Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(b)

Question 1.

Let X={x, y} and Y={u, v}. Write down all the functions that can be defined from X to Y. How many of these are (i) one-one (ii) onto and (ii) one-one and onto?

Solution:

The functions from X = {x, y} to y = {u, v} are:

f1 = {(x, u), (y, v)}

f2 = {(x, v), (y, u)}

f3 = {(x, u), (y, u)}

f4 = {(x, v), (y, v)}

Out of these 4 functions there are:

(i) 2 one-one functions

(ii) 2 onto functions

(iii) 2 one-one and onto function.

Question 2.

Let X and Y be sets containing m and n elements respectively.

(i) What is the total number of functions from X to Y.

(ii) How many functions from X to Y are one-one according as men, m > n and m = n?

Solution:

If | x | = m and | y | = n then

(i) Number of functions =n^{m}

(ii) If m < n then number of one-one functions = ^{n}P_{m}.

If m > n then number of one-one functions = 0

If m = n then number of one-one functions = m!

Question 3.

Examine each of the following functions if it is

(i) injective (ii) surjective, (iii) bijective and (iv) none of the three

(a) f : R → R, f(x) = x²

(b) f : R → [-1, 1], f(x) = sin x

(c) f : R_{+} → R + , f(x) = x + 1/x

where R_{+ }= {x ∈ R : x > 0}

(d) f : R → R, f(x) = x³ + 1

(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)

(f) f : R → R, f(x) = [x] = the greatest integer ≤ x.

(g) f : R → R, f(x) = | x |

(h) f : R → R, f(x) = sgn x

(i) f : R → R, f = id_{R} = the identity function on R.

Solution:

(a) f : R → R, f(x) = x²

for x_{1}, x_{2} ∈ R

Let f(x_{1}) = f(x_{2})

⇒ x_{1}² = x_{2}²

⇒ x_{1 } = ± x_{2}

∴ f is not one-one.

Hence f is not injective or bijective.

Rng f = [0, ∞) ≠ R

∴ f is not surjective.

(b) f : R → [-1, 1], f(x) = sin x

For x_{1 }, x_{2} ∈ R

let f(x_{1}) = f(x_{2}) ⇒ sin x_{1} = sin x_{2}_{ }

⇒ x_{1} = nπ + (- 1)^{n} x_{2}

⇒ x_{1 }= x_{2} (not always)

∴ f is not injective and also not bijective.

But f is onto, as ∀ y ∈ [-1, 1]

there is a x ∈ R such that f(x) = sin x.

i.e., f is surjective.

(c) f : R_{+} → R_{+ }, f(x) = x + \(\frac{1}{x}\)

f(2) = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)

f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)

f(2) = f(\(\frac{1}{2}\))

but 2 ≠ \(\frac{1}{2}\)

∴ f is not injective (one-one).

Again, for 1 ∈ R_{+} (domain)

⇒ there is no x ∈ R_{+}^{(Dom)}

such that x + \(\frac{1}{2}\) = 1

∴ f is not onto.

(d) f : R → R, f(x) = x³ + 1

for x_{1}, x_{2} ∈ R

Let f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} = x_{2}^{3}

⇒ x_{1} = x_{2}

∴ f is injective.

Let f(x) = y ⇒ y = x³ + 1

⇒ x³ = y – 1

⇒ x = (y – 1)^{1/3} which exists ∀ y ∈ R

∴ f is onto.

∴ f is bijective.

(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)

for x_{1}, x_{2} ∈ (-1, 1)

Let f(x_{1}) = f(x_{2}) ⇒ \(\frac{x_1}{1-x_1^2}\) = \(\frac{x_2}{1-x_2^2}\)

⇒x_{1} – x_{1 }x_{2}^{2} = x_{2} – x_{1}^{2} x_{2}

⇒ x_{1} – x_{2} + x_{1}^{2} x_{2} – x_{1 }x_{2}^{2} = 0

⇒ (x_{1} – x_{2}) (1 + x_{1} x_{2}) = 0

⇒ x_{1} = x_{2} (for x_{1} x_{2} ∈ (-1, 1) x_{1} x_{2} ≠ -1)

∴ f is injective.

Again let y = \(\frac{x}{1-x^2}\) ⇒ y – x²y = x

⇒ x²y + x – y = 0

⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

= \(\frac{-1 \pm \sqrt{1+4 y^2}}{2 y}\) ∉ (-1, 1) ∀ y ∈¸.

∴f is surjective.

∴f is bijective.

(f) f : R → R

f(x) = [x]

f(1.2) = f(1.5) =1

∴ f is not injective.

Rng f = Z ⊂ R

∴ f is not surjective.

∴ Hence it is not also bijective.

(g) f : R → R

f(x) = | x |

As f(-1) = f(1) = 1

∴ f is not injective.

Again Rng f = [0, ∞) ⊂ R

⇒ f is not surjective.

Thus f is not bijective.

(h) f : R → R

f(x) = Sgn (x) = \(\left\{\begin{array}{cc}

1, & x>0 \\

0, & x=0 \\

-1, & x<0

\end{array}\right.\)

As f(1) = f(2) = 1

We have f is not injective.

Again Rng f = {- 1, 0, 1} ≠ R

∴ R is not surjective.

⇒ R is not bijective.

(i) f : R → R

f = idx

∴ f(x) = x

for x_{1}, x_{2} ∈ R

Let f(x_{1}) = f(x_{2}) where x_{2} x_{2} ∈ R

⇒ x_{1} = x_{2}

∴ f is one-one.

Again Rng f = R (codomain)

∴ f is onto.

Thus f is a bijective function.

Question 4.

Show that the following functions are injective.

(i) f(x) = sin x on \(\left[0, \frac{\pi}{2}\right]\)

(ii) f(x) = cos x [0, π]

(iii) fix) = log_{a }x on (0, ∞), (a > 0 and a ≠ 1)

(iv) f(x) = a^{x} on R. (a > 0 and a ≠ 1)

Solution:

(i) f(x) = sin x . \(\left[0, \frac{\pi}{2}\right]\)

for α, β ∈ \(\left[0, \frac{\pi}{2}\right]\)

Let f(α) = f(β) ⇒ sin α = sin β

⇒ α = β, as α, β ∈ \(\left(0, \frac{\pi}{2}\right)\) and no other values of α is possible.

∴ f is one-one.

(ii) f(x) = cos x, [0, π]

for α, β ∈ [0, π]

Let f(α) = f(β) ⇒ cos α = cos β

⇒ α = β, since α, β ∈ [0, π] and cos x is +ve in 1st quadrant and -ve in 2nd quadrant.

∴ f is one-one.

(iii) f(x) = log_{a} x [1, ∞]

for α, β ∈ [1, ∞]

Let f(α) = f(β)

⇒ log_{a} α = log_{a} β ⇒ α = β

∴ f is one-one.

(iv) f(x) = a^{x}, (a > 0), x ∈ R

for x_{1}, x_{2} ∈ R

Let f(x_{1}) = f(x_{2}) ⇒ a^{x1} = a^{x2}

⇒ x_{1} = x_{2}

∴ f is one-one.

Question 5.

Show that functions f and g defined by f(x) = 2 log x and g(x) = log x^{2 }are not equal even though log x^{2} = 2 log x.

Solution:

f(x) = 2 log x

g(x) = log x^{2}

Dom f(x) = (0, ∞)

Dom g(x) = R – {0}

As Dom f(x) ≠ Dom g(x) we have f(x) ≠ g(x), though log x^{2} = 2 log x

Question 6.

Give an example of a function which is

(i) Surjective but not injective.

(ii) injective but not surjective.

(iii) neither injective nor surjective.

(iv) bijective

Solution:

(i) f(x) = sin x

from R → [-1, 1]

which is surjective but not injective.

(ii) f(x) = \(\frac{x}{2}\) from Z → R

is injective but not surjective.

(iii) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)

is neither surjective nor injective.

[Refer Q. No. 3(e)].

(iv) f(x) = x^{3} + 1, f : R → R

is bijective.

[Refer No. 3 (d)].

Question 7.

Prove that the following sets are equivalent:

{1, 2, 3, 4, 5, 6,…}

{2, 4, 6, 8, 10,…}

{1, 7, 5, 7, 9,…}

{1, 4, 9, 16, 25,…}

Solution:

Let A = {1, 2, 3, 4, 5, 6, ……..}

B = {2, 4, 6, 8, 10, …….}

C = {1, 3, 5, 7, 9, ……}

D = {1, 4, 9, 16, 25, ……}

Let f : A → B defined as f(x) = 2x

Clearly f is bijective.

There is a one-to-one correspondence between A and B.

⇒ A and B are equivalent.

Let g : A → C defined as g(x) = 2x – 1

Clearly f is bijective.

⇒ There is a one-to-one correspondence between A to C

∴ A and C are equivalent.

Let h : A → D defined as h(x) = x^{2}.

Clearly h is bijective.

⇒ There is a one-to-one correspondence between A to D.

⇒ A and D are equivalent.

Thus A, B, C and D are equivalent.

Question 8.

Let f = {(1, a), (2, b), (3, c), (4, d)} and g = {{a, x), (b, x), (c, y), (d, x)}. Determine gof and fog if possible. Test whether fog = gof.

Solution:

f = {(1, a), (2, b), (3, c), (4, d)}

g = {{a, x), (b, x), (c, y), (d, x)}

gof (1) = g(a) = x

gof (2) = g(b) = x

gof (3) = g(c) = y

gof (4) = g(a) = x

∴ gof = {(1, x), (2, x), (3, y), (4, x)} Here fog is not defined.

Question 9.

Let f = {(1, 3), (2, 4), (3, 7)} and g = {(3, 2), (4, 3), (7, 1)}. Determine gof and fog if possible. Test whether fog = gof.

Solution:

f = {(1, 3), (2, 4), (3, 7)}

g = {(3, 2), (4, 3), (7, 1)}

We have fog (3) = f (g(3)) = f(2) = 4

fog (4) = f (g(4)) = f(3) = 7

fog (7) = f (g(7)) = f(1) = 3

∴ fog = {(3, 4), (4, 7), (7, 3)}

Again gof (1) = g (f(1)) = g(3) = 2,

gof (2) = g (f(2)) = g(4) = 3,

gof (3) = g (f(3)) = g(7) = 1

∴ gof = {(1, 2), (2, 3), (3, 1)}

∴ gof ≠ fog

So the composition of functions is not necessarily commutative.

Question 10.

Let f(x) =√x and g(x) = 1 – x^{2}.

(i) Find natural domains of f and g.

(ii) Compute fog and gof and find their natural domains.

(iii) Find natural domain of h(x) = 1 – x.

(iv) Show that h = gof only on R_{0} = {x ∈ R : x ≥ 0} and not on R.

Solution:

Let f(x) = √x, g(x) = 1 – x^{2}

(i) ∴ Dom f = R_{+} U_{{0}}, Dom g = R

(ii) fog (x) = f (g(x))

= f (1 – x^{2}) = \(\sqrt{1-x^2}\)

∴ fog (x) exists when 1 – x^{2} ≥ 0

⇒ x^{2} ≤ 1 ⇒ -1 ≤ x ≤ 1 i.e., x ∈ [-1, 1]

∴ Dom fog = [-1, 1]

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

= g( √x ) = 1 – ( √x )^{2} = 1 – x

∴ Dom gof = R_{0} = (0, ∞)

(iii) Domain of h(x) = 1 – x is R.

(iv) We have proved in (ii) that gof (x) = 1 – x.

∴ h(x) = gof (x) ⇒ h = gof only when x ∈ R_{0} as dom f is R_{0} = [0, ∞]

Question 11.

Find the composition fog and gof and test whether fog = gof when f and g are functions on R given by the following:

(i) f(x) = x^{3 }+ 1, g(x) = x^{2 }– 2

(ii) f(x) = sin x, g(x) = x^{5}

(iii) f(x) = cos x, g(x) = sin x^{2}

(iv) f(x) = g(x) = (1 – x^{3})^{1/3}

Solution:

(i) f(x) = x^{3 }+ 1, g(x) = x^{2 }– 2

∴ fog (x) = f (g(x)) = f(x^{2 }– 2)

= (x^{2 }– 2)^{3} + 1

gof (x) = g (f(x)) = g(x^{3} + 1)

= (x^{3} + 1)^{2} – 1

fog ≠ gof

(ii) f(x) = sin x, g(x) = x^{5}

∴ fog (x) = f (g(x)) = f(x^{5}) = sin x^{5}

∴ gof (x) = g (f(x))

= g(sin x) = (sin x)^{5} = sin^{5} x

fog ≠ gof

(iii) f(x) = cos x, g(x) = sin x^{2}

∴ fog (x) = f (g(x))

= f( sin x^{2}) = cos (sin x^{2})

and gof (x) = g (f(x)) = g(cos x)

= sin (cos x)^{2} = sin (cos^{2} x)

fog ≠ gof

(iv) f(x) = g(x) = (1 – x^{3})^{1/3}

fog (x) = f (g(x))

= (1 – (g(x))^{3})^{1/3}

= [1 – (1 – x^{3})]^{1/3} = x

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

= [1 – (f(x))^{3}]^{1/3} = x

⇒ fog = gof

Question 12.

(a) Let f be a real function. Show that h(x) = f(x) + f(-x) is always an even function and g(x) = f(x) – f(-x) is always an odd function.

(b) Express each of the following function as the sum of an even function and an odd function:

(i) 1 + x + x^{2 }, (ii) x^{2}, (iii) e^{x}, (iv) e^{x} + sin x

Solution:

(a) We have h(x) = f(x) + f(-x)

∴ h(-x) = f(-x) + f(x) = h(x)

∴ h is always an even function.

Further, g(x) = f(x) – f(-x)

∴ g(-x) = f(-x) – f(x)

= – [f(x) – f(-x)] = – g(x).

∴ g is always an odd function.

(b) (i) Let f(x) = 1 + x + x^{2}

∴ f(-x) = 1 – x + x^{2}

∴ g(x) = \(\frac{f(x)+f(-x)}{2}\)

= \(\frac{1+x+x^2+1-x+x^2}{2}\)

= x^{2} + 1 and

g(-x) = (-x)^{2} + 1 = x^{2} + 1

∴ g is an even function.

h(x) = \(\frac{f(x)-f(-x)}{2}\)

= \(\frac{\left(1+x+x^2\right)-\left(1-x+x^2\right)}{2}\) = x

h(-x) = -x = -h(x)

⇒ h is an odd function.

∴ f(x) = g(x) + f(x)

where g is even and h is odd.

(ii) Let f(x) = x^{2}

So that f(-x) = (-x)^{2} = x^{2}

∴ g(x) = \( \frac{f(x)+f(-x)}{2}\) = \( \frac{x^2+x^2}{2}\) = x^{2}

g(-x) = g(x)

∴ g is an even function.

and h(x) = \( \frac{f(x)-f(-x)}{2}\) = \(\frac{x^2-x^2}{2}\) = 0

h(x) = 0 is both even and odd.

∴ f(x) = g(x) = f(x),

where g is even and h is odd.

(iii) Let f(x) = e^{x}

f(-x) = e^{-x}

g(x) = \( \frac{e^x+e^{-x}}{2}\)

g(-x) = g(x)

g is an even function.

and h(x) = \( \frac{e^x-e^{-x}}{2}\)

h(-x) = \( \frac{e^{-x}-e^x}{2}\) = -h^{2}(x)

⇒ h is an odd function.

∴ f(x) = g(x) + h(x),

where g is even and h is odd.

(iv) Let f(x) = e^{x} + sin x

f(-x) = e^{-x} + sin (-x) = e^{-x} – sin x

∴ g(x) = \( \frac{f(x)+f(-x)}{2}\)

= \( \frac{e^x+\sin x+e^{-x}-\sin x}{2}\)

= \( \frac{e^x+e^{-x}}{2}\) and h(x) = \( \frac{f(x)-f(-x)}{2}\)

= \( \frac{e^x+\sin x-e^{-x}+\sin x}{2}\)

= \( \frac{e^x-e^{-x}+2 \sin x}{2}\)

∴ f(x) = g(x) + h(x)

where g is even and g is odd.

Question 13.

Let X = {1, 2, 3, 4} Determine whether f : X → X defined as given below have inverses.

Find f^{-1} if it exists:

(i) f = {(1, 4), (2, 3), (3, 2), (4, 1)}

(ii) f = {(1, 3), (2, 1), (3, 1), (4, 2)}

(iii) f = {(1, 2), (2, 3), (3, 4), (4, 1)}

(iv) f = {(1, 1), (2, 2), (2, 3), (4, 4)}

(v) f = {(1, 2), (2, 2), (3, 2), (4, 2)}

Solution.

(i) x = {1, 2, 3, 4}

f is bijective. Hence f^{-1 }exists.

f^{-1} = {(4, 1), (3, 2), (2, 3), (1, 4)}

(ii) f(2) = f(3) = 1

⇒ f is not injective

∴ f is not invertible.

(iii) f is bijective. Hence f^{-1} exists.

f^{-1} = {(2,1 ), (3, 2), (4, 3), (1, 4)}

(iv) f is not a function as

f(2) = 2 and f(2) = 3

(v) f is not injective hence not invertible.

Question 14.

Let f : X → Y.

If there exists a map g : Y → X such that gof = id_{x} and fog = id_{y}, then show that

(i) f is bijective and (ii) g = f^{-1}

[Hint: Since id_{x} is a bijective function, gof = id_{x} is bijective. By Theorem 2(iv) f is injective. Similarly fog is bijective ⇒ f is surjective by Theorem 2(iii)]

Solution:

Let f : x → y and g : y – x

where gof = id_{x} and fog = id_{y}

we know that id_{x} and id_{y} are bijective functions.

⇒ gof and fog are both bijective functions.

⇒ f is a bijective function.

(ii) As f is bijective (by (i)) we have f^{-1} exists.

and f^{-1} : y → x where f^{-1}of = id_{x} and fof^{-1} = id_{y}

But g : y → x with gof = id_{x} and fog = id_{y}

∴ g = f^{-1}

Question 15.

Construct an example to show that f(A ∩ B) ≠ f(A) ∩ f(B) where A ∩ B ≠ Ø

Solution:

Let f(x) = cos x.

Let A = \(\left\{0, \frac{\pi}{2}\right\}\), B = \(\left\{\frac{\pi}{2}, 2 \pi\right\}\).

∴ f(A) = \(\left\{\cos 0, \cos \frac{\pi}{2}\right\}\)

= {1, 0} = {0, 1}

∴ f(B) = \(\left\{\cos \frac{\pi}{2}, \cos 2 \pi\right\}\) = {0, 1}

∴ f(A) ∩ f(B) = {0, 1}

Again,

A ∩ B = \(\left\{\frac{\pi}{2}\right\}\) and f(A ∩ B) = cos \(\frac{\pi}{2}\) = {0}

∴ f(A ∩ B) ≠ f(A) ∩ f(B)

Question 16.

Prove that for any f : X → Y, foid_{x} = f = id_{y}of.

Solution:

Let f : X → Y, so that y = f(x), x ∈ X.

∴ foid_{x} = fof^{-1} of (x) = fof^{-1} (f(x))

= f(x) = y (∵ id_{x} = fof^{-1})

Again, (id_{y}of)(x) = (fof^{-1}) of (x)

= (fof^{-1})(y) = f (f^{-1}(y)) = f(x) = y ….(2)

∴ From (1) and (2)

we have foid_{x} = f = id_{y}of

Question 17.

Prove that f : X → Y is surjective iff for all B ⊆ Y, f (f^{-1}(B)) = B.

Solution:

Let f : X → Y is surjective.

i.e. for all y ∈ Y, ∃ a x ∈ X such that

y = f(x).

∴ x = f^{-1}(y) ⇔ f(x)

= f (f^{-1}(y)) ∈ f (f^{-1}(B)).

for y = B ⊂ Y ⇔ y ∈ f (f^{-1}(B)).

∴ y ∈ f (f^{-1}(B)) ⇔ y ∈ B

∴ f (f^{-1}(B)) = B

Question 18.

Prove that f : X → Y is injective iff f (f^{-1}(A)) = A for all A ⊆ X.

Solution:

f : X → Y is injective.

Let x ∈ A ⇔ f(x) ∈ f(A) (∵ f is injective)

⇔ x ∈ f (f^{-1}(A))

∴ A = f (f^{-1}(A)) for all A ⊆ X.

Question 19.

Prove that f : X → Y is injective iff for all subsets A, B of X, f(A ∩ B) = f(A) ∩ f(B).

Solution:

f : X → Y is injective.

Let A and B are subsets of X.

Let f(x) ∈ f(A ∩ B)

⇔ x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B

⇔ f(x) ∈ f(A) ∧ f(x) ∈ f(B) (∵ f is injective)

⇔ f(x) = f(A) ∩ f(B)

∴ f(A ∩ B) = f(A) ∩ f(B)

Conversely, suppose that

f(A ∩ B) = f(A) ∩ f(B)

Let f is not injective.

The if f(x) ∈ f(A ∩ B) ⇔ x ∈ A ∩ B

⇔ x ∈ A ∧ ⇔ x ∈ B

≠ f(x) ∈ f(A) ∧ ⇔ f(x) ∈ f(B)

⇔ f(x) ∈ f(A) ∩ f(B)

∴ f(A ∩ B) = f(A) ∩ f(B) is false.

so f must be injective.

Question 20.

Prove that f : X → Y is surjective iff for all A ⊆ X, (f(A))‘ ⊆ f(A‘), where A‘ denotes the complement of A in X.

Solution:

f : X → Y is surjective.

Then for all y ∈ Y ∃ x ∈ X

such that f(x) = y.

Let y ∈ [f(A)]‘ ⇒ y ∉ f(A)

⇒ f(x) ∉ f(A) ⇒ x ∉ A ⇒ x ∈ A‘

⇒ f(x) ∈ f(A‘) ⇒ y ∈ f(A‘)

∴ [f(A)]‘ ⊂ f(A‘)