CBSE Class XI

Question 1:

Find the radian measures corresponding to the following degree measures:

(i) 25°         (ii) − 47° 30′                  (iii) 240°               (iv) 520°

Answer

(i) 25°

We know that 180° = π radian

(ii) −47° 30′

(iii) 240°

We know that 180° = π radian

(iv) 520°

We know that 180° = π radian

Question 12:

Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Answer

A = {9, 10, 11, 12, 13}

f: A → N is defined as

f(n) = The highest prime factor of n

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

∴ f(9) = The highest prime factor of 9 = 3

f(10) = The highest prime factor of 10 = 5

f(11) = The highest prime factor of 11 = 11

f(12) = The highest prime factor of 12 = 3

f(13) = The highest prime factor of 13 = 13

The range of f is the set of all f(n), where n ∈ A.

∴ Range of f = {3, 5, 11, 13}

Question 11 :

Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.

Answer

The relation f is defined as f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.

Since 2, 6, −2, −6 ∈ Z, (2 × 6, 2 + 6), (−2 × −6, −2 + (−6)) ∈ f

It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and -8. Thus, relation f is not a function.

Question 10:

Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

(i) f is a relation from A to B

(ii) f is a function from A to B.

Justify your answer in each case.

Answer

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It is observed that f is a subset of A × B.

Thus, f is a relation from A to B.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

Question 9:

Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b²}. Are the following true?

(i) (a, a) ∈ R, for all a ∈ N

(ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Justify your answer in each case.

Answer

R = {(a, b): a, b ∈ N and a = b²}

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4.

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3². Now, 3 ≠ 9² = 81; therefore, (3, 9) ∉ N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4²

Now, 9 ≠ 4² = 16; therefore, (9, 4) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question 8:

Let f = {(1, 1), (2, 3), (0, −1), (−1, −3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Answer

f = {(1, 1), (2, 3), (0, −1), (−1, −3)}

f(x) = ax + b

(1, 1) ∈ f

⇒f(1) = 1

⇒a × 1 + b = 1

⇒a + b = 1

(0, −1) ∈ f

⇒f(0) = −1

⇒a × 0 + b = −1

⇒b = −1

On substituting b =−1 in a + b = 1, we obtain a + (−1) = 1

⇒ a = 1 + 1 = 2. Thus, the respective values of a and b are 2 and −1.

Questions 7 :

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x − 3. Find f + g, f – g and f/g

Answer

f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3

(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

∴ (f + g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

∴ (f – g) (x) = –x + 4

Question 6:

Answer

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus, range of f = [0, 1]

Question 5 :

Find the domain and the range of the real function f defined by f (x) = |x −1|

Answer

The given real function is f (x) = |x – 1|.

It is clear that |x – 1| is defined for all real numbers.

∴ Domain of f = R

Also, for x ∈  R, |x – 1| assumes all real numbers.

Hence, the range of f is the set of all non-negative real numbers.

Question 4:

Answer

Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞).

As x ≥ 1 ⇒ (x – 1) ≥ 0

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0, ∞).

Question 3:

Answer

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2. Hence, the domain of f is R − {2, 6}.

Question 2:

Answer

Question 1:

Show that f is a function and g is not a function.

Answer

It is observed that for

0 ≤ x < 3, f(x) = x²

3 < x ≤ 10, f(x) = 3x

Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9

Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique. Thus, the given relation is a function.

It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6

Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

Question 5:

Find the range of each of the following functions.

(i) f(x) = 2 − 3x, x ∈ R, x > 0.

(ii) f(x) = x² + 2, x, is a real number.

(iii) f(x) = x, x is a real number

Answer

f(x) = 2 – 3x, x ∈ R, x > 0

The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

i.e., range of f = (−∞, 2)

Alter:

Let x > 0

⇒ 3x > 0

⇒2 – 3x < 2

⇒f(x) < 2

∴ Range of= (−∞, 2)

(ii) f(x) = x² + 2, x, is a real number

The values of f(x) for various values of real numbers x can be written in the tabular form as

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.

i.e., range of f =[2, ∞)

Alter :

Let x be any real number.

Accordingly,

x² ≥ 0

⇒ x² + 2 ≥ 0 + 2

⇒ x² + 2 ≥ 2

⇒f(x) ≥ 2

∴ Range of f = [2, ∞)

(iii) f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R

Question 4:

Find (i) t (0) (ii) t (28) (iii) t (−10) (iv) The value of C, when t(C) = 212

Answer

(iv) It is given that t(C) = 212

Thus, the value of t, when t(C) = 212, is 100.

Question 3:

A function f is defined by f(x) = 2x − 5. Write down the values of

(i) f(0),        (ii) f(7),       (iii) f(−3)

Answer

The given function is f(x) = 2x − 5.

Therefore,

(i) f(0) = 2 × 0 − 5 = 0 − 5 = −5

(ii) f(7) = 2 × 7 − 5 = 14 − 5 = 9

(iii) f(−3) = 2 × (−3) − 5 = − 6 − 5 = −11

Question 2:

Find the domain and range of the following real function:

(i) f(x) = −|x|

Answer

(i) f(x) = −|x|, x ∈ R

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of f(x) = −|x| is all real numbers except positive real numbers.

∴ The range of f is (−∞, 0].

Since  is defined for all real numbers that are greater than or equal to −3 and less than or equal to 3, the domain of f(x) is {x : −3 ≤ x ≤ 3} or [−3, 3].

For any value of x such that −3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

∴ The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

Question 1:

Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}

Answer

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

Question 9:

Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.

Answer

R = {(a, b): a, b ∈ Z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

∴ Domain of R = Z

Range of R = Z

Question 8:

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer

It is given that A = {x, y, z} and B = {1, 2}.

∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6, the number of subsets of A × B is 2⁶.

Therefore, the number of relations from A to B is 2⁶.