Question

Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.

If \( \mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4 \), then which one of the following is not true?

• A. ƒ(1) - 4ƒ(-1) = 4.
• B. x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.
• C. x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ.
• D. ƒ is an odd function.

Hint:

The limit condition fixes the lowest-order terms of f; critical points at +-1 give the remaining coefficients; then test the nature of extrema.

Answer 1

The Correct Option is B

Step 1: \(\lim_{x\to0}\left(2+\dfrac{f(x)}{x^3}\right)=4\) means \(\dfrac{f(x)}{x^3}\to2\), so f has no terms below \(x^3\) and the \(x^3\) coefficient is 2: \(f(x)=ax^5+bx^4+2x^3\).
Step 2: \(f'(x)=x^2(5ax^2+4bx+6)\). Critical points \(\pm1\): \(5a+4b+6=0\) and \(5a-4b+6=0\Rightarrow b=0,\;a=-\tfrac65\). So \(f(x)=-\tfrac65x^5+2x^3\), an odd function.
Step 3: \(f'(x)=-6x^2(x-1)(x+1)\): f' changes from + to - at x = 1 (maximum) and from - to + at x = -1 (minimum).
Step 4: \(f(1)=\tfrac45,\;f(-1)=-\tfrac45\), so \(f(1)-4f(-1)=\tfrac45+\tfrac{16}5=4\). True.
Step 5: Statement (C) (maxima at 1, minima at -1) is true and (D) is true, while (B) (minima at 1 and maxima at -1) is false.
Answer: (B)