Key Concepts and Formulas

• Limits and Polynomials: If $\lim_{x \to 0} \frac{f(x)}{x^n} = L$ exists and is non-zero, then $f(x)$ has a term $Lx^n$ and all terms of degree less than $n$ are zero.
• Critical Points: A critical point of a function $f(x)$ occurs at $x=c$ if $f'(c) = 0$ or $f'(c)$ is undefined. For polynomials, we only need to consider $f'(c) = 0$.
• First Derivative Test: If $f'(x)$ changes sign from positive to negative at $x=c$, then $f(x)$ has a local maximum at $x=c$. If $f'(x)$ changes sign from negative to positive at $x=c$, then $f(x)$ has a local minimum at $x=c$.

Step-by-Step Solution

Step 1: Determine the form of f(x)

We are given that $\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$. This can be rewritten as $\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^3}}} = 2$. Since the limit exists and is equal to 2, it implies that $f(x)$ must have the form $f(x) = 2x^3 + ax^4 + bx^5$ where $a$ and $b$ are constants. The polynomial cannot have terms of degree less than 3, otherwise the limit would be either 0 or undefined. Also, the coefficient of the $x^3$ term must be 2.

Step 2: Find the first derivative f'(x)

We have $f(x) = 2x^3 + ax^4 + bx^5$. Taking the derivative with respect to x, we get: $f'(x) = 6x^2 + 4ax^3 + 5bx^4$

Step 3: Use the critical point information

We are given that $x = \pm 1$ are critical points of $f(x)$. This means $f'(1) = 0$ and $f'(-1) = 0$. Plugging these values into the expression for $f'(x)$, we get:

$f'(1) = 6(1)^2 + 4a(1)^3 + 5b(1)^4 = 6 + 4a + 5b = 0$ $f'(-1) = 6(-1)^2 + 4a(-1)^3 + 5b(-1)^4 = 6 - 4a + 5b = 0$

Step 4: Solve for a and b

We have a system of two linear equations: $4a + 5b = -6$ $-4a + 5b = -6$

Adding the two equations, we get $10b = -12$, so $b = -\frac{6}{5}$. Substituting this value of $b$ into the first equation: $4a + 5(-\frac{6}{5}) = -6 \implies 4a - 6 = -6 \implies 4a = 0 \implies a = 0$.

Therefore, $a = 0$ and $b = -\frac{6}{5}$.

Step 5: Determine the polynomial f(x)

Substituting the values of $a$ and $b$ into the expression for $f(x)$, we get: $f(x) = 2x^3 + 0x^4 - \frac{6}{5}x^5 = 2x^3 - \frac{6}{5}x^5$

Step 6: Check if f(x) is an odd function

$f(-x) = 2(-x)^3 - \frac{6}{5}(-x)^5 = -2x^3 + \frac{6}{5}x^5 = -(2x^3 - \frac{6}{5}x^5) = -f(x)$. Since $f(-x) = -f(x)$, $f(x)$ is an odd function. Thus, option (D) is true.

Step 7: Find the second derivative f''(x)

$f'(x) = 6x^2 - 6x^4$. Taking the derivative with respect to x, we get: $f''(x) = 12x - 24x^3$

Step 8: Analyze the nature of critical points using the second derivative test

$f''(1) = 12(1) - 24(1)^3 = 12 - 24 = -12 < 0$. Since $f''(1) < 0$, $x = 1$ is a point of local maxima. $f''(-1) = 12(-1) - 24(-1)^3 = -12 + 24 = 12 > 0$. Since $f''(-1) > 0$, $x = -1$ is a point of local minima.

Thus, option (C) is true, and option (B) is false.

Step 9: Evaluate f(1) and f(-1)

$f(1) = 2(1)^3 - \frac{6}{5}(1)^5 = 2 - \frac{6}{5} = \frac{10 - 6}{5} = \frac{4}{5}$ $f(-1) = 2(-1)^3 - \frac{6}{5}(-1)^5 = -2 + \frac{6}{5} = \frac{-10 + 6}{5} = -\frac{4}{5}$

Step 10: Check option (A)

$f(1) - 4f(-1) = \frac{4}{5} - 4(-\frac{4}{5}) = \frac{4}{5} + \frac{16}{5} = \frac{20}{5} = 4$. Thus option (A) is true.

Since we are looking for the statement that is NOT true, and we have found that options (B), (C) and (D) are true, we examine option (A) again.

The question asks which statement is NOT true. We have that option (C) says "x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ". This is consistent with our calculations.

Option (B) says "x = 1 is a point of minima and x = -1 is a point of maxima of ƒ." This contradicts our calculation. So option (B) is false.

Common Mistakes & Tips

• Carefully check the signs when evaluating derivatives and plugging in values. A small sign error can lead to incorrect conclusions.
• Remember that the existence of a limit as $x \to 0$ gives information about the lower-degree terms of the polynomial.
• When dealing with critical points, use the first or second derivative test to determine the nature of the extrema (maxima, minima, or inflection points).

Summary

We found the polynomial $f(x)$ using the limit condition and the critical point information. We then used the first and second derivatives to analyze the nature of the critical points and determine which of the given statements is not true. We concluded that the statement "x = 1 is a point of minima and x = -1 is a point of maxima of ƒ" is not true, because x = 1 is a point of maxima and x = -1 is a point of minima.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).