Problem 217 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(p(x)\) be a function defined on \(R\) such that \(p^{\prime}(x)=\) \(p^{\prime}(1-x)\), for all \(x \in[0,1], p(0)=1\) and \(p(1)=41\) Then \(\int_{0}^{1} p(x) d x\) equal (A) 21 (B) 41 (C) 42 (D) 20

Step-by-Step Solution

Answer

The integral \(\int_{0}^{1} p(x) \, dx = 21\); correct answer option is (A).

Step 1: Analyze the Function Condition

The problem states that for all \(x \in [0, 1]\), the derivative \(p'(x) = p'(1-x)\). This implies that the rate of change of the function is symmetric around \(x = \frac{1}{2}\).

Step 2: Deduce p(x) as a Linear Function

Given the symmetry of the derivative and the boundary conditions \(p(0) = 1\) and \(p(1) = 41\), \(p(x)\) is likely a linear function because any deviation from linearity would generally ruin this symmetry.

Step 3: Assume p(x) as a Linear Function

We assume a linear form: \(p(x) = ax + b\). Using the conditions \(p(0) = 1\) and \(p(1) = 41\), we find: \[ p(0) = a(0) + b = 1 \Rightarrow b = 1 \]\[ p(1) = a(1) + 1 = 41 \Rightarrow a + 1 = 41 \Rightarrow a = 40 \] Thus, \(p(x) = 40x + 1\).

Step 4: Calculate the Integral

Integrate \(p(x) = 40x + 1\) from 0 to 1. The integral is calculated as follows:\[ \int_{0}^{1} p(x) \, dx = \int_{0}^{1} (40x + 1) \, dx \]\[ = \left[ 20x^2 + x \right]_{0}^{1} \]\[ = (20(1)^2 + 1) - (20(0)^2 + 0) \]\[ = 20 + 1 = 21 \]

Step 5: Verify Solution with Options

The computed integral is 21, which matches option (A). We calculated based on normal linear assumptions and constraints provided.

Key Concepts

Linear Functions

Linear functions are one of the simplest types of functions in mathematics. They are defined by equations of the form \(y = ax + b\), where \(a\) and \(b\) are constants. This means that the graph of a linear function is a straight line. Linear functions have a constant rate of change or slope, represented by the coefficient \(a\).

• Slope (\(a\)): This tells us how steep the line is. A larger \(a\) value means steeper lines. If \(a\) is positive, the line rises; if it's negative, the line falls as you move from left to right.
• Y-intercept (\(b\)): This is the point where the line crosses the y-axis. It's the value of \(y\) when \(x = 0\).

In this exercise, the function \(p(x) = 40x + 1\) is a linear function. Here, \(a = 40\) and \(b = 1\). The function starts at 1 when \(x = 0\) and ends at 41 when \(x = 1\), consistently increasing by 40 as \(x\) progresses by 1 unit. The symmetry of its derivative hints towards this linearity.

Symmetry in Derivatives

Symmetry in derivatives is an intriguing concept in calculus. It indicates that a function's derivative has the same behavior around a specific point or interval. In this exercise, the condition \(p'(x) = p'(1-x)\) implies that the rate of change of \(p(x)\) is mirrored about \(x = \frac{1}{2}\).

• This means the increase in \(p(x)\) from 0 to \(\frac{1}{2}\) is balanced by the increase from \(\frac{1}{2}\) to 1.
• Such symmetry often simplifies problems and can suggest underlying properties like linearity.

For instance, if any deviation from linearity occurred, it would disrupt this perfect balance of change across \(x = \frac{1}{2}\). Such symmetry ensures that \(p(x)\) behaves predictably, making calculations like integrals more straightforward.

Definite Integrals

Definite integrals are a powerful tool in calculus, used to find the total accumulation of a quantity or area under a curve over a specific interval. When you calculate \(\int_{a}^{b} p(x) \, dx\), you're summing up all the infinitesimal rectangular slices from \(a\) to \(b\).

• Integral from 0 to 1: In this problem, it calculates the area under the curve \(p(x)\) between \(x = 0\) and \(x = 1\).
• Using limits: The integral \(\int_{0}^{1} (40x + 1) \, dx\) simplifies to computing \([20x^2 + x]_{0}^{1}\). This means you substitute the values of the upper and lower limits into the antiderivative and subtract.

After substituting, the integral evaluates to 21, conforming to the straightforward linear changes in \(p(x)\). Definite integrals like these are essential for determining total values from continuously varying functions, providing critical insights in both mathematics and applied sciences.