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.

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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)

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II.    అంక గణితం (Arithmetic)

1. నిష్పత్తి – అనుపాతం  (Ratio and Proportions)

2. సాదారణ వడ్డీ – బారు వడ్డీ (Simple Interest – Compound Interest)

3. కాలము – దూరం (Time and Distance)

4. కాలము – పని  (Time and Work)

5. లాభము – నష్టము – డిస్కౌంట్   (Profit and Loss – Discount)

6. పన్నులు (Taxes)

III.    సమితులు  ( Sets)

1. సమితి బావన – సమితి భాష (Concept of set – Set language)

2. శూన్య సమితి   (Empty set)

3. ఉప సమితి  (Sub set)

4. పరిమిత – అపరిమిత సమితులు   (Finite and Infinite Sets)

5. సమ సమితులు   (Equality of sets)

6. కార్డినల్ సంఖ్య (Cardinal number of a set)

7. సమితుల పరిక్రియలు  (Set operations)

8. వెన్ చిత్రాలు  (Venn diagrams)

IV.  బీజ గణితం   ( Algebra)

1. బీజ గణిత పరిచయం   (Introduction to Algebra)

2. ఘాతాలు – ఘతాంకాలు    (Exponents – Powers)

3. ప్రత్యేక లబ్ధాలు (Special Products)

4. బహుపదులు   (Polynomials)

5. రేఖీయ సమాసాలు , వాటి గ్రాఫ్లు  (Linear equations , their graphs)

6. వర్గ సమీకరణాలు   (Quadratic Equations)

7. శ్రేడులు   (Progressions)

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V.  రేఖా గణితం    ( Geometry)

1. రేఖా గణిత పరిచయం   (Introduction to Geometry)

2. జ్యామితి అభివృద్ధిలో భారతదేశం యొక్క సహకారం

                  (Contribution of India in the development of geometry )

3. యూక్లిడ్ రేఖా గణితం  (Euclid Geometry)

4. సరళ రేఖలు మరియు కోణాలు  (Lines and Angles)

5. సరూప త్రిభుజాలు   (Similar Triangles)

6.  సర్వ సమాన త్రిభుజాలు  (Congruent Triangles)

7. పైథాగరస్ సిద్దాంతం   (Pythagoras Theorem )

8. వృత్తాలు – వృత్త ధర్మాలు  (Circles – Properties of Circles)

9. వృత్తాల నిర్మాణం     (Construction of circles )

10. త్రిభుజాలు (Triangles)

11.  బహుభుజులు – చతుర్భుజాలు  (Polygons – Quadrilaterals )

12. నిరూపక జ్యామితి    (Coordinate Geometry)

13.  సరళ రేఖలు   (Straight lines)

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VI.  క్షేత్ర మితి   ( Mensuration)

1. పరిధి మరియు వైశాల్యం : ( Perimeter and Area )

     త్రిభుజం(triangle ),చతురస్రం(s quare  ), దీర్ఘ చతురస్రం (rectangle ), వృత్తం (circle ), కంకణం(ring )   

2. ఘనము – దీర్ఘ ఘనము ( Cube  –  Cuboid)

3. స్తూపం   (Cylindre)

4.  శంఖువు  (Cone)

5. గోళము – అర్థ గోళము  (Sphere – Hemi Sphere)

VII.  దత్తాంశ నిర్వహణ    (Data  Handling )

1. దత్తాంశ సేకరణ – వర్గీ కరణ   ( Collection of data – Classification of data )

2.  పౌనఃపున్య పట్టిక   (Frequency Table)

3.  గణన చిహ్నాలు (Tally Marks )

4.  ప ట చిత్రాలు   (Graphs)

5. కేంద్రీయ స్థాన కొలతలు (Measures of Central Tendency)

      అంక గణితం (Arithmetic mean ), మధ్య గతం (Median), బాహులకం ( Mode )

6.  సంభావ్యత  (Probability )

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VIII.  త్రికోణమితి  (  Trigonometry )

1. త్రికోణమితి పరిచయం  ( Introduction of Trigonometry )

2.  త్రికోనమితీయ నిష్పత్తులు  (Trigonometric Ratios)

3.  ప్రత్యేక కోణాల త్రికోనమితీయ నిష్పత్తులు (Specific Trigonometric ratios of angles )

4.  పూరక కోణాల త్రికోనమితీయ నిష్పత్తులు (Trigonometric ratios of complimentary angles )

5.  త్రికోనమితీయ సర్వ సమీకరణాలు  (Trigonometric Identities)

6. త్రికోణమితి అనువర్తనాలు ( Applications  of Trigonometry )

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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 6th Class Maths Concept

TS 10th class maths concept (E/M)

Ts Inter Maths IA Concept

TS 10th Class Maths Concept (T/M)

TS 10th Class Maths Concept (T/M)

Ts Inter Maths IA Concept

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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Inter Maths 1B  Model Papers – II

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Ts Inter Maths 1B Concept

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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 Inter Maths IA Concept

TS 10th Class Maths Concept (T/M)

TS 10th Class Maths Concept (T/M)

Ts Inter Maths IA Concept

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Real Numbers

1. Insert 4 rational numbers between 1 and without using formula

 Sol: we have to insert 4 rational numbers between 1 and  and 1

4 +1 = 5

1 can be written as 

Lcm of (4, 5) = 20

2. The prime factorisation of a natural number (n) is 2³ × 3² × 5² × 7. How many consecutive zeros will it have at the end of it? Justify your answer.

Sol:

 n = 2³ × 3² × 5² × 7

    = 2 × 2² × 3² × 5² × 7

    = 2 × 3² × 7 (5² × 2²)

    = 2 × 3² × 7 × (10)² = 12600

∴ n has 2 consecutive zeros at the end

3. Find the value of

Sol:

4. Write any two irrational numbers between 3 and 4

Sol: 

 5. Find the value of

 6.  Find the HCF and LCM of 90, 144 by using the prime factorisation method

Sol:

7.  Is  rational or irrational?  justify your answer

Sol:

8. Expand

Sol:

9. Find the HCF of 24 and 33 by using division method

Sol:

33 > 24

 33 = 24 × 1 + 9

 24 = 9 × 2 + 6

  9 = 6 × 1 + 3

  6 = 3 × 2 + 0

∴ HCF of 24 and 33 = 3

10. Find the value of

Sol:

11. Ramu says, “if = 0, the value of x is 0”. Do you agree with him? Give reason.

Sol:

1. Write any three numbers of two digits. Find the LCM and HCF for the above numbers by the prime factorisation method.

Sol: let 10, 12, 16 be three two-digit numbers

         Prime factarisation of 10 = 2 × 5 = 2¹× 5¹

         Prime factarisation of 12 = 2 × 2× 3 = 2² × 3

         Prime factarisation of 16 = 2 × 2× 2× 2 = 2⁴

HCF of 10, 12 and 16 = 2¹ = 2

LCM of 10, 12 and 16 = 2⁴ × 3× 5 = 8 × 3× 5 = 120

2. Give an example for each of the following

(i) The product of two irrational numbers is a rational number

(ii) The product of two irrational numbers is an irrational number

Sol:

3. State with reasons which of the following are rational numbers and which are irrational numbers

4. Expand

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TS 10th Chapter 4 Pair of Linear Equations in Two Variables, Parallel, coincident and intersecting lines. Dependent and independents lines

An equation of the form ax + by + c = 0 where a, b, c are real numbers and a² + b² ≠ 0 is called a linear equation in two variables x and y.

Two linear equations in the same variables are called a pair of linear equations in two variables

a₁x + b₁y + c₁ = 0; a₂x + b₂y + c₂ = 0 are the pair of linear equations in two variables in x and y.

Solution of Pair of linear equations in two variables:

For linear equations in two variables, there are infinitely many solutions.

Ex:  x + y = 10

x =7, y = 3; x = 6, y = 4; x = 1, y = 9; x =2, y = 8; x=3, y = 7 like we have infinitely   many solutions.

⇰ For finding exact values of x and y we have to know two linear equations.

⇰ A pair linear equations in two variables solved by four methods

1. Graphical method
2. Substitution method
3. Elimination method
4. Cross – Multiplication method

Solving the pair of linear equations in two variables by using graphical Method:

 1. 2x + y −5 = 0, 3x – 2y − 4 = 0

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines intersect at the point (2, 1)

There is only one solution for this pair of linear equations in two variables.

These equations are known as consistent pair of linear equations and they have a unique solution.

2. 2x – 3y = 15; 4x – 6y = 9

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are parallel

There is no solution for this pair of linear equations in two variables.

These equations are known as inconsistent pair of linear equations and they have no solution.

3. 3x + 4y = 2; 6x + 8y = 4

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are coincide

There are infinitely many solutions for this pair of linear equations in two variables.

These equations are known as consistent pair of linear equations and they have infinitely solution.

Consistent and inconsistent:

If the system of equations has a solution, then they are consistent. If the system of equations has no solution, then they are inconsistent.

The relationship between coefficients and the nature of the equation system:

Examples:

1. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph

    3x – 2y = 2 and 2x + y = 6

The two lines intersect at the point (2, 2)

∴ solutions is x = 2, y = 2

2. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph

    x – 2y = –1 and 2x – y – 4 = 0

The two lines intersect at the point (3, 2)

∴ solutions is x = 3, y = 2

3. Represent the solution of linear equation graphically

     x – 2y = –3 and 2x + y = 4

The two lines intersect at the point (1, 2)

∴ solutions is x = 1, y = 2

Word problems:

1. Neha went to a sale to purchase some points and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts is two less than twice the number of points purchased. Also, the number of number of skirts is four less than four times the number of pants purchased”.

Help her friend to find how many pants and skirts Neha bought.

Sol:

Let the number of points = x

and the number of skirts = y

the number of skirts is two less than twice the number points purchased

⟹ y = 2x – 2

the number of number of skirts is four less than four times the number of pants purchased

⟹ y = 4x – 4

The two lines intersect at the point (1, 0)

∴ solution is x = 1, y = 2

No. of Points = 1 and no. of skirts = 0

2. 10 students of a class X took part in a Mathematics quiz. If the no. of girls is 4 more than the no. of boys, then find the no, of boys and no. of girls who took part in the quiz.

Sol:

let the no. of boys = x

No. of girls = y

Total no. of students = 10

⟹ x + y = 10

The no. of girls is 4 more than the no. of boys

⟹ y = x + 4

⟹ x – y = – 4

The two lines intersect at the point (3, 7)

∴ solution is x = 3, y = 7

No. of boys = 3 and no. of girls = 7

3. Half the perimeter of a rectangular garden, whose length is 4m more than its width is 36 m. Find the dimensions of the garden.

Sol:

let the width of garden = x m.

Length of garden = y m.

Length of garden is 4m mote than its width

⟹ y = x + 4

x – y = – 4

half of the perimeter of rectangular garden is 36 m.

⟹ x + y = 36

⟹ x – y = – 4

The two lines intersect at the point (16, 20)

∴ solution is x = 16, y = 20

Width of garden = 16 m and Length of garden = 20 m

4. The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will be increased by 50 sq units. Find the length and breadth of rectangle.

Sol:

Let the length of rectangle = x units

the breadth of rectangle = y units

The area of rectangle = xy sq units

If the Length of rectangle is reduced by 5 units and breadth is increased by 2 units, then area will be reduced by 80 sq units.

⟹ (x – 5) (y + 2) = xy – 80

⟹ xy + 2x – 5y – 10 = xy – 80

⟹ 2x – 5y = – 70

If length is increased by 10 and breadth decreased by 5 then area will be increased by 50 squnits

⟹ (x + 10) (y – 5) = xy + 50

⟹ xy – 5x + 10y – 50 = xy + 50

⟹ – 5x + 10y = 100

The two lines intersect at the point (40, 30)

∴ solution is x = 40, y = 30

Length of rectangle = 40 units and Breadth of rectangle = 30 units

5. In X class, if three students sit on each bench one student will be left, if four students sit on each bench, one bench will be left. Find the number of students and number of benches in that class.

Sol:

Let the no. of students = x

The no. of benches = y

If three students sit on each bench one student will be left

⟹ 3y = x – 1

⟹ x – 3y = 1

If four students sit on each bench one bench will be left

⟹ 4 (y – 1) = x ⟹ 4y – 4 = x

⟹ x – 4y = – 4

The two lines intersect at the point (16, 5)

∴ solution is x = 16, y = 5

No. of students = 16 and No. of benches = 5

The relationship between coefficients and the nature of the equation system Related Problems:

1. Check whether the following pair of linear equations have a unique solution, infinitely many solutions or no solution

(i) 2x + 3y = 1; 3x – y = 7

(ii) 2x + 3y = 5; 4x + 6y = 10

(iii) 5x + 3y = 4;10x + 6y = 12

Sol:

(i) Given equations are x + 3y = 1; 3x – y = 7

a₁ = 1, b₁ = 3, c₁ = 1; a₂ = 3, b₂ = –1, c₂ = 7

∴ Given equations have unique solution

(ii) Given equations are      2x + 3y = 5; 4x + 6y = 10

a₁ =2, b₁ = 3, c₁ = 5; a₂ = 4, b₂ = 6, c₂ = 10

∴ Given equations have infinite solutions

(iii) Given equations are 5x + 3y = 4;10x + 6y = 12

a₁ =5, b₁ = 3, c₁ = 4; a₂ = 10, b₂ = 6, c₂ = 12

∴ Given equations have no solution

2. For what value of ‘p’ the following pair of equations has a unique solution

      2x + py = – 5 and 3x + 3y = – 6

Sol:

Given equations are 2x + py = – 5, 3x + 3y = – 6

a₁ =2, b₁ = p, c₁ = –5; a₂ = 3, b₂ = 3, c₂ = –6

Above equations have a unique solution

∴ Given pair of linear equations have a unique solution when p ≠ 2

3. For what value of ‘p’ the following pair of equations has a unique solution

      px + 3y –(p – 3)= 0 and 12x + py – p = 0

Sol:

Given equations are px + 3y – (p – 3) = 0, 12x + py – p = 0

a₁ =p, b₁ = 3, c₁ = – (p – 3),; a₂ =12, b₂ = p, c₂ = –p

Above equations have infinitely many solution

∴ Given pair of linear equations have infinitely many solutions when p = ±6

4. For what value of ‘k’ the following pair of equations represent coincident lines.

   3x + 4y + 2 = 0 and 9x + 12y + k = 0

Sol:

Given equations are 3x + 4y + 2 = 0 and 9x + 12y + k = 0

a₁ =3, b₁ = 4, c₁ = 2; a₂ = 9, b₂ = 12, c₂ = k

Above equations are coincident lines

∴ Given pair of linear equations are coincident when k = 6

5. For what value of ‘k’ the following pair of equations represents the parallel lines

        2x – ky + 3 = 0 and 4x + 6y – 5 = 0

Sol:

Given equations are 2x – ky + 3 = 0, 4x + 6y – 5 = 0

a₁ =2, b₁ = –k, c₁ = 3; a₂ = 4, b₂ = 6, c₂ = –5

Above equations represents parallel lines

∴ Given pair of linear equations represents parallel lines when k = – 3

Substitution Method:

In this method, we make one of the variables x or y as the subject from the 1^st equation (or 2^nd equation). Substitute this equation in 2^nd equation (or 1^st equation) and get the value of the variable involved, then by substituting this value in any one of the equations we get the value of second variable.

1. Solve the following pair of equations by using substitution method

    2x – y = 5 and 3x + 2y = 11

Sol:

Given equations are 2x – y = 5 ———— (1)

3x + 2y = 11———– (2)

From (1)

y = 2x – 5

substitute y value in (2)

⟹ 3x + 2(2x – 5) = 11

3x + 4x – 10 = 11

7x = 11 + 10 = 21

x = 3

now y = 2x – 5

⟹ y = 2(3) – 5 = 6 – 5 = 1

∴ the solution is x = 3 and y = 1

2. Solve the following pair of equations by using substitution method

     2x + 3y = 9 and 3x + 4y = 5

Sol:

Given equations are 2x + 3y = 9 ———— (1)

3x + 4y = 5———– (2)

From (1)

3y = 9 – 2x

∴ the solution is x = – 21 and y = 17

3. Solve the following pair of equations by using substitution method

    3x – 5y = – 1 and x – y = – 1

Sol:

Given equations are 3x – 5y = – 1 ———— (1)

x – y = – 1———– (2)

From (2)

y = x + 1

substitute y value in (1)

⟹ 3x – 5(x + 1) = – 1

3x – 5x – 5= – 1

– 2x – 5= – 1

– 2x = – 1 + 5

– 2x = 4 ⟹ x = – 2

now y = x + 1

⟹ y = – 2 + 1 = – 1

∴ the solution is x = – 2 and y = – 1

4. Solve the following pair of equations by using substitution method

     x +  = 6 and 3x –   = 5

Sol:

5. Solve the following pair of equations by using substitution method

      0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

Sol:

Elimination Method:

In this method first we eliminate one of the two variables by equating its coefficients. This give a single equation which can be solved to get the value of the other variable

Steps for Elimination Method:

• Write the both the equations in the form of ax + by = c
• Multiply both the equations by suitable non-zero real number to make the coefficient of one variable (x or y) equal
• If the variable to be eliminated has the same sign in both the equations, subtract one equation from another to get an equation in one variable.
• Solve the equation for the one variable
• Substitute the value of this variable in any one of the given equations and find the value of the eliminated variable.

1. Solve the following pair of equations by using Elimination method

     2x – y = 5 and 3x + 2y = 11

2. Solve the following pair of equations by using Elimination method

     8x + 5y = 9 and 3x + 2y = 4

3. Solve the following pair of equations by using Elimination method

     3x + 4y = 25 and 5x – 6y = –9

4. Solve the following pair of equations by using Elimination method

     2x + y = 5 and 3x – 2y = 4

5. Solve the following pair of equations by using Elimination method

     3x + 2y = 11 and 2x + 4y =4

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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 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

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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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 cos ³ θ – 3cos θ)

   = cos^-1(cos 3θ)

= 3 θ

= 3 cos^-1 x

 = 3  (cos^-1 x)

      = 3

     =

(viii)

(ix)  

(x)

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

Limits

Question 1

Find

Sol:

= 9

Question 2

Compute

Sol:

= a + a = 2a

Question 3

compute  

Sol:

Question 4

Show that = 1and  = –1

Sol:

we know that if x > 0

                            = –x if x < 0

As x → 0+ ⟹ x > 0

⟹  = x

=  1

As x → 0+ ⟹ x < 0

⟹  = –x

  = –1

Question 5

If f (x) = , then find and . Does  exist?

Sol:

Question 6

Show that = –1

Sol:

As x → 2– ⟹ x < 2

   x – 2 < 0

 ⟹   < 0

Question 7

Compute and

Sol:

As x → 2+ ⟹ x > 2

  ⟹  = 2

=                          

As x → 2– ⟹ x < 2

  ⟹  = 1

Question 8

Find

Sol:

Question 9

Compute

Sol:

Question 10

Show that

Sol:

Question 11

Compute

Sol:

Question 12

Compute

Sol:

Question 13

Evaluate

Sol:

let y = x – 1 ⟹ x = y + 1

then as x → 1, y → 0

Question 14

Compute

Sol:

Question 15

Compute

Sol:

Question 16

Compute

Sol:

     As x → ∞,  →0

                      = ∞ (3/4)

                      = ∞

Question 17

Compute

Sol:

We know that – 1 ≤ sin x ≤ 1

x² – 1 ≤ x² – sin x ≤ x² + 1

Question 18

Find

Sol:

Question 19

Find

Sol:

Question 20

Find

Sol:

Question 21

Find

Sol:  

Question 22

Compute

Sol:  

Question 23

Compute

Sol:

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

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

Hyperbolic Functions

Question 1

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh³ x

Sol:

sinh (3x) = sinh (2x + x)

                  = sinh 2x cosh x + cosh 2x sinh x

                 = (2 sinh x cosh x) cosh x + (1 + 2 sinh² x) sinh x

                 = 2sinh x cosh² x + sinh x + 2 sinh³ x

                 = 2 sinh x (1 + sinh² x) + sinh x + 2 sinh³ x

                 = 2 sinh x + 2 sinh³ x+ sinh x + 2 sinh³ x

                 = 3 sinh x + 4 sinh³ x

Question 2

If cosh x = , find the values of (i) cosh 2x and (ii) sinh 2x

Sol:

Given cosh x =

 Cosh 2x = 2 cosh² x – 1

                 = 2.  – 1

                 =

Sinh² 2x = cosh² 2x – 1

                 = – 1

                 =  – 1

       =

  Sinh² 2x   =

Question 3

If cosh x = sec θ then prove that tanh²= tan²

Sol:

tanh²  =

              =

             =

             =

             = tan²

Question 4

If sinh x = 5, then show that x =

Sol:

Given, sinh x = 5

      ⟹ x = sinh^-15    

We know that sinh^-1x =  

        ⟹ x =       

               x  = 

Question 5

Show that tanh^-1 =   log3

Sol:

Given tanh^-1

We know that tanh^-1 x =    

 tanh^-1  =   

                 =    

                 =   log3

Question 6

For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)

Sol:

R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)

=  

= sinh (x + y)

Question 7

For any x∈ R, prove that cosh⁴ x – sinh⁴ x = cosh 2x

Sol:

cosh⁴ x – sinh⁴ x = (cosh² x)² – (sinh² x)²

                                 = (cosh² x + sinh² x) (cosh² x – sinh² x)

                                 = 1. cosh 2x

                                 = cosh 2x

Question 8

Prove that = cosh x + sinh x

Sol:

= cosh x + sinh x

Question 9

If sin hx = ¾ find cosh 2x and sinh 2x.

Sol:

Given sin hx = ¾

We know that cosh² x = 1 + sinh² x

                                           = 1 + (3/4)²

                                           = 1 + 9/16

                                           = 25/16

cos hx = 5/4

cosh 2x = 2cosh² x – 1

                = 2(25/16) – 1

                = 25/8 – 1

                = 17/8

Sinh 2x = 2 sinh x cosh x

                = 2 (3/4) (5/4)

                = 15/8

Question 10

Prove that (cosh x – sinh x) ⁿ = cosh nx – sinh nx

Sol:

∴ (cosh x – sinh x) ⁿ = cosh nx – sinh nx

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Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

Product Of Vectors

Question 1

If a = 6i +2 j +3 k, b = 2i – 9 j+ 6k, then find the angle between the vectors a and b

Sol:

Given vectors are a = 6i +2 j +3 k, b = 2i – 9 j+ 6k

 If θ is the angle between the vectors a and b, then cos θ = 

 a .b = (6i +2 j +3 k). (2i – 9 j+ 6k) = 6(2) + 2 (– 9) + 3(6)

          = 12 – 18 + 18 = 12

              = 7

              = 11

  ⟹ cos θ =

 θ =

Question 2

If a = i +2 j –3 k, b = 3i – j+ 2k, then show that a + b and a – b are perpendicular to each other.

Sol:

Given vectors are a = i +2 j –3 k, b = 3i – j+ 2k

a + b = (i +2 j –3 k) + (3i – j+ 2k) = 4i + j – k

a – b = (i +2 j –3 k) – (3i – j+ 2k) = –2i +3 j – 5k

(a + b). (a – b) = (4i + j – k). (–2i +3 j – 5k)

                            = – 8 + 3 + 5

                           = 0

∴ a + b and a – b is perpendicular to each other.

Question 3

If a and b be non-zero, non-collinear vectors. If , then find the angle between a and b

Sol:

Given

 Squaring on both sides

 (a + b) (a + b) = (a – b) (a – b)

  a² + 2 a. b + b² = a² – 2 a.b + b²

 ⟹ 4 a.b = 0

  a.b = 0

∴ the angle between a and b  is 90⁰

Question 4

If = 11, = 23 and  = 30, then find the angle between the vectors a and b and also find

Sol:

Given = 11,   = 23 and = 30

        = 30

 = 900

 = 900

(11)² – 2 ×11×23 cos θ + (23)² = 900

121 – 506 cos θ + 529   = 900

 650 – 506 cos θ = 900

cos θ =  ⟹ θ =

                = (11)² + 2 ×11×23 cos θ + (23)²

                = 121 + 2 ×11×23 ×  + 529

               = 400

 = 20

Question 5

If a = i – j – k and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude.

Sol:

Given vectors are a = i – j – k and b = 2i – 3j + k

 a.b = (i – j – k). (2i – 3j + k) = 2 + 3 – 1 = 4

   =

 The projection vector of b on a =

                  =  (i – j – k)

  The magnitude of the projection vector =  =

Question 6

If the vectors λ i – 3j + 5k and 2λ i – λ j – k are perpendicular to each other, then find λ

Sol:

let a = λ i – 3j + 5k, b = 2λ i – λ j – k

Given, that a and b are perpendicular to each other

⟹ a.b = 0

(λ i – 3j + 5k). (2λ i – λ j – k) = 0

2 λ² + 3 λ – 5 = 0

2 λ² + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

Question 7

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3i + j + 5k

Sol:

let P (x, y, z) be any point on the plane

 ⟹ OP = xi + yj + zk

 OA = – 2i +j +3k

AP = OP – OA = (xi + yj + zk) – (– 2i +j +3k)

 AP = (x + 2) i + (y – 1) j + (z – 3) k

 AP is perpendicular to the vector 3i + j + 5k

 ⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

  ⟹ 3x + 6 + y – 1 + 5z – 15 = 0

 ∴   3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

Question 8

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

Sol:

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

 Vector equations of the above planes are: r. (2i – 3j – 6k) = 5 and r. (6i + 2j – 9k) = 4

 ⟹ n₁ = 2i – 3j – 6k and n₂ = 6i + 2j – 9k

 If θ is the angle between the planes r. n₁ = d₁ and r. n₂ = d₂, then

 Cos θ =

  ⟹ Cos θ =

 ⟹ θ =

Question 9

a = 2i – j + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle.

Sol:

Given a = 2i – j + k, b = i – 3j – 5k

 If a, b, and c form the sides of a triangle, then a + b + c = 0

 ⟹ a + b = – c

⟹ c = – (a + b)

       = – [(2i – j + k) +( i – 3j – 5k)]

      = – (3i –4 j –4k)

     ∴ c = – 3i +4 j + 4k     

Question 10

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

Sol:

 Let a = 3i – 2j + k and n = 4i + 7j – 4k

 The equation of the plane passing through point A(a) and perpendicular to the vector n is (r – a). n = 0

 ⟹ [r – (3i – 2j + k)]. (4i + 7j – 4k) = 0

 ⟹ r. (4i + 7j – 4k)– [(3i – 2j + k). (4i + 7j – 4k)] = 0

r. (4i + 7j – 4k)– (12 – 14 – 4) = 0

r. (4i + 7j – 4k)– 6 = 0

r. (4i + 7j – 4k) = 6

Question 11

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k

Sol:

The vector which is parallel to the XOY-plane is of the form xi + yj

The vector which is parallel to the XOY-plane and perpendicular to 4i – 3j + k

 is 3i + 4j

 its magnitude =     = 5

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k =

Question 12

If a + b + c = 0, = 3,  = 5 and    = 7, then find the angle between a and b

Sol:

Given, a + b + c = 0,  = 3,  = 5 and    = 7

 a + b = – c

 (a + b)² =

 3² + 5² + 2  cos θ = 7²

  9 + 25 + 2.3.5 cos θ = 49

 34 + 30cos θ = 49

 30cos θ = 49 – 34

 30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

Question 13

If a = 2i – 3j + 5k, b = – i + 4j + 2k, then find a × b and unit vector perpendicular to both a and b

Sol:

 Given, a = 2i – 3j + 5k, b = – i + 4j + 2k             

 a × b =

         = i (–6 – 20) – j (4 + 5) + k (8 – 3)

         = –26i – 9j + 5k

The unit vector perpendicular to both a and b    =  

  =

=

Question 14

If a = i + j + 2k and b = 3i + 5j – k are two sides of a triangle, then find its area.

Sol:

Given, a = i + 2j + 3k and b = 3i + 5j – k