If varphi(x) is a polynomial function an...

If `varphi(x)` is a polynomial function and `varphi^(prime)(x)>varphi(x)AAxgeq1a n dvarphi(1)=0,` then `varphi(x)geq0AAxgeq1` `varphi(x)

`phi(x)ge0forallxge1`

`phi(x)lt0forallxge1`

`phi(x)=0forallxge1`

none of these

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Verified by Experts

The correct Answer is:

Given `phi (x) -phi(x) gt0 forall x ge 1`
or `e^(-x){phi(x) -phi(x)}gt0 forall x ge 1`
or `(d)/(dx)e^(-x)phi(x) gt 0 forall x ge1`
Therefore `e^(-x) phi (x)` is an increasing function `forall x ge 1`
since `phi(x)` is a polynomial
`e^(-x)phi(x)gte^(-1)phi(1)or e^(-x)phi(x)gt0`
`or phi(x) gt0`

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