Question

The value of a(≠0) for which the equation \(\frac{1}{2}(x-2)^2+1=\sin(\frac{a}{x})\) holds is/are

• A. \((4n+1)\pi,n\isin \Z\)
• B. \(2(n-1)\pi,n\isin \Z\)
• C. \(n\pi,n\isin \N\)
• D. \(\frac{n\pi}{2},\isin \N\)
• E. 1

Answer 1

The Correct Option is A

Given:

• Equation: \( \frac{1}{2}(x - 2)^2 + 1 = \sin\left(\frac{a}{x}\right) \)
• \( a \neq 0 \)

Step 1: Minimum LHS value: \[ \frac{1}{2}(x - 2)^2 + 1 \geq 1 \]

Step 2: RHS range: \[ \sin\left(\frac{a}{x}\right) \in [-1, 1] \]

Step 3: Find critical point: \[ x = 2 \implies \sin\left(\frac{a}{2}\right) = 1 \]

Solution: \( a = (4n + 1)\pi \)

Answer 2

The given equation is:

\[ \frac{1}{2}(x - 2)^2 + 1 = \sin\left(\frac{a}{x}\right) \]

Let's analyze the left-hand side (LHS):

\[ f(x) = \frac{1}{2}(x - 2)^2 + 1 \]

This is a parabola with a vertex at \( (2, 1) \). The minimum value of \( f(x) \) is 1, and it increases as \( x \) moves away from 2.

Now let's analyze the right-hand side (RHS):

\[ g(x) = \sin\left(\frac{a}{x}\right) \]

The range of the sine function is \([-1, 1]\).

For the equation to hold, we must have:

\[ 1 \leq \frac{1}{2}(x - 2)^2 + 1 \leq 1 \]

This implies that \( \frac{1}{2}(x - 2)^2 = 0 \), which means \( x = 2 \).

Substituting \( x = 2 \) into the original equation:

\[ \frac{1}{2}(2 - 2)^2 + 1 = \sin\left(\frac{a}{2}\right) \] \[ 1 = \sin\left(\frac{a}{2}\right) \]

This means:

\[ \frac{a}{2} = \frac{\pi}{2} + 2n\pi \quad \text{or} \quad \frac{a}{2} = \frac{3\pi}{2} + 2n\pi, \quad \text{where } n \text{ is an integer.} \]

Therefore:

\[ a = \pi + 4n\pi = (4n + 1)\pi \quad \text{or} \quad a = 3\pi + 4n\pi = (4n + 3)\pi \]

If we let \( m = 4n+1 \) or \( m = 4n+3 \), then the general solution is \( a = m\pi \), where \( m \) is an odd integer. 

This matches the given option \( (4n+1)\pi \), which represents a subset of the general solution (odd multiples of \( \pi \)).