Problem 105 - A Complete Resource Book in Mathematics for JEE Main

Question

Instructions: In the following questions an Assertion (A) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)\) \((\cos 3 x+i \sin 3 x) \ldots(\cos n x+i \sin n x)\) and \(f(1)=1\) then \(f^{\prime \prime}(1)\) is equal to \(-\left(\frac{n(n+1)}{2}\right)^{2}\). Reason: \(f(x)=\cos \frac{n(n-1)}{2} x+i \sin \frac{n(n-1)}{2} x\)

Step-by-Step Solution

Answer

Option C: Assertion is true, but Reason is false.

Step 1: Understanding the Assertion

The assertion states that if we have the function \(f(x) = (\cos x + i \sin x)(\cos 2x + i \sin 2x)\ldots(\cos nx + i \sin nx)\) and \(f(1) = 1\), then the second derivative \(f''(1)\) is equal to \(-\left(\frac{n(n+1)}{2}\right)^2\).

Step 2: Express the Function in Another Form

Recall the identity \(\cos kx + i\sin kx = e^{ikx}\) for complex numbers. Thus, \\[ f(x) = e^{ix} \cdot e^{i2x} \cdot \, \ldots \, \cdot e^{inx} = e^{i(1+2+\ldots+n)x} = e^{i(\frac{n(n+1)}{2})x}. \\] This simplifies the function \( f(x) \) to an exponentiated form.

Step 3: Verify Assertion Using Simplified Form

Since \(f(1) = 1\), we have \(e^{i\left(\frac{n(n+1)}{2}\right)\cdot 1} = 1\). Hence, \(\frac{n(n+1)}{2}\) must be a multiple of \(2\pi\), which is consistent with the original problem. We now need to determine \(f''(1)\).

Step 4: Differentiate the Simplified Function Twice

Differentiate \(f(x) = e^{i\left(\frac{n(n+1)}{2}\right)x}\) with respect to \(x\), giving first: \\[ f'(x) = i\left(\frac{n(n+1)}{2}\right)e^{i\left(\frac{n(n+1)}{2}\right)x}, \\] and second differentiation gives: \\[ f''(x) = \left(i\left(\frac{n(n+1)}{2}\right)\right)^2 e^{i\left(\frac{n(n+1)}{2}\right)x} = -\left(\frac{n(n+1)}{2}\right)^2 e^{i\left(\frac{n(n+1)}{2}\right)x}.\]

Step 5: Evaluate f''(1)

Since \(f(1) = 1\), \(e^{i\left(\frac{n(n+1)}{2}\right)\cdot 1} = 1\) holds. Thus: \\[ f''(1) = -\left(\frac{n(n+1)}{2}\right)^2 \times 1 = -\left(\frac{n(n+1)}{2}\right)^2.\] This confirms the assertion in the original question.

Step 6: Evaluate the Reason

The reason suggests that \(f(x) = \cos\left(\frac{n(n-1)}{2}x\right) + i\sin\left(\frac{n(n-1)}{2}x\right)\), which corresponds to \(e^{i\left(\frac{n(n-1)}{2}\right)x}\). However, our function \(f(x)\) is actually \(e^{i\left(\frac{n(n+1)}{2}\right)x}\). Therefore, the reason is incorrect.

Key Concepts

Euler's Formula

To understand Euler's Formula, let's look at complex numbers and the relationship between trigonometric functions and exponential functions. Euler's Formula states that for any real number \( x \), the equation \( e^{ix} = \cos x + i \sin x \) holds true. This formula elegantly connects the exponential function with the sine and cosine functions, combining them into one powerful equation.

When you see an expression like \( \cos x + i \sin x \), you can replace it with \( e^{ix} \). This substitution can radically simplify many problems in mathematics, particularly those involving complex numbers.

Consider the function \( f(x) = (\cos x + i \sin x)(\cos 2x + i \sin 2x)\ldots(\cos nx + i \sin nx) \). By employing Euler's Formula, we can rewrite this as \( f(x) = e^{ix} \cdot e^{i2x} \cdot \ldots \cdot e^{inx} \), which further simplifies to \( f(x) = e^{i(1+2+\ldots+n)x} \). It's a neat transformation that provides clarity and simplifies differentiation and integration. Keep this formula in your toolbox as it often reveals hidden connections in mathematical expressions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the equation are defined. These identities are fundamental tools in trigonometry, and help in simplifying expressions and solving equations.

Some of the most crucial identities include:

• Pythagorean identities: \( \sin^2 x + \cos^2 x = 1 \)
• Sum and difference identities: \( \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b \) and \( \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b \)
• Double angle formulas: \( \sin 2x = 2\sin x \cos x \) and \( \cos 2x = \cos^2 x - \sin^2 x \)

In the context of our problem, we use the identity \( \cos kx + i \sin kx = e^{ikx} \) to express the function in a simplified exponential form. Such identities are crucial for manipulating and transforming expressions in trigonometric and exponential forms. Understanding these will help you see their utility beyond the basics and allow more advanced applications such as calculating derivatives and integrals of trigonometric functions.

Differentiation of Exponential Functions

Differentiation is a core concept in calculus, and differentiating functions correctly is essential for understanding dynamic changes and patterns. Exponential functions of the form \( e^{ax} \) are particularly important because of their unique properties.

When you differentiate an exponential function \( f(x) = e^{ax} \), you apply the derivative rule which states that the derivative of \( e^{ax} \) with respect to \( x \) is \( a \cdot e^{ax} \). This property becomes even more interesting when dealing with complex numbers, as in our function \( f(x) = e^{i(\frac{n(n+1)}{2})x} \). Here, \( a \) is a complex coefficient \( i(\frac{n(n+1)}{2}) \).

To find the second derivative, \( f''(x) \), differentiate again, which gives:
\[ f'(x) = i\left(\frac{n(n+1)}{2}\right)e^{i\left(\frac{n(n+1)}{2}\right)x} \]
and subsequently:
\[ f''(x) = \left(i\left(\frac{n(n+1)}{2}\right)\right)^2 e^{i\left(\frac{n(n+1)}{2}\right)x} \]
This step demonstrates how exponentiated forms make differentiation straightforward, even when it involves complex numbers. The key takeaway is the straight rule application that retains its elegance and simplicity, mainly due to the power of the base \( e \) in calculus.