Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | January | Q#6

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Question

a.  The polynomial  is given by .

                    i.       Use the Factor Theorem to show that  is a factor of .

                  ii.       Express  in the form  , where a and b are constants.

b. The curve C with equation  , sketched below, crosses the x-axis at the point .

                    i.       Find the gradient of the curve C at the point Q.

                  ii.       Hence find an equation of the tangent to the curve C at the point Q.

                 iii.       Find

                 iv.       Hence find the area of the shaded region bounded by the curve C and the coordinate  axes.

Solution

a.
 

            i.
 

Factor theorem states that if  is a factor of   then;

For the given case  is factor of .

Here  and . Hence;

Hence,  is factor of .


     ii.
 

If  is a polynomial of degree  then  will have exactly  factors, some of which may repeat.

We are given a polynomial of degree ,

Therefore, it will have 03 factors some of which may repeat.

From (a:i) we already have 01 factor  of .

We may divide the given polynomial with any of the already known linear factor(s) to get a quadratic  factor and then factorize the obtained quadratic factor to find the 2nd and 3rd linear  factors.

For the given case;

Already known factor from (a:i) is .

We may divide the given polynomial by this factor .

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Therefore, we get the quadratic factor  for given polynomial.

Hence,   can be written as product of linear and quadratic factors as follows;

b.    

                  i.
 

We are required to find the gradient of the curve at 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; 

We are given equation of the curve as;

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

For gradient of the curve at point Q(2,0), substitute  in derivative of the equation of the curve.

 

                          ii.
 

We are required to find the equation of tangent to the curve at point Q(2,0).

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 as Q(2,0).

We need to find slope of tangent at in order 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;

From (b:i) we have slope (gradient) of the curve at point Q(2,0);

Therefore;

With coordinates of a point on the tangent Q(2,0) and its slope  in hand, we can write equation of the tanget.

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


iii.
 

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

 

                         iv.
 

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

For the given case;

Rule for integration of  is:

From (b:iii) we know that;

Therefore;

Hence;

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