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Let

then
  • a)
    f(x, y) is continuous at origin
  • b)
    f(x, y) has removable discontinuity of (0,0)
  • c)
    f(x, y) is not differentiable at (0, 0)
  • d)
    None of these 
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Letthena)f(x, y) is continuous at originb)f(x, y) has removable discon...
We are given that

Let us approaches (0, 0) along the liney = mx, we get

which depends on m. Thus  does not exists and hence, f(x, y) is discontinuous at origin. Thus, option (a), (b) are incorrect.
Now,
So that, A = 0, B = 0 which does not depends on h and k and

Now, approaching (0, 0) along the line h = mk, we get

which depends on m and hence does not exists. Therefore , f(x, y) is not differentiable at origin. Hence, option (c) is correct.
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Most Upvoted Answer
Letthena)f(x, y) is continuous at originb)f(x, y) has removable discon...
ANSWER :- c
Solution :- Consider an approach along the line y=x; then f(x,y)=x^2/(x^2+x^2) = 1/2
for all x≠0. On the other hand, if we approach (0,0) along the line y=2x
f(x,y)=(2x)^2/[x^2+4x^2]
=2/5
So there are two different paths toward the origin, each giving a different limit. Hence, the limit does not exist.
Or more simply, approach along y=0
Therefore , f(x, y) is not differentiable at origin
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Letthena)f(x, y) is continuous at originb)f(x, y) has removable discontinuity of (0,0)c)f(x, y) is not differentiable at (0, 0)d)None of theseCorrect answer is option 'D'. Can you explain this answer?
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