Question

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

Answer 1

Hint: In this question we first need to form $f'(x)$ using local extreme points at $x = - 1,0,1;$. Then, integrate it to get $f(x)$. After that we use the relation $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, $f(x)$ is a non-zero polynomial of degree four then its derivative $f'(x)$ 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 $f(x)$ are at $x = - 1,0,1;$
We can write $f'(x)$ as
$
   \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 $x$, we get
$
   \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 $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
\[ \Rightarrow f(x){\text{ = a\{ (}}\dfrac{{{x^4}}}{4}) - (\dfrac{{{x^2}}}{2})\} {\text{ + }}c{\text{ }}\]( c = constant ) ------eq.2
It is given that
$S = \{ x \in R:f(x) = f(0)\} $
For getting $f(0)$put “0” in eq,2 ,we get
$f(0) = c{\text{ ----eq.3 }}$
Now, equate eq.2 and eq.3, we get
From given
$
   \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, \[\sqrt 2 , - \sqrt 2 \] .In the solution, 0 is a rational number while \[\sqrt 2 , - \sqrt 2 \] 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 $f'(x)$with the help of local extreme points and $f(x)$ by integrating the expression of $f'(x)$.And remember one thing the difference between rational and irrational number. Rational numbers can be expressed as a ratio of two numbers ($\dfrac{{\text{p}}}{q}$ form) while irrational numbers cannot be expressed as a ratio of two numbers.