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

Inter 1st year Maths Reduced Syllabus |
Click Here |

Inter 2nd year Maths Reduced Syllabus |
Click Here |

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**Errors and Approximations**

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**Differentiation**

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**Question 1**

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**Question 1**

The post Product of Vectors (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>

**QUESTION 1**

Find unit vector in the direction of

The unit vector in the direction of a vector is given by

**QUESTION 2**

Find a vector in the direction of that has a magnitude of 7 units.

The unit vector in the direction of a vector is

The vector having the magnitude 7 and in the direction of is

**QUESTION 3**

Find the unit vector in the direction of the sum of the vectors, **a** = 2**i **+ 2**j **– 5**k **and** b **= 2**i **+** j **+ 3**k**

**Sol** Given vectors are **a** = 2**i **+ 2**j **– 5**k **and** b **= 2**i **+** j **+ 3**k**

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

** ****QUESTION 4**

Write the direction cosines of the vector

**QUESTION 5**

Show that the points whose position vectors are – 2**a **+ 3**b **+ 5**c**, **a **+ 2**b **+ 3**c**, 7** a** – **c** are collinear when **a**,** b**,** c **are non-collinear vectors

**Sol:** Let **OA **= – 2**a **+ 3**b **+ 5**c, OB = a **+ 2**b **+ 3**c**, **OC **= 7** a** – **c**

** AB **= **OB – OA = a **+ 2**b **+ 3**c **– **(**– 2**a **+ 3**b **+ 5**c)**

** AB = **3**a **–** b **– 2**c**

** AC **= **OC – OA **= 7** a** – **c** – **(**– 2**a **+ 3**b **+ 5**c)**

** AC = **9**a **– 3**b **– 6**c =** 3**(**3**a **–** b **– 2**c)**

** AC = **3 **AB**

** **A, B and C are collinear

**QUESTION 6**

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) **AL** and **AM** in terms of** AB** and **AD (**ii) ????,** if AM = **???? **AD – LM**

**Sol:** Given, ABCD is a parallelogram and L and M are middle points of BC and CD

(i) Take A as the origin

M is the midpoint of CD

** = AD **+ ½** AB **(∵ AB = DC)

L is the midpoint of BC

** = AB **+ ½ **AD **((∵ BC = AD)

(ii) **AM = **???? **AD – LM**

**AM + LM= **???? **AD **

** AD + ½ AB + AD + ½ AB – (AB + ½ AD) = ???? AD **

**AD + ½ AB + AD + ½ AB – AB – ½ AD = ???? AD **

3/2 **AD = ???? AD **

∴**???? = 3/2**

**QUESTION 7**

If G is the centroid of the triangle ABC, then show that **OG = ** when, are the position vectors of the vertices of triangle ABC.

**Sol:** **OA** = **a, OB **=** b, OC **=** c and OD **=** d**

** **D is the midpoint of BC

G divides median AD in the ratio 2: 1

**QUESTION 8**

If = , = are collinear vectors, then find m and n.

**Sol:** Given , are collinear vectors

Equating like vectors

2 = 4 λ; 5 = m λ; 1 = n λ

∴ m = 10, n = 2

**QUESTION 9**

Let If ,. Find the unit vector in the direction of **a + b.**

** **The unit vector in the direction of **a + b **=

**QUESTION 10**

If the vectors – 3**i** + 4**j** + λ**k** and μ**i** + 8**j** + 6**k**. are collinear vectors, then find λ and μ.

**Sol:** let **a **= – 3**i** + 4**j** + λ**k**, **b** = μ**i** + 8**j** + 6**k**

** ⟹ **** a **= t**b**

– 3**i** + 4**j** + λ**k** = t (μ**i** + 8**j** + 6**k)**

– 3**i** + 4**j** + λ**k** = μt **i + **8t **j + **6t **k**

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

∴ μ=– 6, λ = 3

**QUESTION 11**

ABCD is a pentagon. If the sum of the vectors **AB**, **AE**, **BC**, **DC**, **ED** and **AC **is ????** AC **then find the value of ????

**Sol:** Given, ABCD is a pentagon

**AB** + **AE** + **BC** + + **DC** + **ED** + **AC **= ????** AC**

(**AB** +** BC) **+ (**AE** + **DC** + **ED**) + **AC = **????** AC**

** AC** + **AC** + **AC = **????** AC**

** 3 AC = **????** AC**

** **????** = **3

**QUESTION 12**

If the position vectors of the points A, B and C are – 2**i** + **j** – **k** and –4**i** + 2**j** + 2**k **and 6**i** – 3**j** – 13**k **respectively and** AB = **????** AC, **then find the value of ????

**Sol:** Given, OA = – 2**i** + **j** – **k **, OB = –4**i** + 2**j** + 2**k **and OC = 6**i** – 3**j** – 13**k **

AB = OB – OA = –4**i** + 2**j** + 2**k **– (– 2**i** + **j** – **k**)

= –4**i** + 2**j** + 2**k **+2**i** – **j** + **k**

= –2**i** + **j** + 3**k **

AC = OC – OA = 6**i** – 3**j** – 13**k **– (– 2**i** + **j** – **k**)

= 6**i** – 3**j** – 13**k **+2**i** – **j** + **k**

= 8**i** –4 **j** –12**k**

** = **– 4 (2**i** + **j** + 3**k)**

**AC = **– 4** AB**

Given** AB = **????** AC**

** **???? = – 1/4

**QUESTION 13**

If **OA** = **i** + **j** +**k,** **AB** = 3**i** – 2**j** + **k**, BC = **i** + 2**j** – 2**k, CD** = 2**i** + **j** +3**k** then find the vector **OD**

**Sol:** Given **OA** = **i** + **j** +**k,** **AB** = 3**i** – 2**j** + **k**, BC = **i** + 2**j** – 2**k, CD** = 2**i** + **j** +3**k**

** OD = OA **+** AB **+** BC **+** CD **

** = i** + **j** +**k **+ 3**i** – 2**j** + **k** + **i** + 2**j** – 2**k **+ 2**i** + **j** +3**k**

** = **7**i** + 2**j** +4**k**

**QUESTION 14**

Let **a** = 2**i** +4 **j** –5 **k**, **b** = **i** + **j+** **k**, **c** = **j** +2 **k**, then find the unit vector in the opposite direction of **a **+ **b** + **c**

**Sol:** Given, **a** = 2**i** +4 **j** –5 **k**, **b** = **i** + **j+** **k**, **c** =** j** +2 **k**

** a **+ **b** + **c **= 2**i** +4 **j** –5 **k** + **i** + **j+** **k **+ **j **+2 **k**

** = **3**i** +6**j** –2**k**

** **The unit vector in the opposite direction of **a **+ **b** + **c **is

**QUESTION 15**

Is the triangle formed by the vectors 3**i** +5**j** +2**k**, **2i** –3**j** –5**k**, **–**5**i** – 2**j** +3**k **

**Sol:** Let **a** =3**i** +5**j** +2**k**, **b** = **2i** –3**j** –5**k**,** c** = **–**5**i** – 2**j** +3**k**

∴ Given vectors form an equilateral triangle.

**QUESTION 16**

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are **a**, **b,** (3**a** – 2**b**) are collinear.

**Sol:** the vector equation of the straight line passing through two points** a**, **b **is

**r** = (1 – t) **a**+ t **b **

** **3**a – **2**b = (**1 – t) **a**+ t **b **

** **Equating like vectors

1 – t = 3 and t = – 2

∴ Given points are collinear.

**QUESTION 17**

OABC is a parallelogram If OA = **a** and OC = **c**, then find the vector equation of the side **BC**

**Sol:** Given, OABC is a parallelogram and OA = **a**, OC = **c**

The vector equation of** BC is** a line which is passing through C(**c**) and parallel to **OA**

⟹ the vector equation of **BC** is r = **c** + t **a**

**QUESTION 18**

If **a**, **b**, **c** are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

**Sol:** Given **OA** = **a**, **OB** = **b**, **OC** = c

D is mid of BC

The equation of AD is

**QUESTION 19**

Find the vector equation of the line passing through the point 2**i** +3**j** +**k **and parallel to the vector 4**i – **2**j **+ 3**k**

Sol: Let **a =**2**i** +3**j** +**k**, **b **= 4**i – **2**j **+ 3**k **

The vector equation of the line passing through **a **and parallel to the vector **b **is **r** = **a **+ t**b**

** r** = 2**i** +3**j** +**k **+ t (4**i – **2**j **+ 3**k)**

= (2 + 4t) **i **+ (3 – 2t) **j** + (1 + 3t)** k**

**QUESTION 20**

Find the vector equation of the plane passing through the points **i – **2**j **+ 5**k**, **– **2**j –k **and **– 3i **+ 5**j**

Sol: The vector equation of the line passing through **a,** **b and c**is **r** = (1 – t – s) a + t**b **+ s**c**

** **⟹ **r** = (1 – t – s) (**i – **2**j **+ 5**k**) + t (**– **2**j –k**) + s (**– 3i **+ 5**j**)

= (1 – t – 4s)** i** + (– 2 – 3t + 7s)** j** + (5 – 6t – 5s)** k**

The post Addition of Vectors (2M Questions &Solutions)|| V.S.A.Q.’S|| first appeared on Basics In Maths.]]>

**QUESTION 1**

If A = , then show that A^{2} = –I

∴ A^{2} = –I

**QUESTION 2**

If A = , and A^{2} = 0, then find the value of k.

A^{2} = 0

8 + 4k = 0, – 2 – k = 0 and –4 + k^{2} = 0

4k = –8; k = –2; k^{2} = 4

k = –2; k = –2; k = ± 2

∴ k =– 2

**QUESTION 3**

Trace of A = 1 – 1 + 1 = 1

**QUESTION 4**

If A = , B = and 2X + A = B, then find X.

**Sol:** Given A = , B = and 2X + A = B

2X = B – A

**QUESTION 5**

Find the additive inverse of A, If A =

Additive inverse of A = – A

**QUESTION 6**

If , then find the values of x, y, z and a.

⟹ x- 1 = 1 – x ; y – 5 = – y ; z = 2 ; 1 + a = 1

⟹ x + x = 1 + 1; y + y = 5; z = 2; a =1– 1

⟹ 2x = 1; 2y = 5; z = 2; a = 0

∴ x = ½ ; y = 5/2; z = 2; a = 0

**QUESTION 7**

Construct 3 × 2 matrix whose elements are defined by a_{ij} =

**Sol:**

Let A= _{ }

a_{11} = 1

a_{22} = 2

a_{31} = 0

**QUESTION 8**

If A = and B = , do AB and BA exist? If they exist, find them. D A and B commutative with respect to multiplication.

**Sol:** Given Matrices are A = B =

Order of A = 2 × 3 and Order of B = 3 × 2

AB and BA exist

AB and BA are not Commutative under Multiplication

**QUESTION 9**

Define Symmetric and Skew Symmetric Matrices

**Sol:**

Symmetric Matrix: Let A be any square matrix, if A^{T} = A, then A is called Symmetric Matrix

Skew Symmetric Matrix: Let A be any square matrix if A^{T} = –A, then A is called Skew Symmetric Matrix

**QUESTION 10**

If A = is a symmetric matrix, then find x.

**Sol:** Given, A = is a symmetric matrix

⟹ A^{T} = A

⟹ x = 6

**QUESTION 11**

If A = is a skew-symmetric matrix, then find x

**Sol:** Given A = is a skew-symmetric matrix

⟹ A^{T} = – A

⟹ x = –x

x+ x = 0 ⟹ 2x = 0

⟹ x = 0

**QUESTION 12**

If A = and B = , then find (A B^{T})^{ T}

**QUESTION 13**

If A = and B = , then find A + B^{T}

**QUESTION 14**

If A = , then show that AA^{T} = A^{T}A = I

∴ AA^{T} = A^{T}A = I

**QUESTION 15**

Find the minor of – 1 and 3 in the matrix

**Sol:** Given Matrix is

**QUESTION 16**

Find the cofactors 0f 2, – 5 in the matrix

**Sol:** Given matrix is

Cofactor of 2 = (–1)^{2 + 2} = –3 + 20 = 17

Cofactor of – 5 = (–1)^{3 + 2} = –1(2 – 5) = –1(–3) = 3

**QUESTION 17**

If ω is a complex cube root of unity, then show that = 0(where 1 + ω+ω^{2} = 0)

R_{1} → R_{1} + R_{2} + R_{3}

**QUESTION 18**

If A = and det A = 45, then find x.

Det A = 45

⟹ 1(3x + 24) – 0 (2x – 20) + 0 (– 12 – 15) = 45

⟹ 3x + 24 = 45

3x = 45 – 24

3x = 21

x = 7

**QUESTION 19**

Find the adjoint and inverse of the following matrices

(i)

(ii)

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

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.

The post The Plane (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

**QUESTION 1**

Show that the points A ( – 4, 9, 6), B (– 1, 6, 6), and C(0, 7, 10) right-angled isosceles triangle.

Sol:

Distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is PQ =

AB^{2} + BC^{2 } = 18 + 18 = 36 = AC^{2}

∴ ∠ B = 90^{0}

points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle

**QUESTION 2**

Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0

Sol:

Let P (x, y, z) be the locus of the point

A (0, y, 0) be any point on Y – axis

B = (1, 2, – 1)

Condition is PA = 3PB

PA^{2} = (3PB)^{2}

z^{2} = 9[x^{2 }– 2x + 1 + y^{2} – 4y + 4 + z^{2} + 2z + 1]

x^{2} + z^{2} = 9x^{2 }– 18x + 9 + 9y^{2} – 36y + 36 +9 z^{2} + 18z + 9

∴ 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 =0

**QUESTION 3**

A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O

Sol:

Let P (x, y, z) be the required point

O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

AP = OP

⟹ AP^{2} =OP^{2}

(x – a )^{2 }+ y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

x^{2} + a^{2} – 2ax + y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

a^{2} – 2ax = 0

a (a – 2x) = 0

a – 2x = 0 (∵ a≠0)

a = 2x ⟹ a/2

Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

**QUESTION 4**

Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear

Sol:

Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)

AB + BC = AC

∴ A B and C are collinear

**QUESTION 5**

Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.

Sol:

Let A = (5, – 1, 7), B = (x, 5, 1)

Given AB = 9

⟹ AB^{2} = 81

(5 – x)^{2} + (– 1 – 5)^{2} + (7 – 1)^{2} = 81

(5 – x)^{2 }+ 36 + 36 = 81

(5 – x)^{2} + 72 = 81

(5 – x)^{2 }= 81 – 72 = 9

(5 – x) = ± 3

5 – x = 3 or 5 – x = – 3

5 – 3 = x or 5 + 3 = x

x = 2 or x = 8

**QUESTION 6**

If the point (1, 2, 3) is changed to the point (2, 3, 1) through the translation of axes. Find a new origin.

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

**QUESTION 7**

By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Sol:

If a point P divides the line segment joining the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) in the ratio, then P =

Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

**QUESTION 8**

Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).

Sol:

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

7 – 2k = k + 1

7 – 1 = k + 2k

6 = 3k

k = 2

∴ P divides AB in the Ratio 1 : 2

**QUESTION 9**

Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.

Sol:

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

k = 4 (k + 1)

2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

**QUESTION 10**

Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)

Sol:

The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

∴ The centroid of the triangle =

**QUESTION 11**

Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)

Sol:

The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is

**QUESTION 12**

Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)

Sol:

Let P be any point on the YZ-plane

P = (0, y, z)

Let P divides AB in the ratio k:1

3k + 2 = 0

3k =– 2

YZ-plane divides AB in the ratio – 2:3

**QUESTION 13**

Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).

Sol:

let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

ABCD is a parallelogram

The midpoint of AC = Midpoint of BD

⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

x = 3, y = 3, z = 1

∴ The fourth vertex D = (3, 3, 1)

**QUESTION 14**

A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.

Sol:

We know that the bisector of ∠BAC divides BC in the ratio AB:AC

= 5:3

If D is a point where the bisector of ∠BAC meets BC

⟹ D divides BC in the ratio 5:3

**QUESTION 15**

If M (α, β, γ) is the midpoint of the line joining the points (x_{1}, y_{1}, z_{1}) and B, then find B

Sol:

Let B (x, y, z) be the required point

M is the midpoint of AB

⟹ 2 α = x + x_{1}; 2 β = y + y_{1}; 2 γ = z + z_{1}

x =2 α – x_{1}; x =2 β – y_{1;} x =2 γ – z_{1}

_{ }∴ B = (2 α – x_{1}, 2 β – y_{1}, 2 γ –)

** QUESTION 16**

If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.

Sol:

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

AB = BC = AC

⟹ ∆ ABC is an equilateral triangle

We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same

= (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

**QUESTION 17**

Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).

Sol:

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

a = 5

b = 4

c = 3

∴ I = (1, 1,0)

**QUESTION 18**

Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P

Sol:

Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m: n.

Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)

P divides AB in the ratio is x_{2} – x : x – x_{2}

= 3 – 5 : 5 – 9

= 1 : 2 (internally)

Let Q be the harmonic conjugate of P

⟹ Q divides AB in the ratio –1: 2

= (–3, –4, –2)

Q (–3, –4, –2) is the harmonic conjugate of P (5, 4, – 6)

**QUESTION 19**

If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.

Sol:

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)

**QUESTION 20**

Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.

Sol:

Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)

∴ the points A (3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear

B divides AB in the ratio is AB:BC = = 1:2

The post Three Dimensional Coordinates (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

**1.** Prove that the points (1, 11), (2, 15), and (– 3, – 5) are collinear, and find the equation of the line containing them.

Sol: Let A (1, 11), B (2, 15), and C (– 3, – 5)

The post Inter Maths – 1B (2m Questions & Solutions) first appeared on Basics In Maths.]]>1.If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x^{2} + 2 then find (i) (gof) (x) (ii) (gof) () (iii) (fof) (x) (iv) go(fof) (0).

Sol: Given f(x) = 4x – 1, g(x) = x^{2} + 2

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

= g (4x – 1)

= (4x – 1)^{2} + 2

= 16x^{2} – 8x + 1 + 2

= 16x^{2} – 8x + 3

= g (a + 1 – 1)

= g(a)

= a^{2} + 2

(iii) (fof) (x) = f (f (x))

= f (4x – 1)

= 4 (4x – 1) – 1

= 16x – 4 – 1

= 16x – 5

(iv) go(fof) (0) = g(fof) (0)

= g (f (f (0)))

= g (f (– 1))

= g (– 4 – 1)

= g (– 5)

= (– 5)^{2} + 2

= 25 + 2 = 27

2. If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x^{2}, then find (i) (3f – 2g) (x) (ii) (fg)(x) (iii) (x) (iv) (f + g+ 2) (x)

Sol: Given f(x) = 2x – 1 and g(x) = x^{2}

(i) (3f – 2g) (x) = 3f(x) – 2 g(x)

= 3(2x – 1) – 2(x^{2})

= 6x – 3 – 2x^{2}

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

= (2x – 1) (x^{2})

= 2x^{3} – 3x^{2}

(iv) (f + g+ 2) (x) = f(x) + g(x) + 2

= 2x – 1 + x^{2} + 2

= x^{2} + 2x + 1

3. If f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}, then find (i) f + g (ii) f – g (iii) 2f +4g (iv) f + 4 (v) fg (vi) f/g (vii) (viii) (ix) f^{2} (x) f^{3}

Sol: Given f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}

Domain of f ∩ Domain of g = {4, 6}

(i) (f + g) (4) = f (4) + g (4)

= 5 – 4 = 1

(f + g) (6) = f (6) + g (6)

=– 4 + 5 = 1

∴ f + g = {(4, 1), (6, 1)}

(ii) (f – g) (4) = f (4) – g (4)

= 5 – (– 4) = 5 + 4 = 9

(f – g) (6) = f (6) – g (6)

= – 4– 5 = – 9

∴ f – g = {(4, 9), (6, – 9}

(iii) (2f +4g) (4) = 2 f (4) + 2 g (4)

= 2(5) + 4 (– 4)

= 10 – 16

=– 6

(2f +4g) (6) = 2 f (6) + 2 g (6)

= 2(– 4) + 4 (5)

= – 8 + 20

=12

∴ (2f +4g) = {(4, – 6), (6, 12)}

(iv) (f + 4) (4) = f (4) + 4 = 5 + 4 = 9

(f + 4) (5) = f (5) + 4 = 6 + 4 = 10

(f + 4) (6) = f (6) + 4 = – 4 + 4 = 0

∴ (f + 4) = {(4, 9), (5, 10), (6, 0)}

(v) fg (4) = f (4) g (4) = (5) (– 4) =– 20

fg (6) = f (6) g (6) = (– 4) (5) =– 20

∴ fg = {(4, – 20), (6, – 20)}

(vi) f/g (4) = f(4)/g(4) = 5/ – 4 = – 5/4

f/g (6) = f(6)/g(6) = – 4/ 5

∴ f/g = {(4, – 5/4), (6, – 4/ 5)}

(ix) f^{2}(4) = (f (4))^{2} = (5)^{2} = 25

f^{2}(5) = (f (5))^{2} = (6)^{2} = 36

f^{2}(6) = (f (6))^{2} = (– 4)^{2} = 16

∴ f^{2} = {(4, 25), (5, 36), (6, 16)}

(x) f^{3}(4) = (f (4))^{3} = (5)^{3} = 125

f^{3}(5) = (f (5))^{3} = (6)^{3} = 216

f^{3}(6) = (f (6))^{3} = (– 4)^{3} = –64

∴ f^{3} = {(4, 125), (5, 216), (6, –64)}

The post Inter Maths-1A (2M Questions & Solutions) first appeared on Basics In Maths.]]>

** Natural numbers:** All the counting numbers starting from 1 are called Natural numbers.

1, 2, 3… Etc.

** Whole numbers:** Whole numbers are the collection of natural numbers including zero.

0, 1, 2, 3 …

** Integers: **integers are the collection of whole numbers and negative numbers.

….,-3, -2, -1, 0, 1, 2, 3,…..

3 + 4 = 7

-2 + 4 = 2

** Subtraction of integers on a number line:-**

6 – 3 = 3

** Multiplication of integers on a number line:-**

2 × 3 ( 2 times of 3) = 6

3 × (- 4 ) ( 3 times of -4) = -12

** Multiplication of two negative integers:**

- To multiply two negative integers, first, we multiply them as whole numbers and put plus sign before the result.
- Multiplication of two negative integers is always negative.

Ex:- -3 × -2 = 6, -10 × -2 = 20 and so on.

** Multiplication of more than two negative integers:**

• If we multiply three negative integers, then the result will be a negative integer.

Ex:- -3 × -4 × -5 = -60, -1× -7 × -4 = -28 and so on.

• If we multiply four negative integers, then the result will be a positive integer.

Ex:- -3 × -4 × -5 × -2 = 120, -1× -7 × -4 × -2 = 56 and so on.

** Note:-**

1. If the no. of negative integers is even, then the result will be positive.

2. If the no. of negative integers is odd, then the result will be negative.

** Division of integers:**

- The division is the inverse of multiplication.
- When we divide negative integer by a positive integer or positive integer by negative integer, we divide them as whole numbers then put negative signs for the quotient.

Ex:- -3 ÷ 1 = 3, 4 ÷ -2 = -2 and so on.

- When we divide a negative integer by a negative integer, we get a positive number as the quotient.

Ex:- -3 ÷ -1 = 3, -4 ÷ -2 = 2 and so on.

** Properties of integers:**

** 1.Closure property:-**

** 2.commutative property:-**

**3.associative property:-**

**Additive identity:-**

1 + 0 = 0 + 1 = 1, 10 + 0 = 0 + 10 = 10

•For any integer ‘a’, a + 0 = 0 + a

•0 is the additive identity.

**Additive inverse:- **

2 + (-2) = (-2) + 2 = 0, 5 + (-5) = (-5) + 5 = 0

•For any integer ‘a’, a+ (-a) = (-a) + a = 0

•Additive inverse of a = -a and additive inverse of (-a) = a

**Multiplicative identity:-**

2 × 1 = 1 × 2 = 2, 5 × 1 = 1 × 5 = 5

•For any integer ‘a’, a × 1 = 1 × a = a

•1 is the multiplicative identity.

**multiplicative inverse:-**

For any integer ‘a’, 1/a × a = a × 1/a = 1

- multiplicative inverse of a = 1/a
- Multiplicative inverse of 1/a = a.

**distributive property:- **

For any three integers a, b and c, a × (b + c) = (a × b) + (a × c).

3 × (2 + 4) = 18

(3 × 2) + (3 × 4) = 6 + 12 = 18

∴ 3 × (2 + 4) = (3 × 2) + (3 × 4).

**Fraction: A fraction** is a number that represents a part of the whole. A group of objects is divided into equal parts, then each part is called a fraction.

** The proper and improper fractions:**

In a **proper fraction,** the numerator is less than the denominator.

Ex: – 1/5, 2/3, and so on.

In an **improper fraction,** the numerator is greater than the denominator.

Ex: – 5/2,11/5 and so on.

**Comparing fractions:**

**Like fractions: – **We have to compare the like fractions with the numerator only because the like fractions have the same denominator. The fraction with the greater numerator is greater and the fraction with the smaller numerator is smaller.

**Unlike fractions: – **

**With the same numerator: **For comparing unlike fractions, we have to compare denominators when the numerator is the same. The fraction with a greater denominator is smaller and the fraction with the smaller denominator is smaller.

**Note: – **To find the equivalent fractions of both the fractions with the same denominator, we have to take the LCM of their denominators.

**Addition of fractions:**

∗ **Like Fractions:**

∗ Unlike** fractions:**

**Subtraction of fractions:**

**∗ Like fractions:**

**Unlike fractions: –** Firs**t,** we have to find the equivalent fraction of given fractions and then subtract them as like fractions

**Multiplication of fractions:**

**Multiplication of fraction by a whole number: –**

Multiplication of numbers means adding repeatedly.

• To multiply a whole number with a proper or improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator the same.

2.**Multiplication of fraction with a fraction: –**

multiplication of two fractions =

**Division of fractions:**

⇒ 6 one-thirds in two wholes

**Reciprocal of fraction: **reciprocal of a fraction is .

**Note:**

- dividing by a fraction is equal to multiplying the number by its reciprocal.
- For dividing a number by mixed fraction, first, convert the mixed fraction into an improper fraction and then solve it.

**1.Division of whole number by a fraction: –**

**2.Division of a fraction by another fraction: –**

** ** **Decimal number or fractional decimal: **

In a decimal number, a dot(.) or a decimal point separates the whole part of the number from the fractional part.

The part right side of the decimal point is called the **decimal part** of the number as it represents a part of 1. The part left to the decimal point is called the **integral part** of the number.

Note: –

- while adding or subtracting decimal numbers, the digits in the same places must be added or subtracted.
- While writing the numbers one below the other, the decimal points must become one below the other. Decimal places made equal by placing zeroes on the right side of the decimal numbers.

**Comparison of decimal numbers:**

while comparing decimal numbers, first we compare the integral parts. If the integral parts are the same, then compare the decimal part.

Ex: – which is bigger: 13.5 or 14.5

Ans: 14.5

Which is bigger: 13.53 or 13. 25

Ans: 13.53

**Multiplication of decimal numbers:**

For example, we multiply 0.1 × 0.1

**Multiplication of decimal numbers by 10, 100, and 1000: –**

• Now Compare 5432 and 4678…

5432 is greater as the digits at the ten thousand place in 5432 is greater than that in 4678.

**Order of numbers:**

arrange the numbers from smallest to the greatest; this order is called Ascending order.

Ex:- 23, 44, 65, 79, 100

arrange the numbers from greatest to the smallest, this order is called Ascending order.

Ex:- 100,79, 65, 33, 23

**Formations of numbers**

• Form the largest and smallest possible numbers using the digits 3, 2, 4, 1 without repetition

• Largest number formed by arranging the given digits in descending order _ 4321.

• Smallest number formed by arranging the given digits in ascending order _ 1234.

• Greatest two-digit number is 99.

• Greatest three-digit number is 999.

• Greatest four-digit number is 9999.

• Place value is the positional notation, which defines the position of a digit.

Ex:- 3458

8 is one place, 5 is tens place, 4 is hundreds place and 3 is thousands place.

**Expanded form**

• It refers to expand the numbers to see the value of each digit.

Ex :- 3458 = 3000 + 400 + 50 + 8

= 3×1000 + 4×100 + 5×10 + 8×1

**• Note:-**

1 hundred = 10 tens

1 thousand = 10 hundreds

1 lakh = 100 thousands = 1000 hundreds

**Place value table for Indian system :**

Example: Represents the number in 6,35,21,892 in place value table

**Place value table for International system :**

Ex:- represents the number in 635,218,924 in place value table

**• Indian system of numeration:-** in the Indian system of numeration we use ones, tens, hundreds, thousands, lakhs and crores. The first comma comes after three digits from the right, the second comma comes two digits latter and the third comma comes after another two digits.E

Ex:- “three crores thirty-five lakh seventeen thousand four hundred thirty” can be written as.3,35,17,430

**• International system of numeration:- **in the International system of numeration we use ones, tens, hundreds, thousands, millions and billions.

Ex:- “ six hundred thirty-five million two hundred eighteen thousand nine hundred twenty-four” can be written as 635,218,924.

Note:-10 millimetres = 1centimeter

100 centimetres = 1 meter

1000 meters = 1 kilometer

1000 milligrams = 1 gram

1000 grams = 1 kilo gram

**Natural numbers:** All the counting numbers starting from 1 are called Natural numbers.

1, 2, 3… Etc.

** Successor and Predecessor: **If we add 1 to any natural number, we get the next number, which is called the Successor. If we subtract 1 from any natural number, we get the previous number, which is called Predecessor.

Ex: – successor of 23 is 24 and predecessor of 32 is 31.

Note:- There is no predecessor of 1 in natural numbers.

**Whole numbers:** Whole numbers are the collection of natural numbers.

0, 1, 2, 3 …

**Representation of whole number on the number line:**

• Draw a line mark a point on it.

• Label it as ‘0’

• Mark as many points at equal distance to the right of 0.

• Label the points as 1, 2, 3, 4, … respectively.

• The distance between any two consecutive points is the unit distance.

** Addition on the number line:**

- The distance between 2 and 4 is 2 units, like as the distance between 2 and 6 is 4 units
- The number on the write is always greater than the number on the left
- The number on the left of any number is always smaller than that number

Addition of the whole number can represent on the number line

Ex:- 3 + 2 = 5

Start from three, we add 3 to 2. We make two jumps to the right of the number line as shown above. We reach at 5.

Subtraction of the whole number can be represented on the number line

Ex :-5 – 3 = 2

Start from 5, we subtract 3 from 5. We make three jumps to the left of the number line shown as above. We reach at 2.

**Multiplication on the number line:**

For multiplying 2 and 3, start from 0, make 2 jumps using 3 units at a time to the right, as you reach to 6. Thus, 2 × 3 =6.

**Properties of whole numbers**

**Closer property: **Two whole numbers are said to be closed if their operation (+, -, ×,÷) is always closed.

**Addition:-**Whole numbers are closed under addition.

Ex: 3, 2 are whole numbers ⟹ 3 + 2 = 5 ( 5 is whole number)

**Subtraction:- **Whole numbers are not closed under subtraction as their difference not always a whole number.

Ex:- 2 – 3 = −1 ( −1 is not a whole number)

**Multiplication:- **Whole numbers are closed under multiplication.

Ex:- 3 × 2 = 6, 6 is a whole number.

**Division:-** Whole numbers are not closed under division, as their division is not always a whole number.

Ex:- 3 ÷ 2 is not a whole number.

**Commutative property: **Two whole numbers are said to be commutative if the result is the same when we change their position.

**Addition:-**Whole numbers are commutative under addition.

Ex: 3, 2 are whole numbers ⟹ 3 + 2 = 5 and 2 + 3 = 5 ( 3 + 2 = 2 + 3).

**Subtraction:- **Whole numbers are not commutative under subtraction.

Ex:- 2 – 3 = −1 and 3 – 2 = 1( 2 −3 ≠ 3 – 2 ).

**Multiplication:- **Whole numbers are commutative under multiplication.

Ex:- 3 × 2 = 6 and 2 ×3 = 6 (3 × 2 = 2 ×3)

**Division:-** Whole numbers are not commutative under division.

Ex:- 3 ÷ 2 ≠ 2 ÷ 3.

**Associative property: **For any three whole numbers a, b and c if (a ⨀ b)⨀ c = a ⨀ (b ⨀ c), then whole numbers are associative under operation ⨀. [ ⨀ = +, –, × and ÷ ].

**Addition:-**Whole numbers are associative under addition.

Ex: ( 3 + 2) + 5 = 10 and 3 + (2 + 5) = 10 ⟹ ( 3 + 2) + 5 = 3 + (2 + 5)

**Subtraction:- **Whole numbers are not associative under subtraction.

Ex:- : ( 3 − 2) − 5 = −4 and 3 − (2 − 5) = 6 ⟹ ( 3 + 2) + 5 ≠3 + (2 + 5)

**Multiplication:- **Whole numbers are associative under multiplication.

Ex:- (3 × 2) ×5 = 30 and 3 ×(2 × 5) = 30 ⟹ (3 × 2) ×5 = 3 ×(2 × 5)

**Division:-** Whole numbers are not associative under division.

Ex:- ( 3 ÷ 2) ÷ 5 ≠3 ÷ (2 ÷ 5).

**Distributive property:**

For any three whole numbers a, b and c, a×(b + c) = (a × b) +( a × c).

**Note :Division by zero is not defining.**

**Identity under addition and multiplication: **

2 +0 = 2, 5 + 0 = 5 and so on.

Thus, 0 is the additive identity.

2 ×1 = 2, 4 × 1 = 4 and so on.

Thus, 1 is a multiplicative identity.

**Patterns:**

- Every number can be arranged as a line. The number 2 is shown as

The number 3 as shown as

- Some numbers can be shown as rectangles. 8 can be shown as

- Some numbers can be arranged as squares. 9 can be shown as

** **

- Some numbers can be shown as triangles.

3 can be shown as 6 can be shown as

**Divisibility Rule:**

The process of checking whether a number is divisible by a given number or not without actual division is called divisibility rule for that number.

**Divisibility by 2:- **a number is divisible by 2 if its once place is either 0, 2, 4, 6 or 8.

Ex:- 26 is divisible by 2. 35 not divisible by 2.

**Divisibility by 3:- **if the sum of the digits of a number is divisible by 3, then that number is divisible by 3.

Ex:- 231 → 2 + 3 +1 =6, 6 is divisible by 3

∴ 231 is divisible by 3

436 → 4 + 3 + 6 = 13, 13 is not divisible by 3

∴ 436 is not divisible by 3.

**Divisibility by 4:- **if the last two digits of a number is divisible by 4, then that number is divisible by 4.

Ex:- 4**36, **36 is divisible by 4 ∴ 436 is divisible by 4

6**23**, 23 is not divisible by 4 ∴ 623 is not divisible by 4.

**Divisibility by 5:- **a number is divisible by 5, if its once place is either 0 or 5.

Ex:- 20, 25 are divisible by 5. 22, 46 are not divisible by 5.

**Divisibility by 6:- **a number is divisible by 6, if it is divisible by both 3 and 2.

Ex:- 242 is divisible by both 2 and 3 ∴ 242 is divisible by 6

232 is divisible by 3 but not 2 ∴ 232 is not divisible by 6

**Divisibility by 8:- **if the last three digits of a number is divisible by 8, then that number is divisible by 8.

Ex:- 4**232**, last three digits 232 are divisible by 8

∴ 4232 is divisible by 8.

**Divisibility by 9:- **if the sum of the digits of a number is divisible by 9, then that number is divisible by 9.

Ex:- 459, 4 + 5 + 9 = 18 → 18 is divisible by 9 ∴ 459 is divisible by 9

532, 5 + 3 + 2 = 10 → 10 is not divisible by 9 ∴ 532 is not divisible by 9.

**Divisibility by 10:- **a number is divisible by 10 if its once place is 0.

Ex:- 20 is divisible by 10. 22, 45 are not divisible by 10.

**Divisibility by 11:- **A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places is either 0 or 11.

Ex:- 6545

Sum of the digits at odd places = 5 + 5 = 10

Sum of the digits at even places = 4 + 6 = 10

Now difference is 10 – 10 = 0

∴ 6545 is divisible by 11.

**Factors:** a number which divides the other number exactly is called a factor of that number.

6 = 1×6

= 2×3 ⟹ factors of 6 are: 1, 2, 3 and 6

Note- 1)1 is a factor of every number.

2) Every number is a factor of itself.

3) Every factor is less than are equal to the given number.

4) Factors of a given number are countable.

**Prime numbers: **The numbers, which have only two factors 1, and itself are called prime numbers.

2, 3, 5, 7, …. Are prime numbers

**Composite numbers: **The number, which has more than two factors are called composite numbers.

4, 6,8,9….. are composite numbers.

- Note: – 1) 1 is neither prime nor composite

2) 2 is the smallest prime number

3) 4 is the smallest composite number.

**Co – prime number: **The number which has no common factor except 1 is called co-prime number.

Ex:- (2, 3), (4,5) ……

**Twin – primes: **If the difference of two prime numbers is 2, then those numbers are called twin prime numbers.

Ex:- (2,3), (3,5), (17,19)…..

**Factorization: **When a number is expressed as the product of its factors, we say that the number has been factorized. The process of finding the factors is called Factorisation.

Ex:- factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24

24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6

**Prime factorisation: **The process of finding the prime factors is called prime factorisation.

Ex:- 24 = 2 × 12

2 × 3 × 4

2 × 3 × 2 × 2

∴ Prime factorisation of 24 is 2 × 2 × 2 × 3.

**Methods of prime factorization: **

**Division method:- **Prime factorization of 12 using the division method,

fallow the procedure.

Start dividing by the least prime factor. Continue division till the resulting number to be divided is 1.

The prime factorization of 12 is 2 × 2 × 3.

**Factor tree method:- **To find the prime factorization of 24, using the factor tree method we proceed as follows:

- Express 24 as a product of two numbers.
- Factorise 4 and 6 further, since they are composite numbers.
- Continue till all factors are prime numbers.
- The prime factorization of 24 is 2 × 2 × 2 × 3.

**Common factors: **Common factors are those numbers, which are factors of all the given numbers.

Ex:- 12, 9

Factors of 12 are: 1, 2, 3, 4, 6 and 12

Factors of 9 are: 1, 3 and 9

∴ Common factors of 12, 9 are 1,3

**Highest Common Factor (H.C.F):- **The highest common factor of two or more numbers is the highest of their common factors. It is also called ad Greatest Common Divisor(G.C.D).

Ex:- H.C.F of 12, 9

Factors of 12 = 1, 2, 3,4, 6, 12

** **Factors of 9 = 1, 3, 9

Common factors of 12, 9 = 1,3

Highest common factor is 3

∴ H.C.F of 12, 9 is 3

The HCF of 9 , 12 can be found by the prime factorization method as f]]>

**• Rational number:** The number, which is written in the form of p/q where p,q are integers q not equal to zero is called rational number. It is denoted by Q.

**• Irrational numbers:- **the number, which is not rational is called irrational number. It is denoted by Q’ or S.

**• Euclid division lemma :- **For any positive integers a and b, then q, r are integers exists uniquely satisfying he rule a = bq + r, 0 ≤ r < b.

**• Prime number :- **The number which has only two factors 1 and itself is called prime number. (2, 3, 5, 7 …. Etc.)

**• Composite number :- **the number which has more than two factors is called composite number. (4, 6, 8, 9, 10,… etc.)

**• Co-prime numbers :- **Two numbers are said to be co-prime numbers, if they have no common factor except 1. [Ex: (1, 2) , (3, 4), (4, 7)…etc.]

**• To find HCF, LCM by using prime factorisation method: H. C****.F**= product of the smallest power of each common prime factors of given numbers. **L.C.M **= product of the greatest power of each prime factor of given numbers.

- In p/q, if prime factorisation of q is in the form
**2**, then p/q is terminating decimal. Otherwise non-terminating repeating decimal.^{m}5^{n} - Decimal numbers with the finite no. of digits is called
Decimal numbers with the infinite no. of digits is called*terminating*decimal. In a decimal, a digit or a sequence of digits in the decimal part keeps repeating itself infinitely. Such decimals are called*non- terminating*decimals .*non- terminating repeating*

• ‘p’ is a prime number and ‘a’ is a positive integer, if p divides a^{2}, then p divides a.

• If a^{x} = N then x =

(i) log (xy) = log(x) + log(y) (ii) log (x/y) = log( x) – log( y) (iii) log (x^{m} ) = m log (x