The funciton f(x)=`x^(2)+(lambda)/(x)`has a

Options

A. minimum at x =2 if `lambda` =16
B. maximum at x =2 if `lambda` =16
C. maximum for no real value of `lambda`
D. point of inflectin at x=1 if `lambda`=-1

Answer

1,3,4

Explanation

`f(x)=2x-(lambda)/(x^(2)),f(x)=0`
`rarr x=((lambda)/(2))^(1//3)`
if `lambda=16,x=2`
Now `f(x)=2+(2lambda)/(x^(3))`
Thus if `lambda=16 f(x)gt0` i.e f(X) has a minimum x=2
Also `f{((lambda)/(2))^(1//3)}=2+(2lambda)/(lambda//2)=2+4gt0`
Hence f(x) has maixmum for no real value of `lambda`
When `lambda =-1 f(x)=0` if x=1 so f(x) has a point of inflection at x=1