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This is a problem from ISI MStat 2019 PSA Problem 16 based on calculating area bounded by the curve.

The functions \(f, g:[0,1] \rightarrow[0,1]\) are given by \( f(x)=\frac{1}{2} x(x+1)\) and \( g(x)=\frac{1}{2} x^{2}(x+1) .\) What is the area enclosed between the graphs of \( f^{-1}\) and \( g^{-1} ?\)

- (A) 1/8
- (B) 1/4
- (C)5/12
- (D) 7/24

Integration

Graph of a function

But try the problem first...

Answer: is (A)

Source

Suggested Reading

ISI MStat 2019 PSA Problem 16

Introduction to Real Analysis by Bertle Sherbert

First hint

Inverse of a function is basically reflection about y=x line .

So , we can get \( f^{-1}\) and \( g^{-1} \) from \( f(x) \) and \( g(x) \) respectively by replacing x by y and y by x .

Let's draw the curves .

Second Hint

This is graph of inverses of f and g when they are defined in \( R \to R \) . But in our problem we should consider one positive x axis and y axis .

Final Step

Therefore area of the curve bounded by the graphs of \( f^{-1}\) and \( g^{-1} \) is \( \int^1_{0}\frac{y^2+y-y^3-y^2}{2} \,dx \) = 1/8 .

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