Problem 67 - A Complete Resource Book in Mathematics for JEE Main 2017

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 choice is (B): continuous for all \(x\) but not differentiable at \(x=2\).

Step 1: Analyze Parametric Equations

The parametric equations are given as \( x = 2t - |t-1| \) and \( y = 2t^2 + t|t| \). We need to determine how these expressions affect the continuity and differentiability of the function \( y = f(x) \).

Step 2: Simplify x in Terms of t

First, rewrite the expression for \( x \):- For \( t \geq 1 \), \( |t-1| = t-1 \), so \( x = 2t - (t - 1) = t + 1 \).- For \( t < 1 \), \( |t-1| = -(t-1) = 1-t \), so \( x = 2t - (1 - t) = 3t - 1 \).

Step 3: Determine Points of Interest for x

From the simplification:- For \( t \geq 1 \), \( x = t + 1 \).- For \( t < 1 \), \( x = 3t - 1 \).The changes in the expression of \( x \) happen at \( t = 1 \). Thus, \( x \) is continuous everywhere but could have a non-differentiable point at \( t=1 \).

Step 4: Express y in Terms of t

The expression for \( y \) is \( y = 2t^2 + t|t| \):- For \( t \geq 0 \), \( y = 2t^2 + t^2 = 3t^2 \).- For \( t < 0 \), \( y = 2t^2 - t^2 = t^2 \).

Step 5: Identify Changes in y Expression

The expression for \( y \) changes at \( t=0 \) and potentially at \( t=1 \) because of the behavior of \( x \). Both these points need attention when checking for differentiability.

Step 6: Check for Differentiability and Continuity

For differentiability, check where \( x \) or \( y \) expressions change abruptly, especially at \( t=0 \) and \( t=1 \). The function is continuous but may not be differentiable at the points where the expressions for \( x \) and \( y \) change due to non-linearity or the presence of absolute values.

Step 7: Evaluate the Options

- The function is continual over \( x \in \mathbb{R} \), but the changes at \( t=0 \) and \( t=1 \) might affect differentiability.- Differentiability fails at \( x=2 \) (for \( t=1 \), \( x=2 \)).- Thus, we look for where differentiability fails, confirming options against provided scenarios.

Key Concepts

Continuity

A function is continuous if, at any point in its domain, the function's limit exists and equals the function's value at that point. For parametric functions like the one given, analyzing continuity involves examining both the parametric expressions for changes that might disrupt the smooth graph flow.
In this context, the function described by parametric equations is continuous over all real numbers. This continuity comes from the fact that the expressions for both x (namely, "2t - |t-1|") and y ("2t^2 + t|t|") are composed of continuous functions themselves. The piecewise nature of the absolute value function may raise concerns, but it does not introduce discontinuities in this case.
Thus, continuity in parametric functions, even with absolute values, is usually not an issue unless there is a division by zero or infinite disjoint intervals. The given function remains uninterrupted, confirming its continuity for all real x.

Differentiability

Differentiability of a function depends on the smoothness of its graph; a function is differentiable at a point if it has a defined tangent line at that point. For parametric equations, this translates to the differentiability of both x(t) and y(t) in terms of t.
In the given exercise, x(t) includes an absolute value function, which can introduce kinks where differentiability fails. Specifically, at t=1, x becomes non-differentiable. Inspection shows that at t=1, x translates to 2, hence the point of non-differentiability for x is validated in terms of x. As y(t) involves a piecewise absolute function too, possible non-differentiability arises at t=0 and t=1, thus x=2 or x=-1 could be problematic for y as well.
Therefore, while the function is continuous everywhere for x in the reals, being non-differentiable at critical points like x=2 due to the nature of absolute value parts makes differentiability tricky.

Absolute Value Function

The absolute value function, written as |a|, outputs the magnitude of a number, turning negative values to positive while leaving positives unchanged. In parametric functions, they often signal changes in function's derivative and sometimes complicate solving differentiability.
For the parametric form here, absolute value shows how |t-1| creates two situations based on t: when t≥1, it equals t-1 and when t<1, it changes to 1-t. This creates diverse pathways for x, signifying significant shifts near t=1 potentially affecting differentiability.
Similarly in y, t|t| factors expose critical switch-overpoints which demand careful inspection for smooth derivations. The bifurcation at t=0, where y transitions, is directly due to this property of absolute value. Understanding these dynamics is crucial in grasping the outcome that certain areas like x=2 lack differentiability.