## 1.

OR

### Answer.

OR

## 2. Write the sum of the order and the degree of the following differential equation:

### Answer.

### Write the sum of the order and the degree of the following differential

### Solution: Order = 2

Degree = 1

Sum = 3

## 3. If πΜ and πΜ are unit vectors, then prove that

|πΜ + πΜ| = 2πππ 0/2 , where π is the angle between them.

### Answer.

If πΜ and πΜ are unit vectors, then prove that

## 4. Find the direction cosines of the following line:

### Answer.

### Find the direction cosines of the following line:

## 5. A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-byone without replacement.

### Answer.

A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement. Solution: Let X be the random variable defined as the number of red balls.

Then X = 0, 1

## 6. Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?

### Answer.

Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack? Solution: The required probability = P((The first is a red jack card and The second is a jack card) or (The first is a red non-jack card and The second is a jack card))

## 7.

### Answer.

## 8. Find the general solution of the following differential equation:

OR

## Find the particular solution of the following differential equation, given that

### Answer.

πππ|πππ πππ£ β πππ‘π£| = βπππ|π₯| + ππππΎ,πΎ > 0 (Here, ππππΎ is an arbitrary

constant.)

β πππ|(πππ πππ£ β πππ‘π£)π₯| = ππππΎ

β |(πππ πππ£ β πππ‘π£)π₯| = πΎ

β (πππ πππ£ β πππ‘π£)π₯ = Β±πΎ

β (πππ ππ π¦/π₯β πππ‘ π¦/π₯) π₯ = πΆ, which is the required general solution.

OR

## 9. If πβ β 0,

βββ πβ. πββ = πβ. πβ, πβ Γ πββ = πβ Γ πβ, then show that πββ = πβ.

### Answer.

### If πβ β 0,

βββ πβ. πββ = πβ. πβ, πβ Γ πββ = πβ Γ πβ, then show that πββ = πβ.

Solution: We have πβ. (πββ β πβ) = 0

β (πββ β πβ) = β0β or πβ β₯ (πββ β πβ)

β πββ = πβ or πβ β₯ (πββ β πβ)

Also, πβ Γ (πββ β πβ) = β0β

β (πββ β πβ) = β0β or πβ β₯ (πββ β πβ)

β πββ = πβ or πβ β₯ (πββ β πβ)

πβ πππ πππ‘ ππ πππ‘β ππππππππππ’πππ π‘π (πββ β πβ) πππ ππππππππ π‘π (πββ β πβ)

Hence, πββ = πβ.

## 10. Find the shortest distance between the following lines:

πβ = (πΜ+ πΜβ πΜ) + π (2πΜ+ πΜ+ πΜ)

πβ = (πΜ+ πΜ+ 2πΜ) + π‘(4πΜ+ 2πΜ+ 2πΜ)

OR

## Find the vector and the cartesian equations of the plane containing the point

πΜ+ 2πΜβ πΜ and parallel to the lines πβ = (πΜ+ 2πΜ+ 2πΜ) + π (2πΜβ 3πΜ+ 2πΜ) and

πβ = (3πΜ+ πΜβ 2πΜ) + π‘(πΜβ 3πΜ+ πΜ)

### Answer.

Find the shortest distance between the following lines:

πβ = (πΜ+ πΜβ πΜ) + π (2πΜ+ πΜ+ πΜ)

πβ = (πΜ+ πΜ+ 2πΜ) + π‘(4πΜ+ 2πΜ+ 2πΜ)

OR

## 11. Evaluate: β« |π₯

3 β 3π₯

2 + 2π₯|ποΏ½

### Answer.

## 12. Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y 2 = x and the x-axis.

OR

## Using integration, find the area of the region

{(π₯, π¦): 0 β€ π¦ β€ β3π₯, π₯

2 + π¦

2 β€ 4}

### Answer.

### Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y 2 = x and the x-axis.

Solution: Solving x + y = 2 and y 2 = x simultaneously, we get the points of intersection as (1, 1) and (4, -2).

OR