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Question

The value of 21(g(x)f(x))dx))dx is

A
2
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B
3
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C
4
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D
4ln2
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Solution

The correct option is B 3
Given y=f(x),y=g(x) are two curves, the equation of tangents to
these curves at points with equal abscissa say (x=α)
are given by
yf(α)=f(α)(xα) & yg(α)=g(α)(xα)...........{1}
These curves (lines) given by {1} intersect at a point on y-axis
yf(α)=αf(α) &
yg(α)=αg(α)
y=f(α)αf(α)
& y=g(α)αg(α)
f(α)αf(α)=g(α)αg(α)
f(α)g(α)=α{f(α)g(α)}
replacing α=x
f(x)g(x)=x{f(x)g(x)}
f(x)g(x)=xddx{f(x)g(x)}
d{f(x)g(x)}f(x)g(x)=dxx
log{f(x)g(x)}=logx+logc=log(xc)
f(x)g(x)=xc (A)
Again, the equations ofthe normals at points with equal abscissa say x=α the two curves are
yf(α)=1f(α)(xα) & yg(α)=1g(α)(xα).......... {2}
These lines intersect at a point on xaxis
y=0
0f(α)=1f(α)(xα) & 0g(α)=1g(α)(xα)
x=α+f(α)f(α) &
x=α+g(α)g(α)
α+f(α)f(α)=α+g(α)g(α)
f(α)f(α)=g(α)g(α) replacing α=x
we get
f(x)f(x)=g(x)g(x) (f(x))2={g(x)}2+k
{f(x)}2{g(x)}2=k
{f(x)+g(x)}{f(x)g(x)}=k
f(x)+g(x)=kcx (using A)(B)
Now solving (A) & (B) we get
f(x)=12(cx+kcx) (C)
and g(x)=12(kcxcx) (D)
Now
Equation (C) passes through (1,1) when x=1,f(x)=1 and
Equation (D) passes through (2,3) when x=1,g(x)=2
2×1=c+kc & 3×2=k2c2c
i.e. 4=c+kc.........{3}
and 6=k2c2c.........{4}
Solving {3} and {4} for k & c we get c=2, k=8
Now putting the value of c,k in f(x) & g(x) i.e. in (C) & (D) we get
f(x)=2xx and g(x)=2x+x
Now, 21(g(x)f(x))dx=21(2x+x(2xx))dx
21(2x)dx=3

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