Let `f:[0,\ oo)->R` be a continuous function such that `f(x)=1-2x+int_0^x e^(x-t)f(t)dt` for all `x in [0,\ oo)` . Then, which of the following statement(s) is (are) TRUE?

Options

A. (a) The curve `y=f(x)` passes through the point (1,2)
B. (b) The curve `y=f(x)` passes through the point (2, -1)
C. (c) the area of the region
`{(x,y) in [0,1]xxR:f(x)le y le sqrt(1-x^(2))} is (pi-2)/4`
D. (d) The area of the region
`{(x,y) in [0,1]xxR:f(x)le y le sqrt(1-x^(2))} is (pi-1)/4`

Answer

(b,c)

Explanation

we have ,
`f(x)=1-2x+int_(0)^(x) e^(x-t)f(t) dt `
On multiplying e^(-x) both sides, we get
`e^(-x)f(x)=e^(-x)-2xe^(-x)+ int_(0)^(x) e^(-t) f(t) dt`
On differentiating both side w.r.t.x, we get
`e^(-x) f'(x)-e^(-x) f(x)=-e^(-x)-2xe^(-x)+2xe^(-x)+e^(-x) f(x)`
`rArr f'(x)-2f(x)=2x-3`
[dividing both sides by e^(-x)]
Let `f(x) = y`
`rArr f'(x)=dy/dx`
`therefore dy/dx-2y=2x-3`
which is linear differential equation of the form
`dy/dx+Py=Q. Here, P=-2 and Q=2x-3.`
Now, `IF = e^(intP dx)=e^(int-2 dx)=e^(-2x)`
`therefore` Solution of the given differential equation is
`y cdot e^(-2x)=int (2x-3)e^(-2x)dx +C`
`y cdot e^(-2x)= (-(2x-3) cdote^(-2x))/2 +2inte^(-2x)/2dx+C`
[by using integration by parts]
`rArr ycdot e^(-2x)= (-(2x-3) e^(-2x))/2-e^(-2x)/2+C`
`rArr y=(1-x)+Ce^(2x)`
On putting x = 0 anfd y = 1, we get
`1=1+C rArr C=0`
`therefore y=1-x`
y=1-x passes through (2,-1)
Now, area of region bounded by curve `y=sqrt(1-x^(2)` and
y=1-x is shows as

`therefore` Area of shaded region
= Area of 1st quadrant of a
ciralc-Area of `Delta OAB`
`=pi/4 (1)^(2)-1/2xx1xx1`
`=pi/4-1/2=(pi-2)/4`
Hencee, options b and c are correct.