Question

A cubic curve $y = f(x)$ passes through the points $(1, -7)$ and $(2, 11)$. If $\frac{dy}{dx} = 6x^2 + kx - 5$, where $k$ is a constant, then $f(x) =$}

• A. $2x^3 + 6x^2 - 5x - 7$
• B. $2x^3 + 3x^2 - 5x - 7$
• C. $2x^3 + 3x^2 - 5x - 4$
• D. $2x^3 + 3x^2 - 5x - 5$

Hint:

Check the options! All options start with $2x^3$ and have $-5x$. Testing point $(1, -7)$ in the options is often faster.
For (B): $2(1) + 3(1) - 5(1) - 7 = -7$. Matches.
For (A): $2 + 6 - 5 - 7 = -4 \neq -7$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We integrate the given derivative $\frac{dy}{dx}$ to find the general equation of the curve, which will contain two unknowns ($k$ and the integration constant $C$). We then use the two given points to solve for these constants.

Step 2: Detailed Explanation:
1. Integrate $\frac{dy}{dx} = 6x^2 + kx - 5$:
\[ f(x) = \int (6x^2 + kx - 5) dx = 2x^3 + \frac{k}{2}x^2 - 5x + C \]
2. Use point $(1, -7)$:
\[ -7 = 2(1)^3 + \frac{k}{2}(1)^2 - 5(1) + C \implies -7 = 2 + \frac{k}{2} - 5 + C \]
\[ \frac{k}{2} + C = -4 \quad \dots \text{(Eqn 1)} \]
3. Use point $(2, 11)$:
\[ 11 = 2(2)^3 + \frac{k}{2}(2)^2 - 5(2) + C \implies 11 = 16 + 2k - 10 + C \]
\[ 2k + C = 5 \quad \dots \text{(Eqn 2)} \]
4. Subtract Eqn 1 from Eqn 2:
\[ (2k - \frac{k}{2}) = 5 - (-4) \implies \frac{3k}{2} = 9 \implies k = 6 \]
5. Find $C$:
\[ 2(6) + C = 5 \implies 12 + C = 5 \implies C = -7 \]
6. The coefficient of $x^2$ is $\frac{k}{2} = \frac{6}{2} = 3$.
Substituting these into the curve equation: $f(x) = 2x^3 + 3x^2 - 5x - 7$.

Step 3: Final Answer:
The equation of the curve is $2x^3 + 3x^2 - 5x - 7$.