A1)

We can simply find the solution as follows,

A2)

The function is well defined at (0,0)

(2) check for the second step,

That means the limit exists at (0,0)

Now check step-3:

So that the function is continuous at origin.

A3)

2.

Here f1 = f2

3. now put y = mx, we get

Here f1 = f2 = f3

Now put y = mx²

4.

Therefore ,

                      F1 = f2 = f3 =f4

We can say that the limit exists with 0.

A4)

To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

                           = cos(y²x + 5x – 8)(y²x + 5x – 8)

                           = (y² + 50)cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

                           = cos(y²x + 5x – 8)(y²x + 5x – 8)

                           = 2xycos(y²x + 5x – 8)

A5)

Here we have,

u =     …………………..(1)

now partially differentiate eq.(1) w.r to x and y , we get

                                               =

Or

                 ………………..(2)

And now,

                                                =

         ………………….(3)

Adding eq. (1) and (3) , we get

=  0

Hence proved.

A6)

Partially differentiating given equation with respect to and x and y then equate them to zero                 

       On solving above we get 

         Also

            Thus we get the pair of values (0,0), (,0) and (0,

        Now, we calculate

        At the point (0,0)

       So function has saddle point at (0,0).

       At the point (

         So the function has maxima at this point (.

        At the point (0,

       So the function has minima at this point (0,.

        At the point (

        So the function has an saddle point at (

A7)

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e. &

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

A8)

Suppose   and

Now taking L.H.S,

Which is

                                                                                               Hence proved.

2.

So that

A9)

Here-                     

Now-

Apply

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

A10)

Here

Now,

And

We know that-

So that-

Now,

            Directional derivative =