TS POLYCET 2023 Question Paper Solutions – Complete Maths Answers PDF
TS POLYCET 2023 Question Paper Solutions – Complete Maths Answers PDF Are you searching for TS POLYCET 2023 Question Paper Solutions? You are in the right place. Practising previous year question papers is one of the smartest ways to score high marks in the TS POLYCET exam. It helps students understand the exam pattern, important topics, and time management. In this post, we provide complete and easy-to-understand solutions for the TS POLYCET 2023 Maths Question Paper, with clear step-by-step explanations. The TS POLYCET 2023 question paper solutions are highly useful for every student preparing for the upcoming exams. Solve the paper sincerely and compare answers with detailed solutions to improve your score. Stay connected for more previous papers, mock tests, and chapter-wise maths solutions. Real Numbers 1. If ‘n’ is a prime number, then √n is (1) Prime number (2) Composite number (3) Rational number (4) irrational number ‘n’ అనేది ఓకే ప్రధాన సంక్య అయిన √n అనేది (1) ప్రధాన సంఖ్య (2) సంయుక్త సంఖ్య (3) అకరణీయ సంఖ్య (4) కరణీయ సంఖ్య Answer: (4) Solution: If ‘n’ is a prime number, then √n is an irrational number 2.Among 1/2, 1/3, 1/4, 1/5 the non-terminating decimal is 1/2, 1/3, 1/4, 1/5 అనే సంఖ్యలలో అంతం కాని దశాంశం (1) 1/2 (2) 1/3 (3) 1/4 (4) 1/5 Answer: (2) Solution: A fraction in its simplest form has a terminating decimal if the prime factors of its denominator are only 2, 5, or both. If the denominator has any other prime factor, it results in a non-terminating repeating decimal. 3. The value of log625(5) l og625(5) యొక్క విలువ (1) 1/2 (2) 1/4 (3) 1/3 (4) 1/5 Answer: (2) Solution: let log625(5) = x 5 = 625x 5 = (54)x 5 = 54x 1 = 4x x = 1/4 4. √2 + √3 is (1) Rational number (2) Irrational number (3) Prime number (4) composite number √2 + √3 is అనునది ఒక (1) అకరణీయ సంఖ్య (2) కరణీయ సంఖ్య (3) ప్రధాన సంఖ్య (4) సంయుక్త సంఖ్య Answer: (2) Solution: √2 + √3 is an irrational number 5. HCF of 7, 8, 9 is 7, 8, 9 ల గా.సా.భా. (1) 9 (2) 7 (3) 1 (4) 2 Answer: (3) Solution: factors of 7 = 1, 7 factors of 8 = 1, 2, 4, 8 factors of 9 = 1, 3, 0039 HCF 0f 7, 8, 9 = 1 Sets 1. If A={P, O, L, Y, T, E,C, H, N, I} and B = {E, X, A, M}then A∩B = A={P, O, L, Y, T, E,C, H, N, I} మరియు B = {E, X, A, M}అయితే, A∩B = (1) {P} (2) {E} (3) {X} (4) {T} Answer: (2) Solution: Given A = {P, O, L, Y, T, E, C, H, N, I} and B = {E, X, A, M} A∩B = {P, O, L, Y, T, E, C, H, N, I}∩{E, X, A, M} = {E} 2. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7} then A – B = A = {1, 2, 3, 4, 5} మరియు B = {4, 5, 6, 7} అయితే, A – B = (1) {1, 2, 3} (2) {3, 4, 5} (3) {5, 6, 7} (4) {2, 3, 4} Answer: (1) Solution: Given A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7} A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3} Polynomials 1. Product of zeroes of polynomial 5×2 – 1 is 5×2 – 1 అనే వర్గ బహుపదీ శూన్యాల లబ్దము (1) 1 (2) ½ (3) 1/5 (4) – 1/5 Answer: (4) Solution: Given polynomial is 5×2 – 1 Product of zeroes = c/a = – 1/5 2. (x + a) is a factor of f(x), if (x + a) అనేది f(x) యొక్క కారణాంకం అయినచో (1) f(a) = 0 (2) f(–a) = 0 (3) f(1/a) = 0 (4) f(– 1/a) = 0 Answer: (2) Solution: Given (x + a) is a factor of f(x),then f(–a) = 0 3. If α, β are the zeroes of the quadratic polynomial ax2 + bx + c, a≠0, then α2 + β2 = ax2 + bx + c, a≠0 అనే వర్గ బహుపది యొక్క శూన్యాలు α, β అయిన α2 + β2 = (1) (b2 + 2ac) (2) (c2 + 2ab) (3) (b2 – 2ac) (4) (c2 – 2ab) Answer: (2) Solution: Given polynomial is ax2 + bx + c, a≠0 α + β = – b/a αβ = c/a α2+ β2 = (α + β)2 – 2αβ α2+ β2 = (– b/a)2 – 2(c/a) α2+ β2 = b2/a2 – 2c/a α2+ β2 = (b2 – 2ac) Linear equations in Two Variables 1. If 5x + py + 8 = 0 and 10x + 15y + 12 = 0 has no solution, then p = 5x + py + 8 = 0 మరియు 10x + 15y + 12 = 0 అను సమీకరణాలకు సాధన లేనిచో, p విలువ (1) 15/2 (2) 13/2 (3) 7/2 (4) 5/2 Answer: (1) Solution: Given 5x + py + 8 = 0 and 10x + 15y + 12 = 0 have no solution a1 = 5; b1 = p; c1 = 8 a2 = 10; b2 = 15; c2 = 12 p = 15/2 2. If ax + b = 0 then x = ax + b = 0 అయిన, x విలువ (1) – a (2) a (3) b/a (4) – b/a Answer: (4) Solution: Given ax + b = 0 ⟹ ax = – b ⟹ x = – b/a 3. The solution of system of equations and is మరియు సమీకరణాల సాధన (1) (1/4, 1/3) (2) (1/3, 1/4) (3) (1/2, 1/3) (4) (1/3, 1/2) Answer: (3) Solution: Given equations are and Let = a and = b ⟹2a + 3b= 13 and 5a – 4b = – 2 a
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