Let $${\text{f}}\left( {{\text{x, y}}} \right) = \frac{{{\text{a}}{{\text{x}}^2} + {\text{b}}{{\text{y}}^2}}}{{{\text{xy}}}},$$ where a and b are constants. If $$\frac{{\partial {\text{f}}}}{{\partial {\text{x}}}} = \frac{{\partial {\text{f}}}}{{\partial {\text{y}}}}$$ at x = 1 and y = 2, then the relation between a and b is
A. $${\text{a}} = \frac{{\text{b}}}{4}$$
B. $${\text{a}} = \frac{{\text{b}}}{2}$$
C. a = 2b
D. a = 4b
Answer: Option D
Related Questions on Calculus
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]

Join The Discussion