Let `p(x)` be a function defined on `R` such that `p'(x)=p'(1-x)` for all `x epsilon[0,1],p(0)=1` and `p(1)=41`.
Then `int_(0)^(1)p(x)dx` is equals to (a)`42` (b)`sqrt(41)` (c)`21` (d)`41`

Options

A. 41
B. 42
C. `sqrt(41)`
D. 21

Answer

D

Explanation

We have,
p'(x)=p'(1-x)
i.e., `(d)/(dx)p(x)=(d)/(d(1-x)){p(1-x)}`
`rArr p(x)=-p(1-x)+C " " `….(i)
Putting x=0 in (i), we get
`p(0)=-p(1)+C rArr 42`
Putting C=42 in (i), we get
`p(x)=-p(1-x)+42" "`......(ii)
Now, `overset(1)underset(0)int p(x)dx=overset(1)underset(0)int p(1-x)dx`
`rArr overset(1)underset(0)int p(x)=overset(1)underset(0)int {42-p(x)}dx`
`rArr 2overset(1)underset(0)int p(x)dx=42`
`rArr overset(1)underset(0)int p(x)dx=21`