Question:

To solve the equation \( x \log x = 1 \), using Newton-Raphson method, the iterative formula and the first approximate \( x_1 \), when \( x_0 = 1 \) is (Assuming \(\log\) is natural logarithm \(\ln\))

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Newton-Raphson: \(x_{n+1} = x_n - f(x_n)/f'(x_n)\).
Remember derivative rules (e.g., product rule, \(\frac{d}{dx}\ln x = 1/x\)).
\(\ln 1 = 0\).
Updated On: Jun 10, 2025
  • \( x_{n+1} = \frac{x_{n}-1}{1+\log x_n} ; x_1 = 0 \)
  • \( x_{n+1} = \frac{x_n + 2x_n \log x_n + 1}{1+\log x_n} ; x_1 = 2 \)
  • \( x_{n+1} = \frac{x_n - 2x_n \log x_n + 1}{1+\log x_n} ; x_1 = 2 \)
  • \( x_{n+1} = \frac{x_n+1}{1+\log x_n} ; x_1 = 2 \)
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The Correct Option is D

Solution and Explanation

Two functions \(f(x)\) and \(g(x)\) are linearly dependent if there exist constants \(c_1, c_2\), not both zero, such that \(c_1f(x) + c_2g(x) = 0\) for all \(x\). This is equivalent to one function being a constant multiple of the other (if neither is identically zero).

Let's examine option (b): \(f(x) = \sin x(4\sin^2 x - 3)\) and \(g(x) = \sin 3x\). 

We know the trigonometric identity for \(\sin 3x\): \(\sin 3x = 3\sin x - 4\sin^3 x\).

Now consider \(f(x)\): \(f(x) = \sin x(4\sin^2 x - 3) = 4\sin^3 x - 3\sin x\).

Comparing \(f(x)\) with \(\sin 3x\): \(f(x) = 4\sin^3 x - 3\sin x = -(3\sin x - 4\sin^3 x) = -\sin 3x\).

So, \(f(x) = -g(x)\), or \(f(x) + g(x) = 0\). Since we can write \(1 \cdot f(x) + 1 \cdot g(x) = 0\), with non-zero constants, the functions are linearly dependent.

Let's briefly check other options:

(a) \(e^x \sin 2x, e^x \cos 2x\): Linearly independent as \(\sin 2x\) and \(\cos 2x\) are independent.

(c) \(\cos x, x \cos x\): If \(c_1 \cos x + c_2 x \cos x = 0 \Rightarrow \cos x (c_1 + c_2 x) = 0\). For this to hold for all \(x\), \(c_1 = 0\) and \(c_2 = 0\). Linearly independent.

(d) \(e^{3x}, (x+1)e^{2x}\): Different exponential growth rates and forms. Linearly independent.

Conclusion: \[ \boxed{f(x) = \sin x(4\sin^2 x - 3), g(x) = \sin 3x} \]

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