Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2017  MayJun  (P29709/21)  Q#8
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Question
The diagram shows the curve with equation
The curve crosses the xaxis at the point P and has a minimum point M.
i. Find the gradient of the curve at the point P.
ii. Find the exact coordinates of the point M.
Solution
i.
We are required to find the gradient of the curve at the point P which is given as xintercept of the curve.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore, first we find expression for gradient of the curve.
We are given that;
Therefore;
If and are functions of , and if , then;
If , then;
Let and ;
First we solve .
Rule for differentiation of is:
Next we solve .
We utilize Chain Rule to differentiate .
If we define , then derivative of is;
If we have and then derivative of is;
Let , then;
Rule for differentiation natural logarithmic function , for is;
Rule for differentiation of is:
Since ;
Hence;
We need gradient of the curve at point P.
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 value of at point P.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
We are given equation of the curve;
Therefore, we substitute to find the value of .
Now we have two options.



Taking antilogarithm of both sides. 


and are inverse functions. The composite function is an identity function, with domain the positive real numbers. Therefore;





Since denoted origin whereas point P is not origin, therefore, for point P .
Hence, gradient of the curve at point P is;
Since
Hence gradient of the curve at point P is 18.
ii.
We are required to find the exact coordinates of point M which is minimum point of the curve.
We know that minimum point of the curve is a stationary point.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
Since it is stationary point of the curve and, hence, gradient of the curve at this point must be ZERO.
We can equate expression for gradient of the curve at this point to ZERO to find the xcoordinate of point M.
Therefore;
Taking antilogarithm of both sides
and are inverse functions. The composite function is an identity function, with
domain the positive real numbers. Therefore;
Hence .
To find the ycoordinate of point M we can substitute in equation of the curve.
for any
Hence coordinates of minimum point M are
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