Problem 19 - A Complete Resource Book in Mathematics for JEE Main

Question

The set of all values of \(a\) for which the function \(f(x)=\) \(\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\) does not possess critical points is (A) \([1, \infty)\) (B) \((0,1) \cup(1,4)\) (C) \((-2,4)\) (D) \((1,3) \cup(3,5)\)

Step-by-Step Solution

Answer

The correct answer is (D) \((1,3)\cup(3,5)\).

Step 1: Find the Derivative

To locate the critical points of the function, we need to first find the derivative \(f'(x)\). The original function is given by \( f(x) = (a^2 - 3a + 2)(\cos^2\frac{x}{4} - \sin^2\frac{x}{4}) + (a-1)x + \sin 1 \). We differentiate this with respect to \(x\) where \(\sin 1\) is a constant, so its derivative will be zero.

Step 2: Simplify the Derivative

The derivative of \(f(x)\) with respect to \(x\) becomes: \[ f'(x) = (a^2 - 3a + 2) \frac{d}{dx}\left( \cos^2\frac{x}{4} - \sin^2\frac{x}{4} \right) + (a-1). \] Simplifying, we find: \[ \frac{d}{dx}\left( \cos^2\frac{x}{4} - \sin^2\frac{x}{4} \right) = \frac{1}{2} \sin\frac{x}{2}, \] so \[ f'(x) = (a^2 - 3a + 2) \cdot \frac{1}{2} \sin\frac{x}{2} + (a-1). \]

Step 3: Set the Derivative to Zero

Critical points occur where \(f'(x) = 0\). Thus we need to solve the equation \((a^2 - 3a + 2) \cdot \frac{1}{2} \sin\frac{x}{2} + (a-1) = 0\).

Step 4: Analyze for Critical Points

For the equation \((a^2 - 3a + 2) \cdot \frac{1}{2} \sin\frac{x}{2} + (a-1) = 0\) to have no solutions (no critical points), the term multiplying \(\sin\frac{x}{2}\) must be zero and \(a-1\) should not equal zero. This gives us two conditions: \(a^2 - 3a + 2 = 0\) and \(a eq 1\).

Step 5: Solve the Quadratic

Solve \(a^2 - 3a + 2 = 0\) by factoring: \((a-1)(a-2) = 0\). This gives \(a = 1\) or \(a = 2\). However, to meet the condition of no critical points, \(a\) must not equal 1.

Step 6: Determine Valid Interval

Since \(a^2 - 3a + 2 = (a-1)(a-2)\), the conditions together imply that \(a = 2\) because \(a = 1\) violates the \(a eq 1\) condition. Since only \(a=2\) meets these criteria, we examine options to find \(a=2\) falls inside solution sets. Therefore, \((1,3) \cup (3,5)\) contains 2.

Key Concepts

Derivative

When working with functions, finding critical points is an essential task, and this often begins by determining the function's derivative. A derivative tells us how a function changes at any point—it measures the rate at which the function's output is changing with respect to changes in the input. Essentially, it gives the "slope" of the function at any given point.
In the exercise, you're asked to consider the function \( f(x) = (a^2 - 3a + 2)(\cos^2 \frac{x}{4} - \sin^2 \frac{x}{4}) + (a-1)x + \sin 1 \). The critical points are found where the derivative \( f'(x) = 0 \).

• This involves differentiating each part of the function. Constants, such as \( \sin 1 \), have a derivative of 0 because they don't change.
• The term \((a^2 - 3a + 2)(\cos^2 \frac{x}{4} - \sin^2 \frac{x}{4})\) requires the product rule and trigonometric derivatives to find how it changes with \( x \).
• After calculating, \( f'(x) = (a^2 - 3a + 2) \cdot \frac{1}{2} \sin\frac{x}{2} + (a-1) \) represents the derivative of the function.

This derivative helps determine where critical points occur, guiding us on how to solve the function further.

Quadratic Equation

A quadratic equation is any equation that can be rearranged in the standard form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. Recognizing and solving quadratic equations is a foundational skill in algebra.
In solving specific problems, the quadratic \( a^2 - 3a + 2 = 0 \) appears in the step-by-step solution. This represents a form you might need to factor to solve for \( a \).

• To solve, consider factoring: write the equation as \((a-1)(a-2) = 0\).
• This gives us the roots \( a = 1 \) or \( a = 2 \), meaning the equation equals zero when \( a \) is either 1 or 2.

Factoring is one of the simplest ways and usually works when the equation can be easily broken down into factors, as was done here for clarity and simplicity.

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) describe properties of angles and have applications ranging from geometry to wave physics. They also appear frequently in calculus.
In the context of this exercise, the expression \( \cos^2 \frac{x}{4} - \sin^2 \frac{x}{4} \) transforms using trigonometric identities. Factoring or differentiating such combinations requires an understanding of fundamental trigonometric derivatives.

• For example, the derivative \( \frac{d}{dx}(\cos^2 \frac{x}{4} - \sin^2 \frac{x}{4}) = \frac{1}{2} \sin \frac{x}{2} \) is found using differentiation rules for trigonometric functions.
• This conversion uses known derivatives: \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).
• Every function in calculus has specific rules for differentiation, which can be remembered with practice.

This knowledge of trigonometric functions enables the pinpoint of critical points by examining where derivatives zero out.