Problem 92 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=x^{4}-8 x^{3}+22 x^{2}-24 x\) and \(g(x)= \begin{cases}\min . f(t) & x \leq t \leq x+1,-1 \leq x \leq 1 \\ x-10 & x>1\end{cases}\) Then, in the interval \([-1, \infty), g(x)\) is (A) continuous for all \(x\) (B) discontinuous at \(x=1\) (C) differentiable for all \(x\) (D) not differentiable at \(x=1\)

Step-by-Step Solution

Answer

(B) discontinuous at \(x = 1\).

Step 1: Understand the Functions

We have two functions, \(f(x) = x^4 - 8x^3 + 22x^2 - 24x\), and \(g(x)\). The function \(g(x)\) is piecewise defined: it evaluates to the minimum of \(f(t)\) over the interval \([x, x+1]\) for \(-1 \leq x \leq 1\), and it equals \(x - 10\) for \(x > 1\).

Step 2: Analyze \(f(x)\) over \([-1, 2]\)

First, find the behavior of \(f(x)\) in the interval \([-1, 2]\), especially because \(g(x)\) is defined in terms of minimum values of \(f(t)\) within \([x, x+1]\). Calculate the derivative \(f'(x) = 4x^3 - 24x^2 + 44x - 24\) and find any critical points by setting \(f'(x) = 0\). Use these points to find potential minima inside \([-1, 2]\).

Step 3: Evaluate \(g(x)\) for \([-1, 1]\)

For \(-1 \leq x \leq 1\), \(g(x) = \min_{t \in [x, x+1]} f(t)\). Calculate \(f(t)\) at critical points within the range as well as at boundary values \(t = x\) and \(t = x+1\). Determine the minimum value for these intervals.

Step 4: Evaluate \(g(x)\) for \(x > 1\)

For \(x > 1\), \(g(x) = x - 10\). This is a linear function and thus continuous and differentiable wherever it's defined (\(x > 1\)). Evaluate the continuity and differentiability of \(g(x)\) at the edge point \(x = 1\).

Step 5: Check Continuity at \(x = 1\)

At \(x = 1\), connect the two different parts of \(g(x)\). For \(x = 1\), calculate \(g(x)\) by using the formula for \(x \leq 1\) and compare it with the value of \(g(x) = x - 10\) for \(x > 1\). \(g(1)\) must be evaluated for continuity using both definitions before and after \(x = 1\).

Step 6: Check Differentiability at \(x = 1\)

Checking differentiability involves ensuring that the left-hand and right-hand derivatives of \(g(x)\) at \(x = 1\) are equal. Calculate the limit of the derivative of \(g(x)\) as \(x\) approaches 1 from both sides.

Key Concepts

Continuity

Continuity is an essential concept in calculus that involves the behavior of functions without any interruptions or breaks in their graphs. For a function to be continuous at a point, the following conditions must be met:

• The function must be defined at the point.
• The function must have a limit as it approaches the point from both sides.
• The value of the function at the point must equal its limit as it approaches that point.

In the provided exercise, we are interested in whether the piecewise function \(g(x)\) is continuous at \(x = 1\). We analyze the behavior of \(g(x)\) by checking both segments of the piecewise definition: \(-1 \leq x \leq 1\) and \(x > 1\). At \(x = 1\), we calculate \(g(1)\) using its definition for \(-1 \leq x \leq 1\), and compare it to the value given by \(x - 10\) for values where \(x > 1\). For continuity, these values must match. By checking both sides, we confirm if the function exhibits any discontinuity at this point.

Differentiability

Differentiability is about the ability of a function to have a derivative at every point in its domain. A function must be continuous to be differentiable, but just being continuous is not enough for differentiability. The derivative essentially gives us the slope of the tangent line to the function at a particular point. To determine differentiability at a point, particularly at \(x = 1\) in our exercise, we examine the left-hand and right-hand derivatives of the function. To be differentiable at \(x = 1\), the derivatives approaching from both sides must be equal. Calculating limits of the derivative of \(g(x)\) as \(x\) approaches 1 from the left involves the interval \(-1 \leq x \leq 1\). For \(x > 1\), \(g(x)\) simplifies to a simple function \(x - 10\), where the derivative is straightforwardly 1. Any mismatch in these derivative values indicates that the function is not differentiable at the point \(x = 1\). This step helps determine if \(g(x)\) is smooth enough to not have any sharp corners or cusps at that point.

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applied to a certain interval of the domain. This can result in different behavioral characteristics within those intervals. In this problem, \(g(x)\) is a piecewise function:

• Within \(-1 \leq x \leq 1\), \(g(x)\) is defined by the minimum value of \(f(t)\) over the interval \([x, x+1]\).
• For \(x > 1\), \(g(x)\) switches to the linear function \(x - 10\).

Each segment can vary significantly, affecting both continuity and differentiability at transition points. Notably, the transition point at \(x = 1\) will require careful consideration, as it connects the two regimes of \(g(x)\). Calculating and comparing the sub-functions' values at this critical point allows us to check for continuity, while analyzing derivatives on either side informs us of differentiability. Understanding the nature of piecewise functions aids in visualizing and solving problems involving transitions between different function types within a single problem scope.