Question

If `phi` (x) is a differentiable real valued function satisfying `phi (x) +2phi le 1`, then it can be adjucted as `e^(2x)phi(x)+2e^(2x)phi(x)lee^(2x) or (d)/(dx)(e^(2)phi(x)-(e^(2x))/(2))le or (d)/(dx)e^(2x)(phi(x)-(1)/(2))le0`
Here `e^(2x)` is called integrating factor which helps in creating single differential coefficeint as shown above. Answer the following question:
If p(1)=0 and `dP(x)/(dx)ltP(x)` for all `xge1` then
A. `P(x) gt 0 forall x gt 1`
B. P(x) is a constant function
C. `P(x) lt 0 forall x gt 1`
D. none of these

Answer 1

Correct Answer - 1
`(dp(x))/(dx)gtp(x)`
or `e^(-x)(dp(x))/(dx)-e^(-x)p(x)gt0`
or `(d)/(dx)p(x)e^(-x)gt0`
thus P(x) `e^(-x)` is and increasing function i.e
`p(x)e^(-x)gt9(1)e^(-1)forallxge1`
or `p(x)e^(-x)gt 0 forall xgt1 or p(x)gt0 forall x gt1`