Thank you very much Sandy.

The question is from a local institute's book.

Could you please elaborate on the solution a little.

Could you please make it more clear by elaborate?

Hey there, I solved it a little differently. First found the vertex (1, 1): x coordinate of the vertex = -b/2a. Then plug in x and solve for y, which will give you the (x,y) coordinate of the vertex. In this case it is (1,1).


Since the point (s, t) is below the vertex (1,1), s<1 and t<1. From here I just picked a number that satisfied s<1 and t<1 and solved. It worked; there may be a better way to do this, but hope this helps a little.

There is a different way to solve this:

Let's assume there is a point on the parabola directly above the given point. The x coordinate will remain the same that is s. But the y coordinate will be something like t+c, where c is a positive number. So the coordinates of the point should be (s,t+c).

As this point is on the parabola it should satisfy the equation of the parabola so, plugin the values. It gives us t+c=2s-s2. Therefore if we subtract c from both sides we can see that t is c less than 2s-s2, as c is a positive number. So B is the correct choice.

Thanks for this explanation. I now understand that the question does not even need you to perform any calculation.

s=x at the same point of the parabola but from the position it is less than y in the equation of the parabola.

Since s=x is constant and t is less than y, then B is the answer.

Given : y = 2x−x^2 , we need to find the point (s,t) which is slightly less than the vertex of parabola.

There we are many ways, you can find the vertex but easiest will be using formula , let vertex (p , q)

for p = -b/2a (formula) , Given : y = 2x−x^2 which is same as y = ax^2 + bx + c
a= -1 and b = 2 so, p = -2/-2 = 1 , put value of p (i.e x) in parabola equation y = 2x−x^2 = 2 -1 = 1

so (p, q) = (1,1) as (s,t) slightly less than vertex take (s,t) = (0.7, 0.7)

t = 0.7 s= 2(0.7) - (0.7)^2 = 0.91

thus B

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