1.

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

Answer»

<P>`P(x)GT 0 forall x gt 1`
P(x) is a constant function
`P(x) lt 0 forall x gt 1`
none of these

Solution :`(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`


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