Question

Let f : \(\R^2 → \R^2\) be defined by f(x, y) = (e^x cos(y), e^x sin(y)). Then the number of points in \(\R^2\) that do NOT lie in the range of f is

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

The Correct Option is B

To solve the problem, we need to analyze the mapping defined by \( f : \R^2 \to \R^2 \) where \( f(x, y) = (e^x \cos(y), e^x \sin(y)) \). This function transforms a point \((x, y)\) in Cartesian coordinates into another point in \(\R^2\).

Let's rewrite \( f(x, y) \) in polar coordinates for simplicity:

\[ u = e^x \cos(y), \quad v = e^x \sin(y) \] 

Notice that these can be rewritten as:

\[ u^2 + v^2 = (e^x \cos(y))^2 + (e^x \sin(y))^2 = e^{2x}(\cos^2(y) + \sin^2(y)) = e^{2x} \]

The term \(\cos^2(y) + \sin^2(y) = 1\) due to the Pythagorean identity.

Therefore, the range of values that \((u, v)\) can take is constrained by \(|(u, v)| = e^x\), which means that \( (u, v) \) must lie on a circle with radius \( e^x > 0 \). Since \( e^x > 0 \) for all real \( x \), this implies that the origin \((0, 0)\) cannot be reached because \( e^x \) can never be zero.

This limits the points that \((u, v)\) can take on \(\R^2\) to all points except the origin. Hence, there is exactly one point \((0, 0)\) that does not lie in the range of \( f \).

Thus, the number of points in \(\R^2\) that do not lie in the range of \( f \) is 1.