AP Polycet (Polytechnic) 2025 Previous Question Paper with Answers And Model Papers With Complete Analysis

AP Polycet (Polytechnic) 2025 Previous Question Paper with Answers And Model Papers With Complete Analysis
AP Polycet (Polytechnic) Previous Year Question Papers And Model Papers:
While preparing for AP Polycet (Polytechnic), candidates must also refer to the previous year question papers of the same. Scoring well in AP Polycet (Polytechnic) and understanding weaknesses and strengths in the respective sections.

AP 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.

  1. Polycet 2025 Mathematics
  2. Polycet 2025 Physics
  3. Polycet 2025 Chemistry


AP POLYCET 2025
Section — A
MATHEMATICS



1) The points (1,5), (2,3) and (-2, -11) form a:

A) triangle
B) parallelogram
C) square
D) They are collinear

View Answer
A) triangle

Explanation:The points (1,5), (2,3), and (-2,-11) form a:
We check if the points are collinear using the slope method:
Slope of (1,5) and (2,3):
m_1 = \frac{3 - 5}{2 - 1} = -2
Slope of (2,3) and (-2,-11):
m_2 = \frac{-11 - 3}{-2 - 2} = \frac{-14}{-4} = 3.5
Since slopes are not equal, points are not collinear.
They form a triangle.
Answer: triangle

2) If 15 cotA = 8, then sinA = ?

A) \dfrac{8}{15}
B) \dfrac{15}{17}
C) \dfrac{17}{15}
D) \dfrac{8}{17}

View Answer
B) \dfrac{15}{17}

Explanation:If 15 cotA = 8, then sinA = ?
Let’s use Pythagoras.
Given:
\cot A = \frac{8}{15} \Rightarrow \text{adjacent} = 8, \text{opposite} = 15
Then, hypotenuse =
\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17
So,
\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{17}
Answer: \dfrac{15}{17}

3) Evaluate \dfrac{2 \tan 30°}{1 + \tan^2 30°}

A) sin 60°
B) tan 60°
C) sin 30°
D) cot 60°

View Answer
B) tan 60°

Explanation:Evaluate \dfrac{2 \tan 30°}{1 + \tan^2 30°}
We use identity:
\tan 2A = \frac{2\tan A}{1 + \tan^2 A}
So this becomes:
\tan 60° = \sqrt{3}
Answer: tan 60°

4) Evaluate (sec A + tan A) (1 – sinA) = ?

A) sin A
B) cos A
C) csc A
D) sec A

View Answer
B) cos A

Explanation:Evaluate (\sec A + \tan A) (1 - \sin A)
Use identities:
\sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A}
Then:
\frac{1 + \sin A}{\cos A} \cdot (1 - \sin A) = \frac{(1 - \sin^2 A)}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A
Answer: cos A

5) Which of the following is true?

A) sin(A + B) = sinA + sin B
B) The value of sin θ increases as θ increases, 0° ≤ θ ≤ 90°
C) The value of cos θ increases as θ increases, 0° ≤ θ ≤ 90°
D) sin θ = cos θ for all values of θ

View Answer
B) The value of sin θ increases as θ increases, 0° ≤ θ ≤ 90°

Explanation:Which of the following is true?
False — sin(A + B) ≠ sinA + sinB
True — sinθ increases from 0 to 1 as θ increases from 0° to 90°
False — cosθ decreases in that interval
False — sin θ = cos θ only when θ = 45°
Answer: The value of sin θ increases as θ increases, 0° ≤ θ ≤ 90°

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