**Chapter 6 Functions Ex 6.1**

## Chapter 6 Functions Ex 6.1

**Question 1.Check if the following relations are functions.(a)**

Solution:

Solution:

Yes.

Reason: Every element of set A has been assigned a unique element in set B.

**(b)Solution:**

No.

Reason: An element of set A has been assigned more than one element from set B.

**(c)Solution:**

No.

Reason:

Not every element of set A has been assigned an image from set B.

**Question 2.Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}(iii) {(1, 3), (4, 1), (2, 2)}(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}Solution:**

(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.

Reason: (2, -1), (2, 2), show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.

Reason: Every element of set A has been assigned a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.

Reason:

3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function

Reason: Every element of set A has been assigned a unique image in set B.

**Question 3.Check if the relation given by the equation represents y as function of x.**

(iii) x^{2} – y = 25

∴ y = x^{2} – 25

∴ For every value of x, there is a unique value of y.

∴ y is a function of x.

(iv) 2y + 10 = 0

∴ y = -5

∴ For every value of x, there is a unique value of y.

∴ y is a function of x.

(v) 3x – 6 = 21

∴ x = 9

∴ x = 9 represents a point on the X-axis.

There is no y involved in the equation.

So the given equation does not represent a function.

**Question 4.**

**Question 5.**

(iv) g(x) = x^{3} – 2x^{2} – 5x + 6

= ( x- 1) (x^{2} – x – 6)

= (x – 1) (x + 2) (x – 3)

g(x) = 0

∴ (x – 1) (x + 2) (x – 3) = 0

∴ x – 1 = 0 or x + 2 = 0 or x – 3 = 0

∴ x = 1, -2, 3

**Question 6.**

**Question 7.**

**Question 8.Find the domain and range of the following functions.(i) f(x) = 7x**

^{2}+ 4x – 1Solution:

f(x) = 7x

^{2}+ 4x – 1

f is defined for all x.

∴ Domain of f = R (i.e., the set of real numbers)

f(x) ≥ 0 … [∵ The value of square root function is non-negative]

∴ Range of f = [0, ∞)

∴ -4 ≤ x ≤ 4

∴ Domain of f = [-4, 4]

Clearly, f(x) ≥ 0 and the value of f(x) would be maximum when the quantity subtracted from 16 is minimum i.e. x = 0

∴ Maximum value of f(x) = √16 = 4

∴ Range of f = [0, 4]

**Question 9.Express the area A of a square as a function of its(a) side s(b) perimeter PSolution:**

**Question 10.Express the area A of a circle as a function of its(i) radius r(ii) diameter d(iii) circumference CSolution:**

(i) Area (A) = πr

^{2}

**Question 11.An open box is made from a square of cardboard of 30 cms side, by cutting squares of length x centimeters from each corner and folding the sides up. Express the volume of the box as a function of x. Also, find its domain.Solution:**

Length of the box = 30 – 2x

Breadth of the box = 30 – 2x

Height of the box = x

Volume = (30 – 2x)

^{2}x, x < 15, x ≠ 15, x > 0

= 4x(15 – x)

^{2}, x ≠ 15, x > 0

Domain (0, 15)

**Question 12.Let f be a subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify?Solution:**

f = {(ab, a + b): a, b ∈ Z}

Let a = 1, b = 1. Then, ab = 1, a + b = 2

∴ (1, 2) ∈ f

Let a = -1, b = -1. Then, ab = 1, a + b = -2

∴ (1, -2) ∈ f

Since (1, 2) ∈ f and (1, -2) ∈ f,

f is not a function as element 1 does not have a unique image.

**Question 13.Check the injectivity and surjectivity of the following functions.(i) f : N → N given by f(x) = x**
f: N → N given by f(x) = x

^{2}Solution:

^{2}

∴ f is injective.

For every y = x

^{2}∈ N, there does not exist x ∈ N.

Example: 7 ∈ N (codomain) for which there is no x in domain N such that x

^{2}= 7

∴ f is not surjective.

**(ii) f : Z → Z given by f(x) = x ^{2}Solution:**

f: Z → Z given by f(x) = x2

∴ f is not injective.

(Example: f(-2) = 4 = f(2). So, -2, 2 have the same image. So, f is not injective.)

Since x

^{2}≥ 0,

f(x) ≥ 0

Therefore all negative integers of codomain are not images under f.

∴ f is not surjective.

**(iii) f : R → R given by f(x) = x ^{2}Solution:**

f : R → R given by f(x) = x

^{2}

∴ f is not injective.

f(x) = x

^{2}≥ 0

Therefore all negative integers of codomain are not images under f.

∴ f is not surjective.

**(iv) f : N → N given by f(x) = x ^{3}Solution:**

f: N → N given by f(x) = x

^{3}

∴ f is injective.

Numbers from codomain which are not cubes of natural numbers are not images under f.

∴ f is not surjective.

**(v) f : R → R given by f(x) = x ^{3}Solution:**
f: R → R given by f(x) = x

^{3}

∴ For every y ∈ R, there is some x ∈ R.

∴ f is surjective.

**Question 14.Show that if f : A → B and g : B → C are one-one, then gof is also one-one.Solution:**

**Question 15.Show that if f : A → B and g : B → C are onto, then gof is also onto.Solution:**

Since g is surjective (onto),

there exists y ∈ B for every z ∈ C such that

g(y) = z …….(i)

Since f is surjective,

there exists x ∈ A for every y ∈ B such that

f(x) = y …….(ii)

(gof) x = g(f(x))

= g(y) ……[From (ii)]

= z …..[From(i)]

i.e., for every z ∈ C, there is x in A such that (gof) x = z

∴ gof is surjective (onto).

**Question 16.**

**Question 17.Express the following exponential equations in logarithmic form:Solution:**

**Question 18.Express the following logarithmic equations in exponential form:**

**Question 19.Find the domain of(i) f(x) = ln (x – 5)(ii) f(x) = log10 (x**

^{2}– 5x + 6)Solution:

(i) f(x) = ln (x – 5)

f is defined, when x – 5 > 0

∴ x > 5

∴ Domain of f = (5, ∞)

(ii) f(x) = log_{10} (x^{2} – 5x + 6)

x^{2} – 5x + 6 = (x – 2) (x – 3)

f is defined, when (x – 2) (x – 3) > 0

∴ x < 2 or x > 3

Solution of (x – a) (x – b) > 0 is x < a or x > b where a < b

∴ Domain of f = (-∞, 2) ∪ (3, ∞)

**Question 20.Write the following expressions as sum or difference of logarithms:Solution:**

Solution:

Solution:

Solution:

**Question 21.Write the following expressions as a single logarithm.(i) 5 log x + 7 log y – log zSolution:**

Solution:

**(iii) ln (x + 2) + ln (x – 2) – 3 ln (x + 5)Solution:**

**Question 22.Given that log 2 = a and log 3 = b, write log √96 terms of a and b.Solution:**

**Question 23.Prove that:**

Solution:

Solution:

**Question 24.**

**Question 25.Solve for x:(i) log 2 + log (x + 3) – log (3x – 5) = log 3Solution:**

Check:

If x = 3 satisfies the given condition, then our answer is correct.

L.H.S. = log 2 + log (x + 3) – log (3x – 5)

= log 2 + log (3 + 3) – log (9 – 5)

= log 2 + log 6 – log 4

= log (2 × 6) – log 4

= log 12/4

= log 3

= R.H.S.

Thus, our answer is correct.

Solution:

∴ x^{2} = 10x + 11

∴ x^{2} – 10x – 11 = 0

∴ (x – 11)(x + 1) = 0

∴ x = 11 or x = -1

But log of a negative numbers does not exist

∴ x ≠ -1

∴ x = 11

Solution:

Solution:

∴ a + a^{2} = 6

∴ a^{2} + a – 6 = 0

∴ (a + 3)(a – 2) = 0

∴ a + 3 = 0 or a – 2 = 0

∴ a = -3 or a = 2

Since 2x = -3 is not possible,

2x = 2 = 21

∴ x = 1

**Question 26.**

**Solution:**

**Question 27.Solution:**

**Question 28.Solution:**