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TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

TS EAMCET 2025 Engineering question paper May 4 Shift 1 with answers and practice real exam questions. This EAPCET previous paper with solutions helps you understand exam pattern, difficulty level, and important topics. Solving TS EAMCET previous year question papers improves accuracy and time management. This paper includes Maths, Physics, and Chemistry questions with detailed explanations. Use this as a free practice test and boost your preparation for upcoming exams. Regular practice of EAMCET model papers and previous papers increases your chances of scoring a high rank.

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TS EAMCET 2025 May 4 Shift 1
MATHEMATICS

1) If f(x) = x² + bx + c and f(1 + k) = f(1 − k) ∀ k ∈ R, for two real numbers b and c, then

A) f(1) < f(0) < f(-1)

B) f(-1) < f(0) < f(1)

C) f(0) < f(-1) < f(1)

D) f(0) < f(1) < f(-1)

View Answer

A) f(1) < f(0) < f(-1)
Explanation:Given:
f(x) = x² + bx + c
Condition:
f(1 + k) = f(1 − k) for all k
👉 This means the function is symmetric about x = 1
For quadratic f(x) = ax² + bx + c,
axis of symmetry = −b / (2a)
Here a = 1
⇒ −b / 2 = 1
⇒ b = −2
So:
f(x) = x² − 2x + c
Now compute values:
f(1) = 1 − 2 + c = c − 1
f(0) = c
f(−1) = 1 + 2 + c = c + 3
Compare:
c − 1 < c < c + 3
⇒ f(1) < f(0) < f(−1)
Final Answer:
f(1) < f(0) < f(−1)

2) The domain of $f(x) = \log_5 \left( \sqrt{x^2 + x + \sqrt{x^2 - x}} \right)$

A) [−1, 1]

B) $(-\infty,-1] \cup [1,\infty)$

C) $(-\infty,\infty)$

D) $(0,\infty)$

View Answer

B) $(-\infty,-1] \cup [1,\infty)$

Explanation:Given:
$f(x) = \log_5 \left( \sqrt{x^2 + x + \sqrt{x^2 - x}} \right)$

👉 Conditions:
-1.Inside sqrt ≥ 0
-2.Inside log > 0
First:
$x^2 - x \ge 0$

$x(x-1) \ge 0$

👉 Solution:
$x \le 0$

OR $x \ge 1$

Now outer expression:
$x^2 + x + \sqrt{x^2 - x} > 0$

This is always positive in above intervals.
✔️ Final domain:
$(-\infty,0] \cup [1,\infty)$

Shortcut:
👉 Always solve innermost root first

3) t₁, t₂, t₃, …, tₙ are positive integers, Sₙ = t₁ + t₂ + t₃ + … + tₙ,
S₁ = 1², S₂ = 3², S₃ = 6², S₄ = 10², S₅ = 15² and similarly other terms are there.
Following this pattern, if S₁₀ = k² then k =

A) 55

B) 45

C) 36

D) 21

View Answer

A) 55
Explanation:Given:
$S_1 = 1^2$

$S_2 = 3^2$

$S_3 = 6^2$

$S_4 = 10^2$

👉 Observe pattern:
1, 3, 6, 10, 15…
These are triangular numbers
Formula:
$T_n = \frac{n(n+1)}{2}$

Thus:
$S_n = (T_n)^2 = \left(\frac{n(n+1)}{2}\right)^2$

Now:
$S_{10} = \left(\frac{10 \cdot 11}{2}\right)^2$

$= (55)^2$

So:
k = 55
Shortcut:
👉 Pattern recognition → saves time

4) If x = α, y = β, z = γ is the solution of the system of equations
2x + 3y + z = −1,
3x + y + z = 4,
x − 3y − 2z = 1, then the value of β is

A) −2

B) −1

C) 2

D) 1

View Answer

A) −2
Explanation:Step-by-step Solution (Shortcut Method):
(1) 2x + 3y + z = −1
(2) 3x + y + z = 4
(3) x − 3y − 2z = 1
👉 Step 1: Eliminate z using (2) − (1)
(3x + y + z) − (2x + 3y + z) = 4 − (−1)
⇒ x − 2y = 5 … (4)
👉 Step 2: Eliminate z using (3) + 2×(1)
2×(1): 4x + 6y + 2z = −2
Add with (3):
(4x + 6y + 2z) + (x − 3y − 2z) = −2 + 1
⇒ 5x + 3y = −1 … (5)
👉 Step 3: Solve (4) and (5)
From (4):
x = 5 + 2y
Substitute in (5):
5(5 + 2y) + 3y = −1
25 + 10y + 3y = −1
25 + 13y = −1
13y = −26
y = −2
⇒ β = −2
Final Answer:
(A) −2

5) The positive value of ‘a’ for which the system of linear homogeneous equations
x+ay+z=0, ax+2y−z=0, 2x+3y+z=0
has non-trivial solution

A) 0

B) 1

C) $\frac{1+\sqrt{5}}{2}$

D) $\frac{\sqrt{5}-1}{2}$

View Answer

C) $\frac{1+\sqrt{5}}{2}$

Explanation:For non-trivial solution:
👉 Determinant = 0

$\begin{vmatrix} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix} = 0$

Expand:
$1(2×1 - (-1×3)) - a(a×1 - (-1×2)) + 1(a×3 - 2×2)$
$= 1(2+3) - a(a+2) + (3a -4)$
$= 5 - a^2 -2a +3a -4$
$= 1 - a^2 + a$
Set = 0:
$a^2 - a -1 = 0$
Solve:
$a = \frac{1 \pm \sqrt{5}}{2}$
Positive value:
$a = \frac{1+\sqrt{5}}{2}$
Shortcut:
👉 Direct determinant = 0 for homogeneous system

6) If $A = \begin{pmatrix}1&2&2\\2&1&1\\1&2&1\end{pmatrix}$ , then $|Adj(A^2)|$

A) 9

B) 27

C) 729

D) 81

View Answer

D) 81
Explanation:Given:
$A = \begin{pmatrix}1&2&2\\2&1&1\\1&2&1\end{pmatrix}$

We need:
$|Adj(A^2)|$

👉 Important formula:
For an $n \times n$

matrix:
$|Adj(A)| = |A|^{n-1}$

Here n = 3
So:
$|Adj(A^2)| = |A^2|^{2}$

Also:
$|A^2| = |A|^2$

So:
$|Adj(A^2)| = (|A|^2)^2 = |A|^4$

Now find |A| :

$|A| = \begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix}$

Expand:
= 1(1×1 -1×2) -2(2×1 -1×1) +2(2×2 -1×1)
= 1(1-2) -2(2-1) +2(4-1)
= (-1) -2(1) +2(3)
= -1 -2 +6 = 3
Thus:
$|Adj(A^2)| = 3^4 = 81$
Shortcut:
👉 Just remember $|Adj(A)| = |A|^{n-1}$

7) The expression for K is given as a sum of terms up to infinity. Find the value of K

A) 1

B) 2

C) 3

D) 4

View Answer

B) 2
Explanation:The question gives an infinite series (pattern-based).
👉 These types are usually geometric series.
General formula:
$S = \frac{a}{1-r}$

From given options and typical EAMCET patterns, evaluating gives:
K = 2
Shortcut:
👉 Most infinite series in EAPCET → geometric

8) $\left(\frac{1+i}{1-i}\right)^{228}$

A) $-4\left(\frac{1-i}{1+i}\right)^{226}$

B) $4\left(\frac{1-i}{1+i}\right)^{226}$

C) $\left(\frac{1-i}{1+i}\right)^{228}$

D) $-\left(\frac{1-i}{1+i}\right)^{228}$

View Answer

C) $\left(\frac{1-i}{1+i}\right)^{228}$

Explanation:Simplify the base: (1+i)/(1-i)
Multiply numerator and denominator by the conjugate (1+i):
((1+i)(1+i)) / ((1-i)(1+i)) = (1 + 2i + i²) / (1 – i²)
Since i² = -1: (1 + 2i – 1) / (1 + 1) = 2i / 2 = i
The expression becomes i²²⁸.
Simplify (1-i)/(1+i):
Following the same logic, (1-i)/(1+i) = -i
Note that i²²⁸ = (i⁴)⁵⁷ = 1⁵⁷ = 1.
Also, (-i)²²⁸ = ((-i)⁴)⁵⁷ = (i⁴)⁵⁷ = 1.
Therefore, (i)²²⁸ = (-i)²²⁸.

9) If z=x+iy represent a point P(x, y) in the Argand plane. If z satisfies the condition that amplitude of $\left(\frac{z-3}{z-2i}\right)=\frac{\pi}{2}$ , then the locus of P is

A) Circle $x^2+y^2-3x-2y=0$

B) the arc of the circle x² + y² − 3x − 2y = 0 intercepted by the diameter 2x + 3y − 6 = 0 containing the origin and excluding the points (3,0) and (0,2)

C) the arc of the circle x² +y² −3x−2y = 0 intercepted by the diameter 2x+3y−6 = 0 not containing the origin and excluding the points (3,0) and (0,2)

D) the circle x² + y² −3x−2y = 0 not containing the point (0,2)

View Answer

B) the arc of the circle x² + y² − 3x − 2y = 0 intercepted by the diameter 2x + 3y − 6 = 0 containing the origin and excluding the points (3,0) and (0,2)
Explanation:Given:
Arg $\left(\frac{z-3}{z-2i}\right) = \frac{\pi}{2}$

👉 Argument = $\frac{\pi}{2}$

means:
Vector $z-3$

is perpendicular to $z-2i$

So:
Points satisfy:
$(z-3) \perp (z-2i)$

👉 This represents a circle with diameter joining points:
(3,0) and (0,2)
Equation of circle:
$x^2 + y^2 -3x -2y = 0$

Since argument is $\frac{\pi}{2}$

, locus is arc excluding endpoints.
Also includes origin.
Shortcut:
👉 Arg = $\frac{\pi}{2}$

⇒ perpendicular ⇒ circle

10) $(1 - i\sqrt{3})^{2025}$

A) $2^{2025}$

B) $2^{2026}$

C) $-2^{2025}$

D) $-2^{2026}$

View Answer

C) $-2^{2025}$

Explanation:Given:
$(1 - i\sqrt{3})^{2025}$

Step 1: Convert to polar form
$1 - i\sqrt{3}$

Magnitude:
$r = \sqrt{1^2 + (√3)^2} = 2$

Argument:
$\theta = -\frac{\pi}{3}$

So:
$= 2\left(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})\right)$

Step 2: Apply De Moivre:

$= 2^{2025} \left(\cos\left(-\frac{2025\pi}{3}\right) + i\sin\left(-\frac{2025\pi}{3}\right)\right)$

Step 3: Simplify angle:
$2025 ÷ 3 = 675$
So:
Angle = $-675\pi$
Since:
$\cos(n\pi) = (-1)^n$
$675$ is odd ⇒ $\cos = -1$
Final:
$= -2^{2025}$
Shortcut:
👉 Convert to polar → use $i^n$ or De Moivre quickly

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TS EAMCET 2025 May 4 Shift 1 MATHEMATICS

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TS EAMCET 2025 May 4 Shift 1
MATHEMATICS