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

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

Key Concepts:

The solution requires algebraic manipulation of the functional equation using trigonometric identities to deduce the form of \( f(x) \). The relevant trigonometric identities are:

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

After establishing \( f(x) \), differentiation and the given condition \( f'(0) = \frac{1}{2} \) are used to find the specific function. The second derivative is then evaluated at the specified point.

Step-by-Step Derivation:

The provided functional equation is:

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

Rearrange the equation to group terms involving \( f(2x + 2y) \) and \( f(2x - 2y) \).

Expanding the equation yields:

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

Isolate terms with \( f(2x + 2y) \) on one side and terms with \( f(2x - 2y) \) on the other:

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

Applying the sine subtraction and addition formulas simplifies the equation to:

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

For \( \sin(x-y) eq 0 \) and \( \sin(x+y) eq 0 \), 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 transforms to:

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

Since this equality holds for arbitrary \( u \) and \( v \), the expression \( \frac{f(2z)}{\sin z} \) must be a constant, denoted by \( k \).

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

To express \( f(x) \) in terms of \( x \), substitute \( t = 2z \), so \( z = t/2 \). This gives the general form of the function:

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

Utilize the condition \( f'(0) = \frac{1}{2} \) to determine the value of \( k \).

Differentiate \( f(x) \) with respect to \( x \):

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

Evaluate \( f'(x) \) at \( x = 0 \):

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

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

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

Therefore, the specific function is \( f(x) = \sin\left(\frac{x}{2}\right) \). Now, compute 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 Calculation and Result:

Calculate the value of \( 24f''\left(\frac{5\pi}{3}\right) \). First, find \( f''\left(\frac{5\pi}{3}\right) \):

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

Since \( \sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):

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

Finally, compute the required expression:

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

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