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 3Given 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 y−f(α)=f′(α)(x−α) & y−g(α)=g′(α)(x−α)...........{1}These curves (lines) given by {1} intersect at a point on y-axis∴y−f(α)=−αf′(α) & y−g(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 y−f(α)=−1f′(α)(x−α) & y−g(α)=−1g′(α)(x−α).......... {2}These lines intersect at a point on x−axis∴y=00−f(α)=−1f′(α)(x−α) & 0−g(α)=−1g′(α)(x−α)⇒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 getf(x)=12(cx+kcx) (C)and g(x)=12(kcx−cx) (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=k2c−2ci.e. 4=c+kc.........{3}and 6=k2c−2c.........{4}Solving {3} and {4} for k & c we get c=−2, k=−8Now putting the value of c,k in f(x) & g(x) i.e. in (C) & (D) we getf(x)=2x−x and g(x)=2x+xNow, ∫21(g(x)−f(x))dx=∫21(2x+x−(2x−x))dx⇒∫21(2x)dx=3

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