Find the locus of points satisfying the equation \( axy + byz = cy \).
Show Hint
- \( zx \)-plane or planes perpendicular to \( zx \)-plane
- Planes perpendicular to x-axis
- Lines perpendicular to \( zx \)-plane
- Lines perpendicular to \( xy \)-plane
The Correct Option is A
Solution and Explanation
\[ axy + byz = cy \] Factoring \( y \): \[ y(ax + bz - c) = 0 \] Step 2: Identifying Locus
Either: \[ y = 0 \quad \Rightarrow zx\text{-plane} \] or \[ ax + bz = c \quad \Rightarrow \text{Planes perpendicular to } zx \text{-plane} \] Thus, the correct answer is \( zx \)-plane or planes perpendicular to \( zx \)-plane.
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Questions Asked in AP EAMCET exam
Given the function:
\[ f(x) = \begin{cases} \frac{(2x^2 - ax +1) - (ax^2 + 3bx + 2)}{x+1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \]
If \( a, b, k \in \mathbb{R} \) and \( f(x) \) is continuous for all \( x \), then the value of \( k \) is:
Given the function:
\[ f(x) = \begin{cases} \frac{2x e^{1/2x} - 3x e^{-1/2x}}{e^{1/2x} + 4e^{-1/2x}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \]
Determine the differentiability of \( f(x) \) at \( x = 0 \).
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