Problem 100 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f(x)=\left(a^{2}-3 a+2\right) \cos \frac{x}{2}+(a-1) x\) possesses critical points, then \(a\) belongs to the interval (A) \((-\infty, 0) \cup(4, \infty)\) (B) \((-\infty, 0] \cup[4, \infty)\) (C) \((-\infty, 0] \cup\\{1\\} \cup[4, \infty)\) (D) None of these

Step-by-Step Solution

Step 1: Understand Critical Points

Critical points of a function occur where the derivative of the function is zero or undefined. Thus, we need to find the derivative of the given function and identify these points.

Step 2: Differentiate the Function

Let's find the derivative of the function \( f(x) = (a^2 - 3a + 2) \cos \left( \frac{x}{2} \right) + (a - 1)x \). Use the product rule and chain rule for differentiation.\[ f'(x) = (a^2 - 3a + 2) \left( -\frac{1}{2} \sin \left( \frac{x}{2} \right) \right) + (a-1) \]

Key Concepts

Derivative of a function

The derivative of a function is a fundamental concept in calculus, representing how a function changes at any given point. Essentially, it measures the rate at which a function's value is changing with respect to changes in the input value, typically denoted as **x**. Derivatives have practical uses in various fields, such as physics and engineering, wherever we need to understand trends and rate of change. The process of finding a derivative is called differentiation. When you differentiate a function, you apply specific rules that transform it into its derivative form. This transformed function indicates changes in the slope or rate of change of the original function across its domain. In our exercise, the function or expression whose derivative we're finding is often represented by notation such as \(f'(x)\) or \(\frac{df}{dx}\) for a function \(f(x)\). A critical aspect of working with derivatives is understanding that a critical point occurs where this derivative equals zero or becomes undefined. In our problem, finding the derivative first helps us identify these critical points.

Chain rule

The chain rule is a powerful tool in calculus used to differentiate compositions of functions. If you have a function nested within another function, the chain rule helps differentiate them efficiently.In mathematical terms, if you have a composite function \(y = g(f(x))\), then the derivative can be found using the chain rule. The chain rule formula is expressed as:\[ \frac{dy}{dx} = \frac{dy}{df} \cdot \frac{df}{dx} \]This means you first differentiate the outer function with respect to the inner function, and then multiply it by the derivative of the inner function with respect to **x**. In our exercise, we are dealing with the cosine function \(\cos\left(\frac{x}{2}\right)\). This is a composition of the cosine function and a linear function \(\frac{x}{2}\), making the chain rule applicable. By using the chain rule, we take the derivative of \(\cos\) which is \(-\sin\), and multiply it by the derivative of \(\frac{x}{2}\), which is \(\frac{1}{2}\). This step is crucial in correctly identifying how each part of the function contributes to its overall rate of change and eventually pinpoints the critical points effectively.

Product rule

The product rule is another key concept in calculus, used when finding the derivative of a product of two functions. Whenever two functions are multiplied together, their combined rate of change involves the product rule. The product rule formula states:- If you have two functions \(u(x)\) and \(v(x)\), their derivative \((uv)'\) is given by:\[ (uv)' = u'v + uv' \]Where \(u'\) is the derivative of \(u\), and \(v'\) is the derivative of \(v\). In our exercise, we used the product rule to differentiate the expression \((a^2 - 3a + 2) \cdot \cos \left( \frac{x}{2} \right)\). Here, \((a^2 - 3a + 2)\) is a constant term (act as your \(u(x)\)) and \(\cos \left( \frac{x}{2} \right)\) is the \(v(x)\) term. In such cases, when one of the functions is a constant, its derivative is zero, simplifying our work with the product rule considerably, as the derivative focuses on the variable function.