MHT CET202311 May 2023Evening ShiftMathematicsTrigonometric EquationsActual
If a and b are positive number such that a > b , then the minimum value of a -b (0 < < 2 ) is
Options
- A1 a^2-b^2
- B1 a^2+b^2
- Ca^2+b^2
- Da^2-b^2
Correct answer
D. a^2-b^2
Step-by-step solution
aligned & let f ( )= a -b & f ^ ( )= a -b ^2 & = ( a - b ) & f ^ ( )=0 ( a - b )=0 & a - b =0 [ As 0 0 [ a is positive and 0 < < 2 ] & aligned ( ) is minimum when = b a . aligned & Minimum value of f( ) & =a ( a a^2-b^2 )-b ( b a^2-b^2 ) [ From (i) ] & = a^2-b^2 a^2-b^2 & = a^2-b^2 aligned