# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2004 | Oct-Nov | (P2-9709/02) | Q#7

** Question**

The diagram shows the curve y = 2e^{x} + 3e^{-2x}.

The curve cuts the y-axis at A.

** i. **Write down the coordinates of A.

** ii. **Find the equation of the tangent to the curve at A, and state the coordinates of the point where this tangent meets the x-axis.

** iii. **Calculate the area of the region bounded by the curve and by the lines x = 0, y = 0 and x = 1, giving your answer correct to 2 significant figures.

**Solution**

** i.
**

We are given that curve with equation cuts the y-axis at point A and we are required to find the coordinates of point A.

It is evident that point A is the y-intercept of the curve.

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

Therefore;

Hence, coordinates of point A are (0,5).

** ii.
**

We are required to find the equation of the tangent to the curve at point A and find coordinates of its x-intercept.

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 have found coordinates of point A on the tangent (and curve) as (0,5).

Therefore, we only need slope of the tangent to write its equation.

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 gradient (slope) of the curve at point A where tangent intersects the curve, then we can find slope of the tangent.

Hence, we need gradient of the curve at point (0,5).

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;

We are given equation of the curve as;

Therefore;

Rule for differentiation of is:

Therefore;

Rule for differentiation natural exponential function , or ;

Rule for differentiation of is:

We are looking for gradient of the curve at point A(0,5).

Therefore;

Therefore slope of the curve at point A(0,5) is;

Hence, slope of the tangent to the curve at this point is;

Now we can write equation of the tangent as follows.

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

Next we are required to find the coordinates of x-intercept of this tangent to the curve at point A.

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 coordinates of x-intercept of the tangent to the curve at point A(0,5) are .

** iii.
**

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

It is evident from the given data, area of the region is bounded by the curve and by the lines x = 0, y = 0 and x = 1, that area is under the curve along x-axis from x=0 to x=1.

Therefore;

Rule for integration of is:

Rule for integration of , or ;

Therefore;

The answer correct to 2 significant figures;

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