Mathematics Test 260

The normal at (bt₁², 2bt₁) on the parabola y² = 4bx meets the parabola again at (bt₂², 2bt₂).

Given: Ax² + By² = 1

As the solution has two constants, the order of the differential equation is 2,

So, our choices (2) and (3) are discarded from the list, only choices (1) and (4) are possible.

The volume of a parallelepiped is given by the scalar triple product of the vectors representing its sides. For vectors A, B, and C, the volume V is given by:

V = |A · (B × C)|

Given vectors: A = 2i - 3j

B = i + j - k

C = 3i - k

Let's compute the cross product B × C first:

B x C = |i j k |
            |1 1 -1|
            |3 0 -1|

Expanding the determinant:

B × C = i[(1)(-1) - (0)(-1)] - j[(1)(-1) - (3)(-1)] + k[(1)(0) - (3)(1)]

= i[-1] - j[-1 + 3] + k[0 - 3]

= -i - 2j - 3k

Now, compute the dot product A · (B × C):

A · (B × C) = (2i - 3j) · (-i - 2j - 3k)

= (2)(-1) + (-3)(-2) + (0)

= -2 + 6 + 0

= 4

Therefore, the volume of the parallelepiped is:

V = |4| = 4

Thus, the correct answer is:

b) 4

Given that, T_p = a + (p − 1)d = q …….(i) and

T_q = a + (q − 1)d = p ……. (ii)

From (i) and (ii), we get d = [−(p − q)] / [(p − q)] = −1

Putting value of d in equation (i), then a = p + q − 1

Now, r^th term is given by A.P. T_r = a + (r − 1)d

= (p + q − 1) + (r − 1) (−1)

= p + q − r