Multiple Choice QuestionsA random variable X has the probability distribution:
| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
0.87
0.77
0.35
0.35
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
37/256
219/ 256
128/256
128/256
If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is
12
8
4
4
The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
x – 2y + 11 = 0
x + 2y + 11 = 0
x + 2y – 11 = 0
x + 2y – 11 = 0
B.
x + 2y + 11 = 0
Equation of the required plane is
(x + y + z – 6) + λ(2x + 3y + z + 5) = 0
i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0
This plane is perpendicular to xy plane whose
equation is z = 0
i.e. 0 . x + 0 . y + z = 0
∴ By condition of perpendicularity
0 .(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0
i.e. λ = –1
∴ Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) =0
or x + 2y + 11 = 0
The curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -
{–2, 2}
{1}
{-2}
{-2}
Switch
