Problem 75 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)\) and \(g(x)\) are differentiable functions for \(0 \leq x \leq 1\) such that \(f(0)=2, g(0)=0, f(1)=6, g(1)=2\), then in the interval \((0,1)\), (A) \(f^{\prime}(x)=0\) for all \(x\) (B) \(f^{\prime}(x)=2 g^{\prime}(x)\) for at least one \(x\) (C) \(f^{\prime}(x)=2 g^{\prime}(x)\) for at most one \(x\) (D) None of these

Step-by-Step Solution

Answer

The solution is option (B).

Step 1: Identify Given Data

We have functions \(f(x)\) and \(g(x)\) with initial and final values: \(f(0)=2\), \(g(0)=0\), \(f(1)=6\), and \(g(1)=2\). These functions are differentiable on the interval \([0, 1]\).

Step 2: Compute Derivative Condition

We will use the Mean Value Theorem (MVT) for differentiable functions. MVT states there exists some \(c\) in \((a, b)\) such that:\[f'(c) = \frac{f(b)-f(a)}{b-a}\] For \(f(x)\) over the interval \([0, 1]\), we have:\[f'(c) = \frac{6 - 2}{1 - 0} = 4\]For \(g(x)\), we have:\[g'(c) = \frac{2 - 0}{1 - 0} = 2\]

Step 3: Analyze Condition \(f'(x) = 2g'(x)\)

Based on the results, for some \(x\) in \((0, 1)\), the conditions from MVT are: \(f'(x) = 4\) and \(g'(x) = 2\).If \(f'(x) = 2g'(x)\), then set \(f'(x) = 4 = 2g'(x) = 2 \,\cdot\, 2 \,\Rightarrow\, 4 = 4\).Thus, \(f'(x) = 2g'(x)\) is satisfied for every \(x\) since this is a constant equality on the interval.

Step 4: Review Answer Options

From the analysis, option (B) \(f^{\prime}(x) = 2g^{\prime}(x)\) for at least one \(x\) holds true, as seen by our calculation (though technically it holds for all \(x\) in the interval). Therefore, option (B) is true.

Key Concepts

Differentiable Functions

In mathematics, a function is said to be differentiable over an interval if it has a well-defined derivative at every point in that interval. This means the function is smooth and continuous without any sharp edges or breaks in the graph.
Differentiable functions have a derivative, which is essentially the slope of the tangent line to the graph of the function at any given point.

• If you imagine the graph of a function as a road, the derivative tells you how steep that road is at each point.
• Differentiability implies continuity, which means that the function does not jump or have gaps.
• In our exercise, both functions \(f(x)\) and \(g(x)\) are differentiable on the interval \([0, 1]\), meaning they both have derivatives at any \(x\) value in this interval.

Derivative

The derivative of a function provides a measure of how the function changes as its input changes. It gives us important information about the behavior of functions.

• For a function \(f(x)\), its derivative, denoted \(f'(x)\), tells us the rate at which \(f(x)\) increases or decreases as \(x\) changes.
• The concept of derivative is critical for the Mean Value Theorem. This theorem states that for a differentiable function \(f(x)\) on an interval \([a, b]\), there is at least one point \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\).
• For function \(f(x)\), this formula calculates how the average rate of change over the interval can also be seen at some point as an instantaneous rate of change.

In our example, applying the theorem gives \(f' = 4\) and \(g' = 2\), allowing us to analyze further.

Interval Analysis

Interval analysis involves examining the behavior of functions over a specified range. It allows us to draw conclusions about how functions behave and predict their values within given bounds.

• In the context of our problem, we analyze the interval \((0,1)\) to understand how \(f(x)\) and \(g(x)\) behave.
• This analysis is crucial for applying the Mean Value Theorem and determining if certain conditions, such as \(f'(x) = 2g'(x)\), hold true at any point within the interval.
• The result shows us that the condition \(f'(x) = 2g'(x)\) is satisfied throughout the entire interval, affirming option (B) in the exercise solution.

By understanding how functions behave in specified ranges, we can make informed predictions and solve complex calculus problems with greater ease.