ISI2019-MMA-28

Question

Consider the functions $f,g:[0,1] \rightarrow [0,1]$ given by

$$f(x)=\frac{1}{2}x(x+1) \text{ and } g(x)=\frac{1}{2}x^2(x+1).$$

Then the area enclosed between the graphs of $f^{-1}$ and $g^{-1}$ is

• A. $1/4$
• B. $1/6$
• C. $1/8$
• D. $1/24$

Answer 1

When you take inverse of a function, it just means you are reflecting it along the line $y = x$ .Reflection does not change the area under the curve. So area between two curves is same as the area between the inverses of two curves. So
Area between two curves = $\int_{0}^{1} g^{-1}(x)-f^{-1}(x) dx = \int_{0}^{1} g(x)-f(x) dx = (1/2)*(1/2 - 1/4) = 1/8$ .
So $C$ is the correct answer

Comments

Yes you are about area is same of curves and it's inverses but you get wrong answer it is 1/4 . 

Check the graph I attached.

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Your answer is incorrect. See the domain and range. It is [0,1] → [0,1]. Do  not bother  about these exams if you are going to use such tools. You will hopefully not be allowed to use these tools in the exam. And even if you could, you will get it wrong.

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Moreover, in the range of [-1,1] the graph is not invertible.

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Bro I know these tool are not used but you see your solution upper limit and lower limit . And I am not see -1 to 0. I attached graph because u understand upper limit and lower limit.