Question

The least natural number ${}^{\prime }{k}^{\prime }$ for which $x+{kx}^{-2}>2\forall xϵ(0,\infty )$ is

Answer 1

$x+k{x}^{-2}$

substituting $x=1$ and $k=1$

we get,

$1+\left(1\right)(1{)}^{-2}=2$ (Value is equal to 2, but the condition is (>2))

So,

substituting $x=1$ and $k=2$

we get,

$1+\left(2\right)(1{)}^{-2}=3$ (Value is greater than 2, satisfies the condition is (>2))

Hence, $k=2$