Relations And Functions

1. Reflexivity:
A relation on is reflexive if for all , i.e., .
Consider . Then: Hence, . Therefore, is not reflexive.
2. Symmetry:
A relation is symmetric if .
Take . Then: But: Hence, is not symmetric.
3. Transitivity:
A relation is transitive if and .
Let . Then: But: This example does not disprove transitivity. Let's try another one.
Let . Then: But: Hence, is not transitive.
Conclusion:
The relation is neither reflexive, nor symmetric, nor transitive.

Given: defined by , where is the greatest integer less than or equal to .
To prove: is neither one-one nor onto.

1. Not One-One (Injective):
A function is one-one if different inputs give different outputs.
Take two real numbers: and
But and
Therefore, is not one-one.
2. Not Onto (Surjective):
A function is onto if every real number has a pre-image in the domain.
The function always gives an integer as output.
For example, there is no real number such that or , since the range of the function is only integers.
Therefore, all real numbers are not covered in the co-domain .
Hence, is not onto.
Conclusion: The greatest integer function is neither one-one nor onto.

Given Information:
1. Turbulence can be severe, moderate, or light, each occurring with equal probabilities:

2. The probability of an airplane reaching late due to: - Severe turbulence: ,
- Moderate turbulence: , - Light turbulence: .

(i) Find the probability that an airplane reached its destination late. Using the law of total probability:

Substitute the values:

Simplify:

Thus:

(ii) If the airplane reached its destination late, find the probability that it was due to moderate turbulence.

Using Bayes' theorem:

Substitute the values:

Simplify:

Thus:

Final Answers: 1. The probability that an airplane reached its destination late is:

2. The probability that the airplane was late due to moderate turbulence is:

A camera is installed on a pole at the height of . It detects a car traveling away from the pole at the speed of . At any point, away from the base of the pole, the angle of elevation of the camera to the car is .

(i) Express in terms of the height of the camera and . From the given setup, we can use the right triangle formed by the height of the pole and the distance of the car from the base.

The tangent of the angle is given by:
Thus:

(ii) Find .
Differentiating with respect to , we use the chain rule:

Simplify :

Substitute:

Simplify further:

(iii) (a) Find the rate of change of angle of elevation with respect to time when the car is away from the pole.
Let and the car's speed be .

The rate of change of the angle of elevation with respect to time is:
From part (ii), .

Substitute :

Now:

(iii) (b) If the rate of change of angle of elevation with respect to time for another car at from the base is , find the speed of the car.

Let the speed of the car be .

From part (ii):

Substitute and :

Solve for :

Final Answers:
1. ,
2. ,
3. (a) ,
(b) Speed of the car is .

Step 1: Reflexivity
For : Thus, , and is reflexive.

Step 2: Symmetry
Let , so: Now, check if : For example: Thus, is not symmetric.

Step 3: Transitivity
Let and , so: Check if : For example: This is irrational, but a counterexample exists for other values. Thus, is not transitive.

Final conclusion
The relation is reflexive but neither symmetric nor transitive.

Step 1: Prove that is one-one
Let , where : Cross-multiply: Simplify: If , this leads to a contradiction. Hence, , proving is one-one.

Step 2: Prove that is onto
Let . Solve : For (since ), exists in . Thus, is onto.

Conclude the function properties
Since is both one-one and onto, is a bijection.

Given Information: The sine function is defined from to , but it is neither one-one nor onto over .
By restricting the domain to an interval, we can define the inverse , which is a one-one and onto function.

(i) If is an interval other than the principal value branch, give an example of one such interval. The principal value branch of the sine function is the interval . Another interval where the sine function is one-one and onto is:

(ii) If is defined from to its principal value branch, find the value of:

Step 1: Evaluate . From the definition of , is the angle in the interval such that: Thus:

Step 2: Evaluate . From the definition of , is the angle in the interval such that: Thus:

Step 3: Compute the difference. Simplify:

(iii) (a) Draw the graph of from to its principal value branch. Step 1: Description of the graph. The graph of is obtained by reflecting the graph of (restricted to ) across the line .

(iii) (b) Find the domain and range of . Step 1: Domain of . For , the argument of must lie within , i.e.:
Simplify:
Multiplying through by (and reversing inequalities): Thus, the domain is:

Step 2: Range of . The range of is . Therefore: Simplify:


Final Answers:
1. Example of an interval other than the principal value branch: .
2. .
3. (a) The graph of is shown above.
(b) The domain of is , and the range is .

Step 1: Simplify .
Using the identity , rewrite the integral:

Step 2: Substitution. Let . Then: The limits of integration change as follows: - When , ,

When , . The integral becomes:

Step 3: Integration by parts. To evaluate , use integration by parts: Then:

Using the formula for integration by parts :

Step 4: Simplify the remaining integral. Simplify : Thus:

Evaluate each term: Substitute back:

Step 5: Final expression.
Substitute back into the integral: Simplify: Step 6: Apply limits of integration.

Evaluate from to : At : At : Thus:

Final Answer:

Step 1: Simplify the denominator and numerator.

1. For the denominator , use the identity :

2. For the numerator , use the identity: Thus, the integral becomes:

Step 2: Substitution for simplification. Let .
Then: The limits of integration change as follows:
When , ,
When , .

Using the substitution , , and , the integral becomes:

Step 3: Simplify the trigonometric terms. Using the identity :

Substitute this back:

Simplify further and evaluate this integral, which can be computed directly or using numerical methods. Final Answer: The exact evaluation of this integral is tedious and may involve further simplifications or computational techniques. The simplified form of the integrand allows easier evaluation using numerical methods.

The integrand involves absolute values. To solve, we analyze the behavior of , , and over the given interval .

Step 1: Break the interval at the critical points , , and . The intervals are:

Step 2: Evaluate the expressions for each interval. - For :

Thus, the integrand becomes: - For : Thus, the integrand becomes:

Step 3: Compute the integral over each sub-interval. 1. For :

Evaluate: 2. For :

Evaluate: Step 4: Add the results.

Final Answer:

The given equation is:

Differentiate both sides with respect to : Using the chain rule:

Rearrange the terms:

Factorize on the left-hand side:

Solve for :

For the given condition , this simplifies to: Hence, it is proved that:

From the given conditions, we have two equations: Solving equations (1) and (2) simultaneously:

From equation (1): .
Substitute this into equation (2):

Substituting into equation (1):
Thus, the function is:
One-one (Injective): A function is one-one if distinct inputs lead to distinct outputs. Since is a linear function with a non-zero slope, it is one-one.
Onto (Surjective): A function is onto if for every element , there exists such that . For , for any , we can solve for , which gives . Hence, the function is onto.

Answer: The function is both one-one and onto. \bigskip

The relation contains pairs of students' roll numbers such that the second roll number is three times the first.
Thus, if is the roll number of a student, then is the roll number of another student. For example, if the roll numbers of the students are , the elements of will be:
Now, let's check if the relation is reflexive, symmetric, and transitive:
1. Reflexive: A relation is reflexive if every element is related to itself. For reflexivity, we would need for all .
Since , it is impossible for , so is not reflexive.
2. Symmetric: A relation is symmetric if whenever , then . Since , there is no corresponding pair where , so is not symmetric.
3. Transitive: A relation is transitive if whenever and , then . Since and , we see that is also in .
Hence, the relation is transitive.

- Assertion (A): The assertion claims that the domain of is .
However, this is incorrect. While is defined for all , the function is only defined for .
Therefore, the domain of is not , but rather .
So, Assertion (A) is incorrect.
- Reason (R): The reason is correct. The domain of the sum of two functions is the intersection of the domains of the individual functions.
The domain of is , and the domain of is . Thus, the domain of is the intersection of these two domains, which is . Thus, Assertion (A) is incorrect, but Reason (R) is correct.

To solve the problem, we are given a relation on :

i.e., the pair if and only if is irrational.

We will check whether is reflexive, symmetric, and transitive.

1. Reflexivity:
A relation is reflexive if for all .
Check: Since is irrational, for all .
Hence, is reflexive.

2. Symmetry:
A relation is symmetric if .
Suppose is irrational.
Check .
Now, the sum of an irrational number and its negative may or may not be irrational:
Example: Let (irrational)
Then (rational)
So , but .
Hence, is not symmetric.

3. Transitivity:
A relation is transitive if and .
Suppose: Add both: Now is irrational, but may or may not be irrational, depending on .
Example: Let
Then: - → irrational - → irrational - But → still irrational Try counterexample: Let
- → rational
So, we can't guarantee transitivity.
More accurately, find a direct counterexample where both pairs are in , but the third is not.
That might be hard to construct exactly, but the structure does not preserve irrationality additively in general.
Thus, no general guarantee of transitivity.
Hence, is not transitive.

Final Answer:
- Reflexive: Yes
- Symmetric: No
- Transitive: No