2025 May 21st EAPCET Engineering Forenoon Practice Test

Test Description

This EAPCET Engineering Practice Test was a previous year EAPCET exam conducted on 2025 May 21st shift 1. This test consists of 160 objective type questions from both Intermediate First Year and Intermediate Second Year Mathematics (80), Physics (40) and Chemistry (40). The questions are given in the same format as in the exam conducted by Andhra Pradesh (AP) EAPCET exam in the year 2025.

This test is completely free, and students can attempt as many times as needed and help students in preparing for EAPCET entrance tests and get admissions into various reputed colleges for Engineering in the state of Andhra Pradesh.

Test Instructions

The test contains 160 multiple-choice questions.
Total time for the test is 3h 00m.
Each question carries equal marks.
There is no negative marking for wrong answers.
Use of calculator is not allowed.
Do not refresh the page during the test.
Make sure you have a stable internet connection.

Quick Test

(1 of 160) The domain and range of a real valued function f(x) = cos x-3 are respectively.

R\{0} and [-1, 1]
R and [-1, 1]
R\{0} and [-4, -2]
R and [-4, -2]

(2 of 160) If

f : R \to R

and

g : R \to R

are two functions defined by

f (x) = 2 x - 3

and

g (x) = 5 x^{2} - 2

, then the least value of the function (gof)(x) is

-2
2
-4
4

(3 of 160) For all

n \in N

, if

1^{3} + 2^{3} + 3^{3} + . . . + n^{3} > x

, then a value of x among the following is

$\frac{n^{2}}{4}$
$n^{2}$
$n^{4}$
$\frac{n^{2} {(n + 1)}^{2}}{4}$

(4 of 160) If A and B are both

3 \times 3

matrices, then which of the following statements are true?

(i) AB = 0 \Rightarrow A = 0 or B = 0

(ii) AB = I_{3} \Rightarrow A^{- 1} = B

(iii) {(A - B)}^{2} = A^{2} - 2 AB + B^{2}

(i) is false and (ii), (iii) are true
(ii) is true and (i), (iii) are false
(i) and (ii) are true, (iii) is false
All are true

(5 of 160) If

A = [\begin{matrix} 1 & - 1 & 2 \\ - 2 & 3 & - 3 \\ 4 & - 4 & 5 \end{matrix}]

is the given matrix and

A^{T}

represents the transpose of A, then

{AA}^{T} - A - A^{T} =

$[\begin{matrix} 4 & 8 & 12 \\ 8 & 16 & - 28 \\ 12 & - 28 & 47 \end{matrix}]$
$[\begin{matrix} 4 & - 8 & 12 \\ - 8 & 16 & - 28 \\ 12 & - 28 & 47 \end{matrix}]$
$[\begin{matrix} 4 & - 8 & 12 \\ - 8 & 16 & 28 \\ 12 & 28 & 47 \end{matrix}]$
$[\begin{matrix} 4 & - 8 & - 12 \\ - 8 & 16 & - 28 \\ - 12 & - 28 & 47 \end{matrix}]$