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Trigonometric Ratios vsaqs questions and solutions

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S designed by the ‘Basics in Maths‘ team.These notes to do help intermediate First-year Maths students. Inter Maths – 1A two marks questions and solutions are very useful in I.P.E examinations. Trigonometric Ratios Up to Transformations Question 1 Find the value of sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10) Sol: sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10) = sin2(π/10) + sin2(π/2 – π/10) + sin2(π/2+ π/10) + sin2(π – π/10) = sin2(π/10) + cos2(π/10) + cos2(π/10) + sin2(π/10) = 1 + 1 = 2 Question 2 If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ Sol: Given sin θ = 4/5 and θ not in the first quadrant ⇒ θ in the second quadrant ⇒ cos θ < 0 cos2θ = 1 – sin2 θ =1 – (4/5)2 = 1 – 16/25 ∴cos θ = – 3/5 (∵cos θ < 0) Question 3 If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ Sol: Given, 3sin θ + 4 cos θ = 5 let 4 sin θ – 3cos θ = x (3sin θ + 4 cos θ )2 + (4 sin θ – 3cos θ)2 = 52 + x2 9 sin2 θ + 16 cos2 θ + 12 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 12sin θ cis θ = 25 + x2 25 sin2 θ + 25 cos2 θ = 25 + x2 25 = 25 + x2 ⇒ x2 = 0 x = 0 ∴ 4 sin θ – 3cos θ = 0 Question 4 If sec θ + tan θ =, find the value of sin θ and determine the quadrant in which θ lies Sol: Given, sec θ + tan θ = ———— (1) We know that sec2 θ – tan2 θ = 1 ⇒ (sec θ + tan θ) (sec θ – tan θ) = 1 sec θ – tan θ = ⇒ sec θ – tan θ = ———— (2) (1) + (2) ⇒ (sec θ + tan θ) + (sec θ – tan θ) = 2sec θ = sec θ = (1) – (2) ⇒ (sec θ + tan θ) – (sec θ – tan θ) = 2 tan θ = ⇒ tan θ = Now sin θ = tan θ ÷ sec θ = Sin θ = Since sec θ positive and tan θ is negative θ lies in the 4th quadrant. Question 5 Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1 Sol: cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16) = cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16) = cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16) = [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4) = 1.1.1.1 =1 Question 6 If cos θ + sin θ = cos θ, then prove that cos θ – sin θ = sin θ Sol: Given, cos θ + sin θ = cos θ Sin θ = cos θ – cos θ = ( – 1) cos θ ( + 1) sin θ = ( + 1) ( – 1) cos θ sin θ + sin θ = cos θ ∴ cos θ – sin θ = sin θ Question 7 Find the value of 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) Sol: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) = 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2)2 = 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ] = 2[1 – 3 sin2 θ cos2 θ] – 3 [1 – 2 sin2 θ cos2 θ] = 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ = – 1 Question 8 If tan 200 = λ, then show that Sol: Given tan 200 = λ = = = = Question 9 If sin α + cosec α = 2, find the value of sinn α + cosecn α, n∈ Z Sol: Given sin α + cosec α = 2 ⇒ sin α + 1/ sin α = 2 ⇒ = 2 sin2 α + 1= 2 sin α sin2 α – 2 sin α + 1= 0 (sin α – 1 )2 = 0 ⇒ sin α – 1 = 0 sin α = 1 ⇒ cosec α = 1 sinn α + cosecn α = (1)n + (1)n =1 + 1 =2 ∴ sinn α + cosecn α = 2 Question 10 Evaluate sin2 + cos2 – tan2 Sol: Question 11 Find the value of sin 3300. cos 1200 + cos 2100. Sin 3000 Sol: sin 3300. cos 1200 + cos 2100. Sin 3000 =sin (3600 – 300). cos (1800 – 600) + cos (1800 + 300). sin (3600 – 600) = (– sin 300). (– cos 600) + (– cos300). (– sin600) = sin 300. cos 600 + cos300. Sin600 = sin (600 + 300) = sin 900 =1 Question 12 Prove that cos4 α + 2 cos2 α = (1 – sin4 α) Sol: cos4 α + 2 cos2 α = cos4 α + 2 cos2 α (1 – cos2 α) = (cos2 α)2 + 2 (1 – sin2 α) (sin2 α) = (1 – sin2 α)2 + 2 sin2 α – 2sin4 α = 1 + sin4 α – 2 sin2 α + 2 sin2 α – 2sin4 α = 1 – sin4 α Question 13 Eliminate θ from x =

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Product of Vectors vsaqs questions and solutions

Product of Vectors (Qns.& Ans) V.S.A.Q.’S

Education / Nayini Satyanarayana Reddy

Product of Vectors (Qns.& Ans) V.S.A.Q.’S Product of Vectors (Qns.& Ans) V.S.A.Q.’S; These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students. 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

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Addition of Vectors vsaqs questions and solutions

Addition of Vectors (Qns.& Ans) V.S.A.Q.’S (2 marks)

Education / Nayini Satyanarayana Reddy

Addition of Vectors Addition of Vectors: These solutions were designed by the ‘Basics in Maths’ team. These notes are to help intermediate First-year Maths students. Inter Maths – 1A two-mark questions and solutions are very useful in IPE examinations. Addition of Vectors QUESTION 1 Find the unit vector in the direction of Sol: Given vector is The unit vector in the direction of a vector is given by QUESTION 2 Find a vector in the direction of a where that has a magnitude of 7 units. Sol: Given vector is 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 = 2i + 2j – 5k and b = 2i + j + 3k Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k QUESTION 4 Write the direction cosines of the vector Sol: Given vector is ∴ Direction cosines are QUESTION 5 Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 a – c are collinear when a, b, c are non-collinear vectors Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 a – cA B = OB – OA = a + 2b + 3c – (– 2a + 3b + 5c) AB = 3a – b – 2c AC = OC – OA = 7 a – c – (– 2a + 3b + 5c) AC = 9a – 3b – 6c = 3(3a – b – 2c) 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) 𝛌, if AM = 𝛌 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 AM = = AD + ½ AB (∵ AB = DC) L is the midpoint of BC AL = = AB + ½ AD ((∵ BC = AD) (ii) AM = 𝛌 AD – LM AM + LM= 𝛌 AD AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD AD + ½ AB + AD + ½ AB – AB – ½ AD = 𝛌 AD 3/2 AD = 𝛌 AD ∴𝛌 = 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 OD = G divides median AD in the ratio 2: 1 OG = ∴ OG = QUESTION 8 If = , = are collinear vectors, then find m and n. Sol: Given , are collinear vectors ⟹ = λ Equating like vectors 2 = 4 λ; 5 = m λ; 1 = n λ λ = 5 = m ⟹ m =10 1 = n ⟹ n = 2 ∴ m = 10, n = 2 QUESTION 9 Let If , . Find the unit vector in the direction of a + b. Sol: Given vectors are and a + b = The unit vector in the direction of a + b = = = QUESTION 10 If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ. Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k ⟹ a = tb – 3i + 4j + λk = t (μi + 8j + 6k) – 3i + 4j + λk = μt i + 8t j + 6t k Equating like vectors – 3 = μt; 4 = 8t, λ = 6t 4 = 8t t = – 3 = μ ⟹μ=– 6 λ = 6 ⟹ λ = 3 ∴ μ=– 6, λ = 3 QUESTION 11 ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌 Sol: Given, ABCD is a pentagon AB + AE + BC + + DC + ED + AC = 𝛌 AC (AB + BC) + (AE + DC + ED) + AC = 𝛌 AC AC + AC + AC = 𝛌 AC 3 AC = 𝛌 AC 𝛌 = 3 QUESTION 12 If the position vectors of the points A, B and C are – 2i + j – k and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌 Sol: Given, OA = – 2i + j – k , OB = –4i + 2j + 2k and OC = 6i – 3j – 13k AB = OB – OA = –4i + 2j + 2k – (– 2i + j – k) = –4i + 2j + 2k +2i – j + k = –2i + j + 3k AC = OC – OA = 6i – 3j – 13k – (– 2i + j – k) = 6i – 3j – 13k +2i – j +

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Matrices 2m Questions( Qns & Solutions) || V.S.A.Q’S||

Education / Nayini Satyanarayana Reddy

Matrices 2m Questions ( Qns & Solutions) || V.S.A.Q’S|| Matrices 2m Questions: Matrices V.S.A.Q’s: This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1A two marks 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. Matrices QUESTION 1 If A = , then show that A2 = –I Sol: Given A = ∴ A2 = –I QUESTION 2 If A = , and A2 = 0, then find the value of k. Sol: Given A = and A2 = 0 ⟹ A. A =0 ⟹ = 0 ⟹ = 0 8 + 4k = 0, – 2 – k = 0 and –4 + k2 = 0 4k = –8; k = –2; k2 = 4 k = –2; k = –2; k = ± 2 ∴ k =– 2 QUESTION 3 Find the Trace of A, If A = Sol: Given A = 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 2X = – = = X = ∴ X = QUESTION 5 Find the additive inverse of A, If A = Sol: Given A = Additive inverse of A = – A = – = QUESTION 6 If , then find the values of x, y, z and a. Sol: Given ⟹ 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 aij = Sol: Let A= a11 = a11 = 1 a12 = a12 = a21 = a21 = a22 = a22 = 2 a31 = a31 = 0 a32 = a32 = ∴ A = QUESTION 8 If A = and B = , do AB and BA exist? If they exist, find them. BA and AB 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 = BA = 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 AT = A, then A is called Symmetric Matrix Skew Symmetric Matrix: Let A be any square matrix if AT = –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 ⟹ AT = A ⟹ x = 6 QUESTION 11 If A = is a skew-symmetric matrix, then find x Sol: Given A = is a skew-symmetric matrix ⟹ AT = – A ⟹ x = –x x+ x = 0 ⟹ 2x = 0 ⟹ x = 0 QUESTION 12 If A = and B = , then find (A BT) T Sol: Given A = B = BT = (A BT) = = (A BT) T = QUESTION 13 If A = and B = , then find A + BT Sol: Given A = and B = BT = A + BT = + QUESTION 14 If A = , then show that AAT = ATA = I Sol: Given A = AT = AAT = = = ATA = = = ∴ AAT = ATA = I QUESTION 15 Find the minor of – 1 and 3 in the matrix Sol: Given Matrix is minor of – 1 = = 0 + 15 = 15 minor of 3 = = – 4 + 0 = – 4 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+ω2 = 0) Given matrix is R1 → R1 + R2 + R3 = 0 (∵ 1 + ω+ω2 = 0) QUESTION 18 If A = and det A = 45, then find x. Sol: Given A = Det A = 45 ⟹ = 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) A = Adj A = A-1 = = ∴ A-1 = (ii) A = Adj A = A-1 = ∴ A-1 = QUESTION 20 Find the inverse of (abc ≠ 0) Sol: Let A = Det A = a (bc – 0) – 0(0 – 0) + 0(0

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The Plane 2m Questions & Solutions || V.S.A.Q’S||

Education / Nayini Satyanarayana Reddy

The Plane 2m Questions & Solutions || V.S.A.Q’S|| The Plane 2m Questions: The Plane: These solutions were designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations. 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 Plane Question 1 Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, – 5). Sol: The plane passes through A and is perpendicular to OA, then the line segment OA is normal to the plane. Dr’s of OA = (2, 3, – 5) The equation of the plane passing through the point (x1, y1, z1) and dr’s (a, b, c) is a(x – x1) + b (y – y1) + c (z – z1) = 0 ⟹ 2(x – 2) + 3 (y – 3) – 5 (z + 5) = 0 2x – 4 + 3y – 9 – 5z – 25 = 0 2x + 3y – 5z – 38 = 0 Question 2 Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) Sol: The equation of the plane passing through the points (x1, y1, z1) (x2, y2, z2) (x3, y3, z3) is The plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) is =0 ⟹ = 0 x (30 – 20) – (y + 1) (20 – 6) + (z + 1) (40 – 18) = 0 x (10) – (y + 1) (14) + (z + 1) (22) = 0 10x – 14y – 14 + 22z + 22 = 0 10x – 14y + 22z + 8 = 0 2(5x – 7y + 11z + 4) = 0 ∴ the equation of the plane is 5x – 7y + 11z + 4 = 0 Question 3 Find the equation to the plane parallel to the ZX-plane and passing through (0, 4, 4). Sol: Equation of ZX-plane is y = 0 The equation of the plane parallel to the ZX-plane is y = k But it is passing through (0, 4, 4) ⟹ y = 4 Question 4 Find the equation to the plane passing through the point (α, β, γ) and parallel to the plane axe + by + cz + d = 0. Sol: The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0 But it is passing through the point (α, β, γ) a α + b β + c γ + k = 0 ⟹ k = – a α – b β – c γ The equation of the plane is ax + by + cz – a α – b β – c γ = 0 ⟹ a(x – α) + b (y – β)+ c (z – γ) = 0 The Plane 2m Questions Question 5 Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7. Sol: If θ is the angle between the planes a1 x + b1 y + c1 z + d1 = 0 and a2x + b2 y + c2 z + d2 = 0, then cos θ = Cos θ = = = Cos θ = cos 600 θ = 600 = Question 6 Reduce the equation x + 2y – 2z – 9 = 0 to the normal form and hence find the dc’s of the normal to the plane. Sol: Given plane is x + 2y – 2z – 9 = 0 x + 2y – 2z = 9 dividing into both sides by the normal form is dc’s of the normal to the plane are Question 7 Suppose a plane makes intercepts 2, 3, 4 on X, Y, Z axes respectively. Find the equation of the plane in the intercept form. Sol: Given a = 2, b = 3, c = 4 The equation of the line in the intercept form is ⟹ Question 8 Express x – 3y + 2z = 9 in the intercept form Sol: Given plane is x – 3y + 2z = 9 ⟹ ⟹ It is in the form of a = 9, b = – 3, c = 9/2 The Plane 2m Questions Question 9 Find the direction cosine of the normal to the plane x + 2y + 2z – 4 = 0. Sol: Given plane is x + 2y + 2z – 4 = 0 We know that Dr’s of the normal to the plane ax + by + cz + d = 0 are (a, b, c) ⟹ dc’s of the normal to the plane = ⟹ dr’s of the normal to the plane x + 2y + 2z – 4 = 0 are (1, 2, 2) ⟹ dc’s of the normal to the plane are = Question 10 Find the midpoint of the line joining the points (1, 2, 3) and (–2, 4, 2) Sol: Given points are A (1, 2, 3), B (–2, 4, 2) The midpoint of AB = = = Visit my Youtube Channel: Click on Below Logo

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3D Coordinates vsaqs questions and solutions

Three Dimensional Coordinates (Q’s & Ans) || V.S.A.Q’S||

Education / Nayini Satyanarayana Reddy

Three Dimensional Coordinates || V.S.A.Q’S|| Three-Dimensional Coordinates: These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations. 3D Coordinates 1. Show that the points A (– 4, 9, 6), B (– 1, 6, 6), and C (0, 7, 10) form a right-angled isosceles triangle. Sol: Distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is PQ = ∴ The points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle. Three-Dimensional Coordinates 2. Show that the locus of the point whose distance from the Y–axis is thrice its distance from (1, 2, – 1) is 8x 2 + 9y2 + 8 z2 – 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 PA2 = (3PB)2 PA2= 9 PB2 ⟹ x2 + z2 = 9[(x – 1)2 + (y – 2)2 + (z + 1)2] x2 + z2 = 9[x2 – 2x + 1 + y2 – 4y + 4 + z2 + 2z + 1] x2 + z2 = 9×2 – 18x + 9 + 9y2 – 36y + 36 +9 z2 + 18z + 9 ∴ 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0 3. A, B, and 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 that 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 ⟹ AP2 =OP2 (x – a )2 + y2 + z2 = x2 + y2 + z2 x2 + a2 – 2ax + y2 + z2 = x2 + y2 + z2 a2 – 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) 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) 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 ⟹ AB2 = 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 6.If the point (1, 2, 3) is changed to 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) 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 (x1, y1, z1), (x2, y2, z2) in the ratio, then Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3 Three-Dimensional Coordinates 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 9. Using the section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), and 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 2 – 4k = 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 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 (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is 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 tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3,

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Straight Lines vsaqs questions and solutions

Straight Lines 2m Questions & Solutions || V.S.A.Q’S||

Education / Nayini Satyanarayana Reddy

Straight Lines 2m Questions & Solutions || V.S.A.Q’S|| Straight Lines 2m Questions:This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students. 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. Straight Lines QUESTION1. 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 slope of the line segment joining the points (x1, y1) and (x2, y2) is Slope of AB = = 4 Slope of BC = = = 4 QUESTION2. Find the condition for the points (a, 0), (h, k), and (0, b) to be collinear. Sol: Let A (a, 0), B (h, k) and C (0, b) The slope of the line segment joining the points (x1, y1) and (x2, y2) is Given that A, B, and C are collinear points The slope of AB = The slope of BC ⟹ ⟹ ⟹ – hk = (h – a) ( b – k) – hk = hb – hk – ab + ak ⟹ 0 = hb + ak – ab ⟹ hb + ak = ab or QUESTION3. Find the equations of the straight lines which makes intercepts whose sum is sum is 5 and product is 6. Sol: The equation of the line in the intercept form is Given that, a + b = 5 and ab = 6 ⟹ b = 5 – a a(5 – a) = 6 5a – a2 = 6 a2 – 5a + 6 = 0 a – 3a – 2a + 6 = 0 a (a – 3) – 2(a – 3) = 0 (a – 3) (a – 2) = 0 a = 3 or a = 2 case (i) if a = 3 ⟹ b = 2 ⟹ 2x + 3y – 6 = 0 case (ii) if a = 2 ⟹ b = 3 = 1 ⟹ 3x + 2y – 6 = 0 QUESTION4. Find the equation of the straight line which makes an angle 1350 with the positive X – axis measured countered clockwise and passing through the point (– 2, 3). Sol: Slope of the line m = tan 1350 = – 1 The point is (– 2, 3) The equation of the straight line in slope point form is (y – y1) = m (x – x1) The equation of the line passing through the point (– 2, 3) with slope – 1 is y – 3 = – 1 (x + 2) y – 3 =– x – 2 x + y – 1 = 0 QUESTION5. Find the equation of the straight line passing through the points (1, – 2) and (– 2, 3). Sol: Given points are (1, – 2), (– 2, 3) The equation of the straight line in two points form is (y – y1) = (x – x1) The equation of required straight line is (y + 2) = (x – 1) (y + 2) = (x – 1) – 3 (y + 2) = 5 (x – 1) – 3y – 6 = 5x – 5 5x + 3y + 1 = 0 QUESTION6. Find the slopes of the line x + y = 0 and x – y = 0 Sol: The slope of the line ax + by + c = 0 is The slope of the line x + y = 0 is = – 1 The slope of the line x – y = 0 is = 1 QUESTION7. Find the angle which the straight-line y = x – 4 makes with the Y-axis. Sol: Given equation is y = x – 4 Compare with y = mx + c m = ⟹ tan θ = θ = Angle made by the line with X-axis is Angle made by the line with Y-axis is QUESTION8. Find the equation of the reflection of the line x = 1 in the Y-axis. Sol: Given equation is x = 1 Reflection about the Y-axis is x =– 1 Required equation of the line is x + 1 = 0 QUESTION9. Write the equations of the straight lines parallel to X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis. Sol: (i) The equation of the straight line parallel to X-axis which is at a distance of 3 units above the X-axis is y = 3 ⟹ y – 3 = 0 (ii) The equation of the straight line parallel to X-axis which is at a distance of 4 units below the X-axis is y = – 4 ⟹ y + 4 = 0 QUESTION10. Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it. Sol: (i) The equation of the straight line parallel to the Y-axis which is at a distance of 2 units from the Y-axis to the right of it is x = 2 ⟹ x – 2

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Functions vsaqs questions and solutions - 1

Functions 2M Questions &Solutions || V.S.A.Q.’S||

Education / Nayini Satyanarayana Reddy

Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| Functions (2M Questions &Solutions): This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1A 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. Functions QUESTION 1 Find the Domain of the following real-valued functions. (i) f(x) = Sol: given f(x) = It is defined when 6x – x2 – 5 ≠ 0 ⇒ x2 – 6x + 5 ≠ 0 x2 – 5x – x + 5 ≠ 0 x (x – 5) –1(x – 5) ≠ 0 (x – 5) (x – 1) ≠ 0 x ≠ 5 or x ≠ 1 ∴ domain = R – {1, 5} (ii) f(x) = Sol: Given f(x) = It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0 ⇒ x ≥ –3, x ≤ 3 and x ≠ 0 ⇒ –3≤ x, x ≤ 3 and x ≠ 0 ⇒ –3≤ x ≤ 3 and x ≠ 0 x ∈ [–3, 3] – {0} ∴ domain = [–3, 3] – {0} (iii) f(x) = Sol: Given f(x) = It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0 ⇒ x ≥ –2, x < 1 and x ≠ 0 x ∈ [–2, ∞) ∩ (– ∞, 1) – {0} ⇒ x ∈ [–2, 1) – {0} ∴ domain = [–2, 1) – {0} (iv) f(x) = Sol: Given f(x) = It is defined when 4x – x2 ≥ 0 ⇒ x2 – 4x ≤ 0 x (x – 4) ≤ 0 (x – 0) (x – 4) ≤ 0 x ∈ [0, 4] ∴ domain = [0, 4] (v) f(x) = log (x2 – 4x + 3) Sol: Given f(x) = log (x2 – 4x + 3) It is defined when x2 – 4x + 3 > 0 ⇒ x2 – 3x – x + 3 > 0 x (x – 3) –1(x – 3) > 0 (x – 3) (x – 1) > 0 x ∈ (–∞, 1) ∪ (3, ∞) x ∈ R – [1, 3] ∴ domain = R – [1, 3] (vi) f(x) = Sol: Given f(x) = It is defined when x2 – 1 ≥ 0 and x2 – 3x + 2 > 0 (x + 1)(x – 1) ≥ 0 and x2 – 2x – x + 2 > 0 (x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0 x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞) ∴ domain = R – (–1, 2] (vii) f(x) = Sol: Given f(x) = It is defined when – x > 0 ⇒ > x ⇒ x < 0 ∴ domain = (–∞, 0) (viii) f(x) = Sol: Given f(x) = It is defined when +x ≠0 ⇒ ≠ – x ⇒ x > 0 ∴ domain = (0, ∞) QUESTION 2 If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x2 + 2 then find (i) (gof) (x) (ii) (gof) () (iii) (fof) (x) (iv) go(fof) (0). Sol: Given f(x) = 4x – 1, g(x) = x2 + 2 (i) (gof) (x) = g (f (x)) = g (4x – 1) = (4x – 1)2 + 2 = 16×2 – 8x + 1 + 2 = 16×2 – 8x + 3 (ii) (gof) ()= (g (f ())

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TS VII CLASS MATHS CONCEPT FEATURE IMAGE

TS 7th Class Maths Concept

Education, TS State VI - IX / Nayini Satyanarayana Reddy

7th Class Maths Concept 7th Class Maths TS 6th Maths Concept TS 7th Maths Concept TS 8th Maths Concept TS 9th Maths Concept TS 10th Maths Concept Studying maths in VII class successfully means that children take responsibility for their learning and learn to apply the concepts to solve problems. 7th Class Maths concepts were designed by the ‘Basic in Maths’ team. These notes to do help students fall in love with mathematics and overcome fear. 1. INTEGERS 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,….. Integers on a number line: Operations on integers: addition of integers: 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. The 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 a negative integer by a positive integer or a positive integer by a 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). 2. FRACTIONS, DECIMALS AND RATIONAL NUMBERS 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. Ex: , and so on 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 a smaller denominator is smaller. Ex: – and so on. 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: Ex: Unlike fractions: – First, we have to find the equivalent fraction of given fractions and then subtract them as like fractions Ex: Multiplication of fractions: Multiplication of fraction by a whole number: – Multiplication of numbers means adding repeatedly. Ex: – • 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: Ex: – 2 ÷ ⇒ 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

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

Education, TS State VI - IX / Nayini Satyanarayana Reddy

6th maths notes|| TS 6 th class Maths Concept 6th Maths TS 6th Maths Concept TS 7th Maths Concept TS 8th Maths Concept TS 9th Maths Concept TS 10thMaths Concept Studying maths in the 6th class successfully meaning that children take responsibility for their own learning and learn to apply the concepts to solve problems. This note is designed by the ‘Basics in Maths’ team. These notes to do help students fall in love with mathematics and overcome fear. 1. KNOWING OUR NUMBERS • Number: A number is a mathematical object used to count and measure.1,2,3…….etc. Comparing numbers: • We can compare the numbers by counting the digits in the numbers. • 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: • Ascending Order: – arrange the numbers from smallest to the greatest; this order is called Ascending order. Ex:- 23, 44, 65, 79, 100 • Descending Order: – 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 • 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 6th Maths Reading and Writing the numbers 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 Use of commas: • 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 2. WHOLE NUMBERS 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 on the number line: 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:-

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