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The function f(x)=x^2+lambda/x has a min...

The function `f(x)=x^2+lambda/x` has a minimum at `x=2iflambda=16` maximum at `x=2iflambda=16` maximum for no real value of `lambda` point of inflection at `x=1iflambda=-1`

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

Text Solution

Verified by Experts

The correct Answer is:
1,3,4

`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
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