Question

Let \(f\) and \(g\) be real valued functions defined on interval \((-1,1)\) such that \(g^{\prime \prime}(x)\) is continuous, \(g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq 0\), and \(f(x)=g(x) \sin x\). STATEMENT-1: \(\lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0) .\) and STATEMENT-2: \(f^{\prime}(0)=g(0)\). (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

(C) STATEMENT-1 is True, STATEMENT-2 is False. By computing both limits and derivatives, we find that STATEMENT-1 is consistent with \(f''(0)\), but STATEMENT-2 incorrectly describes \(f'(0)\).

Step-by-step solution

Understand and Evaluate STATEMENT-1

To evaluate the first statement, let's find the limit as x approaches 0 of the expression inside the square brackets. The limit definition and the L'Hôpital's Rule may be used to find the limit of a difficult indeterminate form, such as '0/0'. Here, we need to examine \( \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x] \).

Apply L'Hôpital's Rule for STATEMENT-1

Rewrite the limit using the identity \( \cot x = \frac{1}{\tan x} \) and \( \operatorname{cosec} x = \frac{1}{\sin x} \). Then, apply L'Hôpital's Rule twice to \(g(x) \cot x - g(0) \operatorname{cosec} x\) as x approaches 0 to determine the value of the limit.

Compute the second derivative of \(f(x)\) at x = 0

We find \(f'(x)\) by applying the product rule to \(f(x) = g(x) \sin x\). Then, we compute \(f''(x)\) by differentiating \(f'(x)\) again and evaluate \(f''(0)\).

Evaluate STATEMENT-2

To evaluate the second statement, we find \(f'(x)\) by differentiating \(f(x) = g(x) \sin x\). We then evaluate \(f'(0)\) to see if it is equal to \(g(0)\).

Assess the validity of STATEMENT-1

After evaluating the limit and the second derivative, compare them to establish the validity of the first statement. If the limit and \(f''(0)\) are equal, the statement is true.

Assess the validity and relevance of STATEMENT-2

If \(f'(0)\) equals \(g(0)\), then STATEMENT-2 is true. But to be a correct explanation for STATEMENT-1, it must logically explain why STATEMENT-1 is true. Assess whether the first derivative at 0 has any impact on the statement regarding the second derivative.

Key concepts

L'Hôpital's Rule

In calculus, L'H\text{\textup{o}}pital's Rule is a powerful tool used to evaluate the limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with a limit that yields an indeterminate form, applying L'H\text{\textup{o}}pital's Rule can simplify the process of finding that limit.

This rule states that if the functions \( f(x) \) and \( g(x) \) are differentiable and approach 0 or \(\pm\infty\) as \( x \) approaches a certain value, and if the limit of their derivatives, \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) exists or is \(\pm\infty\), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \).

In the provided exercise, we use L'H\text{\textup{o}}pital's Rule to evaluate the limit of \( g(x) \cot x - g(0) \operatorname{cosec} x \) as \( x \) approaches 0. The original expression is indeterminate at \( x = 0 \) since both \(\cot x\) and \(\operatorname{cosec} x\) involve division by zero. By reworking the expression and differentiating the numerator and denominator accordingly, L'H\text{\textup{o}}pital's Rule allows us to find a determinate limit.

Product Rule of Differentiation

The Product Rule of Differentiation is a fundamental theorem in calculus that allows us to find the derivative of a product of two functions. \( f \) and \( g \) are functions of \( x \), then the derivative of their product \( f(x) \cdot g(x) \) can be calculated as follows: \( (f \cdot g)'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \).

The product rule is particularly useful when dealing with functions that are multiplied together, as is the case in the exercise with \( f(x) = g(x) \sin x \). To find \( f'(x) \), we need to take the derivative of both \( g(x) \) and \( \sin x \) and then apply the product rule. This provides us with the first derivative of \( f \) which can be further differentiated to find higher-order derivatives such as \( f''(x) \), necessary for evaluating the given statements about \( f \).

Understanding the product rule is essential for correctly applying the derivative to solve problems involving the multiplication of functions, which is frequently encountered in calculus.

Second Derivative Test

The Second Derivative Test is used in calculus to determine the concavity of a function at a given point and to identify potential local maxima or minima. It involves taking the second derivative of a function \( f \) and evaluating it at a specific point \( x_0 \).

To apply this test, you first find the critical points of the function, which are where the first derivative \( f'(x) \) is zero or undefined. Once you have a critical point, you evaluate the second derivative \( f''(x) \) at that point. If \( f''(x_0) > 0 \) at a critical point \( x_0 \), the function has a local minimum there because the function is concave up. Conversely, if \( f''(x_0) < 0 \) at that point, there is a local maximum because the function is concave down. If \( f''(x_0) = 0 \) the test is inconclusive.

In the context of the exercise, we use the second derivative test to evaluate \( f''(0) \). This informs us about the concavity of \( f \) at \( x = 0 \) but also directly relates to the first statement that involves the limit of a related expression. Understanding the behavior of the function's second derivative is critical to solving and confirming the validity of the statements provided.