Question

If \[ \frac{d}{dx} \left(\frac{1 + x^2 + x^4}{1 + x + x^2}\right) = ax + b, \] then \( (a,b) \) is:

• A. \( (-1,2) \)
• B. \( (-2,1) \)
• C. \( (2,-1) \)
• D. \( (1,2) \)

Hint:

For differentiating fractions, use the quotient rule: \( \frac{f'g - fg'}{g^2} \).

Answer 1

The Correct Option is C

Step 1: Differentiating using quotient rule Using the quotient rule: \[ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'g - fg'}{g^2}. \] Let \( f(x) = 1 + x^2 + x^4 \), \( g(x) = 1 + x + x^2 \). Computing derivatives: \[ f'(x) = 2x + 4x^3, \quad g'(x) = 1 + 2x. \] Applying the quotient rule and simplifying: \[ \frac{(2x + 4x^3)(1 + x + x^2) - (1 + x^2 + x^4)(1 + 2x)}{(1 + x + x^2)^2}. \] Simplifying, we get: \[ 2x - 1. \] Thus, \( a = 2 \), \( b = -1 \).

Answer 2

Given:
\[ y = \frac{1 + x^2 + x^4}{1 + x + x^2} \]

We need to find:
\[ \frac{dy}{dx} = a x + b \] and then find \( (a, b) \).

Step 1: Let numerator \( u = 1 + x^2 + x^4 \) and denominator \( v = 1 + x + x^2 \).
Compute derivatives:
\[ u' = 2x + 4x^3 \] \[ v' = 1 + 2x \]

Step 2: Use the quotient rule:
\[ \frac{dy}{dx} = \frac{u' v - u v'}{v^2} \]

Step 3: Calculate numerator of derivative:
\[ u' v = (2x + 4x^3)(1 + x + x^2) \]
Multiply:
\[ = (2x)(1 + x + x^2) + (4x^3)(1 + x + x^2) \] \[ = 2x + 2x^2 + 2x^3 + 4x^3 + 4x^4 + 4x^5 = 2x + 2x^2 + 6x^3 + 4x^4 + 4x^5 \]

\[ u v' = (1 + x^2 + x^4)(1 + 2x) \] \[ = (1)(1 + 2x) + x^2 (1 + 2x) + x^4 (1 + 2x) \] \[ = 1 + 2x + x^2 + 2x^3 + x^4 + 2x^5 \]

Step 4: Subtract:
\[ u' v - u v' = (2x + 2x^2 + 6x^3 + 4x^4 + 4x^5) - (1 + 2x + x^2 + 2x^3 + x^4 + 2x^5) \] \[ = 2x - 2x + 2x^2 - x^2 + 6x^3 - 2x^3 + 4x^4 - x^4 + 4x^5 - 2x^5 - 1 \] \[ = 0 + x^2 + 4x^3 + 3x^4 + 2x^5 - 1 \] or reordered:
\[ = -1 + x^2 + 4x^3 + 3x^4 + 2x^5 \]

Step 5: The denominator is:
\[ v^2 = (1 + x + x^2)^2 = 1 + 2x + 3x^2 + 2x^3 + x^4 \]

Step 6: Since the problem states:
\[ \frac{dy}{dx} = a x + b, \] the derivative must be a linear polynomial.
So the complicated fraction simplifies to \( a x + b \).

Step 7: To find \( a \) and \( b \), substitute convenient values of \( x \) into the derivative formula and equate.
Calculate \( \frac{dy}{dx} \) at \( x=0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-1 + 0 + 0 + 0 + 0}{1} = -1 \] This equals \( a \cdot 0 + b = b \), so
\[ b = -1 \]

Step 8: Calculate \( \frac{dy}{dx} \) at \( x=1 \):
Numerator:
\[ -1 + 1 + 4 + 3 + 2 = 9 \]
Denominator:
\[ 1 + 2 + 3 + 2 + 1 = 9 \]
So:
\[ \frac{dy}{dx}\bigg|_{x=1} = \frac{9}{9} = 1 \] This equals \( a \cdot 1 + b = a -1 \), so
\[ 1 = a - 1 \implies a = 2 \]

Therefore,
\[ (a, b) = (2, -1) \]