⁻¹ x- 4 ) 4 &0 ( ⁻¹ x- 4 )^2 9 16 ^2 aligned Statement II is true aligned & ( ⁻¹ x )^3+ ( ⁻¹ x )^3=a ^3 & ( ⁻¹ x+ ⁻¹ x ) [ ( ⁻¹ x+ ⁻¹ x )^2- . & .3 ⁻¹ x ⁻¹ x ]=a ^3 & ^2 4 -3 ⁻¹ x ⁻¹ x=2 a ^2 & ⁻¹ x ( 2 - ⁻¹ x )= ^2 12 (1-8 a) & ( ⁻¹ x- 4 )^2= ^2 12 (8 a-1)+ ^2 16 & ( ⁻¹ x- 4 )^2= ^2 48 (32 a-1) aligned Putting this value in equation (1) 0 ^2 48 (32 a-1) 9 16 ^2 aligned & 0 32 a-1 27 & 1 32 a 7 8 aligned Statement-I is also true.