RBSE Class 12 Maths Solutions Chapter 1 Relations and Functions Ex 1.3

RBSE Class 12 Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.

Let f: {1, 3, 4} {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Answer:
According to question.
f = {(1, 2), (3, 5), (4, 1)}
and g = {1, 3), (2, 3), (5, 1)}
then f(1) = 2, f(3) = 5, f(4) = 1
and g(1) = 3, g(2) = 3, g(5) = 1
Now, (gof)(1) = g(f(1)) = g(2) = 3
(gof) (3) = g(f(3)) = g(5) = 1
(gof)(4) = g(f(4)) = g(1) = 3
Thus, gof = {(1, 3), (3, 1), (4, 3)}

Question 2.
Let f, g and h be functions from R to R. Show that:
(i) (f + g) oh = foh + goh
(ii) (f . g) oh = (foh).(goh)
Answer:
Given functions are f, g, h, R → R.
(i) ∴ (f + g)oh (x) = (f + g) (h(x))
= f(h(x)) + g(h(x))
= (foh)(x) + (goh)(x)
Thus (f + g)oh = foh + goh

(ii) (f.g)oh(x) = (f.g)(h(x))
= f(h(x)).g(h(x))
= {(foh)(x)}.{goh(x)}
= {(goh) > (goh)} (x)
Thus (f.g)oh = (foh).(goh)
Hence Proved.

Question 3.
Find gof and fog, if
(i) f(x) = |x| and g(x) = |5x - 2|
(ii) f(x) = 8x3 and g(x) = x1/3
Answer:
(i) Given, f(x) = |x| and g(x) = | 5x - 2 |
Then gof(x) = g(f(x)) = g(|x|)
= |5| x |- 2
Now fog(x) = f(g(x)) = f(|5x - 2|)
= | | 5x - 2 | |

(ii) Given, f(x) = 8x3 and g(x) = x1/3
then gof(x) = g(f(x)) = g( 8x3)
= (8x3)1/3
= [(2x)3]1/3
= 2x
Now fog(x) = f(g(x)) = f(x1/3)
= 8(x1/3)3 = 8x

Question 4.
If f(x) = $\frac{(4x+3)}{(6x-4)}$, x ≠ $\frac{2}{3}$, show that fof(x) = x, for all x ≠ $\frac{2}{3}$. What is the inverse of f? 
Answer:

Note: We can find inverse of /in the following way, also.
Let y = $\frac{4x+3}{6x-4}$
⇒ 6xy - 4y = 4x + 3
⇒ x(6y - 4) = 4y + 3
⇒ x = $\frac{4y+3}{6y-4}$
= f(y) (By definition of function)
∴ x = f(y)
⇒ f-1 (x) = y = $\frac{4x+3}{6x-3}$
∴ f-1 (x) = $\frac{4x+3}{6x-4}$
Thus, f is inverse of itself.

Question 5.
State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Answer:
f = {1, 2, 3, 4} → {10}
where f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Inverse, of function f does not exist since it is not one-one onto it is many-one onto.
f(1) = 10, f(2) = 10, f(3) = 10, f(4) = 10

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Answer:
g = {5, 6, 7, 8} → {1, 2, 3, 4}
where g = {(5, 4), (6, 3), (7,4), (8, 2)}
Here, g(5) = 4, g(6) = 3, g(7) = 4, g(8) = 2
Function is not one-one onto.
It is many one into.
Element 1 of set {1, 2, 3, 4} is not image of any element of set {5, 6, 7, 8}.
Thus, function g is not one-one onto.
Thus, g will not have inverse function.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answer:
h = {2,3, 4, 5} → {7, 9,11,13}
where h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, h(2) = 7, h(3) = 9, h(4) = 11, h(5) = 13
Thus, each element of set {2, 3, 4, 5} has unique image in set {7, 9, 11, 13} and there is no element in set {7, 9,11,13}
which has no image of element of set {2,3,4,5}.
Thus, function h is one-one onto.
∴ Inverse of function h is
h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
where h-1: {7, 9, 11, 13} → {2, 3, 4, 5}

Question 6.
Show that f: [- 1, 1] → R, given by f(x) = $\frac{x}{(x+2)}$ is one-one. Find the inverse of the function f: [- 1, 1] → Range f.
(Hint: For y ∈ Range f, y = f(x) = $\frac{x}{(x+2)}$, or some x in [- 1, 1], i.e., x = $\frac{2y}{1-y}$)
Answer:
Let x1, x2 ∈ [- 1, 1] then
f(x1) = f(x2) ⇒ $\frac{x_1}{x_1+2}$ = $\frac{x_2}{x_2+2}$
⇒ x1(x2 + 2) = x2(x1 + 2)
⇒ x1x2 + 2x1 = x2x1 + 2x2
⇒ 2x1 = 2x2 ⇒ x1 = x2
Thus, function f: [- 1,1] → R is one-one.
Let y ∈ (range f) is any arbitrary element and y = f(x), where x ∈ [- 1, 1],
Then f(x) = y ⇒ $\frac{x}{x+2}$ = y
⇒ y(x + 2) = x
⇒ yx + 2y = x
⇒ yx - x = - 2y
⇒ x(y - 1) = - 2y
⇒ x = $\frac{-2y}{y-1}$ ⇒ = $\frac{2y}{1-y}$, y ≠ 1 ...... (i)
Thus, range of f = R - {1}
Again x = $\frac{2y}{1-y}$, y ≠ 1

So, f is onto.
Since, f is one-one onto.
f:[- 1, 1] → R - {1}
Thus, f will be invertible.
Then fof-1(x) = f(f-1(x)) = x
⇒ $\frac{f^{-1}\left(x\right)}{f^{-1}\left(x\right)+2}$ = x (by definition of f)
⇒ f-1(x) = xf-1 (x) + 2x
⇒ f-1(x) - xf-1(x) = 2x
⇒ (1 - x)f-1(x) = 2x
⇒ f-1(x) = $\frac{2x}{1-x}$
Thus, inverse of f-1: R - {1} → [-1, 1] is
⇒ f-1(x) = $\frac{2x}{1-x}$ (∵ x ≠ 1)

Question 7.
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Answer:
f(x) = 4x + 3 and f: R → R
Let x1, x2 ∈ R (domain), then
f(x1) = f(x2) ⇒ 4x1 + 3 = 4x2 + 3
⇒ 4x1 = 4x2 ⇒ x1 = x2
Thus, f is one-one.
Again, let y ∈ R (co-domain) is an arbitrary element then in set R (domain) x will be such element for which
f(x) = y
Now y = f(x) ⇒ y = 4x + 3
⇒ y - 3 = 4x ⇒ x = 6 R (domain)
⇒ f(x) = f$\left(\frac{y-3}{4}\right)$ = 4$\left(\frac{y-3}{4}\right)$ + 3
= y - 3 + 3 = y
f$\left(\frac{y-3}{4}\right)$ = y ∈ R (co-domain)
∴ Function is onto.
∵ Function is one-one onto;
so, f is invertible.
Now, for inverse f-1 of f
fof-1 (x) = x
⇒ f(f-1(x)) = x
⇒ 4(f-1(x)) + 3 = x
⇒ 4 (f-1(x)) = x - 3
⇒ f-1(x) = $\frac{x-3}{4}$
Thus, inverse of f, f-1(x): R → R is such that
f-1(x) = $\frac{x-3}{4}$

Question 8.
Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f-1 of given by f-1 (y) = $\sqrt{y-4}$, where R+ is the set of all non-negative real numbers. 
Answer:
Let x1, x2 ∈ R, then
f(x1) = f(x2) ⇒ x12 + 4 = x22 + 4
x12 = x22 ⇒ x1 = x2
Thus, f is one-one.
Again, let y ∈ [4, ∞) is an arbitrary element then x ∈ R+ will be such element for which y = f(x)
Now y = f(x)
⇒ y = x2 + 4
⇒ x2 = y - 4
⇒ x = $\sqrt{y-4}$, y ≥ 4
For each y ≥ 4, $\sqrt{y-4}$ ∈ R+
Now f(x) = f($\sqrt{y-4}$)
= ($\sqrt{y-4}$)2 + 4
= y - 4 + 4
= y (By definition of f)
∴ f($\sqrt{y-4}$) = y
Thus, f is onto.
Now, f is one-one onto then it will be invertible.
Thus, f is invertible.
Now, for f-1, inverse of f
fof-1(x) = x
⇒ f(f-1(x)) = x
⇒ [f-1(x)]2 + 4 = x (By definition of f)
⇒ (f-1(x))2 = x - 4
⇒ f-1(x) = $\sqrt{x-4}$
⇒ f-1(y) = $\sqrt{y-4}$
Note: f($\sqrt{y-4}$) = y
∴ f-1 = $\sqrt{y-4}$
Which is inverse function of f.

Question 9.
Consider f: R+ → [- 5, ∞) given by f(x) = 9x2 + 6x - 5.
Show that f is invertible with f-1 (y) = $\left(\frac{\sqrt{(y+6)}-1}{3}\right)$
Answer:
Let x1, x2 ∈ R+ then
f(x1) = f(x2)
9x12 + 6x1 - 5 = 9x22 + 6x2 - 5
⇒ 9x12 + 6x + 1 - 6 = 9x22 + 6x2 + 1 - 6
⇒ (3x1 + 1)2 - 6 = (3x2 + 1)2 - 6
⇒ 3x1 + 1 = 3x2 + 1 [∵ x1, x2 ∈ R+]
⇒ 3x1 = 3x2
⇒ x1 = x2
Thus, f is one-one.
Again, let y ∈ [- 5, ∞) is any arbitrary element then x e R+ will be such element for which y = f(x), then
y = f(x)
⇒ y = 9x2 + 6x - 5
⇒ y = 9x2 + 6x + 1 - 1 - 5
⇒ y = (3x + 1)2 - 6
⇒ (3x + 1)2 = y + 6

Thus, f is onto.
Since, f is one-one onto so f is invertible.
From (i), f-1(y) = $\left(\frac{\sqrt{y+6}-1}{3}\right)$
or
fof-1(x) = x (∵ f-1, inverse of function f)
⇒ f(f-1(x)) = x
⇒ 9(f-1(x))2 + 6f-1(x) - 5 = x
⇒ 9(f-1(x))2 + 6f-1(x) + 1 - 1 - 5 = x
⇒ (3f-1(x) + 1)2 = x + 6
⇒ 3f-1(x) + 1 = $\sqrt{x+6}$
⇒ 3f-1(x) = $\sqrt{x+6}$ - 1
⇒ f-1 (x) = $\left(\frac{\sqrt{x+6}-1}{3}\right)$
Thus f-1 (y) = $\left(\frac{\sqrt{(y+6)}-1}{3}\right)$
where f-1 (y), is inverse of f(y).
Hence Proved.

Question 10.
Let f: X → Y be an invertible function. Show that/has unique inverse. (Hint: Suppose^ andg2 are two inverses off. Then for all y ∈ Y, fog1(y) = IY(y) = fog2(y). Use one-one ness of f).
Answer:
Given that f: X → Y, is invertible so it will be one- one onto.
Let g1, g2 are two inverse of f.
then fog1(y) = Iy(y) ......... (i)
and fog2(y) = Iy(y) ......... (ii)
From (i) and (ii), we get
fog1(y) = fog2(y)
(Since, Iy(y) will be one not two )
⇒ f(g1(y)) = f(g2(y))
⇒ g1(y) = g2(y)
⇒ g1 = g2 [Since f is one-one]
Thus f(x1) = f(x2)
⇒ x1 = x2
i.e., inverse of f is unique.
Hence Proved.

Question 11.
Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f-1 and show that (f'-1)-1 = f.
Answer:
f: {1, 2, 3} → {a, b, c}, f(1) = a, f(2) = b, f(3) = c.
Writing function f in ordered pair form:
f = {(1, a), (2, b), (3, c)}
Clearly, each element of set {1, 2, 3} has distinct image in set {a, b, c} and there is no element in set {a, b, c) which is not the image of any element of set {1, 2, 3}. Thus, function f is one-one onto. Hence, function f is invertible.
Let inverse of f is f-1.
Then in f = {(1, a), (2, b), (3, c)} by interchanging the elements f-1 will be obtained.
Thus, f-1 = {(a, 1), (b, 2), (c, 3)}
i.e., f-1: {a, b, c} → {1, 2, 3} will be defined as:
f-1(a) = 1, f-1(b) = 2, f-1(c) = 3
Since, function f is one-one onto so its inverse f-1 be one-one onto then inverse of f-1 will be (f-1)-1.
Thus, by interchanging the elements of ordered pair in
f-1 = {(a, 1), (b, 2), (c, 3)} we will get (f-1)-1.
Thus, (f-1)-1 = {(1, a), (2, b), (3, c)} = f (By definition of f)
Here (f -1)-1(1) = a, (f-1)-1 (2) = b and (f-1)-1(3) = c, which is definition of f.
Hence Proved.

Question 12.
Let f: X → Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1)-1 = f.
Answer:
It is given that f: X → Y is invertible. We have to prove that inverse of f-1 is f i.e.
(f-1)-1 = f
Here, to show f-1of = IX and fof-1 = Iy will be sufficient for(f-1)-1 = f.
Clearly f: X → Y is one-one onto.
∴ f-1: Y → X is one-one onto.
Let x ∈ X and f(x) = y then f-1(y) = x
∴ (f-1of) = f-1(f(x))
= f-1(y) [∵ f(x) = y]
= x = Ix(x)

where Ix(x) is identity function.
∴ f-1of = Ix
Again, function f is onto, then x ∈ X will be such that f(x) = y, then f-1(y) = x
∴ (fof-1) (y) = f(f-1(y)) = f(x)
= y (∵ f-1(y) = x)
= IY(y)
Thus, f-1of = IX and fof-1 = IY
∴ (f-1)-1 = f
Hence Proved.

IInd Method: Since, we know that inverse of any bijective function is bijective.
∴ f-1: Y → X will also be bijective
Now, f-1 : Y → X is bijective, then (f-1)-1: X → Y will also be bijective
Let x ∈ X is an arbitrary element.
∴ (f-1)-1 (x) = f(x), ∀ x ∈ A
Thus, (f-1)-1 = f
Hence Proved.

Question 13.
If f: R → R be given by f(x) = ${(3-x^3)}^{\frac{1}{3}}$ , then fof(x) is:
(A) $x^{\frac{1}{3}}$
(B) x3
(C) x
(D) (3 - x3)
Answer:
Given f(x) = (3 - x3)1/3,f: R → R
Now, fof(x) = f(f(x))
= f[(3 - x3)1/3]
= [{3 - {3 - x3)1/3}3]1/3
= [3 - (3 - x3)]1/3
= (3 - 3 + x3)1/3
= (x3)1/3
= x
Thus, option (C) is correct.

Question 14.
Let f: R - $\{-\frac{4}{3}\}$ → R be a function defined as f(x) = $\frac{4x}{3x+4}$. The inverse of f is the map g: Range f → R - $\{-\frac{4}{3}\}$ given by.
(A) g(y) = $\frac{3y}{3-4y}$
(B) g(y) = $\frac{4y}{1-3y}$
(C) g(y) = $\frac{4y}{34-4}$
(D) g(y) = $\frac{4y}{3-4y}$
Answer:

Thus, option (B) is correct.

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