If P is a 3 x 3 matrix such that PT = 2P + I, where pT is the transpose of P and I is 3 x 3 identity, then there exists a column matrix X = such that PX is equal to
X
2X
- X
If the tangent at (1, 1) on y = x(2 - x)2 meets the curve again at P, then P is
(4, 4)
(- 1, 2)
(9/4, 3/8)
None of these
C.
(9/4, 3/8)
We have,
y2 = x(2 - x)2 ...(i)
y2 = x3 - 4x2 + 4x
On differentiating both sides w.r.t. x we get
The equation of the tangent at (1, 1) is
y - 1 = - 1/2 (x - 1)
On solving Eq. (i) and (ii), we get x = 9/4 and y = 3/8
Hence, the coordinates of Pare (9/4, 3/8).
If Then, the value of the pair (a, b) for which f(x) cannot be continuous at x = 1, is
(2, 0)
(1, - 1)
(4, 2)
(1, 1)
The differential equation has the solution
x = y(log(x) + C)
y = x(log(y) + C)
x = (y + C)log(x)
y = (x + C)log(y)
The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 succeses is
28/256
128/256
37/256