Problem 24 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

Let \(f(x)\) and \(g(x)\) be differentiable for \(0 \leq x \leq 1\), such that \(f(0)=0, g(0)=0, f(1)=6\). Let there exist a real number \(c\) in \((0,1)\) such that \(f(c)=\) \(2 g^{\prime}(c)\), then the value of \(g(1)\) must be [Pb. CET-1991] (a) 1 (b) 3 (c) \(-2\) (d) \(-1\)

Step-by-Step Solution

Answer

The value of \( g(1) \) is 3, option (b).

Step 1: Understanding the Given Information

We have two functions, \( f(x) \) and \( g(x) \), both differentiable for \( 0 \leq x \leq 1 \). We know the initial values: \( f(0) = 0 \), \( g(0) = 0 \), and the endpoint value for \( f(x) \): \( f(1) = 6 \). Additionally, there is a point \( c \) in \( (0,1) \) such that \( f(c) = 2g'(c) \). We need to find \( g(1) \).

Step 2: Apply the Mean Value Theorem to f(x)

Since \( f(x) \) is differentiable on \( (0,1) \) and continuous on \( [0,1] \), by the Mean Value Theorem, there exists some number \( c_1 \) in \( (0,1) \) such that \[ f'(c_1) = \frac{f(1) - f(0)}{1 - 0} = \frac{6 - 0}{1} = 6. \]

Step 3: Consider the Condition on c

According to the problem, there exists a \( c \) such that \( f(c) = 2g'(c) \). Assuming \( c = c_1 \), we have \[ f(c_1) = 2g'(c_1). \] Using the Mean Value Theorem again, we get \[ f'(c_1) = 6 = 2g'(c_1). \] Solving for \( g'(c_1) \) gives \[ g'(c_1) = 3. \]

Step 4: Estimate g(1) Using g'(c_1)

Assuming a linear behavior of \( g(x) \) for simplicity, since \( g(0) = 0 \) and the derivative \( g'(c_1) = 3 \), integration gives \[ g(1) - g(0) = g'(c_1) \cdot (1 - 0) = 3 \cdot 1 = 3. \]Thus, \( g(1) = 3. \)

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental principle in calculus. It connects the average rate of change of a function to its instantaneous rate of change at some point within the interval. In simple terms, if you have a continuous function over a closed interval and differentiable on the open interval, there is at least one point where the tangent to the curve is parallel to the secant line connecting the endpoints of the interval.

For example, with a function like \( f(x) \) that meets these conditions on the interval \([0, 1]\), the MVT guarantees that there exists some \( c \) in \( (0, 1) \) such that:

• \( f'(c) = \frac{f(1) - f(0)}{1 - 0} \)

In our problem, applying this to \( f(x) \), we find a \( c \) such that \( f'(c) = 6 \). This is because the change from \( f(0) \) to \( f(1) \) is 6 over that interval.

Function Differentiation

Differentiating a function involves computing its derivative, which gives us the rate at which the function changes with respect to its input. In this problem, \( g'(x) \) is the derivative of \( g(x) \) and represents the rate of change of \( g(x) \) at any point \( x \) in its domain.

The problem introduces a condition where \( f(c) = 2g'(c) \) at some point \( c \) in \( (0,1) \). This serves to connect the behavior of these two functions. By assuming \( c \) as the same point determined by the MVT for \( f(x) \), it turns out by calculation that \( g'(c) = \frac{f'(c)}{2} \). Since \( f'(c) = 6 \) from earlier, we can solve for:

• \( g'(c) = 3 \)

Calculus Problem Solving

Solving calculus problems often involves applying known theorems, like the Mean Value Theorem, and using properties of derivatives to find unknown values, just as we found \( g(1) \) in this problem.

Once we know that \( g'(c) = 3 \), we can make an assumption about the behavior of \( g(x) \). If we assume \( g(x) \) behaves linearly, the derivative \( g'(x) \) remains constant over the interval. The constant derivative implies a consistent slope, indicating:

• \( g(x) = 3x \) over the interval \([0, 1]\)

Given that \( g(0) = 0 \), integrating the constant slope from 0 to 1:

• \( g(1) - g(0) = g'(1) \times (1 - 0) = 3 \cdot 1 \)

Thus, using these steps, general problem-solving methods simplify the process to find the value of \( g(1) = 3 \).