Problem 91 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=x^{3}-x^{2}+x+1\) and \(g(x)=\left\\{\begin{array}{cc}\max \cdot f(t) & 0 \leq t \leq x \text { for } 0 \leq x \leq 1 \\ 3-x & 1

Step-by-Step Solution

Answer

\( g(x) \) is continuous at all \( x \) but not differentiable at \( x=1 \). (D) applies.

Step 1: Understanding the Problem

We need to analyze the continuity and differentiability of the function \( g(x) \) in the interval \([0,2]\). \( g(x) \) is given as a piecewise function, where \( g(x) = \max \cdot f(t) \) for \( 0 \leq x \leq 1 \) and \( g(x) = 3-x \) for \( 1

Step 2: Evaluate g(x) for 0 ≤ x ≤ 1

For \( 0 \leq x \leq 1 \), \( g(x) \) is defined as the maximum of \( f(t) \) where \( 0 \leq t \leq x \). We need to find \( f(t) \) from \( x = 0 \) to \( x = 1 \) and identify its maximum value during this interval.

Step 3: Finding f(t) and Maximum on [0, 1]

Calculate \( f(t) = t^3 - t^2 + t + 1 \). Critical points are found by setting \( f'(t) = 3t^2 - 2t + 1 = 0 \) and analyzing within \( t \in [0, 1] \). Solving \( f'(t) = 0 \), we observe behavior without specific roots (as discriminant \(< 0\)). Evaluate \( f(t) \) at boundaries and compare: \[ f(0)=1, f(1)=2 \]. The maximum within \([0, 1]\) is 2 at \( t = 1 \). Hence, \( g(x) = 2 \) for \( 0 \leq x \leq 1 \).

Step 4: Evaluate g(x) for 1 < x ≤ 2

For \( 1 < x \leq 2 \), function is \( g(x) = 3-x \). Calculate limits as \( x \to 1^+ \) and substitutions for closure existence check.

Step 5: Check Continuity at x = 1

Ensure continuity at transition \( x=1 \). Calculate \( \lim_{x \to 1^-} g(x) = 2 \) and \( \lim_{x \to 1^+} g(x) = 3-1 = 2 \). Thus, \( g(x) \) continuous at \( x=1 \) since limits and function value at \( x=1 \) match.

Step 6: Examine Differentiability at x = 1

Check piecewise results for differentiability. For \( 0

Key Concepts

Piecewise Functions

Piecewise functions are a type of function that have different expressions based on the input value, usually categorized by specific intervals. In the problem we examined, the function \( g(x) \) is an excellent example of this kind of function. The function is defined as follows:

• For \(0 \leq x \leq 1\), \( g(x) = \max \cdot f(t)\) where \( f(t) = t^3 - t^2 + t + 1 \).
• For \(1 < x \leq 2\), \( g(x) = 3 - x \).

This means within the first interval from \( 0 \) to \( 1 \), the function takes the maximum value of \( f(t) \) as \( t \) ranges from \( 0 \) to \( x \). Meanwhile, in the second interval from \( 1 \) to \( 2 \), \( g(x) \) simplifies to a linear expression, \( 3 - x \). Piecewise functions often reflect real-world situations where a rule or behavior changes over different intervals. Understanding them involves analyzing each segment and its implications separately.

Continuity

Continuity of a function at a point means that the function has no breaks, jumps, or holes at that point. A continuous function smoothly connects without any sudden changes. Checking for continuity, especially in piecewise functions, involves several steps. For the function \( g(x) \) at \( x = 1 \), we calculated:

• \( \lim_{x \to 1^-} g(x) = 2 \)
• \( \lim_{x \to 1^+} g(x) = 2 \)
• \( g(1) = 2 \)

The fact that the limit from the left (\(x\) approaching 1 from the negative side) and the limit from the right (\(x\) approaching 1 from the positive side) are equal, and that both are equal to the value of the function at \( x = 1 \), establishes that \( g(x) \) is continuous at \( x = 1 \). Continuity across the whole interval \([0, 2]\) for \( g(x) \) implies similar checks must hold for each point in the interval.

Differentiability

Differentiability refers to whether a function can have a derivative at all points in its domain. A function must be continuous to be differentiable, but being continuous doesn't automatically imply differentiability. For our piecewise function \( g(x) \) between \( 0 \) and \( 1 \), the derivative is \( 0 \), indicating a constant function since it represents the maximum value of \( f(t) \). Between \( 1 \) and \( 2 \), the derivative of \( 3-x \) is \( -1 \). The abrupt change of the derivative's value at \( x = 1 \) causes \( g(x) \) to not be differentiable at this point, despite being continuous. Differentiability requires not just continuity, but also matching of the derivative slopes from either side of the evaluated point. Thus, \( g(x) \) presents an ideal study of how piecewise definitions can affect derivative behavior at transition points.

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Problem 90

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Problem 92

Other exercises in this chapter

Problem 89

Let \(f^{\prime \prime}(x)\) be continuous at \(x=0\). If \(\lim _{x \rightarrow 0} \frac{2 f(x)-3 a f(2 x)+b f(8 x)}{\sin ^{2} x}\) exists and \(f(0) \neq 0\),

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Problem 90

If \(f(x)=\left\\{\begin{aligned} x-3, & x

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Problem 92

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 inter

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Problem 93

If \(f(x)=[\tan x]+\sqrt{\tan x-[\tan x]}, 0 \leq x

View solution

Key Concepts

Piecewise Functions

Piecewise functions are a type of function that have different expressions based on the input value, usually categorized by specific intervals. In the problem we examined, the function \( g(x) \) is an excellent example of this kind of function. The function is defined as follows:

• For \(0 \leq x \leq 1\), \( g(x) = \max \cdot f(t)\) where \( f(t) = t^3 - t^2 + t + 1 \).
• For \(1 < x \leq 2\), \( g(x) = 3 - x \).

This means within the first interval from \( 0 \) to \( 1 \), the function takes the maximum value of \( f(t) \) as \( t \) ranges from \( 0 \) to \( x \). Meanwhile, in the second interval from \( 1 \) to \( 2 \), \( g(x) \) simplifies to a linear expression, \( 3 - x \). Piecewise functions often reflect real-world situations where a rule or behavior changes over different intervals. Understanding them involves analyzing each segment and its implications separately.

Continuity

Continuity of a function at a point means that the function has no breaks, jumps, or holes at that point. A continuous function smoothly connects without any sudden changes. Checking for continuity, especially in piecewise functions, involves several steps. For the function \( g(x) \) at \( x = 1 \), we calculated:

• \( \lim_{x \to 1^-} g(x) = 2 \)
• \( \lim_{x \to 1^+} g(x) = 2 \)
• \( g(1) = 2 \)

The fact that the limit from the left (\(x\) approaching 1 from the negative side) and the limit from the right (\(x\) approaching 1 from the positive side) are equal, and that both are equal to the value of the function at \( x = 1 \), establishes that \( g(x) \) is continuous at \( x = 1 \). Continuity across the whole interval \([0, 2]\) for \( g(x) \) implies similar checks must hold for each point in the interval.

Differentiability

Differentiability refers to whether a function can have a derivative at all points in its domain. A function must be continuous to be differentiable, but being continuous doesn't automatically imply differentiability. For our piecewise function \( g(x) \) between \( 0 \) and \( 1 \), the derivative is \( 0 \), indicating a constant function since it represents the maximum value of \( f(t) \). Between \( 1 \) and \( 2 \), the derivative of \( 3-x \) is \( -1 \). The abrupt change of the derivative's value at \( x = 1 \) causes \( g(x) \) to not be differentiable at this point, despite being continuous. Differentiability requires not just continuity, but also matching of the derivative slopes from either side of the evaluated point. Thus, \( g(x) \) presents an ideal study of how piecewise definitions can affect derivative behavior at transition points.