Question

If $y = (tan^{-1} x)^2, show\ that\ (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$

Answer 1

Mutually exclusive means that none of the 3 events can happen simultaneously. As a result, the probability of any of the events happening is just the sum of the individual probabilities.

Exhaustive means that they are the only events that can happen.

Using those two facts, we know that:

P(A or B or C) = P(A) + P(B) + P(C) = 1

Let s express each in terms of P(B). First fact, we know P(A) = 2P(B), so:

2P(B) + P(B) + P(C) = 1

Next we know that 3P(C) = 2P(B), so we could also say P(C) = (2/3)P(B), so:

2P(B) + P(B) + (2/3)P(B) = 1

To simplify things, let s just use x = P(B):

2x + x + (2/3)x = 1

Combine like terms:

3x + (2/3)x = 1

Multiply both sides by 3:

9x + 2x = 3

11x = 3

x = 3/11

Answer:

P(B) = 3/11

Double-check:

P(A) = 6/11

P(B) = 3/11

P(C) = 2/11

P(A) = 2P(B) = 2*3/11

P(A) = 3P(C) = 3*2/11

Answer:

P(B) = 3/11