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Question: If \(f(x)\) is a non-zero polynomial of degree four, having local extreme points at \(x = - 1,0,1;\)...

If f(x)f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1,0,1;x = - 1,0,1; then the set S=x∈R:f(x)=f(0)S = \\{ x \in R:f(x) = f(0)\\}
Contains exactly:
A.{\text{A}}{\text{.}}Four irrational numbers
B.{\text{B}}{\text{.}}Two irrational number and one rational number
C.{\text{C}}{\text{.}}Four rational numbers
D. {\text{D}}{\text{. }}Two irrational and two rational numbers

Explanation

Solution

In this question we first need to form f′(x)f'(x) using local extreme points at x=−1,0,1;x = - 1,0,1;. Then, integrate it to get f(x)f(x). After that we use the relation S=x∈R:f(x)=f(0)S = \\{ x \in R:f(x) = f(0)\\} to get the required result.

Complete step-by-step answer:
A local extremum (extreme) of a function is the point at which a maximum or minimum value of the function in some open interval containing the point is obtained.
Since, f(x)f(x) is a non-zero polynomial of degree four then its derivative f′(x)f'(x) is polynomial of degree three because on differentiation of any function its degree will decrease by one. It is given that local extreme points of f(x)f(x) are at x=−1,0,1;x = - 1,0,1;
We can write f′(x)f'(x) as
⇒f′(x)=a(x+1)(x−0)(x−1) ⇒f′(x)=a(x)(x2−1) ⇒f′(x)=a(x3−x) ………..eq.1   \Rightarrow f'(x) = a(x + 1)(x - 0)(x - 1) \\\ \Rightarrow f'(x) = a(x)({x^2} - 1) \\\ \Rightarrow f'(x) = a({x^3} - x){\text{ ………..eq.1 }} \\\
On integration of eq.1 with respect to xx, we get
⇒∫f′(x)dx=∫a(x3−x)dx ⇒∫f′(x)dx=a∫(x3−x)dx  \Rightarrow \int {f'(x)dx = \int {a({x^3} - x)dx} } \\\ \Rightarrow \int {f'(x)dx = a\int {({x^3} - x)dx} } \\\
Now, using the formula of integration ∫xndx=xn+1n+1\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}
⇒f(x) = a (x44)−(x22) + c \Rightarrow f(x){\text{ = a\\{ (}}\dfrac{{{x^4}}}{4}) - (\dfrac{{{x^2}}}{2})\\} {\text{ + }}c{\text{ }}( c = constant ) ------eq.2
It is given that
S=x∈R:f(x)=f(0)S = \\{ x \in R:f(x) = f(0)\\}
For getting f(0)f(0)put “0” in eq,2 ,we get
f(0)=c —-eq.3 f(0) = c{\text{ ----eq.3 }}
Now, equate eq.2 and eq.3, we get
From given
⇒f(x)=f(0)  ⇒ a (x44)−(x22) + c = c ⇒ a (x44)−(x22) = 0 ⇒  (x44)−(x22) = 0 ⇒ x22 (x22)−1 = 0 ⇒ x2(x2−2) = 0 ⇒ x=0,2,−2  \Rightarrow f(x) = f(0){\text{ }} \\\ \Rightarrow {\text{ a\\{ (}}\dfrac{{{x^4}}}{4}) - (\dfrac{{{x^2}}}{2})\\} {\text{ + }}c{\text{ }} = {\text{ }}c \\\ \Rightarrow {\text{ a\\{ (}}\dfrac{{{x^4}}}{4}) - (\dfrac{{{x^2}}}{2})\\} {\text{ }} = {\text{ 0}} \\\ \Rightarrow {\text{ \\{ (}}\dfrac{{{x^4}}}{4}) - (\dfrac{{{x^2}}}{2})\\} {\text{ }} = {\text{ 0}} \\\ \Rightarrow {\text{ }}\dfrac{{{x^2}}}{2}{\text{\\{ (}}\dfrac{{{x^2}}}{2}) - 1\\} {\text{ }} = {\text{ 0}} \\\ \Rightarrow {\text{ }}{x^2}{\text{(}}{x^2} - 2){\text{ }} = {\text{ 0}} \\\ \Rightarrow {\text{ }}x = 0,\sqrt 2 , - \sqrt 2 \\\

Therefore, by using the given conditions we get three solutions as 0, 2,−2\sqrt 2 , - \sqrt 2 .In the solution, 0 is a rational number while 2,−2\sqrt 2 , - \sqrt 2 are irrational numbers. So there are two irrational numbers and one rational number.

So, the correct answer is “Option B”.

Note: Whenever you get this type of question the key concept of solving is to find f′(x)f'(x)with the help of local extreme points and f(x)f(x) by integrating the expression of f′(x)f'(x).And remember one thing the difference between rational and irrational number. Rational numbers can be expressed as a ratio of two numbers (pq\dfrac{{\text{p}}}{q} form) while irrational numbers cannot be expressed as a ratio of two numbers.