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 are given the derivative of a function. We can find the original function \(f(x)\) by integrating the derivative. The integration will introduce a constant \(C\). We use the two given points to solve for both \(k\) and \(C\).
Step 2: Key Formula or Approach:
Integrate: \(y = \int (6x^2 + kx - 5) dx\).
Substitute \((x, y) = (1, -7)\) to get Equation 1.
Substitute \((x, y) = (2, 11)\) to get Equation 2.
Solve the system for \(k\) and \(C\).
Step 3: Detailed Explanation:
Find the general equation of the curve:
\[ y = \int (6x^2 + kx - 5) dx \] \[ y = 2x^3 + \frac{k}{2}x^2 - 5x + C \] Use point (1, -7):
\[ -7 = 2(1)^3 + \frac{k}{2}(1)^2 - 5(1) + C \] \[ -7 = 2 + \frac{k}{2} - 5 + C \] \[ -7 = -3 + \frac{k}{2} + C \implies \frac{k}{2} + C = -4 \implies k + 2C = -8 \quad \text{--- (Eq 1)} \] Use point (2, 11):
\[ 11 = 2(2)^3 + \frac{k}{2}(2)^2 - 5(2) + C \] \[ 11 = 16 + 2k - 10 + C \] \[ 11 = 6 + 2k + C \implies 2k + C = 5 \quad \text{--- (Eq 2)} \] Solve the system. From Eq 2, \(C = 5 - 2k\). Substitute into Eq 1:
\[ k + 2(5 - 2k) = -8 \] \[ k + 10 - 4k = -8 \] \[ -3k = -18 \implies k = 6 \] Find \(C\):
\[ C = 5 - 2(6) = 5 - 12 = -7 \] Substitute \(k\) and \(C\) back into the general equation:
\[ f(x) = 2x^3 + \frac{6}{2}x^2 - 5x - 7 \] \[ f(x) = 2x^3 + 3x^2 - 5x - 7 \] Step 4: Final Answer:
The function is \(2x^3 + 3x^2 - 5x - 7\).