Problem 97 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a real valued function defined on an open internal \(I \subset R\). If \(x_{0} \in I\), then we define \(g\) with domain \(I-\left\\{x_{0}\right\\}\) by setting \(g(x)=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}\), for all \(x \in I-\left\\{x_{0}\right\\} .\) If \(\lim _{x \rightarrow x_{*}} g(x)\) exists and is finite, we denote it by \(f^{\prime}\left(x_{0}\right)\) and say that \(f\) is derivable at \(x_{0} \cdot f^{\prime}\left(x_{0}\right)\) is called the derivative of \(f\) at \(x_{0}\). If \(\lim _{x \rightarrow x_{i}^{\prime}} g(x)\) exists and is finite, we denote it by \(R f^{\prime}\left(x_{0}\right)\) and say that \(f\) is derivable from right at \(x_{0}\). If \(\lim _{x \rightarrow x_{0}^{-}} g(x)\) exists and is finite, we denote it by \(L f^{\prime}\left(x_{0}\right)\) and say that \(f\) is derivable from left at \(x_{0}\). It is obvious that \(f\) is derivable at \(x_{0}\) iff \(L f^{\prime}\left(x_{0}\right)\) and \(R f^{\prime}\left(x_{0}\right)\) both exist and are equal. Also, if this condition be satisfied, then the common value is nothing else but \(f^{\prime}\left(x_{0}\right)\). Let \(f(x)=\cos x\) and \(g(x)=[x+2]\), where \([.]\) denotes the greatest integer function. Then, (gof) \(\left(\frac{\pi}{2}\right)\) is (A) 1 (B) 0 (C) \(-1\) (D) does not exist

Step-by-Step Solution

Answer

The value of \((g \circ f)\left(\frac{\pi}{2}\right)\) is 2, so none of the given options are correct.

Step 1: Interpret the Problem

We want to find the value of \((g \circ f)\left(\frac{\pi}{2}\right)\), where \(f(x) = \cos x\) and \(g(x) = [x+2]\). Here, \([x]\) denotes the greatest integer function, also known as the floor function.

Step 2: Evaluate \(f\left(\frac{\pi}{2}\right)\)

To evaluate \((g \circ f)\left(\frac{\pi}{2}\right)\), we first need to find \(f\left(\frac{\pi}{2}\right)\). Here, \(f(x) = \cos x\). Therefore, \(f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\).

Step 3: Evaluate \(g(f(\frac{\pi}{2})) = g(0)\)

The next step is to evaluate \(g(0)\), where \(g(x) = [x+2]\). We substitute \(x = 0\) into the function: \(g(0) = [0 + 2] = [2]\). The greatest integer less than or equal to 2 is 2, so \(g(0) = 2\).

Step 4: Conclude the Value of \((g \circ f)\left(\frac{\pi}{2}\right)\)

From our calculations: \((g \circ f)\left(\frac{\pi}{2}\right) = g(0) = 2\). Since this option is not among (A) 1, (B) 0, (C) -1, or (D) does not exist, none of the given options is correct based on our calculations.

Key Concepts

Function Composition

Function composition is a process where two functions are combined to create a new function. Imagine you have two functions, say \(f(x)\) and \(g(x)\). When you "compose" these functions, you create a new function \((g \circ f)(x)\), which means you're inserting the output of \(f(x)\) into the input of \(g(x)\). In simpler terms, you're doing \(f\) first, then \(g\) on whatever result \(f\) gave you.

• The operation looks like this: \((g \circ f)(x) = g(f(x))\).
• It's like feeding the result of \(f(x)\) directly to \(g(x)\).
• Order is crucial: \((g \circ f)(x)\) is typically not the same as \((f \circ g)(x)\).

In the exercise, we first evaluate \(f\) at \(\frac{\pi}{2}\) to get \(f(\frac{\pi}{2}) = 0\). We then use this result to find \(g(0)\). This process helps us understand how combining functions can lead to an entirely different result, which in our case was 2.To fully master function composition, practice with various functions to see how the outcomes differ based on the order in which functions are applied.

Greatest Integer Function

The greatest integer function, often represented as \([x]\), is also known as the floor function. It works by taking a real number and rounding it down to the nearest integer. This function takes each number and finds the greatest integer that is less than or equal to the given number.

• For any integer, \([x] = x\).
• For non-integers, if \(x\) is between \(1.2\) and \(1.9\), then \([x] = 1\).
• Rounding down happens regardless of whether the decimal is closer to the next integer.

In our solution, we defined \(g(x) = [x + 2]\). Once we calculated the nested function output, \(g\), at 0, it became \([0 + 2]\), which equals 2. Understanding how the greatest integer function rounds down is crucial, especially when combining it with other functions in calculus for determining exact values.

Limits and Continuity

Limits are a fundamental concept in calculus that describe how a function behaves as its input approaches a certain point. Continuity, closely tied with limits, refers to a function's smoothness — a continuous function has no sudden jumps or gaps.When finding the derivative in calculus, as in the original exercise, we're often looking at forms of limits:

• The derivative is defined as \(f'(x_0) = \lim_{x \to x_0}\frac{f(x) - f(x_0)}{x - x_0}\).
• If this limit exists and is the same for both left and right, the function is said to be continuous and derivable at this point.

In the exercise, determining the limit of \(g(x)\) when creating the derivative is central to the concept of continuity. Understanding these foundational ideas of limits helps establish the continuity and smoothness of functions, and forms the bedrock of differentiating functions. By breaking down and studying limits, students can pave the way to mastering more complex calculus topics.