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Question

The area bounded by the curve y=f(x), y-axis and the normal to the curve at P is

A
1a(a12+ea)
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B
1a(a+12+ea)
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C
1a(a12+ea)
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D
1a(a+12+ea)
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Solution

The correct option is A 1a(a12+ea)
Fig.(1)
Given:
curve y=f(x) passes through the point P(1,1) ;
Normal at the point P(1,1) is a(y1)+(x1)=0 ;
slope of tangent dydxy (ordinate of point )
dydx=ky where k is a proportionality constant.
dyy=k.dx
integrating both the sides
dyy=k.dx
ln(y)=kx+ln(c) where ln(c) is a integral constant
ln(yc)=kx
yc=ekx
y=c.ekx
curve y=c.ekx passes through point P(1,1)
1=c.ek.1
1=c.ek
c=ek
on putting this value of c into the equation of curve
y=ek.ekx
y=ek(x1)
Slope of Normal at point P(1,1)
N(1,1) = dxdy
N(1,1) = 1(dydx)
N(1,1) =1ddx(ek(x1))
N(1,1) =1k.e(11)
N(1,1) =1k.e(0)
N(1,1) = 1k ......(1)
equation of Normal at point P(1,1) is a(y1)+(x1)=0
slope of normal will be dydx of the equation of normal at point P(1,1)
a(y1)+(x1)=0
a(dydx0)+(10)=0
dydx=1a
So N(1,1) = 1a ......(2)
from equation (1) and (2)
1k=1a
a=k
Now equation of curve y=ea(x1)
equation of normal a(y1)+(x1)=0
y=ea(x1)
at x=0, y=ea
at x=1, y=1
a(y1)+(x1)=0
at x=0, y=(1+1a)
at x=1, y=1
Area bounded by the curve y=f(x) , yaxis and the normal to the curve at point P(1,1)
Total area = shaded area(1) + shaded area(2); (shown in Fig. 1)
Now, shaded area(1) = area of triangle having coordinates (0,1), (1,1), (0,1+ 1a)
= 12×(base)×(height)
= 12×(10)×(1+1a1)
= 12×(1)×(1a)
= 12a
so shaded area(1) =12a
y=ea(x1)
ln(y)=a(x1)
x= 1aln(y)+1
Now,
shaded area(2) = 1eaxdy
= 1ea ( 1aln(y)+1) dy
=1a1ealn(y)dy + 1ea1dy
= 1a×[y.ln(y)y]1ea + [y]1ea
= 1a[1+aea+ea] + [1ea]
= (11a+eaa)
so shaded area(2) = (11a+eaa)
Total area = shaded area(1) + shaded area(2)
= (12a) + (11a+eaa)
= (112a+eaa)
= 1a(a12+ea)
Hence the correct answer is 1a(a12+ea) is correct.

747612_306603_ans_93a923ea23024950b826988e98b0fbac.png

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