Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/13) | Q#10
Question
The function is defined by
for
.
i. Find and
and hence verify that the function
has a minimum value at
.
The points and
lie on the curve
, as shown in the diagram.
i. Find the distance AB.
ii. Find, showing all necessary working, the area of the shaded region.
Solution
i.
We are given;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
A stationary point on the curve
is the point where gradient of the curve is equal to zero;
If the given function has a minimum (stationary) value at
then
at
.
We have found above that for given function ;
At ;
Therefore, function has a stationary value at
.
Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2nd derivative of the curve.
We substitute of the stationary point
in the expression of 2nd derivative of the curve and evaluate it;
If or
then stationary point (or its value) is minimum.
If or
then stationary point (or its value) is maximum.
We have found above that for given function ;
At ;
Since , function
has a minimum value at
.
ii.
Expression to find distance between two given points and
is:
Therefore for points and
;
iii.
It is evident from the diagram that;
To find the area of region under the curve , we need to integrate the curve from point
to
along x-axis.
We are given equation of the curve as follows;
However, we do not have equation of line but using points and
we can write equation of line AB.
Two-Point form of the equation of the line is;
Therefore;
Now we can find area of shaded region.
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
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