Problem 162 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(y=f(x)\) has a second order derivative \(f^{\prime \prime}(x)=6(x-1)\). If its graph passes through the point \((2,1)\) and at that point the tangent to the graph is \(y=\) \(3 x-5\), then the function is (A) \((x-1)^{2}\) (B) \((x-1)^{3}\) (C) \((x+1)^{3}\) (D) \((x+1)^{2}\)

Step-by-Step Solution

Answer

The function is \((x-1)^{3}\).

Step 1: Identify the Second Derivative

We start with the given second derivative of the function: \( f''(x) = 6(x-1) \). This is the information provided related to the rate of change of slope of our function.

Step 2: Integrate to Find the First Derivative

To find the first derivative, \( f'(x) \), we integrate the second derivative: \[ \int 6(x-1) \, dx. \]This yields:\[ f'(x) = 3(x-1)^2 + C_1, \]where \( C_1 \) is the constant of integration.

Step 3: Use the Tangent Line Information

We know the slope of the tangent at \( x=2 \) is 3. Therefore, \( f'(2) = 3 \). Plugging \( x=2 \) into our expression for \( f'(x) \),\[ 3 = 3(2-1)^2 + C_1, \] \[ 3 = 3 \cdot 1 + C_1. \]This simplifies to \( C_1 = 0 \), so \[ f'(x) = 3(x-1)^2. \]

Step 4: Integrate to Find the Original Function

Next, we find \( f(x) \) by integrating the first derivative: \[ \int 3(x-1)^2 \, dx. \]Perform the integration: \[ f(x) = (x-1)^3 + C_2, \]where \( C_2 \) is another constant.

Step 5: Use the Given Point to Find the Constant

The graph passes through \((2,1)\), meaning \( f(2) = 1 \). Substitute \( x=2 \) and \( f(x)=1 \) into the function:\[ 1 = (2-1)^3 + C_2, \]\[ 1 = 1 + C_2. \]Solving this gives \( C_2 = 0 \), so \[ f(x) = (x-1)^3. \]

Key Concepts

Second Derivative

The second derivative of a function is a fundamental concept in calculus. When we talk about the second derivative, we are referring to the derivative of the derivative of a function. In other words, it tells us how the rate of change of a function's slope is varying. For the exercise problem, the second derivative function is given as \( f''(x) = 6(x-1) \).

• This indicates that the rate at which the slope of the tangent line is changing depends on \(x\).
• The second derivative can indicate the concavity of the function's graph. If \( f''(x) > 0 \), the function is concave up at that interval, and if \( f''(x) < 0 \), it is concave down.

Understanding how changes affect the slope is crucial for predicting the behavior of the function and its graph. It can help identify points of inflection where the concavity changes.

Integration

Integration can be considered the reverse process of differentiation. To find the original function from its derivative, we integrate. This exercise involves integrating the second derivative to find the first derivative, and then integrating again to find the original function.
Steps to integrate include:

• Identify the function to integrate, such as \( 6(x-1) \) in the example.
• Perform the integration: \( \int 6(x-1) \, dx = 3(x-1)^2 + C_1 \), where \( C_1 \) is the constant of integration.
  
• Use any additional conditions, like the slope of a tangent or points on the graph, to find constants.

Integration allows us to rebuild the function step by step, using constants derived from given conditions such as points the function passes through or the slope of a tangent line.

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. The slope of the tangent line is given by the first derivative at that point. In this problem, we use the tangent line equation \( y = 3x - 5 \) at the point \((2, 1)\).

• Knowing the tangent's slope helps us set \( f'(2) = 3 \), which assists in calculating integration constants.
• Tangent lines are useful for approximating values of the function near the point where the tangent is drawn.
• The information a tangent provides can simplify solving equations by using straight lines instead of potentially more complex curves.

Analyzing the tangent line allows you to confirm if your integrated function aligns with known behavior at specific points, makes connections between different calculus concepts, and ensures you are on the right path with the equation of the function you are integrating toward.