A geometry problem by Soumava Pal

Geometry Level 3

Consider the following problems:

Let $G$ be the number of points of discontinuity of $f(x)=\lfloor x^2\rfloor -\lfloor x \rfloor^2$ in the interval $(0,1.6)$ .

Let $E$ be the number of solutions to $x=99\sin(\pi x)$ .

Find $G+E-GE$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1.

2 solutions

Soumava Pal
May 19, 2016

Plotting the graph (or a little bit of investigation of the function f(x) ) shows that $G=1$ because the graph is discontinuous at all points where | $x$ | $=\sqrt{n}$ where $n$ is a non-negative integer, except for $n=1$ . So in the given interval it is discontinuous at $0<\sqrt{2}<1.6$ only.

Now $G-1=0$ implies $(G-1)(H-1)=0$

so that $G+H-GH=1$ .

Also if we want, we can find the value of $H$ by considering that $x=0$ is a solution to the given equation. Also the number of positive solutions is equal to the number of negative solutions which can be identified from the shape of the graph only. Also from the fact that $y=x$ and $y=99sin(\pi x)$ both are odd, if $x_0$ is a solution then $-x_0$ is a solution.

The number of positive solutions is 99, which can be found from sketching the graph of the two functions itself.

Exactly... I didn't bothered to find E as it was not required at all since it gets cancelled in the process since G=1.... Although nice soln.. (+1)

Rishabh Jain - 5 years ago

Yeah, but i am getting 100 solutions!

Dipanjan Chowdhury - 5 years ago

The number of solutions is the number of times the line y = x 99 meets the graph of y = sin(x). This can occur only for x belongs to [-99; 99] because sin(x) has range [-1; 1]. Also sin(x) is periodic with period 2. For x >=0, the two graphs meet twice in each cycle of sin(x), both intersections occurring in the first half of the cycle. There are 50 such half-cycles from x = 0 to x = 99, over intervals [0; 1]; [2; 3]; : : : ; [98; 99]. So there are 100 non-negative solutions. Similarly there are 100 solutions<= 0 because both graphs are odd. Since x = 0 is counted twice, the total number of solutions is 100 + 100 - 1 = 199. so E=99/199?

rajdeep brahma - 4 years, 2 months ago

Ankan Dutta
May 21, 2016

No need to calculate the second equation as its trivial to see E=1.. which cancels G n thus E.. 😉 N if u want to calculate G.. it's 4*((99-1)/2) -1.. one gets reduced for a root which we calculated twice 0.

A geometry problem by Soumava Pal

The answer is 1.

2 solutions

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