Problem 47 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\prod_{k=1}^{n}(\cos (2 k-1) x+i \sin (2 k-1) x)\), then \((\operatorname{Re} f(x))^{\prime \prime}+i(\operatorname{Im} f(x))^{\prime \prime}\) is equal to (A) \(n^{2} f(x)\) (B) \(-n^{4} f(x)\) (C) \(-n^{2} f(x)\) (D) \(n^{4} f(x)\)

Step-by-Step Solution

Answer

The correct answer is (C) \(-n^{4} f(x)\).

Step 1: Understand the function

The function is given as \(f(x) = \prod_{k=1}^{n}(\cos((2k-1)x) + i\sin((2k-1)x))\). This is a product of complex exponential terms as each term can be expressed as an exponential: \(e^{i(2k-1)x}\).

Step 2: Simplify using Euler's formula

Using Euler's formula, \(e^{i\theta} = \cos\theta + i\sin\theta\), each term \(\cos((2k-1)x) + i\sin((2k-1)x)\) can be rewritten as \(e^{i(2k-1)x}\). Hence, \(f(x) = \prod_{k=1}^n e^{i(2k-1)x} = e^{i\sum_{k=1}^n (2k-1)x}\).

Step 3: Sum the exponents

Compute the sum of the exponents: \(\sum_{k=1}^n (2k-1) = 1 + 3 + 5 + \ldots + (2n-1)\). This is an arithmetic series with \(n\) terms, first term 1 and last term \(2n-1\).

Step 4: Calculate arithmetic series

The sum of the first \(n\) odd numbers is given by \(n^2\). Therefore, \(\sum_{k=1}^n (2k-1) = n^2\). Thus, \(f(x) = e^{in^2x}\).

Step 5: Differentiate the real and imaginary parts

The real part is \(\operatorname{Re}(f(x)) = \cos(n^2 x)\) and the imaginary part is \(\operatorname{Im}(f(x)) = \sin(n^2 x)\). Compute the second derivatives: \(\left(\operatorname{Re}f(x)\right)'' = -n^4\cos(n^2 x)\) and \(\left(\operatorname{Im}f(x)\right)'' = -n^4\sin(n^2 x)\).

Step 6: Combine derivatives

The expression \((\operatorname{Re} f(x))'' + i(\operatorname{Im} f(x))''\) therefore becomes \[-n^4\cos(n^2 x) - in^4\sin(n^2 x) = -n^4(\cos(n^2 x) + i\sin(n^2 x)) = -n^4 f(x)\].

Step 7: Final Answer

By comparing with the options, the expression matches option C with \(-n^4 f(x)\).

Key Concepts

Euler's Formula

Euler's formula is a key concept when dealing with complex functions. It states that for any real number \( \theta \), the exponential function can be expressed as a sum of trigonometric functions: \( e^{i\theta} = \cos \theta + i\sin \theta \). This formula elegantly links the complex exponential function with sine and cosine functions.

In the context of the problem, Euler's formula helps us convert complex forms into exponential ones. Each term in the product \(f(x) = \prod_{k=1}^{n}(\cos((2k-1)x) + i\sin((2k-1)x))\) can be rewritten using Euler's formula as \(e^{i(2k-1)x}\).

This conversion simplifies the function into a product of exponential terms, \(\prod_{k=1}^n e^{i(2k-1)x} = e^{i\sum_{k=1}^n (2k-1)x}\), which is much easier to work with in later steps.

Complex Exponential

A complex exponential function takes the form \(e^{ix}\), where \(i\) is the imaginary unit. It's widely used since it simplifies the manipulation of trigonometric identities. Rather than dealing with separate sine and cosine terms, using complex exponentials allows for a more straightforward analysis.

In the exercise, after applying Euler's formula, we deal with \(f(x) = e^{in^2x}\). This is because the sum of the exponents in the exponential terms gives us \(n^2\) due to the properties of arithmetic series.

By writing \(f(x)\) in this exponential form, we manage to compress the original multi-term problem into a single exponential term, making differentiation much simpler.

Differentiation of Trigonometric Functions

Differentiation involves finding how a function changes at any point, and for trigonometric functions, the process is well established. The derivatives of sine and cosine functions are given by:

• \(\frac{d}{dx}(\cos x) = -\sin x\)
• \(\frac{d}{dx}(\sin x) = \cos x\)
• Second derivatives: \(\frac{d^2}{dx^2}(\cos x) = -\cos x\), and \(\frac{d^2}{dx^2}(\sin x) = -\sin x\)

In our problem, we need the second derivatives:\(\left(\operatorname{Re}f(x)\right)'' = -n^4\cos(n^2 x)\) and \(\left(\operatorname{Im}f(x)\right)'' = -n^4\sin(n^2 x)\). This requires using the chain rule, multiplying by the derivative of the inside function \(n^2\).

Combining these second derivatives gives us the expression required in the exercise: \(-n^4(\cos(n^2 x) + i\sin(n^2 x)) = -n^4 f(x)\). This highlights how differentiation simplifies the mathematics of trigonometric functions.