Problem 67 - A Complete Resource Book in Mathematics for JEE Main

Question

The function \(y=f(x)\), defined parametrically as \(x=2 t-|t-1|\) and \(y=2 t^{2}+t|t|\), is (A) continuous and differentiable for \(x \in R\) (B) continuous for \(x \in R\) and differentiable for \(x \in\) \(R-\\{2\\}\) (C) continuous for \(x \in R\) and differentiable for \(x \in\) \(R-\\{-1,2\\}\) (D) None of these

Step-by-Step Solution

Answer

The correct option is (C).

Step 1: Understand the Parametric Equations

The function is defined using parameters. We have given \(x = 2t - |t-1|\) and \(y = 2t^2 + t|t|\). To check for continuity and differentiability, first understand how these parameters behave for different real values of \(t\). Particularly, handle the absolute value expressions and how they change at \(t = 1\) and \(t = 0\).

Step 2: Simplify the Parametric Expressions

For \(x = 2t - |t-1|\):- When \(t \geq 1\), \(|t-1| = t-1\), so \(x = 2t - (t-1) = t+1\).- When \(t < 1\), \(|t-1| = 1-t\), so \(x = 2t - (1-t) = 3t-1\).For \(y = 2t^2 + t|t|\):- When \(t \geq 0\), \(|t| = t\), so \(y = 2t^2 + t^2 = 3t^2\).- When \(t < 0\), \(|t| = -t\), so \(y = 2t^2 - t^2 = t^2\).

Step 3: Analyze the Function for Continuity

Both \(x(t)\) and \(y(t)\) are polynomials, which are continuous for all real \(t\). Therefore, the function \(y = f(x)\) is continuous over \(x \in \mathbb{R}\) since the ranges of these \(t\) provide continuous mappings for \(x\).

Step 4: Explore Differentiability

Check for differentiability of \(x(t)\) and \(y(t)\) by differentiating:For \(x(t)\), find \(dx/dt\). In the expression for \(x\), there might be a potential discontinuity at \(t=1\) because of the absolute value:- \(dx/dt = 1\) for \(t \geq 1\)- \(dx/dt = 3 \) for \(t < 1\)For \(y(t) = 3t^2\) when \(t \geq 0\) and \(y(t) = t^2\) otherwise:- \(dy/dt = 6t\) when \(t \geq 0\)- \(dy/dt = 2t\) otherwiseFrom the \(t=0\) and \(t=1\) reflections, ensure \(dy/dx\) is undefined there, implicating differentiability excludes certain points by examining those closely.

Step 5: Conclusion for Options

Now, plug any excluded points from differentiability into the options.- (A): Is not valid since it's not differentiable for all \(x \in \mathbb{R}\).- (B): Incorrect as differentiability is affected at points where changes around \(t\) (potentially near zero).- (C): Correct, as differentiable points are all \(x\) excluding points affected, namely \(-1, 2\).- (D): Negation of other valid options.

Key Concepts

Continuity

When discussing parametric equations like those given in the exercise, continuity is a key concept. A function is continuous if there are no breaks or jumps in its graph. For our exercise, both parametric definitions, like the one for \( x(t) = 2t - |t-1| \), need to be continuous over real values of \( t \).
For different values of \( t \), the absolute value impacts the expression: when \( t \geq 1 \), the expression becomes \( x = t + 1 \), and when \( t < 1 \), it becomes \( x = 3t - 1 \). Since these are linear expressions, they're continuous over their respective intervals.
Likewise, \( y(t) = 2t^2 + t|t| \) is combined into simple polynomial expressions for different \( t \)'s: \( y = 3t^2 \) for \( t \geq 0 \) and \( y = t^2 \) for \( t < 0 \). Polynomials are inherently continuous, so \( y(t) \) is continuous as well.

• Thus, the function described by these parametric equations provides a continuous mapping for all real numbers \( t \).
• The continuity translates to the function \( y = f(x) \) being continuous over all real numbers \( x \).

Differentiability

Differentiability indicates whether a function has a defined slope at every point, meaning it must be smooth without any sharp turns or corners.
For parametric equations, checking differentiability often involves calculating derivatives. We take \( dx/dt \) and \( dy/dt \), and use them to find \( dy/dx \).

For \( x(t) \), depending on the value of \( t \) and due to the presence of an absolute value, the derivative changes:

• For \( t \geq 1 \), \( dx/dt = 1 \).
• For \(t < 1\), \( dx/dt = 3 \).

These reflect that there's a potential point of non-differentiability at \( t = 1 \).

For \( y(t) \),

• When \( t \geq 0 \), \( dy/dt = 6t \).
• When \( t < 0 \), \( dy/dt = 2t \).

It shows that \( y'(t) \) changes form as \( t \) switches between positive and negative.
Assessing \( dy/dx \), we notice undefined behavior at \( t = 0 \) and \( t = 1 \) (due to division in \( dy/dx \)), implying the function is not differentiable at these \( x \) points.

• This results in exclusions from the differentiable domain for points in \( x \) corresponding to \( t = -1, 2 \).

Absolute Value Functions

Absolute value functions are crucial in parametric equations since they affect how expressions behave at certain points. An absolute value, like in \( |t-1| \) from our exercise, modifies the expression based on the condition of its argument.
If \( t \geq 1 \), then \( |t-1| = t-1 \), otherwise, it becomes \( 1-t \) when \( t < 1 \). This leads to different linear expressions for different intervals of \( t \).

• This behavior creates potential points of non-differentiability or non-smoothness in the function as slope may suddenly change.
• It's essential to analyze these critical points (where the behavior changes) to understand the function's behavior entirely.

Using absolute values in functions contributes to piecewise definitions. Each piece is continuous on its interval, but transitions between pieces must be carefully checked, especially for continuity and differentiability.