MAHARASHTRA

XII (12) HSC

XI (11) FYJC
X (10) SSC

## 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.

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as
= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. ## Question 2:Let f, g and h be functions from R to R. Show that To prove:     ## Question 3:Find gof and fog, if(i) (ii) (i)  (ii)  ## Question 4:If , show that f o f(x) = x, for all . What is the inverse of f?

It is given that . Hence, the given function f is invertible and the inverse of f is f itself.

## Question 5:State with reason whether following functions have inverse(i) f: {1, 2, 3, 4} → {10} withf = {(1, 10), (2, 10), (3, 10), (4, 10)}(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} withg = {(5, 4), (6, 3), (7, 4), (8, 2)}(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} withh = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
is not one-one.
Hence, function does not have an inverse.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.
is not one-one,
Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.

## Question 6:Show that f: [−1, 1] → R, given by is one-one. Find the inverse of the function f: [−1, 1] → Range f.(Hint: For y ∈Range f, y = , for some x in [−1, 1], i.e., )

f: [−1, 1] → R is given as Let f(x) = f(y). ∴ f is a one-one function.
It is clear that f: [−1, 1] → Range f is onto.
∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [−1, 1] → Range exists.
Let g: Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f
Since f: [−1, 1] → Range f is onto, we have: Now, let us define g: Range f → [−1, 1] as gof = and fo  f−1 = g
⇒ ## Question 7:Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

fR → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y). ∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3. Therefore, for any ∈ R, there exists such that ∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define gR→ R by .  Hence, f is invertible and the inverse of f is given by ## Question 8:Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by , where R+ is the set of all non-negative real numbers.

fR+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y). ∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4. Therefore, for any ∈ R, there exists such that .
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → Rby,   Hence, f is invertible and the inverse of f is given by ## Question 9:Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with .

fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6− 5. f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) → R+ as We now have:   and Hence, f is invertible and the inverse of f is given by ## Question 10:Let f: X → Y be an invertible function. Show that f has unique inverse.(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,fog1(y) = IY(y) = fog2(y). Use one-one ness of f).

Let fX → Y be an invertible function.
Also, suppose f has two inverses (say ).
Then, for all y ∈Y, we have: Hence, f has a unique inverse.

## 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.

Function f: {1, 2, 3} → {abc} is given by,
f(1) = af(2) = b, and f(3) = c
If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:  and , where X = {1, 2, 3} and Y= {abc}.
Thus, the inverse of exists and f−1 = g.
f−1: {abc} → {1, 2, 3} is given by,
f−1(a) = 1, f−1(b) = 2, f-1(c) = 3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {abc} as
h(1) = ah(2) = bh(3) = c, then we have:  , where X = {1, 2, 3} and Y = {abc}.
Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.
It can be noted that h = f.
Hence, (f−1)−1 = f.

## 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.

Let fX → Y be an invertible function.
Then, there exists a function gY → X such that gof = IXand fo= IY
Here, f−1 = g
Now, gof = IXand fo= IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.

## Question 13:If f: R → R be given by , then fof(x) is(A) (B) x3(C) x(D) (3 − x3)

fR → R is given as . The correct answer is C.

## Question 14:Let be a function defined as . The inverse of f is map g: Range (A) (B) (C) (D) It is given that Let y be an arbitrary element of Range f.
Then, there exists x ∈ such that  Let us define g: Range as Now,    Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map g: Range , which is given by The correct answer is B.

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SUBJECTS

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION
SSC MATHS I PAPER SOLUTION
SSC MATHS II PAPER SOLUTION
SSC SCIENCE I PAPER SOLUTION
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SSC ENGLISH PAPER SOLUTION
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2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

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SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

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Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

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Important-formula

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