# 2022

## Mathematical Indunction (M.I) Exerxise wise Solutions

Mathematical Indunction: It is a technique for proving results or establishing statements for natural numbers. This part illustrates the method through a variety of examples.

Definition:
Mathematical Induction  is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.

The technique involves steps to prove a statement, as stated below −
Let P(n) or S(n) be the given statement

Step 1: For n =1
we get LHS = RHS
then P(n) is true for n= 1
Step 2: Let us assume thet P(n) is true for n =k
Step 3: We have to Prove P(n) is true for n= k + 1

Laplace:

Laplace was a mathematecian and astronomer whose work was pivotal to the development of mathematical astronomy. His most outstanding work was done in the fields of celestial mechonics, probability, differential equations, and geodesy. His five volume work on celestial mechonics earned him the title of the Newton of France.

“Analysis and natural philosophy owe their most important discoveries to this fruitful means, which is called indunction” – Pierr Simon de Laplace

### Exercise 2(a)

#### Using Mathematical Indunction, Prove each of the following statement for all n ∈ N.

##### 1. 12 + 22 + 32 + …… + n2=

Let p(n) be the given statement that

12 + 22 + 32 + …… + n2=

For n= 1

LHS = 12 = 1

RHS = =  = 1

LHS = RHS

P(n) is true for n = 1

Let us assume that P(n) is true for n = k

i.e., 12 + 22 + 32 + …… + k2=  ………… (1)

for n = k + 1

add (k +1)2 on both sides of (1)

12 + 22 + 32 + …… + k2 + (k +1)2 =

P(n) is true for n = k+ 1

∴ By the principle of  M.I.  P(n) is true for all n ∈ N

∴ 12 + 22 + 32 + …… + n2=

##### 2.  2.3 + 3.4 + 4.5 + ……… up to n terms =

First factors of given series are: 2, 3, 4, 5, …

a = 2, d = 1

an = a + (n – 1) d

= 2 + (n – 1) (1)

= 2 + n – 1

= n + 1

Second factors of given series are: 3, 4, 5,…

a = 3, d = 1

an = 3 + (n – 1) d

= 3 + (n – 1) (1)

= 3 + n – 1

= n + 2

nth term of given series is (n + 1) (n + 2)

let P(n) be the given statement that

2.3 + 3.4 + 4.5 + ……… + (n + 1) (n + 2) =

For n = 1

LHS = 2.3 = 6

RHS =  =  =  = 6

LHS = RHS

P(n) is true for n = 1

Let us assume that P(n) is true for n = k

i.e., 2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) =  ………… (1)

for n = k + 1

add (k + 2) (k + 3) on both sides of (1)

2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) + (k + 2) (k + 3)

=  + (k + 2) (k + 3)

P(n) is true for n = k+ 1

∴ By the principle of  M.I.

P(n) is true for all n ∈ N

∴ 2.3 + 3.4 + 4.5 + ……… up to n terms =

###### 3 .

Sol:

let P(n) be the given statement that

For n = 1

LHS =  =

RHS =  =  =

LHS = RHS

P (n) is true for n = 1

Let us assume that P(n) is true for n = k

………… (1)

For n = k + 1

Add on both sides of (1)

P (n) is true for n = k + 1

∴ By the principle of M.I.   P(n) is true for all n ∈ N

∴

###### 4. 43 + 83 + 123 + … up to n terms = 16 n2 (n + 1)2

Sol:

let P(n) be the given statement that

4, 8, 12, … are in AP

a = 4, d = 4

an = a + (n – 1) d

= 4 + (n – 1)4

= 4 + 4n – 4

= 4n

nth term of given series is (4n)3

let P(n) be the given statement that

43 + 83 + 123 + … + (4n)3= 16 n2 (n + 1)2

For n = 1

LHS = 43 = 63

RHS = 16 (1)2 (1 + 1)2 = 16 × 4 = 64

LHS = RHS

P (n) is true for n = 1

Let us assume that P(n) is true for n = k

43 + 83 + 123 + … + (4k)3= 16 k2 (k + 1)2 ………… (1)

For n = k + 1

Add [4 (k + 1)]3 on both sides of (1)

43 + 83 + 123 + … + (4k)3= 16 k2 (k + 1)2 + [4 (k + 1)]3

= 16 k2 (k + 1)2 + 64 (k + 1)3

= 16 (k + 1)2 [k2 + 4 (k + 1)]

= 16 (k + 1)2 [k2 + 4 k + 4]

= 16 (k + 1)2 (k + 2)2

= 16 (k + 1)2 ( + 1)2

P (n) is true for n = k + 1

∴ By the principle of Mathematical induction

P(n) is true for all n ∈ N

##### 5. a + (a + d) + (a + 2d) + …. up to n terms =

Sol:

Given series is a + (a + d) + (a + 2d) + …. up to n terms =

nth term of the given series is a + (n – 1) d

let P(n) be the given statement that

a + (a + d) + (a + 2d) + …. +[a + (n – 1) d] =

for n = 1

LHS = a;     RHS = = a

LHS = RHS

P(n) is true for n = 1

Let us assume that p(n) is true for n = k

a + (a + d) + (a + 2d) + …. + [a + (k – 1) d] = ……. (1)

add (a + k d) on both sides of (1)

a + (a + d) + (a + 2d) + …. [a + (k – 1) d] + [a + kd] = + [a + kd]

P (n) is true for n = k + 1

∴ By the principle of Mathematical induction

P(n) is true for all n ∈ N

Mathematical Induction

## Functions Exercise 1c Solutions

Functions Exercise 1c

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1c Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

## Functions Exercise 1c

### I.

#### 1.Find the domains of the following real valued functions

(i)

f (x) =

given function is f (x) =

f (x) is defined when (x2 – 1) (x + 3) ≠ 0

⟹ (x2 – 1) ≠ 0 or (x + 3) ≠ 0

⟹ (x + 1) (x – 1) ≠ 0 or (x + 3) ≠ 0

⟹ x ≠ 1, x ≠ – 1 or x ≠ – 3

∴ Domain of f(x) is R – {– 1, – 3, 1}

###### (ii)

f (x) =

Given function is f (x) =

f (x) is defined when (x – 1) (x – 2) (x – 3) ≠ 0

⟹ x ≠ 1, x ≠ 2 or x ≠ 3

∴ Domain of f(x) is R – {1, 2, 3}

(iii)

f (x) =

Given function is f (x) =

f (x) is defined when 2 – x > 0 and 2 – x ≠ 1

⟹ 2 > x  and  2 – 1 ≠ x

⟹ 2 > x  and  x ≠ 1

∴ Domain of f(x) is (– ∞, 2) – {1}

(iv)

f (x) =

Given function is f (x) =

f (x) is defined when x ∈ R

∴ Domain of f(x) is R

Functions Exercise 1c

(v)

f (x) =

Given function is f (x) =

f (x) is defined when 4x – x2 ≥ 0

⟹ x (4 – x) ≥ 0

⟹ x (x – 4) ≤ 0

⟹ (x – 0) (x – 4) ≤ 0

⟹ x ∈ [0, 4]

∴ Domain of f(x) is [0, 4]

(vi)

f (x) =

Given function is f (x) =

f (x) is defined when 1 – x2 > 0

⟹ x2 – 1 < 0

⟹ (x – 1) (x + 1) < 0

⟹ x ∈ (– 1, 1)

∴ Domain of f(x) is (– 1, 1)

(vii)

f (x) =

Given function is f (x) =

f (x) is defined when x + 1≠ 0

⟹ x ≠ – 1

∴ Domain of f(x) is R – {– 1}

(viii)

f(x) =

Given function is f (x) =

f (x) is defined when x2 – 25 ≥ 0

⟹ (x – 5) (x + 5) ≥ 0

⟹ x ∈ (–∞, –5] ∪ [5, ∞)

⟹ x ∈ R – (– 5, 5)

∴ Domain of f(x) is R – (– 5, 5)

Functions Exercise 1c

(ix)

f(x) =

Given function is f (x) =

f (x) is defined when x – [x] ≥ 0

⟹ x ≥ [x]

⟹ x ∈ R

∴ Domain of f(x) is R

(x)

f(x) =

Given function is f (x) =

f (x) is defined when [x] – x ≥ 0

⟹ [x] ≥ x

⟹ x ∈ Z

∴ Domain of f(x) is Z

Functions Exercise 1c

##### 2. find the ranges of the following real valued functions

(i)

f(x) =

Given function is f (x) =

Let y =

⟹  |4 – x2| = ey

∵ ey > 0 ∀ y ∈ R

∴ Range of f(x) is R

(ii)

f(x) =

Given function is f (x) =

f (x) is defined when [x] – x ≥ 0

⟹ [x] ≥ x

⟹ x ∈ Z

Domain of f(x) is Z

Range of f = {0}

(iii)

f(x) =

Given function is f (x) =

f (x) is defined when x ∈ R

Domain of f(x) is R

For x ∈ R   [x] is an integer

Since sin nπ = 0, ∀ n ∈ z

⟹ sin π[x] = 0

∴ Range of f = {0}

(iv)

f (x) =

Given function is f (x) =

f (x) is defined when x – 2 ≠ 0

⟹ x ≠ 2

Domain of f(x) is R – {2}

Let y =

=

= x + 2

If x = 2 ⟹ y = 2 + 2 = 4

∴ Range of f(x) is R – {4}

Functions Exercise 1c

(v)

f (x) =

let y =

y2 = 9 + x2

x2 = y2 – 9

x =

it is defined when y2 – 9 ≥ 0

⟹ (y – 3) (y + 3) ≥ 0

y ∈ (– ∞, – 3] ∪ [3, ∞)

but y = ≥ 0

∴ Range of f(x) is [3, ∞)

##### find

Sol:

Given f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2

(i)

(3f – 2g) (x) = 3 f(x) – 2g (x)

= 3 (2x – 1) – 2(x2)

= 6x – 3 – 2x2

= – 2x2 + 6x – 3

∴ (3f – 2g) (x) =– 2x2 + 6x – 3

(ii)

(fg) (x) = f (x) g (x)

= (2x – 1) (x2)

= 2x3 + x2

∴ (fg) (x) = 2x3 + x2

(iii)

##### 4. If f = {(1, 2), (2, – 3) (3, – 1)} then find (i) 2f (ii) (fog) 2 + f iii) f2 (iv)

Sol:

Given f = {(1, 2), (2, – 3) (3, – 1)}

(i)

(2f) (1) = 2 f (1) = 2 × 2 = 4

(2f) (2) = 2 f (2) = 2 × – 3 = – 6

(2f) (3) = 2 f (3) = 2 × – 1 = – 2

∴ 2f = {(1, 4), (2, – 6) (3, – 2)}

(ii)

(2 + f) (1) = 2 + f (1) = 2 + 2 = 4

(2 + f) (2) = 2 + f (2) = 2 + (– 3) = – 1

(2 + f) (3) = 2 + f (3) = 2 + (– 1) = 1

∴ 2 + f = {(1, 4), (2, – 1) (3, 1)}

(iii)

(f2) (1) = [f (1))]2 = 22 = 4

(f2) (2) = [f (2))]2 = (– 3)2 = 9

(f2) (3) = [f (1))]2 = (– 1)2 = 1

∴ f2= {(1, 4), (2, 9) (3, 1)}

(iv)

### II.

#### 1.Find the domain of the following real valued functions

(i)

f (x) =

f(x) is defined when x2 – 3x + 2 ≥ 0

x2 – 2x – x + 2 ≥ 0

x (x – 2) – 1(x – 2) ≥ 0

(x – 1) (x – 2) ≥ 0

x ∈ (– ∞, 1] ∪ [2, ∞)

∴ Domain of f(x) is R – (1, 2)

(ii)

f(x) = log (x2 – 4x + 3)

f(x) is defined when x2 – 4x + 3 > 0

x2 – 3x – x + 3 > 0

x (x – 3) – 1(x – 3) > 0

(x – 1) (x – 3) > 0

x ∈ (– ∞, 1) ∪ (3, ∞)

∴ Domain of f(x) is R – [1, 3]

(iii)

f(x) =

f(x) is defined when 2 + x ≥ 0, 2 – x ≥ 0  and x ≠ 0

x ≥ – 2, x ≤ 2 and x ≠ 0

– 2 ≤ x ≤ 2 and x ≠ 0

x ∈ [–2, 2] – {0}

∴ Domain of f(x) is [–2, 2] – {0}

(iv)

f(x) =

f(x) is defined in two cases as follows:

case (i) 4 – x2 ≥ 0 and [x] + 2 > 0

x2 – 4 ≤ 0 and [x] > –2

(x – 2) (x + 2) ≤ 0 and [x] > –2

x ∈ [– 2, 2] and x ∈ [– 1, ∞)

x ∈ [– 1, 2]

case (ii) 4 – x2 ≤ 0 and [x] + 2 < 0

x2 – 4 ≥ 0 and [x] < –2

(x – 2) (x + 2) ≥ 0 and [x] < –2

x ∈ (– ∞, –2] ∪ [2, ∞) and x ∈ (–∞, –2)

x ∈ (–∞, –2)

from case (i) and case (ii)

x ∈ (–∞, –2) ∪ [– 1, 2]

∴ Domain of f(x) is (–∞, –2) ∪ [– 1, 2]

(v)

f(x) =

f(x) is defined when ≥ 0 and x – x2 > o

x – x2 ≥ (0.3)0   and x2 – x < 0

x – x2 ≥ 1 and x (x – 1) < 0

x2 –x + 1 ≤ 0 and (x – 0) (x – 1) < 0

it is true for all x ∈ R and x ∈ (0, 1)

∴ domain of f(x) =  R∩ (0, 1) = (0, 1)

(vi)

f(x) =

f(x) is defined when x +|x| ≠ 0

|x| ≠ – x

|x| ≠ – x

⟹|x| = x

⟹ x > 0

x ∈ (0, ∞)

∴ Domain of f(x) is (0, ∞)

#### 2. Prove that the real valued function is an even function

Sol:

Given f(x) =

#### 3. Find the domain and range of the following functions

(i)

f (x) =

Given f(x) =

Since [x] is an integer

sin π[x] = tan π[x] = 0 ∀ x ∈ R

∴ domain of f(x) is R

and

since tan π[x] = 0

Range of f(x) = {0}

(ii)

f(x) =

Given f (x) =

It is defined when 2 – 3x ≠ 0

⟹ 2 ≠ 3x

⟹ x ≠ 2/3

∴ Domain of f (x) = R – {2/3}

Let y = f(x)

y =

⟹ y (2 – 3x) = x

2y – 3xy = x

2y = x + 3xy

2y = x (1 + 3y)

⟹ x =

It is defined when 1 + 3y ≠ 0

1 ≠ –3y

y ≠ – 1/3

∴ Range of f (x) = R – {– 1/3}

(iii)

f(x) = |x| + | 1 + x|

Given function is f (x) = |x| + |1 + x|

f (x) is defined for all x ∈ R

∴ domain of f(x) = R

f (– 3) = |– 3| + |1 – 3|

=|– 3| + |– 2|

= 3 + 2 = 5

f (– 2) = |– 2| + |1 – 2|

=|– 2| + |– 1|

= 2 + 1 = 3

f (– 1) = |– 1| + |1 – 1|

=|– 1| + |0|

= 1 + 0 = 1

f (0) = |0| + |1 + 0| = 1

f (1) = |1| + |1 + 1|

= 1 + |2|

= 1 + 2 = 3

f (2) = |2| + |1 + 2|

= |2| + |3|

= 2 + 3 = 5

f (3) = |3| + |1 + 3|

= |3| + |4|

= 3 + 4 = 7

∴ Range of f(x) = [1, ∞)

## Functions Exercise 1b Solutions

Functions Exercise 1b

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1b Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

## Functions Exercise 1b

Exercise 1(b) Solutions

### I.

##### 1.If f (x) = ex and g(x) = , then show that f og = gof and f-1 = g-1

Sol:

Given     f (x) = ex and g(x) =

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

= f ( )

=

= x ————- (1)

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

= g (ex)

=

= x

= x ————- (2)

From (1) and (2)  f og = gof

let y = f(x)

x = f-1(y)

y = ex

x =

f-1 (x) =

let g(x) = z

x = g-1(z)

z =

x = ez

g-1 (x)  = ex

##### 2. If f (y) = ,  g (y) = then show that fog(y) = y

Sol:

Given f (y) = ,  g (y) =

Now

fog(y) = f(g(y))

∴ fog(y) = y

##### (i)  (fog) (x)    (ii) (gof) (x)       (iii) (fof) (0)      (iv) go(fof)(3)

Sol:

Given f: R ⟶ R is, g: R ⟶ R is defined by f(x) = 2x2 + 3 ang g (x) = 3x – 2

(i)  (fog) (x) = f(g(x))

= f(3x – 2)

= 2 (3x – 2)2 + 3

= 2 (9x2 – 12x + 4) + 3

= 18 x2 – 24x + 8 + 3

= 18 x2 – 24x + 11

∴ (fog) (x) = 18 x2 – 24x + 11

(ii)  (gof) (x) = g (f (x))

= g (2x2 + 3)

= 3(2x2 + 3) – 2

= 6x2 + 9 – 2

= 6 x2 + 7

∴ (gof) (x) = 6 x2 + 7

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(iii)  (fof) (0) = f (f (0))

= f (2(0)2 + 3)

= f (2(0) + 3)

= f (3)

= 2 (3)2 + 3

= 2 (9) + 3

= 18 + 3 = 21

(iv)  go(fof) (3) = go (f (f (3)))

= go (f (2 (3)2 + 3))

= go (f (21))

= g (f (21))

= g (2 (21)2 + 3))

= g (2 (441) + 3))

= g (882 + 3)

= g (885)

= 3 (885) – 2

= 2655 – 2

= 2653

∴ go(fof) (3) = 2653

Functions Exercise 1b

Ts Inter Maths IA Concept

##### (i) (fof) (x2 + 1)    (ii) (fog) (2)        (iii) (gof) (2a – 3)

Sol:

Given f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1

(i) (fof) (x2 + 1) = f (f (x2 + 1))

= f (3 (x2 + 1)– 1)

= f (3x2 + 3– 1)

= f (3x2 + 2)

= 3(3x2 + 2) – 1

= 9x2 + 6 – 1

= 9x2 + 5

(ii) (fog) (2) = f (g (2))

= f (22 + 1)

=f (4 + 1)

= f (5)

= 3(5) – 1

= 15 – 1

= 14

(iii) (gof) (2a – 3) = g (f (2a – 3))

= g (3(2a – 3) – 1)

= g (6a – 9 – 1)

= g (6a – 10)

= (6a – 10)2 + 1

= 362 – 120a + 100 + 1

= 362 – 120a + 101

##### 5. If f(x) = and g(x) = for all x ∈ (0, ∞) then find (gof) (x)

Sol:

Given f(x) =  and g(x) =  for all x ∈ (0, ∞)

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

= g ( )

=

∴ (gof) (x) =

##### 6. If f(x) = 2x – 1 and g(x) = for all x ∈ R then find (gof) (x)

Sol:

Given f(x) = 2x – 1 and g(x) =  for all x ∈ R

∴ gof(x) = x

##### 7. If f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)

Sol:

Given f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R

(fo(goh)) (x) = fo (g (h (x))

= fo g(2x)

= f (g(2x))

= f((2x)2)

= f(4x2)

= 2

∴ (fo(goh)) (x) = 2

Functions Exercise 1b Solutions

#### 8. Find the inverse of the following functions

(i) a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b, (a ≠ 0)

Given a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b

Let y = f(x)

x = f-1(y)

now y = ax + b

ax = y – b

x =

f-1(y) =

∴   f-1(x) =

(ii) f: R ⟶ (0, ∞) defined by f(x) = 5x

Let y = f(x)

x = f-1(y)

now y = 5x

x =

f-1(y) =

∴   f-1(x) =

(iii) f: (0, ∞) ⟶ R defined by f(x) =

Let y = f(x)

x = f-1(y)

now y =

x = 2y

f-1(y) = 2y

∴   f-1(x) = 2x

##### 9. If f(x) = 1 + x + x2 + … for , then show that f-1 (x) =

Sol:

Given, f(x) = 1 + x + x2 + … for

1 + x + x2 + … is an infinite G.P

a = 1, r = x

S =

=

Now

f (x) =

Let y = f(x)

x = f-1(y)

now y =

1 – x =

x = 1 –

=

f-1(y) =

∴   f-1(x) =

##### 10 . If f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1) then find f-1 (x)

Sol:

Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1)

Let y = f(x)

x = f-1(y)

now y = 2x (x – 1)

= x (x – 1)

x2 – x =

x2 – x –   = 0

Functions Exercise 1b Solutions

### II.

#### 1. If f (x) =   , x ≠ ± 1, then verify (fof-1) (x) = x

Sol:

Given f (x) =  , x ≠ ± 1

Let y = f(x)

x = f-1(y)

now y =

y (x + 1) = x – 1

xy + y = x – 1

1 + y = x – xy

1 + y = x (1 – y)

x =

f-1(y) =

∴   f-1(x) =

Now

##### 2.  If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)-1 = f-1og-1

Sol:

Given A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, p), (β, r), (γ, p)}

A = {1, 2, 3}, B = {α, β, γ}

f: A ⟶ B; and f = {(1, α), (2, γ), (3, β)}

⟹ f (1) = α; f (2) = γ; f (3) = β

⟹    Every element of set A has a unique image in set B

f is injection (one – one)

range of f = codomain of f

⟹ f is surjection (on to)

∴ f is bijective

B = {α, β, γ}, C = {p, q, r}

g: B ⟶ C is defined by g = {(α, q), (β, r), (γ, p)}

⟹ g (α) =q; g (β) = r; g (γ) = p

⟹    Every element of set B has a unique image in set C

g is injection (one – one)

range of g = codomain of g

⟹ g is surjection (on to)

∴ g is bijective

Now

f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}

gof = {(1, q), (2, p), (3, r)

(gof)-1 = {(q, 1), (p, 2), (r, 3)} ———– (1)

f-1 = {(α, 1), (γ, 2), (β, 3)

g-1 = {(q, α), (r, β), (p, γ)}

f-1og-1 = {(q, 1), (p, 2), (r, 3)} ———– (2)

From (1) and (2)

(gof)-1 = f-1og-1

##### (i) (gof-1) (2)           (ii) (gof) (x – 1)

Sol:

Given f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x2 + 1

Let y = f(x)

x = f-1(y)

now y = 3x – 2

3x = y + 2

x =

f-1(y) =

f-1(x) =

∴ (gof-1) (2)   =

(ii)  (gof) (x – 1) = g (f (x – 1))

= g (3(x – 1) – 2)

= g (3x – 3 – 2)

= g (3x – 5)

= (3x – 5)2 + 1

= 9x2 – 30x + 25 + 1

= 9x2 – 30x + 26

∴(gof) (x – 1) = 9x2 – 30x + 26

##### 4. Let f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)-1 = f-1og-1

Sol:

Given f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}

⟹ f -1 = {(a, 1), (c, 2), (d, 4), (b, 3)} and g = {(a, 2), (b, 4), (c, 1), (d, 3)}

Now gof = {(1, 2), (2, 1), (4, 3), (3, 4)}

(gof)-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(1)

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

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

from (1) and (2)

(gof)-1 = f-1og-1

##### 5. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5 then find (fog)-1 (x)

Sol:

Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5

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

= f (x3 + 5)

= 2 (x3 + 5) – 3

= 2 x3 + 10 – 3 = 2x3 + 7

Let y = (fog) (x)

⟹ x = (fog)-1(y)

y = 2x3 + 7

2x3 = y – 7

x3 =

x =

(fog)-1(y)=

(fog)-1(x)=

##### 6. Let f(x) = x2, g(x) = 2x then solve the equation (fog) (x) = (gof) (x)

Sol:

Given f(x) = x2, g(x) = 2x

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

= f (2x)

= (2x)2

= 22x

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

= g (x2)

=

Now

(fog) (x) = (gof) (x)

⟹ 22x =

2x = x2   [∵ if am = an , then m = n]

x2 – 2x = 0

⟹ x (x – 2) = 0

⟹x = 0 or x – 2 = 0

∴ x = 0 or x = 2

##### 7.  If f (x) = , x ≠ ±1 then find (fofof) (x) and (fofofof) (x)

Sol:

Given f (x) = , x ≠ ±1

 PDF Files || Inter Maths 1A &1B || (New)6th maths notes|| TS 6 th class Maths ConceptTS 10th class maths concept (E/M)Ts Inter Maths IA ConceptTS 10th Class Maths Concept (T/M)

## Functions Exercise 1a Solutions

Functions Exercise 1a

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1a Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

### Exercise 1a

#### I.

##### 1. If the function f is defined by

then find the values of (i) f (3)    (ii) f (0)      (iii) f (– 1.5)       (iv) f (2) + f (– 2)       (v) f (– 5 )

Sol:

Given

Domain of f(x) is (– 3, ∞)

(i) f (3)

3 lies in the interval x > 1

⟹ f(x) = x + 2

f(3) = 3 + 2 = 5

∴ f (3) = 5

(ii) f (0)

0 lies in interval – 1 ≤ x ≤ 1

⟹f(x) = 2

∴ f (0) = 2

(iii) f (– 1.5)

– 1.5 lies in interval – 3 < x < – 1

⟹f(x) = x – 1

f (– 1.5) = – 1.5 – 1 = – 2.5

∴ f (– 1.5) = – 2.5

(iv) f (2) + f (– 2)

2 lies in the interval x > 1

⟹ f (x) = x + 2

f (3) = 2 + 2 = 4

f (2) = 4

– 2 lies in interval – 3 < x < – 1

⟹f(x) = x – 1

f (– 2) = – 2 – 1 = – 3

f (– 2) = – 2 – 1 = – 3

now f (2) + f (– 2) = 4 – 3 = 1

∴ f (2) + f (– 2) = 1

(v) f (– 5)

since domain of f(x) is (– 3, ∞)

f (– 5) is not defined

##### 2. If f: R – {0} ⟶ R is defined by f(x) = , then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) =

f (1/x)  =

Now

f (x) + f (1/x)  =

∴ f (x) + f (1/x)  = 0

##### 3. If f: R ⟶ R is defined by f(x) = , then show that f (tan θ) = cos 2θ

Sol:

Given f: R ⟶ R is defined by f(x) =

f (tan θ) =

= cos 2θ

##### 4. If f: R – {±1} ⟶ R is defined by f(x) = , then show that f  = 2 f (x)

Sol:

Given f: R – {±1} ⟶ R is defined by f(x) =

##### 5. If A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection (onto function) defined by f(x) = x2 + x + 1, then find B

Sol:

Given A = {– 2, – 1, 0, 1, 2} and   f: A ⟶ B is a surjection defined by f(x) = x2 + x + 1

f(– 2) = (–2)2 + (–2) + 1

= 4 – 2 + 1 = 3

f(– 1) = (–1)2 + (–1) + 1

= 1 – 1 + 1 = 1

f(0) = (0)2 + (0) + 1

= 0 + 0 + 1 = 1

f(1) = (1)2 + (1) + 1

=1 +1 + 1 = 3

f( 2) = (2)2 + (2) + 1

= 4 + 2 + 1 = 4

∴ B = {1, 3, 7}

TS 10th class maths concept (E/M)

##### 6. If A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = , then find range of f.

Sol:

Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =

##### 7. If f (x + y) = f (xy) ∀ x, y ∈ R, then prove that f is a constant function

Sol:

Given f (x + y) = f (xy) ∀ x, y ∈ R

Let x = 0 and y = 0

f (0 + 0) = f (0 × 0) = f (0)

f (1) = f (0 + 1)

= f (0 × 1)

= f (0)

f (2) = f (1 + 1)

= f (1 × 1)

= f (1)

= f (0)

f (3) = f (1 + 2)

= f (1 × 2)

= f (2)

= f(0)

Similarly, f(4) = 0

f(5) = 0

and so on.

∴ f is a constant function

II.

##### 1. If A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2, g(x) = x3, which of the following are surjections?

(i) f: A ⟶ A                 (ii) g: A ⟶ A

Sol:

###### (i) Given A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2

A = {– 1, 0, 1}; f: A ⟶ A

f (x) = x2

f (– 1) = (– 1)2 = 1

f (0) = (0)2 = 0

f (1) = (1)2 = 1

∵ range is not equal to co domain of f

f is nor a surjection

###### (ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x3

A = {– 1, 0, 1}; g: A ⟶ A

g (x) = x3

g (– 1) = (– 1)3 = – 1

g (0) = (0)3 = 0

g (1) = (1)3 = 1

∵ range is equal to co domain of f

f is a surjection

#### 2. Which if the following are injection, surjection or bijection? Justify your answer

##### (i) f: R ⟶ R defined by f(x) =

let x1, x2 ∈ R

f(x1) = f(x2)

2x1 + 1 = 2x2 + 1

2x1 = 2x2

x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y =

3y = 2x + 1

3y – 1 = 2x

⟹ x =  ∈ R

Now

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (ii) f: R ⟶ (0, ∞) defined by f(x) = 2x

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y = 2x

x =   ∈ (0, ∞)

Now f(x) = 2x

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (iii) f: (0, ∞) ⟶ R defined by f(x) =

let x1, x2 ∈ (0, ∞)

f(x1) = f(x2)

x1 = x2

x1, x2 ∈ (0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y =

x = ey ∈ (0, ∞)

Now f(x) =

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (iv) f: [0, ∞) ⟶ [0, ∞) defined by f(x) = x2

let x1, x2 ∈ [0, ∞)

f(x1) = f(x2)

x1 = x2 [∵ x1, x2 ∈ [0, ∞)]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y = x2

x =  ∈ [0, ∞)

Now f(x) = x2

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (v) f: R ⟶ [0, ∞) defined by f(x) = x2

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = ± x2 [∵ x1, x2 ∈ R]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

Let y = f(x)

y = x2

x = ∈ R

Now f(x) = x2

=

= y

∴ f is surjection

f is not an injection but surjection

∴ f is not a bijection

##### (vi) f: R ⟶ R defined by f(x) = x2

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = ± x2 [∵ x1, x2 ∈ R]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

f (1) = 12 = 1

f (– 1) = (–1)2 = 1

here ‘– 1’ has no pre image

∴ f is not a surjection

f is not an injection and not surjection

∴ f is not a bijection

Ts Inter Maths IA Concept

hai

#### 3. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}? If this is given by the formula g(x) = ax + b, then find a and b

Sol:

Given A = {1, 2, 3, 4}, B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

g (1) = 1; g(2) = 3; g(3) = 5 ; g(4) = 7

here, every element of set A has a unique image in set B

∴ g: A ⟶ B is a function

And also given g(x) = ax + b

g (1) = 1

⟹ a (1) + b = 1

a + b = 1

b = 1 – a _______________   (1)

g (2) = 3

a (2) + b = 3

2a + b = 3

2a + 1 – a = 3 (from (1))

a+ 1 = 3

a = 2

b = 1 – 2

b = – 1

∴ a = 2, b = – 1

#### 4. If the function f: R ⟶ R defined by f(x) = , then show that f (x + y) + f (x – y) = 2 f (x) f (y).

Sol:

Given the function f: R ⟶ R defined by f(x) =

= 2 f (x) f (y)

5. If the function f: R ⟶ R defined by f(x) = , then show that f (1 – x) = 1 – f (x)

and hence reduce the value of

Sol:

Given the function f: R ⟶ R defined by f(x) =

∴ f (1 – x) = 1 – f(x)

#### 6. If the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection, then find a and b

Sol:

Given the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection

Case(i)

If f(– 1) = 0 and f(1) = 2

a (– 1) + b = 0

– a + b = 0

b = a ————(1)

and

a (1) + b = 2

a + b = 2

a + a = 2         [ from (1)]

2a = 2

a = 1

b = 1

Case (ii)

If f(– 1) = 2 and f(1) = 0

a (– 1) + b = 2

– a + b =

b = 2 + a ————(2)

and

a (1) + b = 0

a + b = 0

a + 2 + a = 0         [ from (2)]

2a + 2 = 0

2a = – 2

a = – 1

b = 1

From Case(i) and  Case (ii) a = ±1 and b = 1

##### 7. If f(x) = cos (log x), then show that = 0

Sol

Given f(x) = cos (log x)

##### =

cos (log x) cos (log y) – [cos (log x) cos (log y) – sin (log x) sin (log x)

+ cos (log x) cos (log y) + sin (log x) sin (log x)]

= cos (log x) cos (log y) – [2cos (log x) cos (log y)]

= cos (log x) cos (log y) – cos (log x) cos (log y)

∴    = 0

## polycet solved previous qp 2020 TS Polycet || Solved Previous Question Papers 2020 Math

polycet solved previous qp 2020

The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology.

The Main Subjects in this Exam are Maths, Physics, Chemistry and Biology. Here we are providing Previous Maths  Papers Questions and Solutions.

The syllabus of  Maths, Physics, Chemistry and Biology for POLYCET-2022 is the same as that of SSC Examination conducted by the Board of Secondary Education, Telangana.

TS Polycet || Solved Previous Question Papers 2021 Mathematics gives an idea becuase,  it is very helpful to solve the problems in POLYCET entence  examination

TS VI – IX Maths Concepts

### polycet solved previous qp 2020

#### Chapter 1: Real Numbers

1.If 7 divides a2 then

a2  ను 7 భాగించినచో

(1)  7 divides a (a ను 7 భాగిస్తుంది)

(2) 7 divides ( ను 7 భాగిస్తుంది)

(3) a divide 7 (7 ను a భాగిస్తుంది)

(4) none (ఏదీ కాదు)

2. In the formula , which of the following is true?

అయిన, ఈ క్రింది వాటిలో ఏది సత్యము.

(1)  x > 0, y > 0, a = 1

(2) x < 0, y < 0, a = 1

(3)  a > 0, y > 0, x = 1

(4) x > 0, y > 0, a ≠ 1

` `

##### 3. 5 =____________

` `

##### (3)  5                  (4) 2

` `

###### (3) c2       (4) None(ఏది కాదు)

` `

TS 10th class maths concept (E/M)

TS 6th Class Maths Concept

TS 10th Class Maths Concept (T/M)

## Chapter 2: Sets

1.Cardinal number of set A = {P, O, L, Y, T, E, C, H, N, I, Q} , B = {P, O, L, Y, C, E, T, 2020}, then B – A =

A = {P, O, L, Y, T, E, C, H, N, I, Q} , B = {P, O, L, Y, C, E, T, 2020}, అయిన B – A =

(1)  {20}

(2) {2020}

(3) {40}

` `

##### 2. If A = {a} and B = {a, b} , C =  {a, b, c}, then A ∩ B ∩ C  =

A = {a} and B = {a, b} , C =  {a, b, c}, అయిన A ∩ B ∩ C  =

(1) {a}                  (2)  {b}

(3) {c}                  (4) {2021}

` `

Ts Inter Maths IA Concept

TS 10th class maths concept (E/M)

TS 6th Class Maths Concept

### Chapter 3: Polynomials

##### (3) 2                    (4) 3

` `

` `

### Chapter 4: Linear equations in Two Variables

` `

##### 7x + 5y = 12 మరియు 5x – 7y = – 2 సమీకరణాల సాధన ఈ క్రిది వానీలో దేనికి సమాన కాదు

` `

3. If  then (x, y) =

అయిన,(x, y) =

(1) (2019, 2020)         (2) (2020, 2019)

(3)  (2019, 2019)       (4) (2020, 2020)

` `

` `

` `

##### x + y = , x – y = 0, అయిన x =

` `

TS 6th Class Maths Concept

TS 10th class maths concept (E/M)

Ts Inter Maths IA Concept

1.If the roots of 2x2 + kx + 3 = 0 are real and equal, then k =

2x2 + kx + 3 = 0 యొక్క మూలాలు వాస్తవాలు మరియు సమానాలు అయిన k విలువ =

(1)± 6                       (2) ± 4

(2)± 2                      (4) ± 5

` `

###### 2. 8x2 – 6x – 9 = 0= ______

(1)(2x – 3) (x – 3)         (2) (2x – 3) (x +1)

(3) (2x + 1) (x – 1)            (4) (2x – 3) (4x + 3)

` `

##### 4. Roots of  5x2 – 8x = 4 are

5x2 – 8x = 4   యొక్క మూలాలు

(1) 2,                           (2) 1,

(3)   2,                               (4) 2, 7

` `

TS 6th Class Maths Concept

TS 10th class maths concept (E/M)

#### Chapter 6: Progressions

###### 1. 1, -1/2, 1/4…. Are in G.P, then find 8th term

1, -1/2, 1/4….  అనే గుణ శ్రేడిలోని ఎనిమిదవ పదం

(1) 1/128                    (2) 1/64

(3) – 1/128                 (4) –1/64

` `

##### 2. 4, 7, 10… are in AP, the sum 15 terms is____

4, 7, 10… A.P లో ఉన్నచో 15 పదాల మొత్తం ____

(1) 385                   (2) 475

(3)  375                   (4) 325

` `

###### 3. 10th term of AP: 13, 8, 3, – 2, …. is

13, 8, 3, – 2, …. అను అంకశ్రేడిలోని 10 వ పదం

(1) – 32                  (2) – 23

(3) 30                       (4) – 30

` `

###### 4. Which term of GP:  is 729?

అనే గుణ శ్రేడిలో  729 ఎన్నవ పదం?

(1) 10             (2) 12

(3) 14               (4) 16

` `

#### Chapter 7: Coordinater Geometry

##### 1.If the slope of the line through (2, – 7) and (x, 5) is 3 then x =_________

(2, – 7), (x, 5) ల గుండా పోవు రేఖ వాలు 3 అయిన x యొక్క విలువ _____

(1) 4                 (2) 5

(3) 6                   (4) 7

` `

###### 2. If (8, 1), (k, – 4), (2, – 5) are collinear, then k = ______

(8, 1), (k, – 4), (2, – 5) లు సరేఖీయాలైన k యొక్క విలువ ______

(1) 4                   (2) 3

(3) 2                      (4) 1

` `

##### 3. The point (2, – 3) divides the line segment joining the points (– 1, 3), (4, – 7) in the ratio__

(– 1, 3), (4, – 7) బిందువులతో ఏర్పడు రేఖా ఖండాన్ని (2, – 3) బిందువు విభజించు నిష్పత్తి___

(1) 3: 2              (2) 2 : 3

(3) 8 : 1                         (4) 1 : 4

` `

###### 4. The centroid of the triangle whose vertices are (3, – 5), (–7, 4), (10, –2) is

(3, – 5), (–7, 4), (10, –2) లు శీర్శాలుగా గల త్రిభుజం యొక్క గురుత్వ కేంద్రం____

(1) (1, 1)                  (2) (1, – 2)

(3) (–2, 1)               (4) (2, –1)

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## ts polycet 2021 || Solved Previous Question Papers 2021 Maths

The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology.

The Main Subjects in this Exam are Mathematics, Physics, Chemistry and Biology. Here we are providing Previous Maths  Papers Questions and Solutions.

The syllabus of  Mathematics, Physics, Chemistry and Biology for POLYCET-2022 is the same as that of SSC Examination conducted by the Board of Secondary Education, Telangaana.

### ts polycet 2021

TS Polycet || Solved Previous Question Papers 2021 Maths gives an idea to solve the problems in POLYCET entence  examination

### ts polycet 2021

#### Chapter 1: Real Numbers

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##### (1) 3               (2) 0           (3)  2          (4)   1

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###### అనునది____________

(1) Natural numbers (సహజ సంఖ్య)

(2) Irrational number (కరణీయ సంఖ్య)

(3) Rational Number (అకరణీయ సంఖ్య)

(4)   An Integer (పూర్ణ సంఖ్య)

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##### (1)  1               (2) 2              (3) 3              (4) 4

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## Power point Presentations

Power point presentations

A communication device that relays a topic to an audience in the form of a slide show, a demonstration, a lecture or speech where words and pictures intend to complement each other.

### Different Types of  Power Point Presentations:

1) Informative Presentations
2) Instructive Presentations
3) Persuasive Presentations
4) Motivational Presentations
5) Decision-making Presentations
6) Progress Presentations

## TS TET syllabus 2022 For Mathametic Paper 2

TS Tet 2022 syllabus For Mathametic Paper 2 is very useful for writing candidates for TS TET. Knowing the syllabus will make the practice of the candidates easier.

### I.    సంఖ్యా వ్యవస్థ ( Number System)

1 . ప్రధాన సంఖ్యలు మరియు సంయుక్త సంఖ్యలు (Prime and Composite numbers)

2. భాజనీయత సూత్రాలు (Test of divisibility Rules)

3.  సంఖ్యల రకాలు (types of numbers)

4. వాస్తవ సంఖ్యలు (Real numbers)

5. భిన్నాలు మరియు దశాంశ భిన్నాలు (Fractions and Decimal Fractions)

6. క .సా .గు .  మరియు గ . సా .భా. యూక్లిడ్ భాగహార పద్ధతి (LCM and HCF – Euclid division Lemma)

7. వర్గాలు – వర్గ మూలాలు (Squares – Square roots)

8. ఘనాలు  – ఘన మూలాలు (CubsCube roots

9. సంఖ్యల అమరిక మరియు సంఖ్యల ఫజిల్ (Numbers Pattern and Numbers puzzle)

10. యూక్లిడ్ భాజనీయత సూత్రం  (Euclid Division lemma)

11. సంవర్గమానము (Logarithms)

 Ts Inter Maths IA Concept Ts Inter Maths 1B Concept TS 10th Class Maths Concept (T/M) TS Inter 2nd Year Maths 2A Concept TS Inter 2nd Year Maths 2B Concept TS 6th Class Maths Concept TS 10th class maths concept (E/M)

## TS Intermediate First Year Maths Question Papers 2022 PDF

As per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Model papers  2022 to do help the intermediate First-year Maths students.

TS Inter Maths 1A and 1B Modelpapers  2022  as per reduced syllabus are very useful in IPE examinations.

 Ts Inter Maths IA Concept Ts Inter Maths 1B Concept TS 10th Class Maths Concept (T/M) TS Inter 2nd Year Maths 2A Concept TS Inter 2nd Year Maths 2B Concept TS 6th Class Maths Concept TS 10th class maths concept (E/M)

## Inter Maths 1A Model Papers

TS Inter Maths 1A Model papers  2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Model papers  2022 to do help the intermediate First-year Maths students.

Ts Inter Maths IA Concept

#### Inter Maths 1A  Model Papers – I

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Ts Inter Maths 1A Concept

### Inter Maths 1A  Model Papers – II

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### Inter Maths 1A  Model Papers – III

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#### Inter Maths 1A  Model Papers – IV

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##### Inter Maths 1B Model Papers

TS Inter Maths 1B Model papers  2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Model papers  2022 to do help the intermediate First-year Maths students.

TS 6th Class Maths Concept

### TS Inter 1st Year Maths Model Papers PDF – 2022

#### Inter Maths 1B  Model Papers – I

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#### Inter Maths 1B  Model Papers – II

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#### Inter Maths 1B  Model Papers – III

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### Inter Maths 1B  Model Papers – IV

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 Ts Inter Maths IA Concept Ts Inter Maths 1B Concept TS 10th Class Maths Concept (T/M) TS Inter 2nd Year Maths 2A Concept TS Inter 2nd Year Maths 2B Concept TS 6th Class Maths Concept TS 10th class maths concept (E/M)

## 10 Maths ICSE Banking MCQS With Answers

ICSE Banking MCQS

A) ₹ 200

B) ₹ 600

C) ₹  500

D) ₹ 300

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A) ₹ 7865

B) ₹ 12875

C) ₹875

D) ₹13865

### 3.  Meena opens a recurring deposit account in a bank and deposits ₹ 600  for 20 months, find the maturity value of this account if the bank pays interest at the rate of 10% per annum

A) ₹ 78650

B) ₹ 2350

C) ₹12350

D) ₹13050

Ts Inter Maths IA Concept