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

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)

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

B) B
Explanation:If the roots of (ax^2 + bx + c = 0) are (sin α) and (cos α), then (1 + 2 = ?)
Given roots: (sin α + cos α = –) 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:
\[ 1 + 2\frac{c}{a} = \frac{b^2}{a^2} \]

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:
\[ \frac{a_1}{a_2} ≠q \frac{b_1}{b_2} ⇒ \frac{3}{k} ≠q \frac{-2}{2} ⇒ \frac{3}{k} ≠q -1 ⇒ k ≠q -3 \]

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

C) 13
Explanation:Distance between (0, 0) and (5, 12) is:
\[ \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

D) 6
Explanation:Slope of the line through (x, 5) and (5, 2) is 3. Find (x):
\[ \text{Slope} = \frac{2 – 5}{5 – x} = 3 ⇒ \frac{-3}{5 – x} = 3 ⇒ -3 = 15 – 3x ⇒ x = 6 \]

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°

A) obtuse
Explanation:The angle in the minor segment is:
An angle in the minor segment is obtuse(greater than (90°).

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°).