Question

Let \[p\left( x \right)\] be a real polynomial of least degree which has a local maximum at \[x = 1\] and a local minimum at \[x = 3\] . lf \[p\left( 1 \right) = 6\] and \[p\left( 3 \right) = 2\] , then \[p{\text{ }}\prime \left( 0 \right)\] is
A. 9
B. 13
C. 11
D. 12

Answer 1

Hint: If a polynomial of lowest degree p has zeros at \[x = x{}_1,x{}_2, \ldots ,x{}_n\] , then the polynomial can be written in the factored form: \[f(x) = a{(x - {x_1})^{{p_1}}}{(x - {x_2})^{{p_2}}} \cdots {(x - {x_n})^{{p_n}}}\] where the powers on each factor can be determined by the behaviour of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x-intercept.
We will assume two factors from local maximum and local minimum of a polynomial of least degree and integrate it. Then try to solve the polynomial for x=1 and x=3 and find equations in terms of k and c, one is constant for polynomial and other is constant of integration. With the value of k we will find \[p'\left( x \right)\] and then putting x=0, we will find the value of \[p{\text{ }}\prime \left( 0 \right)\] .

Complete step-by-step solution:
Given that \[p\left( x \right)\] be a real polynomial of least degree
So, \[p'\left( x \right)\] is a real polynomial of least degree.
It has two extremes: local maximum at \[x = 1\] and local minimum at \[x = 3\] . As Maxima and minima are derived from derivatives do we will assume (x-1) and (x-3) factors of p(x) derivative..
Hence,
 \[p'\left( x \right) = k\left( {x - 1} \right)\left( {x - 3} \right)\]
Open the brackets and multiply.
 \[ \Rightarrow p'\left( x \right) = k\left( {{x^2} - 4x + 3} \right)\]
Then by integrating both side with respect to x
We get
 \[ \Rightarrow p(x) = k\left( {\dfrac{{{x^3}}}{3} - 2{x^2} + 3x} \right) + c\]
Now taking into consideration, if \[p\left( 1 \right) = 6\] and \[p\left( 3 \right) = 2\] , we will substitute these values in the above equation.
Given that \[p\left( 1 \right) = 6\] , i.e. substituting the value of x=1
 \[ \Rightarrow p(1) = k\left( {\dfrac{{{1^3}}}{3} - 2{{(1)}^2} + 3(1)} \right) + c = 6\]
Now simplifying the equation, we get
 \[ \Rightarrow \dfrac{4}{3}k + c = 6................(1)\]
Given that \[p\left( 3 \right) = 2\] i.e. substituting the value of x=3
 \[ \Rightarrow p(3) = k\left( {\dfrac{{{3^3}}}{3} - 2{{(3)}^2} + 3(3)} \right) + c = 2\]
Now simplifying the equation, we get
 \[ \Rightarrow c = 2\]
Now substituting the value of c in the equation (1)
 \[ \Rightarrow \dfrac{4}{3}k + 2 = 6 \Rightarrow k = 3\]
So, by using the equation \[ \Rightarrow p'\left( x \right) = k\left( {{x^2} - 4x + 3} \right)\] and substituting the value of k our polynomial becomes:-
 \[p'\left( x \right) = 3\left( {x - 1} \right)\left( {x - 3} \right)\]
We need to find the value of \[p{\text{ }}\prime \left( 0 \right)\] i.e. substituting the value of x=0
 \[
   \Rightarrow p'\left( 0 \right) = 3\left( {0 - 1} \right)\left( {0 - 3} \right) \\
   \Rightarrow p'\left( 0 \right) = 9
 \]

So, option (A) is the correct answer.

Note: A local maximum or local minimum at x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a. If a function has a local maximum at a, then \[f\left( a \right) \geqslant f\left( x \right)\] for all x in an open interval around x = a.