TS Polycet (Polytechnic) 2021 Previous Question Paper with Answers And Model Papers With Complete Analysis

21)Which of the following equation is not a linear equation

A) 2+3x=y-5

B) 3-x=y2+7

C) x+3y=2y-x

D) 5x+2y=0

View Answer

B) 3-x=y2+7
Explanation:Which of the following equations is not a linear equation?
(2 + 3x = y – 5) → Linear(Degree 1)
(3 – x = y2 + 7) → Not linear(Degree 2 in (y))
(x + 3y = 2y – x) → Linear(Degree 1)
(5x + 2y = 0) → Linear(Degree 1)
22)If x2+kx+1=0 has a root x=1 then k =

A) 1

B) 2

C) 3

D) -2

View Answer

D) -2
Explanation:If (x2 + kx + 1 = 0) has a root (x = 1), then (k = ?)
Substitute (x = 1):
12 + k(1) + 1 = 0 ⇒ 1 + k + 1 = 0 ⇒ k = -2
23)If the roots of the quadratic equation ax2 + bx+c=0 are sinα and cosα, then 1+2\frac ca = …….

A) \frac{a^2}{b^2}

B) \frac{b^2}{a^2}

C) a2

D) b2

View Answer

B) \frac{b^2}{a^2}
Explanation:If the roots of (ax^2 + bx + c = 0) are (sin α) and (cos α), then (1 + 2\frac{c}{a} = ?)
Given roots: (sin α + cos α = –\frac{b}{a}) and (sin α cos α = $\frac{c}{a}).
Using the identity:
sin2 α + cos2 α = 1 ⇒ (sin α + cos α)2 – 2 sin α cos α = 1
Substitute:
\[ \left(-\frac{b}{a}\right)^2 – 2\left(\frac{c}{a}\right) = 1 ⇒ \frac{b^2}{a^2} – \frac{2c}{a} = 1 \] Rearrange:
\[ 1 + 2\frac{c}{a} = \frac{b^2}{a^2} \]
24)If the system of equation 3x-2y-7=0 and kx+2y+11=0 has unique solution then …….

A) k≠3

B) k≠-3

C) k=3

D) k=-3

View Answer

B) k≠-3
Explanation:If the system (3x – 2y – 7 = 0) and (kx + 2y + 11 = 0) has a unique solution, then:
For a unique solution, the lines must not be parallel:
\[ \frac{a_1}{a_2} ≠q \frac{b_1}{b_2} ⇒ \frac{3}{k} ≠q \frac{-2}{2} ⇒ \frac{3}{k} ≠q -1 ⇒ k ≠q -3 \]
25)If 7x-5y=2 and 3x+y=4, then x=

A) 3

B) -3

C) 1

D) 2

View Answer

C) 1
Explanation:If (7x – 5y = 2) and (3x + y = 4), then (x = ?)
From the second equation: (y = 4 – 3x).
Substitute into the first equation:
7x – 5(4 – 3x) = 2
⇒ 7x – 20 + 15x = 2
⇒ 22x = 22
⇒ x = 1
26)The distance between the points (0, 0) and (5,12) is

A) 11

B) 12

C) 13

D) 14

View Answer

C) 13
Explanation:Distance between (0, 0) and (5, 12) is:
\[ \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
27)If the slope of the line through (x,5) and (5,2) is 3, then the value of x is

A) 3

B) 4

C) 5

D) 6

View Answer

D) 6
Explanation:Slope of the line through (x, 5) and (5, 2) is 3. Find (x):
\[ \text{Slope} = \frac{2 – 5}{5 – x} = 3 ⇒ \frac{-3}{5 – x} = 3 ⇒ -3 = 15 – 3x ⇒ x = 6 \]
28)If △ABC~△PQR, ∠A=32°, ∠R=65° then ∠B=

A) 93°

B) 83°

C) 73°

D) 63°

View Answer

B) 83°
Explanation:If (ΔABC ~ ΔPQR), (∠A = 32°), (∠R = 65°), then (∠B = ?)
Corresponding angles are equal, so (∠C = ∠R = 65°).
Sum of angles in (ΔABC):
∠B = 180° – ∠A – ∠C = 180° – 32° – 65° = 83°
29)The angle in the minor segment is

A) obtuse

B) acute

C) 90°

D) None of these

View Answer

A) obtuse
Explanation:The angle in the minor segment is:
An angle in the minor segment is obtuse(greater than (90°).
30)In the figure ∠BAO=30°, ∠BCO=40° then ∠AOC=

A) 100°

B) 120°

C) 140°

D) 150°

View Answer

C) 140°
Explanation:In the figure, (∠BAO = 30°), (∠BCO = 40°). Then (∠AOC = ?)
From the figure (assuming (O) is the center of the circle):
– (∠AOC = 2(∠ABC) (Central angle theorem).
– (∠ABC = ∠BAO + ∠BCO = 30° + 40° = 70°).
– Thus, (∠AOC = 2 × 70° = 140°).
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