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Let f:RtoR be a differentiable function ...

Let `f:RtoR` be a differentiable function such that `f(x)=x^(2)+int_(0)^(x)e^(-t)f(x-t)dt`.
`y=f(x)` is

A

The curve y=f(x) passes through the point (1,2)

B

The curve y=f(x) passes through the point (2,-1)

C

The area of the region `{(x,y) in [0,1]xxR:f(x)leylesqrt(1-x^(2))}" is "(pi-2)/(4)`

D

The area of the region `{(x,y)in [0,1]xxR:f(x)leylesqrt(1-x^(2))}" is "(pi-1)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`f(x)=1-2x+overset(x)underset(0)inte^(x-1)f(t)dt`
`rArr" "f(x)=1-2x+e^(x)overset(x)underset(0)inte^(-t)f(t)dt" ...(i)"`
Differentiating w.r.t. x, we get
`f'(x)=-2+e^(x)e^(-x)f(x)+e^(x)overset(x)underset(0)inte^(-t)f(t)dt" ...(ii)"`
Subtracting (i) from (ii), we get
`f'(x)-f(x)=-2+f(x)-1+2x`
`rArr" "f'(x)-2f(x)=2x-3`
This is linear differential equation.
`I.F. =e^(int-2dx)=e^(-2x)`
Therefore, solution is
`ycdote^(-2x)=int(2x-3)e^(-2x)dx+c`
`rArr" "ycdote^(-2x)=(2x-3)cdot(e^(-2x))/(-2)-int2xx(e^(-2x))/(-2)dx+c`
`rArr" "ycdote^(-2x)=-((2x-3)e^(-2x))/(2)-(e^(-2x))/(2)+c`
`rArr" "ycdote^(-2x)=(1-x)e^(-2x)+c`
`rArr" "y=(1-x)+c.e^(2x)" ...(iii)"`
Putting x=0 in (i), we get f(0)=1.
So, from above equation, we have
`1=1+c rArr c=0`
Now, (iii) reduces to y=1-x, which passes through point (2,-1).
`"Now,"f(x)leylesqrt(1-x^(2))`
`rArr" "1-xleylesqrt(1-x^(2))`
`1-xley` represents the region above the line `1-x=y and yle sqrt(1-x^(2))` represents the region inside the semicircle (lying above x-axis) `x^(2)+y^(2)=1.`
So, common region is as the shaded one in the following figure :

Area of the shaded region
=Area of quarter of the circle - Area of triangle OAB
`=(1)/(4)xxpi(1)^(2)-(1)/(2)xx1xx1=(pi-2)/(4)`
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