Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | May-Jun | (P1-9709/13) | Q#11

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Question

The function  is defined for . It is given that  has a minimum value when  and that .

(i)          Find .

It is now given that ,  and  are the first three terms respectively of an arithmetic progression.

(ii)        Find the value of .

(iii)       Find , and hence find the minimum value of .

Solution

(i)
 

To find the derivative of the curve from second derivative we can integrate the second derivative of the curve.

We are given second derivative of the curve as;

Therefore;

Rule for integration of  is:

We are given that  has a minimum value when . The minimum value of a function is its value  at a point which is stationary in general and minimum point in particular.

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

Therefore;

Hence at ;

Hence;

(ii)
 

We are given that the first, second and third terms of a arithmetic progression are respectively ,  and .

Let’s find first two terms of the Arithmetic Progression (A.P).

Hence, first three terms of the Arithmetic Progression (A.P).

Expression for difference  in Arithmetic Progression (A.P) is:

Therefore for the given Arithmetic Progression (A.P);

We can also find the difference   between 2nd and 3rd terms as follows;

As we are given that  and we have also found above that , we can write as;

(iii)        

We can find equation of the curve from its derivative through integration;

We have found in (i) that;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

We have found in (ii) that , which means we can write the above equation as;

Hence;

A stationary value is the maximum or minimum value of a function.

Stationary value of a curve (function) occurs at stationary point on the curve.

We are given that  has a minimum value when .

By substituting x-coordinate of stationary point on the curve in the equation of the function gives the  stationary value of that function.

Therefore, minimum value of the above found function can be found by substituting  in ;

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