Problem 12 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f(x)=3 \cos |x|-6 a x+b\) increases for all \(x \in R\), then the range of values of \(a\) is given by (A) \(a>-\frac{1}{2}\) (B) \(a<-\frac{1}{2}\) (C) \(a \leq b\) (D) \(a \geq b\)

Step-by-Step Solution

Answer

(B) \(a < -\frac{1}{2}\)

Step 1: Understand the Condition for Increasing Functions

A function is said to be increasing for all \(x\) if its first derivative is non-negative for all \(x \in \mathbb{R}\). Let's find the derivative of the function \(f(x) = 3\cos|x| - 6ax + b\).

Step 2: Find the Derivative of the Function

The derivative of \(3\cos|x|\) involves differentiating \(|x|\), which results in being \(-3\sin|x|\cdot(\text{sgn}(x))\) there is a jump at \(x=0\). For differentiating \(|x|\), we replace it with \(-3\sin x \) using piecewise methods for \(x>0\) and \(x<0\). The derivative of \(-6ax\) is \(-6a\), and \(b\) disappears after differentiation.Therefore, \(f'(x) = -3\sin|x|\cdot(\text{sgn}(x))-6a\).

Step 3: Set Condition for Non-Negative Derivative

For the function to be increasing for all \(x\), its derivative must be non-negative: \[-3\sin|x|\cdot(\text{sgn}(x))-6a \geq 0\].

Step 4: Simplify the Inequality Considering All x

As \(\sin|x|\) varies between -1 and 1, the term \(-3\sin|x|\cdot(\text{sgn}(x))\) becomes at maximum 3 and at minimum -3 because of the sign. We consider when \(-3\sin|x| = -3\) to address worst-case scenario:\[-3 - 6a \geq 0 \implies a \leq -\frac{1}{2}.\]

Step 5: Conclude the Range of Values for a Consistent with Choices

Considering the analysis of the derivative, we found that \(a \leq -\frac{1}{2}\) for the function to be increasing. Hence, the range for \(a\) aligns with option (B).

Key Concepts

Derivatives

A derivative is a fundamental concept in calculus, measuring how a function changes at any given point, essentially giving us the slope of the function. When considering increasing functions, their first derivative plays a pivotal role.

• If the first derivative of a function is positive, the function is increasing.
• If it is zero, the function is constant at that particular point.
• If it is negative, the function is decreasing.

For the given exercise, by deriving the function \[ f(x) = 3 \cos |x| - 6ax + b \]we use rules of differentiation. We find the first derivative as\[ f'(x) = -3 \sin |x| \cdot (\text{sgn}(x)) - 6a \]Understanding each component of this derivative helps in knowing how the function's slope behaves across the real number line.

Inequalities

Inequalities provide insights into the relationships between different quantities and are especially useful in calculus when examining the behavior of derivatives. In this exercise, we're examining the inequality associated with ensuring the function is increasing:\[ -3 \sin |x| \cdot (\text{sgn}(x)) - 6a \geq 0 \]This inequality is solved by considering the behavior of the trigonometric part and constant changes:

• The trigonometric part \(-3 \sin |x|\) reaches values between \(-3\) and \(3\) as \(|x|\) grows and oscillates.
• Triple the sine function can shift from a high of \(+3\) to a low of \(-3\), affecting the inequality substantially.

For the purpose of our solution, we consider the maximum possible negative value:\[ -3 - 6a \geq 0 \]Solving this inequality simplifies to: \[ a \leq -\frac{1}{2} \]This condition ensures that the derivative of our function is non-negative for all values of \(x\), ensuring the function is increasing.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in understanding oscillatory motions, periodicity, and waveforms. In calculus, these functions often appear in problems involving angles and periodic behavior.For the function in this exercise, \( 3 \cos |x| \),the absolute value operation and the cosine function together mean that as \(x\) shifts, the output oscillates while maintaining a consistent positive magnitude. Key insights include:

• The cosine of an absolute value \(|x|\) means the graph oscillates in a symmetric fashion across the y-axis.
• The derivative, \(-3 \sin |x| \cdot (\text{sgn}(x))\), introduces a sine component based on the signum function \((\text{sgn}(x))\). This controls the direction of sinusoidal change and is crucial for determining the slope at different points.

For our problem, this means assessing where the sine component is most impactful in shifting the equation towards its limits to guide the derivative inequality solution.