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Question

Let f:R+R be a differentiable function such that f(x)=e(x1)ln(xe)+x1f(x) dx. Then

A
the equation of normal at x=1 is
x+(1+e)y+(e2+e+1)=0.
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B
the equation of tangent at x=1 is x(1e)y1=0.
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C
the equation of tangent at x=1 is x(1+e)y1=0.
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D
the equation of normal at x=1 is
x+(1+e)y(e2+e+1)=0.
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Solution

The correct option is D the equation of normal at x=1 is
x+(1+e)y(e2+e+1)=0.
Given, f(x)=y=e(x1)ln(xe)+x1f(x) dx
Differentiating w.r.t. x, we get
dydx=(x1)xln(xe)+y
dydxy=1xlnx
I.F.=e1dx=ex
Now, the general solution is
yex=(1xlnx)exdx+C
yex=exxdx(lnx)exdx+C
yex=exlnx+(ex)lnxlnxexdx+C
yex=exlnx+C
y=lnx+Cex=f(x)

since f(1)=e, therefore C=1
f(x)=lnx+ex
f(x)=1x+ex
Slope of tangent at point (1,e) is (1+e).
and slope of normal at point (1,e) is 11+e.

Equation of tangent:
ye=(1+e)(x1)
(1+e)xy1=0

Equation of normal:
ye=11+e(x1)
(1+e)(ye)=x+1
x+(1+e)y(e2+e+1)=0

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