AP Inter 1. Year Maths 1a question paper 2025 with reply key, download pdf

AP Inter 1. Year Mathematics 1a Exam 2025: Candidates will find a detailed examination analysis on this page for the mathematics exam of the mathematics -1a examination of AP. The exam ended at 12 noon. This analysis includes the overall degree of difficulty of the paper and the level of difficulty of each section – section A, section B and section C C.

It will also offer an idea of ​​the expected good score and help the students to evaluate their performance. In addition, the ratings and feedback by students are recorded and updated here. These checks provide valuable insights into the question paper and help the students to better understand the exam.

AP Inter 1. Year Mathematics 1a question paper 2025

Download the question paper for the 1st year AP Inter Maths 1a 2025 test here:

AP Inter 1. Year Mathematics 1a Examination Analysis 2025

Here is the AP Inter 1st year Mathematics 1a examination analysis 2025 in a table information:

AP Inter 1. Year Maths 1a Replica 2025

The response key for the mathematics examination of AP Inter 1st year is now available. The students can check the correct answers and rate their performance in the exam. Check the answers and solutions below:

1. Find the domain of the real estimated function f (x) = 1/√ (1 – x²).

Answer: The domain of the function will be [−1,1].

2.Wenn a = {1,2,3,4} and f: a → r is a function that is defined by f (x) = (x² – X + 1)/ (x + 1) = 1 and then find the area of ​​F.

Solution: Given a = {1,2,3,4}

f: a → r so that f (x) = (x² – X + 1)/(X + 1)

f (1) = (1² – 1 + 1)/ (1 + 1) = 1/2

f (2) = (2² – 2+1)/(2+1) = (4 – 2+1)/3 = 3/3 = 1

f (3) = (3²−3+1)/(3+1) = (9–3+1)/4 = 7/4

f (4) = (4²–4+1)/(4+1) = (16 – 4+1)/5 = 13/5

∴ area of ​​f = {1/2,1,7/4.13/5}

3. Find the CO factors of elements 2 and -5 in the matrix,

Answer:

Co-factor of 2 = 17

CO -factor of -5 = 3

4.

Answer:

x = 2, y = 2, z = 5 and a = 5

5. Find a vector in the direction of Vector A = I – 2J with size 7 units.

Answer:

The required vector is:

S.

Answer:

The required vector equation is r = (1 – t) (2i + J + 3k) + T (-4i + 3j – K)

8. Find the maximum and minimal value of 3 Sinx – 4COSX.

Answer:

Maximum value = 5

Minimum value = -5

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