Problem 60 - A Complete Resource Book in Mathematics for JEE Main 2017

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: Understand the function

The given function is \(y = f(x) = \int_{0}^{x} 2|t| dt\). The task is to determine the tangent lines to this function that are parallel to the line \(y = x\), which is the bisector of the first quadrant angle. This means we need derivatives where the slope equals \(1\).

Step 2: Consider the absolute value

Since the function involves \(|t|\), we consider two cases for \(t\): \(t \geq 0\) and \(t < 0\). For \(x \geq 0\), \(|t| = t\), and for \(x < 0\), \(|t| = -t\). Thus, the integral becomes a piecewise function.

Step 3: Solve the integral for the positive case

For \(x \geq 0\), \(f(x) = \int_{0}^{x} 2t dt = [t^2]_{0}^{x} = x^2\).

Step 4: Solve the integral for the negative case

For \(x < 0\), \(f(x) = \int_{0}^{x} -2t dt = [-t^2]_{0}^{x} = x^2\) (after substituting limits properly). Hence, the function is \(y = x^2\) for both cases.

Step 5: Differentiate the function

Differentiate \(f(x) = x^2\) with respect to \(x\). The derivative is \(f'(x) = 2x\).

Step 6: Find where the derivative equals the slope of the bisector

Set \(f'(x) = 2x = 1\) (since we want the slope of the tangent to be \(1\), equal to the slope of \(y = x\)). Solving for \(x\), we get \(x = \frac{1}{2}\).

Step 7: Calculate the y-coordinate of the tangent point

Substituting \(x = \frac{1}{2}\) back into the function \(y = x^2\), we get \(y = (\frac{1}{2})^2 = \frac{1}{4}\).

Step 8: Write the equation of the tangent line

The equation of the tangent line with a slope of 1 through the point \((\frac{1}{2}, \frac{1}{4})\) is \(y - \frac{1}{4} = 1(x - \frac{1}{2})\). Simplifying, we find \(y = x - \frac{1}{4}\).

Step 9: Determine the symmetrical tangent line

Since the function is symmetric around the y-axis, another tangent point at \(x = -\frac{1}{2}\) will have \(y = (\frac{1}{2})^2 = \frac{1}{4}\). The symmetrical tangent line equation would thus be \(y - \frac{1}{4} = 1(x + \frac{1}{2})\), which simplifies to \(y = x + \frac{3}{4}\).

Step 10: Match the options

Checking the options, \(y = x - \frac{1}{4}\) is consistent with option A (\(y = x \pm \frac{1}{4}\)), since it accounts for both lines.

Key Concepts

Definite Integrals

In integral calculus, a definite integral evaluates the accumulation of quantities, such as areas under a curve, within a specific interval. The given exercise involves calculating a definite integral from 0 to x. The expression \( y = f(x) = \int_{0}^{x} 2|t| \, dt \) represents the area under the curve \( 2|t| \) between the limits of 0 and x. This integral is particularly interesting because it includes an absolute value function, \(|t|\), which requires us to consider it as a piecewise function.
The limits of integration, 0 to x, specify the interval over which we calculate the area. By evaluating this definite integral for different intervals defined by \( |t| \), we can determine the function \( y = f(x) \) and its behavior. Understanding how derivatives are used in conjunction with definite integrals can also help identify points where the tangent to the curve has specific characteristics, such as a particular slope, which is key in solving such problems.

Piecewise Functions

Piecewise functions are defined by different expressions depending on the interval of the input variable. In our exercise, we see the function \( 2|t| \) treated differently when considering its integral for different ranges of t.

• For \( x \geq 0 \): \( |t| = t \) and thus the function becomes \( 2t \), and its integral results in \( f(x) = x^2 \).
• For \( x < 0 \): \( |t| = -t \) leading the function to be \( -2t \), but upon integration from 0 to x, it evaluates to \( x^2 \) (thanks to properly positioning limits as well as symmetry in the graph of \( y = |t| \)).

Understanding the structure of piecewise functions like this helps us form the complete definition over the real number line. This approach simplifies what might otherwise be a complex function into manageable parts, each with its own characteristics. By recognizing patterns and integrating within these separate intervals, we can thoroughly solve problems involving absolute values and piecewise behaviors.

Equation of Tangent Line

The equation of a tangent line to a curve at a point gives a straight line that just "touches" the curve at that point. The derivative of a function at a particular point gives the slope of this tangent line.
In the case of our problem, we found that the derivative \( f'(x) = 2x \) for the integrad \( y = x^2 \). To find a line parallel to the bisector of the first quadrant angle (i.e., the line \( y = x \)), we need where \( f'(x) = 1 \) since the slope of \( y = x \) is also 1. Solving this gives \( x = \frac{1}{2} \).
At \( x = \frac{1}{2} \), substituting into \( f(x) \) provides the y-coordinate: \( y = (\frac{1}{2})^2 = \frac{1}{4} \). Thus, the tangent line passing through \( (\frac{1}{2}, \frac{1}{4}) \) is \( y = x - \frac{1}{4} \).
On the symmetrically opposite side at \( x = -\frac{1}{2} \), the same derivation follows, leading to another valid tangent line \( y = x + \frac{3}{4} \). Recognizing and deriving equations of tangent lines is essential in calculus to make predictions and understand curve behaviors across its domain.