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 =nm
(ii) If m < n then number of one-one functions = nPm.
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 = idR = the identity function on R.
Solution:
(a) f : R → R, f(x) = x²
for x1, x2 ∈ R
Let f(x1) = f(x2)
⇒ x1² = x2²
⇒ x1 = ± x2
∴ 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 x1 , x2 ∈ R
let f(x1) = f(x2) ⇒ sin x1 = sin x2
⇒ x1 = nπ + (- 1)n x2
⇒ x1 = x2 (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 x1, x2 ∈ R
Let f(x1) = f(x2)
⇒ x13 = x23
⇒ x1 = x2
∴ 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 x1, x2 ∈ (-1, 1)
Let f(x1) = f(x2) ⇒ \(\frac{x_1}{1-x_1^2}\) = \(\frac{x_2}{1-x_2^2}\)
⇒x1 – x1 x22 = x2 – x12 x2
⇒ x1 – x2 + x12 x2 – x1 x22 = 0
⇒ (x1 – x2) (1 + x1 x2) = 0
⇒ x1 = x2 (for x1 x2 ∈ (-1, 1) x1 x2 ≠ -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 x1, x2 ∈ R
Let f(x1) = f(x2) where x2 x2 ∈ R
⇒ x1 = x2
∴ 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) = loga x on (0, ∞), (a > 0 and a ≠ 1)
(iv) f(x) = ax 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) = loga x [1, ∞]
for α, β ∈ [1, ∞]
Let f(α) = f(β)
⇒ loga α = loga β ⇒ α = β
∴ f is one-one.
(iv) f(x) = ax, (a > 0), x ∈ R
for x1, x2 ∈ R
Let f(x1) = f(x2) ⇒ ax1 = ax2
⇒ x1 = x2
∴ f is one-one.
Question 5.
Show that functions f and g defined by f(x) = 2 log x and g(x) = log x2 are not equal even though log x2 = 2 log x.
Solution:
f(x) = 2 log x
g(x) = log x2
Dom f(x) = (0, ∞)
Dom g(x) = R – {0}
As Dom f(x) ≠ Dom g(x) we have f(x) ≠ g(x), though log x2 = 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) = x3 + 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) = x2.
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 – x2.
(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 R0 = {x ∈ R : x ≥ 0} and not on R.
Solution:
Let f(x) = √x, g(x) = 1 – x2
(i) ∴ Dom f = R+ U{0}, Dom g = R
(ii) fog (x) = f (g(x))
= f (1 – x2) = \(\sqrt{1-x^2}\)
∴ fog (x) exists when 1 – x2 ≥ 0
⇒ x2 ≤ 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 = R0 = (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 ∈ R0 as dom f is R0 = [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) = x3 + 1, g(x) = x2 – 2
(ii) f(x) = sin x, g(x) = x5
(iii) f(x) = cos x, g(x) = sin x2
(iv) f(x) = g(x) = (1 – x3)1/3
Solution:
(i) f(x) = x3 + 1, g(x) = x2 – 2
∴ fog (x) = f (g(x)) = f(x2 – 2)
= (x2 – 2)3 + 1
gof (x) = g (f(x)) = g(x3 + 1)
= (x3 + 1)2 – 1
fog ≠ gof
(ii) f(x) = sin x, g(x) = x5
∴ fog (x) = f (g(x)) = f(x5) = sin x5
∴ gof (x) = g (f(x))
= g(sin x) = (sin x)5 = sin5 x
fog ≠ gof
(iii) f(x) = cos x, g(x) = sin x2
∴ fog (x) = f (g(x))
= f( sin x2) = cos (sin x2)
and gof (x) = g (f(x)) = g(cos x)
= sin (cos x)2 = sin (cos2 x)
fog ≠ gof
(iv) f(x) = g(x) = (1 – x3)1/3
fog (x) = f (g(x))
= (1 – (g(x))3)1/3
= [1 – (1 – x3)]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 + x2 , (ii) x2, (iii) ex, (iv) ex + 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 + x2
∴ f(-x) = 1 – x + x2
∴ g(x) = \(\frac{f(x)+f(-x)}{2}\)
= \(\frac{1+x+x^2+1-x+x^2}{2}\)
= x2 + 1 and
g(-x) = (-x)2 + 1 = x2 + 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) = x2
So that f(-x) = (-x)2 = x2
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\) = \( \frac{x^2+x^2}{2}\) = x2
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) = ex
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) = \( \frac{e^{-x}-e^x}{2}\) = -h2(x)
⇒ h is an odd function.
∴ f(x) = g(x) + h(x),
where g is even and h is odd.
(iv) Let f(x) = ex + 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 = idx and fog = idy, then show that
(i) f is bijective and (ii) g = f-1
[Hint: Since idx is a bijective function, gof = idx 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 = idx and fog = idy
we know that idx and idy 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-1of = idx and fof-1 = idy
But g : y → x with gof = idx and fog = idy
∴ 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, foidx = f = idyof.
Solution:
Let f : X → Y, so that y = f(x), x ∈ X.
∴ foidx = fof-1 of (x) = fof-1 (f(x))
= f(x) = y (∵ idx = fof-1)
Again, (idyof)(x) = (fof-1) of (x)
= (fof-1)(y) = f (f-1(y)) = f(x) = y ….(2)
∴ From (1) and (2)
we have foidx = f = idyof
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‘)