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

Inter 1st year Maths Reduced Syllabus |
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Inter 2nd year Maths Reduced Syllabus |
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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: –**

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