Can it be proved that the vectors u = 3i + xj and v=(x+1)i + xj, can never be perpendicular, no matter what value is given to x.

We use vector theory here. From vector theory we know that
two vectors u and v, can be perpendicular only if the dot product of the two vectors is
zero.

So we have to first find the dot products of the
vectors given.

We have u = 3i + xj and v=(x+1)i + xj. The
dot product of the two vectors is u.v = (3i+xj) .
((x+1)i+xj)

=> 3(x+1) + x^2 = x^2+3x +
3

=>  (x + 3/2)^2 - (3/2)^2 +
3

=> (x+3/2)^2 +3/4

Now
(x+3/2)^2 +3/4 is always greater than or equal to 3/4 no matter what x
is.

Thus we prove that the dot product of u and v is
greater than or equal to 3/4.

Therefore,  as
u dot v can never be equal to zero, we prove that u and v can never be perpendicular, no
matter what the value of x is.