Question

Let \(f(x)=2+\cos x\) for all real \(x\). STATEMENT-1: For each real \(t\), there exists a point \(c\) in \([t, t+\pi]\) such that $$ f^{\prime}(c)=0 $$ because STATEMENT-2: \(f(t)=f(t+2 \pi)\) for each real \(t\). (A) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement- 1 is True, Statement- 2 is False (D) Statement- 1 is False, Statement- 2 is True

Short answer

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. The correct choice is (B).

Step-by-step solution

Analyze Statement-1

To verify the truth of Statement-1, find the derivative of the function, which is \( f'(x) = -\text{sin}(x) \). According to the Intermediate Value Theorem, if \(f'(t)\) and \(f'(t + \text{π}) \) have different signs, then there must be at least one point \(c\) in the interval \([t, t+ \text{π}]\) where \(f'(c) = 0\). We can check this by evaluating \(f'(t)\) and \(f'(t + \text{π})\) to see if they have different signs.

Evaluate the Derivative at Endpoints

Evaluate the derivative function at the two endpoints \(t\) and \(t + \text{π}\): \(f'(t) = -\text{sin}(t) \) and \(f'(t + \text{π}) = -\text{sin}(t + \text{π})\). Since \(\text{sin}(x)\) is a periodic function with period \(2\text{π}\), it is an odd function, so \(\text{sin}(t + \text{π}) = - \text{sin}(t)\). This means that \(f'(t)\) and \(f'(t + \text{π})\) have opposite signs unless sin(t) is zero.

Check the Intermediate Value Theorem Conditions

Since \(f'(t)\) and \(f'(t + \text{π})\) are of opposite signs for all values of \(t\) except when \(\text{sin}(t) = 0\) (in which case there would also be a root at that point), the conditions of the Intermediate Value Theorem are satisfied. Therefore, there exists at least one point \(c\) in the interval \([t, t + \text{π}]\) where \(f'(c) = 0\), which proves that Statement-1 is true.

Analyze Statement-2

Statement-2 states that \(f(t) = f(t+2\text{π})\) for each real \(t\). This can be verified by noting that the cosine function is a periodic function with a period of \(2\text{π}\), meaning \(\text{cos}(x) = \text{cos}(x + 2\text{π})\) for all \(x\). Since \(f(x)\) is defined as \(2 + \text{cos}(x)\), this property directly transfers to \(f(x)\), making Statement-2 true.

Determine if Statement-2 is a Correct Explanation for Statement-1

While Statement-2 is true, it does not provide the reason for the existence of a point where \(f'(c) = 0\) in the interval \([t, t + \text{π}]\). The fact that \(f(x)\) is periodic with a period of \(2\text{π}\) is unrelated to the behavior of its derivative over the interval of length \( \text{π} \). The correct explanation involves the Intermediate Value Theorem and the properties of the sine function, \text{sin}(x), the derivative of \(f(x)\). Therefore, Statement-2 is not the correct explanation for Statement-1.

Key concepts

Derivative of Cosine

Understanding the derivative of the cosine function is a fundamental concept in calculus and plays a crucial role in analyzing the behavior of trigonometric functions within a given interval. When we differentiate the cosine function, we get the negative sine function. Formally, if we have a function such as \( f(x) = \text{cos}(x) \), then the derivative, denoted as \( f'(x) \), is \( -\text{sin}(x) \).

This derivative helps us determine the rate at which the cosine function's value is changing at any point on its curve. If we consider a function like \( f(x) = 2 + \text{cos}(x) \), then by differentiating it, we obtain \( f'(x) = -\text{sin}(x) \). Understanding the sine and cosine functions' derivatives is pivotal for solving many problems in calculus, including finding the points where the function has horizontal tangents (where the derivative is zero). This knowledge is essential for students preparing for rigorous exams, such as the JEE Advanced, where calculus plays a prominent role.

Periodicity of Trigonometric Functions

Trigonometric functions such as sine and cosine exhibit periodic behavior, which means that these functions repeat their values in regular intervals, known as their periods. For the cosine function, the period is \(2\text{π} \), which implies that \( \text{cos}(x) = \text{cos}(x + 2\text{π}n) \) for any integer \( n \). This property is known as the periodicity of the cosine function.

Periodicity is a crucial factor when analyzing the behavior of trigonometric functions over different intervals. For example, when solving an exercise such as identifying if there is a point \( c \) in \( [t, t+\text{π}] \) where the derivative of the function \( f(x) = 2 + \text{cos}(x) \) equals zero, knowing the periodic nature of the cosine function can help us understand that \( f(t) \) will have the same value as \( f(t+2\text{π}) \). Consequently, periodicity is indispensable when considering the long-term behavior of trigonometric functions, which is also a critical aspect in the JEE Advanced mathematics syllabus.

JEE Advanced Mathematics

The JEE Advanced exam is known for its challenging mathematics section, which tests a wide range of topics, including algebra, calculus, and trigonometry. Understanding concepts such as the derivative of cosine and the periodicity of trigonometric functions is essential for success in this examination.

The exam's questions often integrate multiple concepts, such as using the Intermediate Value Theorem in conjunction with the properties of the cosine function's derivative to prove that a function has certain behaviors within an interval. For students, grasping these concepts not only aids in solving textbook exercises but also builds the deeper understanding necessary for tackling the complex problems presented in the JEE Advanced mathematics section. Clear, methodical problem-solving steps and solid theoretical knowledge are key to mastering the types of questions encountered in the exam.