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TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis

TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis
TS Polycet (Polytechnic) Previous Year Question Papers And Model Papers:
While preparing for TS Polycet (Polytechnic), candidates must also refer to the previous year question papers of the same. Scoring well in TS Polycet (Polytechnic) and understanding weaknesses and strengths in the respective sections.

TS Polycet (Polytechnic) Previous Year Question Papers can be found on this page in PDF format. Students taking the exam to get into some of the best Polytechnic colleges/institutes in the state of Andhra Pradesh may practice these papers to get a clear idea of the structure of the exam, marking scheme, important topics, etc.

TS Polycet Mock Test

Section — I
MATHEMATICS

1)10^th term of A.P.: 13, 8, 3, -2,…………… is ___________

A) -32

B) -23

C) 30

D) -30

View Answer

A) -32
Explanation:13, 8, 3,-2…….. A.P. Here a = 13, d = 8 — 13 = —5 N^th term a_n = a + (n – 1) d ∴ 10^th term a₁₀ = 13 + (10 – 1) (-5) = 13 + 9 (-5) = 13-45 = -32

2)The centroid of the triangle whose vertices are (3, -5), (-7, 4), (10, -2) is ____________

A) (1, 1)

B) (1,-2)

C) (-2,1)

D) (2,-1)

View Answer

D) (2,-1)
Explanation:The centroid of the Ale formed by the points A (x₁,y₁), B (x₂,y₂), C (x₃,y₃) is G = ( $\\frac{x1+x2+x3}3,\\frac{y1+y2+y3}3$

) G = ( $\\frac{3-7+10}3,\\frac{-5+4-2}3$

) G = ( $\\frac63,\\frac{-3}3$

) = (2,-1)

3)If the roots of 2 x²+ kx + 3 = 0 are real and equal, then k =

A) ±6√2

B) ±4

C) ±2√6

D) ±5

View Answer

C) ±2√6
Explanation:Given quadratic equation is 2 x²+ kx + 3 = 0 Comparing with a x² + bx + c = 0 a = 2, b = k, c = 3 Roots are real and equal, then b² – 4ac = 0 => (k)² – 4(2)(3) = 0 => k² – 24 = 0 k² = 24 => k = ± √24 = ±2 √6

4)Product of the roots of √3 x² – 2x -√3 =0 is

A) 1

B) -1

C) 2/√3

D) -2/√3

View Answer

B) -1
Explanation:Given quadratic equation is √3 x²-2x-√3=0 Comparing with a x² + bx + c = 0 a = √3, b = —2, c = — √3 Product of the roots = c/a = -√3/√3 = -1

5)8 x²-6x-9 = ________

A) (2x-3)(x-3)

B) (2x-3)(x + 1)

C) (2x+1) (x- 1)

D) (2x – 3) (4x + 3)

View Answer

D) (2x – 3) (4x + 3)
Explanation:8 x²-6x-9 = 8 x²– 12x + 6x-9 = 4x(2x-3) + 3 (2x-3) = (2x – 3) (4x + 3)

6)1,-1/2,1/4,…………………. are in G.P., then find 8th term.

A) 1/128

B) 1/64

C) -1/128

D) -1/64

View Answer

C) -1/128
Explanation:1,-1/2,1/4,………………G.P. Here s = 1, r=-1/2/1 = -1/2, n = 8 A_n = a r^n-1 ∴ A₈ =1(-1/2)^8-1 = (-1/2)⁷ = -1/128

7)Which term of the G.P.√3,3,3√3 ,…………………… is 729?

A) 10

B) 12

C) 14

D) 16

View Answer

B) 12
Explanation:√3,3,3√3……………..G.P. Here a = √3, r = 3/√3 = √3, a_n = 729 a_n = a. r^n-1 729 = √3 (√3)^n-1 729 = (√3)ⁿ => 729 = 3^n/2 => 3⁶ = 3^n/2 => 6 = n/2 => n = 12

8)Roots of 5 x²– 8x = 4 are:

A) 2,-2/5

B) 1,8/5

C) 2,1/5

D) 2,7

View Answer

A) 2,-2/5
Explanation:5 x²-8x = 4 =>5 x²-8x-4 = 0 =>5 x²-10x + 2x-4 = 0 =>5x(x-2) + 2(x-2) = 0 => (x -2) (5x + 2) = 0 => x – 2 = 0 (or) 5x + 2 = 0 => x = 2 (or) x = -2/5

9)5=

A) 7^log₁₂5

B) 7^log₅7

C) 7^log₇5

D) None

View Answer

C) 7^log₇5
Explanation:5 = 7^log75 [∴a^log_am = m]

10)log_x2 + log_x2² + log_x2³+………..+ log_x2ⁿ = $\\frac{n(n+1)}2$ , then x =

A) n

B) 1

C) 5

D) 2

View Answer

D) 2
Explanation:log_x2 + log_x2² + log_x2³+………..+ log_x2ⁿ = n(n+1)/2 log_x2 + 2 log_x2 + 3 log_x2 +…………………+n log_x2 = n(n+1)/2 log_x2 (1+2+3 +……………..+n) = n(n+1)/2 log_x2 [n(n+1)/2] = n(n+1)/2 log_x2 = 1 => x¹ = 2 => x = 2

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TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis

Section — I
MATHEMATICS

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Section — I MATHEMATICS

1 thought on “TS Polycet (Polytechnic) 2020 Previous Question Paper with Answers And Model Papers With Complete Analysis”

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Section — I
MATHEMATICS