# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/11) | Q#10

Question

A curve has equation  and it is given that . The point A is the only point  on the curve at which the gradient is 1.

i.       Find the x-coordinate of A.

ii.       Given that the curve also passes through the point (4,10), find the y-coordinate of A, giving  your answer as a fraction.

Solution

i.

We are given that;

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Therefore;

We are also given that point A is the only point on the curve at which the gradient is 1.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

Therefore;

Hence;

Now we have two options.

Two possible values of  imply that there are two points on the curve, one at each value of , where  curve has gradient -1.

But we are given that point A is the only point on the curve at which the gradient is 1.

We can see that substitution of both values of  gives;

 For For Hence,  is not valid option.

Therefore;

ii.

We are required to find the y-coordinate of point A.

To find y-coordinate of a point  on the curve, we substitute value of x-coordinate of the point     on the curve in the equation of the curve.

We have already found x-coordinate of point A in (i).

Therefore, we need equation of the curve.

We are given that;

We can find equation of the curve from its derivative through integration;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and    in the equation obtained from integration of the derivative of the curve i.e. .

We are given that the curve passes through the point (4,10).

Therefore, substituting  and  in the above derived equation of the curve;

Hence, equation of the curve is;

Now we can find y-coordinate of point A by substituting x-coordinate of point A  in equation of  the curve;