# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Feb-Mar | (P1-9709/12) | Q#10

**Question**

The diagram shows part of the curve , which touches the x-axis at the point P. The point Q(3,4) lies on the curve and the tangent to the curve at Q crosses the x-axis at R.

** i. **State the x-coordinate of P.

Showing all necessary working, find by calculation

**
ii. **the x-coordinate of R,

**iii. **the area of the shaded region PQR.

**Solution**

i.

It is evident that point P is the x-intercept of the given curve. Therefore, we are required to find the x-coordinate of the x-intercept of the curve with equation given as;

The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

Hence, substituting in given equation of the curve;

ii.

It is evident that point R is the x-intercept of the tangent to the curve at point Q(3,4). Therefore, we are required to find the x-coordinate of the x-intercept of the tangent to the curve.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

Therefore, we need equation of the tangent to the curve at point Q(3,4).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).

We already have coordinates of a point on the tangent; Q(3,4). Therefore, we need slope of the tangent to write its equation.

Let’s find slope of the tangent.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;

Therefore, if we find the slope of the curve at point Q(3,4), we can find slope of tangent to the curve at this point.

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, we need derivative of the curve to find its slope at point Q(3,4).

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

We have equation of the curve given as;

Rule for differentiation of is:

Now we can find slope of the curve at point Q(3,4). Substituting in above equation;

Now we can fins slope of tangent to the curve at point Q(3,4).

With coordinates of point (3,4) on the tangent and slope of the tangent , we can write equation of the tangent.

Point-Slope form of the equation of the line is;

Finally, with equation of tangent at hand we can find x-coordinate of the x-intercept of tangent to the curve.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

Substituting in equation of the tangent;

iii.

It is evident from the diagram that;

Let’s first find area under the curve.

To find the area of region under the curve , we need to integrate the curve from point to along x-axis.

We have equation of the curve given as;

Therefore;

Rule for integration of is:

Now we find area under line.

To find the area of region under the curve , we need to integrate the curve from point to along x-axis.

We have found equation of the line as;

Therefore;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Finally, we can find area of shaded region;

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