Problem 60 - A Complete Resource Book in Mathematics for JEE Main

Question

. For \(y=f(x)=\int_{0}^{x} 2|t| d t\), the tangent lines parallel to the bisector of the first quadrant angle are (A) \(y=x \pm \frac{1}{4}\) (B) \(y=x \pm \frac{3}{2}\) (C) \(y=x \pm \frac{1}{2}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (A) \(y=x\pm\frac{1}{4}\).

Step 1: Compute the Integral

The function is defined as \(y = f(x) = \int_{0}^{x} 2|t| dt\). We must consider the absolute value to determine the integral in parts, depending on whether \(x\) is positive or negative. Thus:\[ y = \begin{cases} \int_{0}^{x} 2t \, dt & \text{for } x \geq 0 \int_{0}^{x} -2t \, dt & \text{for } x < 0 \end{cases} \]For \(x \geq 0\), the integral becomes \[ y = \left. t^2 \right|_0^x = x^2. \]For \(x < 0\), the integral becomes \[ y = -\left. t^2 \right|_0^x = -x^2. \]Thus, the function is piecewise defined as \(y = x^2\) for \(x \geq 0\) and \(y = -x^2\) for \(x < 0\).

Step 2: Compute the Derivative

To find the slope of the tangent line, compute the derivative of \(y = f(x)\). For \(x \geq 0\), \(f(x) = x^2\) gives \(f'(x) = 2x\). For \(x < 0\), \(f(x) = -x^2\) gives \(f'(x) = -2x\).

Step 3: Identify the Slope for the Tangent Line

The problem states that the tangent lines are parallel to the bisector of the first quadrant angle, which has a slope of 1 (since it's \(y = x\)). Thus, we set \(f'(x) = 1\) for \(x \geq 0\) and \(f'(x) = -1\) for \(x < 0\).

Step 4: Solve for Points of Tangency

Find \(x\) such that \(f'(x) = 1\) for positive \(x\), which gives \(2x = 1\) resulting in \(x = \frac{1}{2}\). For negative \(x\), solve \(-2x = 1\), resulting in \(x = -\frac{1}{2}\).

Step 5: Equation of the Tangent Lines

For \(x = \frac{1}{2}\), the point is \((\frac{1}{2}, \frac{1}{4})\). The equation of the tangent line is \(y - \frac{1}{4} = 1(x - \frac{1}{2})\), simplifying to \(y = x - \frac{1}{4}\). Similarly, for \(x = -\frac{1}{2}\), the point is \((-\frac{1}{2}, \frac{1}{4})\) and the equation is \(y - \frac{1}{4} = 1(x + \frac{1}{2})\), or \(y = x + \frac{1}{4}\). These equations match (A).

Key Concepts

Definite Integrals

In calculus, a definite integral is a way to calculate the area under a curve, specifically over a certain interval. This concept is denoted with the integral sign and has upper and lower limits. For example, given the function from the exercise, we want to evaluate the integral of \(2|t|\) from 0 to \(x\). This requires us to break the problem into parts because of the absolute value function involved.

• When \(x \geq 0\), the function inside the integral simplifies to \(2t\), resulting in \(y = x^2\).
• When \(x < 0\), it simplifies to \(-2t\), leading to \(y = -x^2\).

This division of the integral is crucial in understanding that the definite integral not only involves calculation but also understanding parts of the curve's behavior under the given interval.

Piecewise Functions

Piecewise functions are functions that have different expressions based on the input value. In the example, the function \(y = f(x)\) is defined differently for positive and negative \(x\).

• For \(x \geq 0\), the function is expressed as \(x^2\).
• For \(x < 0\), it becomes \(-x^2\). This means that the graph of this function looks different on either side of the y-axis.

This approach allows us to deal with situations where the behavior of a function changes at certain points, making it versatile for modeling real-world scenarios.

Tangent Lines

Tangent lines touch a curve at just one point. They represent the instantaneous rate of change or slope of the function at that point. In the context of this exercise, we need to find tangent lines that are parallel to a reference line, in this case, the bisector \(y = x\), which has a slope of 1.
To find these points:

• The derivative (or slope) \(f'(x)\) of \(y = f(x)\) must equal 1 for the tangent line to be parallel.
• For \(x \geq 0\), solve \(2x = 1\), giving \(x = \frac{1}{2}\).
• For \(x < 0\), solve \(-2x = 1\), giving \(x = -\frac{1}{2}\).

This calculation finds the exact points where these tangent lines touch the curve, making them parallel to \(y = x\).

Derivatives

Derivatives represent the rate at which a function is changing at any point, essentially giving us the slope of the tangent line to the function at that point. Calculating derivatives is a fundamental task in calculus.
For the piecewise function discussed:

• The derivative \(f'(x) = 2x\) when \(x \geq 0\).
• The derivative \(f'(x) = -2x\) when \(x < 0\).

These expressions show how the slope varies for different portions of the function. Understanding how to derive these expressions is key in determining how steeply functions rise or fall, which in turn helps in sketching graphs and solving optimization problems.