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Let
  • a)
    f(x, y) is continuous at origin
  • b)
    f(x, y) is differentiable at origin
  • c)
    fx(0,0) ≠ 0
  • d)
     fy(0 ,0) ≠ 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Leta)f(x, y) is continuous at originb)f(x, y) is differentiable at ori...
We are given that 
Let us take ε > 0 and x2 + y2 ≠ 0.
Now consider,

We know that, 
or equivalently

Thus,
| f(x ,y )-0 | ≤ | y | = 0- |x| + 1 • |y|
take 
Therefore, for given ε > 0, there exists δ > 0 such that
|f(x, y) - 0 | < ε. whenever 0<|x|<δ,
0< |y |< δ
Hence, 
Thus, f(x,y) is continuous at (0, 0).
Hence, option (a) is correct.
Now,
 f(0 + h ,0 + k) - f(0,0) = f(h , k) - f(*0,0)
= 0 • h + 0 • k + • 
So that A = 0, B = 0 which does not depends on h and k and

Now approaching along h = mk, we get 

which depends on m. Therefore does not exists. Hence, f(x, y) is not differentiable at origin. Hence, option (b) is incorrect.
Next,
Hence, option (c) is incorrect.
And,

Hence, option (d) is incorrect.
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Most Upvoted Answer
Leta)f(x, y) is continuous at originb)f(x, y) is differentiable at ori...
We are given that 
Let us take ε > 0 and x2 + y2 ≠ 0.
Now consider,

We know that, 
or equivalently

Thus,
| f(x ,y )-0 | ≤ | y | = 0- |x| + 1 • |y|
take 
Therefore, for given ε > 0, there exists δ > 0 such that
|f(x, y) - 0 | < ε. whenever 0<|x|<δ,
0< |y |< δ
Hence, 
Thus, f(x,y) is continuous at (0, 0).
Hence, option (a) is correct.
Now,
 f(0 + h ,0 + k) - f(0,0) = f(h , k) - f(*0,0)
= 0 • h + 0 • k + • 
So that A = 0, B = 0 which does not depends on h and k and

Now approaching along h = mk, we get 

which depends on m. Therefore does not exists. Hence, f(x, y) is not differentiable at origin. Hence, option (b) is incorrect.
Next,
Hence, option (c) is incorrect.
And,

Hence, option (d) is incorrect.
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Leta)f(x, y) is continuous at originb)f(x, y) is differentiable at originc)fx(0,0) ≠ 0d)fy(0 ,0) ≠ 0Correct answer is option 'A'. Can you explain this answer?
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