March, 2021 | Basics In Maths

AS Tutorial For Math Classes / YouTube new no. 1 Channel

By [email protected] Reddy / March 18, 2021 / Education

AS Tutorial తెలుగులో

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In ASTutorialతెలుగులో I will share with you the concepts of Maths in inter.
AS Tutorialతెలుగులో Provides Concepts and Problems for Inter Maths Students In Telugu.

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Simple Equations | LHS to RHS and RHS to LHS, RULE ||AS Tutorial తెలుగులో || 2021|| : https://youtu.be/2Vkb_jYx_YM

Trigonometry// ALL-SILVER-TEA-CUPS Rule ||ASTutorialతెలుగులో || 2021|| :https://youtu.be/UqwfCBsJVMY

‘Locus’ Concept and Problems For Inter First Year Maths 1B||AS Tutorial తెలుగులో || 2021||(PART – 1):https://youtu.be/RAtxfld4ywQ

‘Locus’ Concept and Problems For Inter First Year Maths 1B||AS Tutorial తెలుగులో || 2021||(PART – 2): https://youtu.be/E6IaE09YsbM

Trigonometry Introduction, Sides of Right Triangle and Trigonometry Ratios |ASTutorialతెలుగులో 2021:https://youtu.be/Elt7Uy_yfLU

Trigonometry || How To Prove Specific Angles In Trigonometry| || AS Tutorialతెలుగులో|| || 2021|: https://youtu.be/Fh4Vc66swQQ

Trigonometry| How to remember Specific Angles Table in Trigonometry |ASTutorialతెలుగులో || 2021||: https://youtu.be/D3CZ1ijq7OM

Trigonometry| Trigonometric Identities In Trigonometry|ASTutorialతెలుగులో || 2021|| :https://youtu.be/r1NZyagpN90

‘Transformation Of Axes’ Concept and Problems For Inter Maths 1B||AS Tutorial తెలుగులో || 2021|| P 1: https://youtu.be/9qJJuZzKtJ8

‘Transformation Of Axes’ Concept and Problems For Inter Maths 1B||AS Tutorial తెలుగులో || 2021|| P 2 :https://youtu.be/lAfuMQ1YU3c

Playing With Numbers||Divisibility Rules In Telugu ||AS Tutorial తెలుగులో || || 2021|| : https://youtu.be/8I1izylDH_o

Playing With Numbers||Even| Odd|Prime|Composite|Co-Prime &Twin- Prime Numbers|| AS Tutorial తెలుగులో : https://youtu.be/TZeleb6vd1I

Logarithms|| Product Rule| Quotient Rule| and Power Rule||AS Tutorial తెలుగులో || || 2021||: https://youtu.be/ssYCXUc0Hf8

Periodic Functions| Trigonometric Ratios Up To Transformations ||AS Tutorial తెలుగులో || || 2021|| : https://youtu.be/aeOEDgQpVu0

TS TET Child Developmentt & Pedagogi Part 2: https://youtu.be/qyB5pImDg8c
TS TET Child Developmentt & Pedagogi Part 3: https://youtu.be/m3ui0jFIgmc
TS TET Child Developmentt & Pedagogi Part 4: https://youtu.be/sULWOJctF8k
TS TET Child Developmentt & Pedagogi Part 5: https://youtu.be/J-huGDv7cUs

TS TET 2022, Mathematics Methods, paper 2, Practice Bits p – 1: https://youtu.be/HPBN6WdgbBc

TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 3:https://youtu.be/DZzgHljDklA

TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 4:https://youtu.be/2Bdu_dWIH7U

TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 5: https://youtu.be/INWN_23gXDc

TS TET – 2022: VI Class Maths, Chapter 1, మనసంఖ్యలను తెలుసుకుందాం , Practice Bits:https://youtu.be/wVMzcpNWyPc

TS TET – 2022: VI Class Maths, Chapter 1, మనసంఖ్యలను తెలుసుకుందాం , Practice Bits: https://youtu.be/3-1MEbi6z9E

TS TET – 2022: VI Class Maths, Chapter 2,పూర్ణాంకాలు , Practice Bits: https://youtu.be/Z7w4p2naP7Y

TS TET 2022: VI Class Maths, Chapter 2, Wole Numbers , Practice Bits: https://youtu.be/I2BKs3YBLT8

TS TET – 2022: X Class Maths, Chapter 11, Trigonometry , Practice Bits: https://youtu.be/5xwtndJLhPs

TS TET – 2022: X Class Maths, Chapter 11, త్రికోణమితి , Practice Bits:https://youtu.be/oHbOZPTeWuo

TS TET 2022 , X Class Maths, Chapter 1,Real Numbers , Practice Bits, Practice Bits: https://youtu.be/iUWXfxF3pz4

TS TET 2022,10 th Maths, chapter 1, Real numbres – Logorithms: https://youtu.be/6HoIeoATswY

TS TET 2022, Paper 2, 10 th Maths, chapter 2, SETS, Practicew bits : https://youtu.be/VYypC4xC6Jc
TS TET 2022, Paper 2, 10 th Maths, chapter 14, Statistics, Practicew bits: https://youtu.be/RA6jDQQaRXw

Matrices Basic Concepts: https://youtu.be/_Gqg07ZquN0
Trace of Matrix, Addition of Matrices, Equality of Matrices: https://youtu.be/Hycj-UITfD8
Multiplication of two matrices P – 1:https://youtu.be/rMp96I6ag8s
Multiplication of two matrices P – 2:https://youtu.be/Axu5BbGKkbs
Multiplication of two matrices P – 3:https://youtu.be/1aSVztz9TII
transose of matrices: https://youtu.be/D8nhjYQJqm0
Smmetric and Skew Symmetric Matrices:https://youtu.be/JUv2RT35DaA

Adjoint and Inverse of Matrices P – 1:https://youtu.be/mPnkSkxKaxo

Adjoint and Inverse of Matrices P – 2:https://youtu.be/4fpvFTNpXC0

Adjoint and Inverse of Matrices P – 3:https://youtu.be/Ge-jHQRF4SM

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Reduced Syllabus 2021 Telangana BIE Maths

By [email protected] Reddy / March 3, 2021 / Education

Reduced Syllabus 2021 Telangana BIE Maths

Reduced Syllabus 2021

This content is designed by the ‘Basics in Maths‘ team.

Telangana BIE Maths Reduced Syllabus(2021) very useful I.P.E exam.

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

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

TS 10th class maths concept (E/M)

Inter 1st year Maths Reduced Syllabus	Click Here
Inter 2nd year Maths Reduced Syllabus	Click Here

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Errors and Approximations || V.S.A.Q’S||

By [email protected] Reddy / March 3, 2021 / Education

Errors and Approximations || V.S.A.Q’S||

Errors and approximations: These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

Errors and Approximations

Question 1

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

(i) y = f(x) = x² + x at x = 10 when ∆x = 0.1.

Sol:

Given y = f(x) = x² + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

= f (10 + 0.1) – f (10)

= f (10.1) – f (10)

= (10. 1)² + 10.1 – (10² + 10)

= 102.01 + 10.1 – 100 – 10

= 112.11 – 110

= 2.11

dy = f’ (x) ∆x

= (2x + 1) (0.1)

= [2(10) + 1] (0.1)

= 21 × 0.1

= 2.1

(ii) y = cos x at x = 60⁰ with ∆x = 1⁰ (1⁰ = 0.0174 radians)

Sol:

Given y = cos x, x = 60⁰ and ∆x = 1⁰

∆y = f (x + ∆x) – f (x)

= cos (60⁰ + 1⁰) – cos 60⁰

= cos (61⁰) – 0.5

= 0.4848 – 0.5

= – 0.0152

dy = f’ (x) ∆x

= – sin x (1⁰)

= – sin 60⁰ × 0.0174

=– 0.8660 × 0.0174

= – 0.0150

(iii) y = x² + 3x + 6, x = 10 with ∆x = 0.01

Sol:

y = x² + 3x + 6

∆y = f (x + ∆x) – f (x)

= f (10 + 0.01) – f (10)

= f (10.01) – f (10)

= (10.01)² + 3 (10.01) + 6 – (10² + 3 (10) + 6)

= 100. 2001 + 30.03 + 6 – 100 – 30 – 6

=130. 2301 – 130

= 0.2301

dy = f’ (x) ∆x

= (2x + 3 + 0) (0.01)

= (2× 10 + 3) (0.01)

= 23 × 0.01

= 0.23

(iv) y = , x = 8 and ∆x =0.02

Sol:

Question 2

The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.

Sol:

Let x be the side of the square and the area be A

Area of the square A = x²

x = 3 and ∆x = 0.01

∆A = 2x × ∆x

= 2(3) (0.01)

= 6 × 0.01

= 0.06

Question 3

If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.

Sol:

Let x be the side of the square and A be its area

= 2

A = x²

∆A = 2x × ∆x

The approximate percentage error in area A

= 2 × 2 =4

PDF Files || Inter Mathematics 1A ands 1B

Question 4

From the following. Find the approximations

(i)

Sol:

Let f(x) = , where x = 1000 and ∆x =– 1

f’ (x) =

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= 10 – 0. 0033

= 9.9967

(ii)

Sol:

(iii)

Sol:

(iv) Sin 62⁰

Sol:

Let f(x) = sin x, where x = 60⁰ and ∆x =2⁰

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 60⁰ + cos x (2⁰)

= sin 60⁰+ cos 60⁰ (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

Question 5

The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.

Sol:

Given r = 14 cm and ∆r =0.02cm

Surface area of sphere =A = 4π r²

∆A = 8π r ∆r

= 8 ×3.14× 14 × 0.02

= 7.0336

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Differentiation(2m Q & S) || V.S.A.Q’S||

By [email protected] Reddy / March 2, 2021 / Education

Differentiation(2m Q & S) || V.S.A.Q’S||

Differentiation

This content was designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

Differentiation

Question 1

Find f’ (x) for the following functions

(i) f(x) = (ax + b) (x > -b/a)

Sol:

Given f(x) = (ax + b)ⁿ

f’ (x) = n (ax + b)^{n – 1} (ax + b)

= n (ax + b) ^{n – 1}a

= an (ax + b) ^{n –} 1

(ii) f(x) = x² 2^x log x

Sol:

Given f(x) = x² 2^x log x

f’ (x) = (x²) 2^x log x + x² (2^x) log x + x² 2^x (log x).

= 2×2^x log x +x² 2^x log a log x + x² 2^x (1/x)

= x 2^x[log x² + x log x log 2 + 1]

(iii) f(x) = (x > 0)

Sol:

Given f(x) =

f’ (x) = . log 7 (x³ + 3x)

= log 7 (3x² + 3)

=3 (x² + 1) log 7

(iv) f(x) = log (sec x + tan x)

Sol:

Given, f(x) = log (sec x + tan x)

f’ (x) = (sec x + tan x)

= (sec² x + sec x tan x)

= sec x (sec x + tan x)

= sec x

Question 2

Find the derivative of the following functions

(i) f(x) = e^x (x² + 1)

Sol:

Given f(x) = e^x (x² + 1)

f’ (x) = e^x (x² + 1) + (x² + 1) (e^x)

= e^x (2x + 0) + (x² + 1) e^x

= e^x (x² + 2x + 1)

= e^x (x + 1)²

(ii)

Let y =

(iii) cos (log x + e^x)

(iv) x = tan (e^-y)

e^-y = tan^-1 x

(v) cos [log (cot x)]

(vi) sin[tan^-1(e^x)]

(vii) cos^-1(4x³ – 3x)

let y = cos^-1(4x³ – 3x)

put x = cos θ ⟹ θ = cos^-1 x

y = cos^-1(4 cos³ θ – 3cos θ)

= cos^-1(cos 3θ)

= 3 θ

= 3 cos^-1 x

= 3 (cos^-1 x)

= 3

(viii)

(ix)

(x)

Differentiation

Question 3

Find f’ (x), If f(x) = (x³ + 6 x² + 12x – 13)¹⁰⁰.

Sol:

Given f(x) = (x³ + 6 x² + 12x – 13)¹⁰⁰

f’ (x) = 100(x³ + 6 x² + 12x – 13)⁹⁹ (x³ + 6 x² + 12x – 13)

= 100(x³ + 6 x² + 12x – 13)⁹⁹ (3x² + 12 x + 12 – 0)

=100(x³ + 6 x² + 12x – 13)⁹⁹ 3 (x² + 4 x + 4)

= 300 (x + 2)² (x³ + 6 x² + 12x – 13)⁹⁹

Question 4

If f(x) = 1 + x + x² + x³ + …. + x¹⁰⁰, then find f’ (1).

Sol:

Given f(x) = 1 + x + x² + x³ + …. + x¹⁰⁰

f’(x) = 0 + 1 + 2x + 3 x² + … 100 x⁹⁹

f’(1) = 1 + 2 + 3 + … + 100

= 50 × 101

= 5050

Question 5

From the following functions. Find their derivatives.

Question 6

If y = , find

Sol:

Given y =

Question 7

If y = log (cosh 2x), find

Sol:

Given y = log (cosh 2x)

Question 8

If x = a cos³ t, y = a sin³ t, find

Sol:

Given If x = a cos³ t, y = a sin³ t

Question 9

Differentiate f(x) with respect to g(x) for the following.

(i) f(x) = e^x, g(x) =

f’ (x) = e^x and g’ (x) =

derivative of f(x) with respect to g(x) =

(ii )

put x = tan θ ⟹ θ = tan^-1 x

Question 10

if y = then prove that

Sol:

Given y =

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Practice Makes Maths Perfect

March 2021

AS Tutorial For Math Classes / YouTube new no. 1 Channel

AS Tutorial తెలుగులో

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Reduced Syllabus 2021 Telangana BIE Maths

Reduced Syllabus 2021 Telangana BIE Maths

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Errors and Approximations || V.S.A.Q’S||

Errors and Approximations || V.S.A.Q’S||

Errors and Approximations

Visit my Youtube Channel: Click on Below Logo

Differentiation(2m Q & S) || V.S.A.Q’S||

Differentiation(2m Q & S) || V.S.A.Q’S||

Differentiation

Question 1

Question 2

Differentiation

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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Archive

March 2021

AS Tutorial తెలుగులో

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Reduced Syllabus 2021 Telangana BIE Maths

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Errors and Approximations || V.S.A.Q’S||

Errors and Approximations

Visit my Youtube Channel: Click on Below Logo

Differentiation(2m Q & S) || V.S.A.Q’S||

Differentiation

Question 1

Question 2

Differentiation

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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