KCET2009MathematicsApplication of Derivatives
The minimum value of 27^ 2 x 81^ 2 x is
Options
- A-5
- B1 5
- C1 243
- D1 27
Correct answer
C. 1 243
Step-by-step solution
Let f(x)=27^ 2 x 81^ 2 x =3^ 3 2 x+4 2 x =3^ 5 ( 3 5 2 x+ 4 5 2 x ) Let 3 5 = 4 5 = then f(x)=3^ 5( 2 x+ 2 x) =3^ 5( ( +2 x)) For minimum value of given function, ( +2 x) will be minimum, ie, ( +2 x )=-1 f ( x )=3^ 5(-1) = 1 243