Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)\) be a function defined by \(f(x)=\) \(\int_{1}^{x} x\left(x^{2}-3 x+2\right) d x, 1 \leq x \leq 3\), then the range of \(f(x)\) is (A) \(\left[-\frac{1}{4}, 2\right]\) (B) \(\left[-\frac{1}{4}, 4\right]\) (C) \([0,2]\) (D) None of these

Step-by-Step Solution

Answer

The range of \( f(x) \) is \([-\frac{1}{4}, 2]\).

Step 1: Identify the Function

First, we identify the function given in the problem: \( f(x) = \int_{1}^{x} x\left(x^{2} - 3x + 2\right) \, dx \). This expression describes the function \( f(x) \) as the definite integral from 1 to \( x \) of the function \( x(x^2 - 3x + 2) \).

Step 2: Simplify the Integrand

We simplify the expression inside the integral. The integrand is \( x(x^2 - 3x + 2) \). Expand this to get \( x^3 - 3x^2 + 2x \).

Step 3: Compute the Integral

Now we compute the integral of the polynomial \( x^3 - 3x^2 + 2x \). The integral becomes \( \frac{x^4}{4} - x^3 + x^2 \).

Step 4: Apply Definite Limits

Substitute the limits 1 and \( x \) in the antiderivative: \( f(x) = \left(\frac{x^4}{4} - x^3 + x^2\right) - \left(\frac{1^4}{4} - 1^3 + 1^2\right) \). This simplifies to: \( \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4} + 1 - 1 \). Simplifying further, \( f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4} \).

Step 5: Determine the Range of f(x)

Now, we need to determine the range of \( f(x) \) over the interval \( 1 \leq x \leq 3 \). This involves finding the values of \( f(x) \) at the boundaries and checking for any extreme points. Calculate \( f(1) = -\frac{1}{4} \) and \( f(3) = 0 \). Check the derivative \( f'(x) = x^3 - 3x^2 + 2x \) to find any critical points in the interval. Solve \( x(x-1)(x-2) = 0 \), giving critical points at \( x = 1, 2 \). Evaluate \( f(2) = 2 \).

Step 6: Choose the Correct Range

From the values calculated, \( f(x) \) takes values from \(-\frac{1}{4}\) to 2 on the interval \( 1 \leq x \leq 3 \). This matches the range given in option (A).

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus used to compute the accumulation of quantities, typically representing the area under a curve or between curves over a specific interval.

• Boundaries: In the case of the exercise, the definite integral is taken from 1 to some value of \(x\) within \([1, 3]\). These are your boundaries or limits of integration.
• Expression Understanding: The given function \(f(x)\) in the exercise is defined as the definite integral of the polynomial \(x(x^2 - 3x + 2)\). This means we accumulate all the tiny pieces of the area under the curve from 1 up to \(x\).

To solve the integral:- First, express the integral in terms of a simplified polynomial so it can be easily integrated.- Perform the integration process step-by-step to get a general antiderivative.- Finally, apply the Fundamental Theorem of Calculus by substituting the upper and lower bounds into the antiderivative to get a particular solution, \(f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4}\). This function describes the accumulated area as \(x\) varies over its interval, leading to insights about the function's range.

Polynomial Integration

The process of polynomial integration involves integrating terms like \(x^n\) individually and combining them to find the integral of the entire polynomial.

• Power Rule: The power rule states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). This rule is fundamental when transforming individual terms of a polynomial integrand.
• Breaking Down the Polynomial: In our case, the integrand \(x^3 - 3x^2 + 2x\) is simplified by applying the power rule to each term separately:
  • \(\int x^3 \, dx = \frac{x^4}{4}\)
  • \(\int -3x^2 \, dx = -\frac{3x^3}{3} = -x^3\)
  • \(\int 2x \, dx = x^2\)
  Combine these to form the antiderivative \(\frac{x^4}{4} - x^3 + x^2 + C\).

Pay close attention to each step in polynomial integration, as correctly handling the coefficients and signs will ensure the resulting function is accurate.

Critical Points Analysis

Critical points of a function are where its derivative is zero or undefined, indicating potential maximums, minimums, or inflection points. Analyzing these points helps determine the range of a function, especially when dealing with integrals over an interval.

• Derivative Calculation: For the given function \(f(x)\), the derivative is the original polynomial \(x^3 - 3x^2 + 2x\). Setting \(f'(x)\) equal to zero yields the equation \(x(x-1)(x-2) = 0\).
• Solving for Critical Points: Solve \(x(x-1)(x-2) = 0\) to find the critical values of \(x\), which are \(x = 0, 1, 2\). But relevant within the given interval are \(x = 1\) and \(x = 2\).
• Function Evaluation at Critical Points: Evaluate \(f(x)\) at \(x = 1, 2, \) and other boundaries to find the range:
  • \(f(1) = -\frac{1}{4}\)
  • \(f(2) = 2\)
  • Looking at these values, the range of \(f(x)\) over the interval is \([ -\frac{1}{4}, 2 ]\).

Boundaries, along with values at critical points, provide important insights into how the function behaves across the given domain. This method ensures comprehensive understanding and validation of the solution.