Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)

If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is:

• A. -2
• B. –3
• C. 2
• D. 1

Hint:

When higher derivatives are involved, look for patterns in trigonometric identities and test values at standard angles.

Answer 1

The Correct Option is B

We are given a functional equation for a twice differentiable function \( f: \mathbb{R} \to \mathbb{R} \) and a condition on its first derivative at \( x=0 \). We need to find the value of the expression \( 24 f''\left(\frac{5\pi}{3}\right) \).

Concept Used:

The solution involves simplifying the given functional equation using trigonometric identities to determine the form of the function \( f(x) \). The key trigonometric identities are:

\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \] \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

Once the form of \( f(x) \) is established, we use differentiation and the given initial condition \( f'(0) = \frac{1}{2} \) to find the specific function. Finally, we compute its second derivative at the required point.

Step-by-Step Solution:

The given functional equation is:

\[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)) \]

We start by rearranging the terms to group \( f(2x + 2y) \) and \( f(2x - 2y) \). 

Expanding both sides, we get:

\[ f(2x + 2y)\sin x \cos y - f(2x - 2y)\sin x \cos y = f(2x + 2y)\cos x \sin y + f(2x - 2y)\cos x \sin y \]

Now, we bring terms with \( f(2x + 2y) \) to one side and terms with \( f(2x - 2y) \) to the other side:

\[ f(2x + 2y)(\sin x \cos y - \cos x \sin y) = f(2x - 2y)(\sin x \cos y + \cos x \sin y) \]

Using the sine addition and subtraction formulas, this simplifies to:

\[ f(2x + 2y)\sin(x - y) = f(2x - 2y)\sin(x + y) \]

For values of \(x, y\) where \( \sin(x-y) \neq 0 \) and \( \sin(x+y) \neq 0 \), we can write:

\[ \frac{f(2x + 2y)}{\sin(x + y)} = \frac{f(2x - 2y)}{\sin(x - y)} \]

Let \( u = x + y \) and \( v = x - y \). 

The equation becomes:

\[ \frac{f(2u)}{\sin u} = \frac{f(2v)}{\sin v} \]

Since this relation holds for arbitrary \( u \) and \( v \), it implies that the expression \( \frac{f(2z)}{\sin z} \) must be a constant, let's call it \( k \).

\[ \frac{f(2z)}{\sin z} = k \implies f(2z) = k \sin z \]

To find the form of \( f(x) \), we can substitute \( t = 2z \), which gives \( z = t/2 \). Thus, the function is of the form:

\[ f(t) = k \sin\left(\frac{t}{2}\right) \quad \text{or} \quad f(x) = k \sin\left(\frac{x}{2}\right) \]

Next, we use the given condition \( f'(0) = \frac{1}{2} \) to find the constant \( k \). 

First, we differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = \frac{d}{dx}\left(k \sin\left(\frac{x}{2}\right)\right) = k \cos\left(\frac{x}{2}\right) \cdot \frac{1}{2} = \frac{k}{2}\cos\left(\frac{x}{2}\right) \]

Now, we substitute \( x = 0 \):

\[ f'(0) = \frac{k}{2}\cos(0) = \frac{k}{2} \]

Given \( f'(0) = \frac{1}{2} \), we have:

\[ \frac{k}{2} = \frac{1}{2} \implies k = 1 \]

So, the specific function is \( f(x) = \sin\left(\frac{x}{2}\right) \). Now we need to find its second derivative.

\[ f'(x) = \frac{1}{2}\cos\left(\frac{x}{2}\right) \] \[ f''(x) = \frac{d}{dx}\left(\frac{1}{2}\cos\left(\frac{x}{2}\right)\right) = \frac{1}{2}\left(-\sin\left(\frac{x}{2}\right)\right) \cdot \frac{1}{2} = -\frac{1}{4}\sin\left(\frac{x}{2}\right) \]

Final Computation & Result:

We need to calculate the value of \( 24f''\left(\frac{5\pi}{3}\right) \). First, let's evaluate \( f''\left(\frac{5\pi}{3}\right) \):

\[ f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4}\sin\left(\frac{1}{2} \cdot \frac{5\pi}{3}\right) = -\frac{1}{4}\sin\left(\frac{5\pi}{6}\right) \]

We know that \( \sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \). 

Substituting this value:

\[ f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{8} \]

Finally, we compute the required expression:

\[ 24f''\left(\frac{5\pi}{3}\right) = 24 \times \left(-\frac{1}{8}\right) = -3 \]

Hence, the value of \( 24f''\left(\frac{5\pi}{3}\right) \) is -3.

Answer 2

Using substitution and matching with function properties, the fourth derivative is modeled as: \[ f^{(4)}(x) = -\frac{1}{4} \sin \left( \frac{x}{2} \right) \] Then: \[ 24f^{(4)}\left(\frac{5\pi}{3}\right) = 24 \cdot \left( -\frac{1}{4} \cdot \sin\left(\frac{5\pi}{6}\right) \right) = 24 \cdot \left( -\frac{1}{4} \cdot \frac{1}{2} \right) = -3 \]