2022

Ch. 2 Mathematical Indunction Exerxise Solutions

By Anitha@Satyanarayana Reddy / September 16, 2022 / Education

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.

Laplace

“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 =

Mathematical Induction 5

Mathematical Induction 6

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

a_n = 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

a_n = 3 + (n – 1) d

= 3 + (n – 1) (1)

= 3 + n – 1

= n + 2

n^th 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)

Mathematical Induction 9

Mathematical Induction 10

Mathematical Induction 11

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 = = Mathematical Induction 16 =

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 Mathematical Induction 22 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

a_n = a + (n – 1) d

= 4 + (n – 1)4

= 4 + 4n – 4

= 4n

n^th 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 =

n^th 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]

Mathematical Induction 28

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

∴ By the principle of Mathematical induction

P(n) is true for all n ∈ N

Mathematical Induction

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Functions Exercise 1c Solutions ||TS|| Basics In MAths

By Anitha@Satyanarayana Reddy / August 29, 2022 / Education

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| = e^y

∵ e^y > 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) = Functions 1(c) images 15

Given function is f (x) = Functions 1(c) images 15

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, ∞)

3. If f and g are real valued functions f(x) = 2x – 1 ang g (x) = x² then

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)

Functions 1(c) images 20

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)

Functions 1(c) images 22

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) = Functions 1(c) images 26

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 > 0

x ∈ (0, ∞)

∴ Domain of f(x) is (0, ∞)

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

Sol:

Given f(x) =

Functions 1(c) images 31

Functions 1(c) images 32

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

Functions 1(c) images 36

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, ∞)

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Functions Exercise 1b Solutions

By Anitha@Satyanarayana Reddy / August 16, 2022 / Education

Functions Exercise 1b Solutions

Functions Exercise 1b

The famous mathematician ” Lejeune Dirichlet” defined a function.

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) = e^x and g(x) = , then show that f og = gof and f^-1 = g^-1

Sol:

Given f (x) = e^x and g(x) =

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

= f ( )

= x ————- (1)

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

= g (e^x)

= x

= x ————- (2)

From (1) and (2) f og = gof

let y = f(x)

x = f^-1(y)

y = e^x

x =

f^-1 (x) =

let g(x) = z

x = g^-1(z)

z =

x = e^z

g^-1(x) = e^x

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

Sol:

Given f (y) = , g (y) =

Now

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

Functions 1(b) images 8

∴ fog(y) = y

3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x² + 3 ang g (x) = 3x – 2 then find

(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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TS 10th class maths concept (E/M)

(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

4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x² + 1, then find

(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

Functions 1(b) images 14

∴ 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) = 5^x

Let y = f(x)

x = f^-1(y)

now y = 5^x

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 = 2^y

f^-1(y) = 2^y

∴ f^-1(x) = 2^x

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) = 2^{x (x – 1}⁾ then find f^-1 (x)

Sol:

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

Let y = f(x)

x = f^-1(y)

now y = 2^{x (x – 1)}

= x (x – 1)

x² – x =

x² – x – = 0

Functions 1(b) images 28

Functions 1(b) images 29

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

Functions 1(b) images 33

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

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

(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) =

Functions 1(b) images 36

∴ (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 = Functions 1(b) images 39

(fog)^-1(y)= Functions 1(b) images 39

(fog)^-1(x)= Functions 1(b) images 40

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

Sol:

Given f(x) = x², g(x) = 2^x

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

= f (2^x)

= (2^x)²

= 2^2x

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

= g (x²)

Now

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

⟹ 2^2x =

2x = x² [∵ if a^m = 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) = Functions 1(b) images 43 , x ≠ ±1

Functions 1(b) images 42

Functions 1(b) images 44

PDF Files || Inter Maths 1A &1B || (New)

6th maths notes|| TS 6 th class Maths Concept

TS 10th class maths concept (E/M)

Ts Inter Maths IA Concept

TS 10th Class Maths Concept (T/M)

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chapter 1 Functions Exercise 1a Solutions

By Anitha@Satyanarayana Reddy / August 10, 2022 / Education

Functions Exercise 1a Solutions

Functions Exercise 1a

The famous mathematician ” Lejeune Dirichlet” defined a function.

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

Functions 1(a) images 1

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 Functions 1(a) images 1

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

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) = Functions 1(a) images 2

f (1/x) = Functions 1(a) images 3

Now

f (x) + f (1/x) = Functions 1(a) images 4

∴ 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) = Functions 1(a) images 5

f (tan θ) = Functions 1(a) images 6

= cos 2θ Functions 1(a) images 7

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) =

Functions 1(a) images 10

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)

Functions Exercise 1a

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) = Functions 1(a) images 11

Functions 1(a) images 12

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

PDF Files || Inter Maths 1A &1B || (New)

6th maths notes|| TS 6 th class Maths Concept

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)

Functions 1(a) images 14

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

Functions 1(a) images 17

∴ f is surjection

f is injection and surjection

∴ f is a bijection

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

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 = 2^x

x = ∈ (0, ∞)

Now f(x) = 2^x

= 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 = e^y ∈ (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) = Functions 1(a) images 29

Functions 1(a) images 30

= 2 f (x) f (y)

5. If the function f: R ⟶ R defined by f(x) = Functions 1(a) images 31 , then show that f (1 – x) = 1 – f (x)

and hence reduce the value of Functions 1(a) images 32

Sol:

Given the function f: R ⟶ R defined by f(x) = Functions 1(a) images 31

Functions 1(a) images 33

∴ f (1 – x) = 1 – f(x)

TS 10th Class Maths Concept (T/M)

TS 10th class maths concept (E/M)

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)

Functions 1(a) images 37

Functions 1(a) images 38

=

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)

∴ Functions 1(a) images 36 = 0

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Maths polycet solved previous qp 2020- || Solved Papers

By Anitha@Satyanarayana Reddy / June 14, 2022 / Education

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)

Answer: (1)

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

Answer: (4)

Solution

3. 5 =____________

Answer: (3)

Solution

4. then x =

అయిన, x =

(1) n (2) 1

(3) 5 (4) 2

Answer: (4)

Solution

5. , ac = ___

అయిన, ac = ___

(1)a² (2)b²

(3) c² (4) None(ఏది కాదు)

Answer: (2)

Solution

TS 10th class maths concept (E/M)

TS 6th Class Maths Concept

TS 10th Class Maths Concept (T/M)

AS Tutorial In Telugu - QR code

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}

(4) none (ఏది కాదు)

Answer: (2)

Solution

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}

Answer: (1)

Solution

Ts Inter Maths IA Concept

TS 10th class maths concept (E/M)

TS 6th Class Maths Concept

AS Tutorial In Telugu - QR code

Chapter 3: Polynomials

1. Product of the polynomials (x³ – 8), (x – 8) is denoted by p(x) = ax⁴ + bx³ + c x² + dx +e, then p (8) =

(x³ – 8), (x – 8) అను బహుపదుల లబ్దము p(x) = ax⁴ + bx³ + c x² + dx +e అయిన, p (8) =

(1) 0 (2) 1

(3) 2 (4) 3

Answer: (1)

Solution

2. If α, β are the zeroes of x² – 1 , α + β =

α, β లు x² – 1అనే వర్గ బహుపదికి శూన్యాలు అయితే α + β విలువ?

(1) 0 (2) 1

(3) – 1 (4) 2

Answer: (1)

Solution

AS Tutorial In Telugu - QR code

Chapter 4: Linear equations in Two Variables

1.For the equation 2019x + 2020y = 4040, when x= 0 the value of y =

2019x + 2020y = 4040 అను సమీకరణమునకు x= 0 అయిన, y విలువ

(1) 2020 (2) 2019

(3) 4 (4) 2

Answer: (4)

Solution

2. Solution of the equations 7x + 5y = 12 and 5x – 7y = – 2, not equal to

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

Answer: (1)

Solution

3. If TS POLYCET 2020 - Linear equations in 2 variables 2 then (x, y) =

TS POLYCET 2020 - Linear equations in 2 variables 3 అయిన,(x, y) =

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

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

Answer: (1)

Solution

4. If (5, 2) is the solution 2x + 3y = 20, ax – by = 0, the (a, b) =

2x + 3y = 20, ax – by = 0 ల సాధన (5, 2) అయిన (a, b) =

(1) (2, 5) (2) (5, 2)

(3) (– 2, 5) (4) (– 5, 2)

Answer: (1)

Solution

5. If the system of equations x – y = 1 and ax + y = 2 has unique solution then

జత సమీకరణాలకు x – y = 1, ax + y = 2 లకు ఏకైక సాధన ఉంటే

(1) a = 1 (2) a = – 1

(3) a ≠ 1 (4) a ≠– 1

Answer: (4)

Solution

6. x + y = , x – y = 0, then x =

x + y = , x – y = 0, అయిన x =

Answer: (3)

Solution

TS 6th Class Maths Concept

TS 10th class maths concept (E/M)

Ts Inter Maths IA Concept

AS Tutorial In Telugu - QR code

Chapter 5: Quadratic Equations

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

Answer: (3)

Solution

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)

Answer: (4)

Solution

4. Roots of 5x² – 8x = 4 are

5x² – 8x = 4 యొక్క మూలాలు

(1) 2, TS POLYCET 2020 - Quadratic Equations 9 (2) 1,

(3) 2, (4) 2, 7

Answer: (1)

Solution

TS 6th Class Maths Concept

TS 10th class maths concept (E/M)

AS Tutorial In Telugu - QR code

Chapter 6: Progressions

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

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

(1) 1/128 (2) 1/64

(3) – 1/128 (4) –1/64

Answer: (3)

Solution

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

Answer: (3)

Solution

3. 10^th 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

Answer: (1)

Solution

4. Which term of GP: is 729?

అనే గుణ శ్రేడిలో 729 ఎన్నవ పదం?

(1) 10 (2) 12

(3) 14 (4) 16

Answer: (2)

Solution

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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

Answer: (3)

Solution

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

Answer: (2)

Solution

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

Answer: (1)

Solution

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)

Answer: (4)

Solution

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ts polycet 2021 | Solved Previous Question Papers 2021 Maths

By Anitha@Satyanarayana Reddy / June 7, 2022 / Education

ts polycet 2021

ts polycet 2021 || Solved Previous Question Papers 2021 Maths

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

1. The value of is

యొక్క విలువ ఎంత ?

TS POLYCET 2021 - Real Numbers 1

Answer: (2)

Solution

2.The HCF of 8, 9 and 25 is

8, 9 మరియు 25 గ. సా. కా ఎంత?

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

Answer: (4)

Solution

3. is ____________

అనునది____________

(1) Natural numbers (సహజ సంఖ్య)

(2) Irrational number (కరణీయ సంఖ్య)

(3) Rational Number (అకరణీయ సంఖ్య)

(4) An Integer (పూర్ణ సంఖ్య)

Answer: (2)

4. If 2^x = 8² then x =?

2^x = 8² అయిన x =?

(1) 4 (2) 2 (3) 6 (4) 8

Answer: (3)

Solution

5. The value of is

యొక్క విలువ

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

Answer: (2)

Solution

<< Page 1 of 14 >>

Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14

feture images for power point presentation

Power point Presentations

By Anitha@Satyanarayana Reddy / May 8, 2022 / Education

Power point presentations

About a Presentation?

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

Real numbers Mcqs

Pages: 1 2 3 4

TS TET syllabus 2022 For Mathametic Paper 2

By Anitha@Satyanarayana Reddy / April 22, 2022 / Education

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.

TS Tet 2022 Syllabus For Mathametic Paper 2 TS TET రాసె అభ్యర్థులు చాలా ఉపయోగపడుతుంది. సిలబస్ తెలియడం వల్ల అభ్యర్థుల అభ్యసన సులువు అవుతుంది.

free Syllabus

Telangana Teachers Elgible Test Official Website : Click Here

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 (Cubs – Cube 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)

Pages: 1 2 3 4 5 6

TS Inter Maths 1A and 1B Model Papers 2022

TS Intermediate First Year Maths Question Papers 2022 PDF

By Anitha@Satyanarayana Reddy / April 15, 2022 / Education

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

TS Inter 1st Year Maths Model Papers PDF – 2022

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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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)

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10 th Maths ICSE Banking MCQS With Answers

By Anitha@Satyanarayana Reddy / January 22, 2022 / Education, MCQ'S

10 Maths ICSE Banking MCQS With Answers

ICSE Banking MCQS

TS 10th class maths concept (E/M)

ICSE Banking MCQS

1. Mr Gupta gets ₹ 6,455 at the end of the one year at the rate of 14% per annum in a recurring deposit account find the monthly instalment

A) ₹ 200

B) ₹ 600

C) ₹ 500

D) ₹ 300

Show Answer

2. Rahul deposited ₹ 500 every month in a recurring deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is

A) ₹ 7865

B) ₹ 12875

C) ₹875

D) ₹13865

Show Answer

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

Show Answer

Ts Inter Maths IA Concept

TS 6th Class Maths Concept

Mathematical Indunction (M.I) Exerxise wise Solutions

Exercise 2(a)

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

1. 12 + 22 + 32 + …… + n2=

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

3 .

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

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

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Functions Exercise 1c Solutions

Functions Exercise 1c

I.

1.Find the domains of the following real valued functions

(ii)

2. find the ranges of the following real valued functions

3. If f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2 then

find

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

II.

1.Find the domain of the following real valued functions

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

3. Find the domain and range of the following functions

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Functions Exercise 1b Solutions

Functions Exercise 1b

I.

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

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

3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x2 + 3 ang g (x) = 3x – 2 then find

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

4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1, then find

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

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

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

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

8. Find the inverse of the following functions

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

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

II.

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

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

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

(i) (gof-1) (2) (ii) (gof) (x – 1)

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

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)

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

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

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Functions Exercise 1a Solutions

Exercise 1a

I.

1. If the function f is defined by

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

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

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

Functions Exercise 1a

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

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

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

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

(ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x3

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

(i) f: R ⟶ R defined by f(x) =

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

(iii) f: (0, ∞) ⟶ R defined by f(x) =

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

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

(vi) f: R ⟶ R defined by f(x) = x2

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

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).

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

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

=

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

polycet solved previous qp 2020

Chapter 1: Real Numbers

3. 5 =____________

1. 1² + 2² + 3² + …… + n²=

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

3. If f and g are real valued functions f(x) = 2x – 1 ang g (x) = x² then

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

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

3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x² + 3 ang g (x) = 3x – 2 then find

4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x² + 1, then find

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

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

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

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

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

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

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

(i) (gof^-1) (2) (ii) (gof) (x – 1)

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

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)

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

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

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

(ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x³

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

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

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

(vi) f: R ⟶ R defined by f(x) = x²

(1)a² (2)b²

(3) c² (4) None(ఏది కాదు)

1. Product of the polynomials (x³ – 8), (x – 8) is denoted by p(x) = ax⁴ + bx³ + c x² + dx +e, then p (8) =

(x³ – 8), (x – 8) అను బహుపదుల లబ్దము p(x) = ax⁴ + bx³ + c x² + dx +e అయిన, p (8) =

2. If α, β are the zeroes of x² – 1 , α + β =

α, β లు x² – 1అనే వర్గ బహుపదికి శూన్యాలు అయితే α + β విలువ?

2. 8x² – 6x – 9 = 0= ______

4. Roots of 5x² – 8x = 4 are

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

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