Question

Let \( f(x) \) be a polynomial of degree \( 5 \) having extreme values at \( x = 1 \) and \( x = -1 \). If \[ \lim_{x \to 0} \frac{f(x)}{x^3} = -5, \] then the value of \( f(B) - f(-2) \) is

• A. 110
• B. 112
• C. 115
• D. 118

Hint:

When a limit of the form \( \lim_{x\to 0}\frac{f(x)}{x^3} \) is finite, the polynomial must not contain terms of degree less than \( 3 \). Then use the derivative condition for maxima or minima.

Answer 1

The Correct Option is B

Step 1: Use the given limit.
Since \[ \lim_{x \to 0} \frac{f(x)}{x^3} = -5, \] the polynomial \( f(x) \) must have \( x^3 \) as a factor near \( x=0 \), and the coefficient of \( x^3 \) must be \(-5\).
Because \( f(x) \) is a polynomial of degree \( 5 \), let us write \[ f(x)=ax^5+bx^4-5x^3. \] There are no constant, linear, or quadratic terms, otherwise the limit would not be finite and equal to \(-5\).

Step 2: Use the condition of extreme values at \( x=1 \) and \( x=-1 \).
If \( f(x) \) has extreme values at \( x=1 \) and \( x=-1 \), then \[ f'(A)=0 \quad \text{and} \quad f'(-1)=0. \] Now differentiate: \[ f'(x)=5ax^4+4bx^3-15x^2. \] So, \[ f'(A)=5a+4b-15=0 \] and \[ f'(-1)=5a-4b-15=0. \]
Step 3: Solve for \( a \) and \( b \).
Adding the two equations: \[ (5a+4b-15)+(5a-4b-15)=0 \] \[ 10a-30=0 \] \[ a=3. \] Substituting into \[ 5a+4b-15=0, \] we get \[ 15+4b-15=0 \] \[ 4b=0 \] \[ b=0. \] Hence, \[ f(x)=3x^5-5x^3. \]
Step 4: Calculate \( f(B)-f(-2) \).
First, \[ f(B)=3(B)^5-5(B)^3=3\cdot 32-5\cdot 8=96-40=56. \] Now, \[ f(-2)=3(-2)^5-5(-2)^3=3(-32)-5(-8)=-96+40=-56. \] Therefore, \[ f(B)-f(-2)=56-(-56)=112. \] Final Answer: \( 112 \)