Chapter 6 Functions Ex 6.1 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Ex 6.1

(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.

(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.

(ii) x + y²= 9
∴ y²= 9 – x
∴ y = ±
∴ For one value of x, there are two values of y.
∴ y is not a function of x.

(iii) x²– y = 25
∴ y = x²– 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.

(ii) f (-3) = (-3)²– 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) f() =
=
=
=

(iv) f(x + 1) = (x + 1)²– 3(x + 1) + 1
= x²+ 2x + 1 – 3x – 3 + 1
= x²– x – 1

(v) f(-x) = (-x)2 – 3(-x) + 1 = x2 + 3x + 1

(vi)
=
=
= h + 1

(ii) g(x) =
g(x) = 0
= 0
∴ 18 – 2x²= 0
∴ x²= 9
∴ x = ±3

(iii) g(x) = 6x²+ x – 2
g(x) = 0
∴ 6x²+ x – 2 = 0
∴ (2x – 1) (3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = or x =

(iv) g(x) = x³– 2x²– 5x + 6
= ( x- 1) (x²– 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

(ii) f(x) = √x – 3, g(x) = 5 – x
f(x) = g(x)
∴ √x – 3 = 5 – x
∴ √x = 5 – x + 3
∴ √x = 8 – x
on squaring, we get
x = 64 + x²– 16x
∴ x²– 17x + 64 = 0
∴ x =
∴ x =
∴ x =

(ii) g(x) =
Solution:
g(x) =
Function g is defined everywhere except at x = 2.
∴ Domain of g = R – {2}
Let y = g(x) =
∴ (x – 2) y = x + 4
∴ x(y – 1) = 4 + 2y
∴ For every y, we can find x, except for y = 1.
∴ y = 1 ∉ range of function g
∴ Range of g = R – {1}

(iii) h(x) =
Solution:
h(x) = , x ≠ -5
For x = -5, function h is not defined.
∴ x + 5 > 0 for function h to be well defined.
∴ x > -5
∴ Domain of h = (-5, ∞)
Let y =
∴ y > 0
Range of h = (0, ∞) or R+

(iv) f(x) =
Solution:
f(x) =
f is defined for all real x and the values of f(x) ∈ R
∴ Domain of f = R, Range of f = R

(v) f(x) =
Solution:
f(x) =
For f to be defined,
(x – 2)(5 – x) ≥ 0
∴ (x – 2)(x – 5) ≤ 0
∴ 2 ≤ x ≤ 5 ……[∵ The solution of (x – a) (x – b) ≤ 0 is a ≤ x ≤ b, for a < b]
∴ Domain of f = [2, 5]
(x – 2) (5 – x) = -x²+ 7x – 10
=
=
∴
Range of f = [0, ]

(vi) f(x) =
Solution:
f(x) =
For f to be defined,
≥ 0, 7 – x ≠ 0
∴ ≤ 0 and x ≠ 7
∴ 3 ≤ x < 7
Let a < b, ≤ 0 ⇒ a ≤ x < b
∴ Domain of f = [3, 7)
f(x) ≥ 0 … [∵ The value of square root function is non-negative]
∴ Range of f = [0, ∞)

(vii) f(x) =
Solution:
f(x) =
For f to be defined,
16 – x²≥ 0
∴ x²≤ 16
∴ -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]

(ii) Diameter (d) = 2r
∴ r =
∴ Area (A) = πr²=

(iii) Circumference (C) = 2πr
∴ r =
Area (A) = πr²=
∴ A =

(ii) f : Z → Z given by f(x) = x²
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 x²≥ 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) = x²
Solution:
f : R → R given by f(x) = x²

∴ f is not injective.
f(x) = x²≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iv) f : N → N given by f(x) = x³
Solution:
f: N → N given by f(x) = x³

∴ 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) = x³
Solution:
f: R → R given by f(x) = x³

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

(ii) f(x) = log₁₀(x²– 5x + 6)
x²– 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, ∞)

(b)
Solution:

(c)
Solution:

(d)
Solution:

(ii) log(x – 1) + log(x)
Solution:

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

(b)
Solution:

(c)
Solution: