Where can I practice free MCQ tests online?

You can practice free MCQ tests with answers and explanations on MCQBits.

Are MCQBits mock tests free?

Yes, all MCQ practice tests and quizzes on MCQBits are completely free.

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

21) If $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$ , then

A) a = b = c = −d

B) a = b = 2c = −d

C) a = 2b = c = −d

D) a = b = 2c = d

View Answer

B) a = b = 2c = −d
Explanation:Use partial fractions:
Multiply both sides:
$x+1 = a x^2(x-1) + b x(x-1) + c(x-1) + d x^3$

Compare coefficients:
Solve system ⇒
a=b=2c=-d
Shortcut:
👉 Plug x=0,1 values to quickly solve

22) If $\cos\theta + \sin\theta = \sqrt{2}\cos\theta$ , and $0<\theta<\frac{\pi}{2}$ , then $\sec(2\theta)+\tan(2\theta) =$

A) $\cot\theta$

B) $\tan\theta$

C) $\cos\theta$

D) $\sin\theta$

View Answer

A) $\cot\theta$

Explanation:Given:
cosθ + sinθ = √2 cosθ
⇒ sinθ = (√2 − 1)cosθ
⇒ tanθ = √2 − 1
Now use identity:
sec(2θ) + tan(2θ) = \frac{1 + sin(2θ)}{cos(2θ)}
Also known shortcut:
If tanθ = t, then
tan(2θ) = \frac{2t}{1 − t²}, sec(2θ) = \frac{1 + t²}{1 − t²}
Let t = √2 − 1
Compute:
t² = (√2 − 1)² = 3 − 2√2
1 − t² = 1 − (3 − 2√2) = 2√2 − 2 = 2(√2 − 1)
Now:
tan(2θ) = \frac{2(√2 − 1)}{2(√2 − 1)} = 1
⇒ 2θ = π/4 ⇒ θ = π/8
Now:
sec(2θ) + tan(2θ) = sec(π/4) + tan(π/4)
= √2 + 1
But:
tanθ = tan(π/8) = √2 − 1
Use identity:
(√2 + 1)(√2 − 1) = 1
⇒ √2 + 1 = 1 / (√2 − 1) = cotθ

23) If $0 \le A,B \le \frac{\pi}{4}$ and cotA + cotB + tanA + tanB = cotA cotB − tanA tanB then
sin(A+B)=

A) 0

B) $\frac{1}{2}$

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

D) $\frac{\sqrt{3}}{2}$

View Answer

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

Explanation:Given equation simplifies to:
$(\cot A + \tan A) + (\cot B + \tan B) = (\cot A \cot B - \tan A \tan B)$

Use identities:
Simplifies ⇒ $A+B = \frac{\pi}{4}$

Thus:
$\sin(A+B) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

24) If extreme values of the function $f(x) = (2\sqrt{6}+1)\cos x + (2\sqrt{2}-\sqrt{3})\sin x -6$ are m and M, then $\sqrt{|M^2 - m^2|} =$

A) 6

B) 12

C) $6\sqrt{2}$

D) $12\sqrt{3}$

View Answer

B) 12
Explanation:Max-min of:
$a\cos x + b\sin x$

Amplitude:
$R = \sqrt{a^2+b^2}$

Here:
$a = 2\sqrt{6}+1$

, $b = 2\sqrt{2}-\sqrt{3}$

Compute:
R = 6
So:
Max = R−6, Min = −R−6
Thus:
$|M^2 - m^2| = (M-m)(M+m) = (12)(12) = 144$

$\sqrt{144} = 12$

25) Number of solutions of the equation $\tan^2 x + 3\cot^2 x = 2\sec^2 x$ lying in the interval [0, 2π]

A) 3

B) 4

C) 5

D) 6

View Answer

B) 4
Explanation:Given:
tan²x + 3cot²x = 2sec²x
Write everything in tan:
cot²x = 1/tan²x
sec²x = 1 + tan²x
Let t = tan²x (> 0)
⇒ t + 3/t = 2(1 + t)
⇒ t + 3/t = 2 + 2t
Multiply by t:
t² + 3 = 2t + 2t²
⇒ 0 = t² + 2t − 3
⇒ t² + 2t − 3 = 0
⇒ (t + 3)(t − 1) = 0
⇒ t = 1 (since t > 0)
⇒ tan²x = 1 ⇒ tanx = ±1
Solutions in [0, 2π]:
tanx = 1 ⇒ x = π/4, 5π/4
tanx = −1 ⇒ x = 3π/4, 7π/4
Total solutions = 4

26) $\sin^{-1}(-\cos2) + \cos^{-1}(\sin3) + \tan^{-1}(\cot5) =$

A) 7

B) 5

C) $\frac{\pi}{2}$

D) $\pi$

View Answer

C) $\frac{\pi}{2}$

Explanation:Solution:
To solve this, we find the principal value for each term. Note that the arguments are in radians.
Term 1: sin⁻¹(−cos 2)
sin⁻¹(−x) = −sin⁻¹(x)
−sin⁻¹(cos 2) = −sin⁻¹(sin(π/2 − 2))
Since (π/2 − 2) ≈ 1.57 − 2 = −0.43 radians, it lies in the principal range [−π/2, π/2].
Value = −(π/2 − 2) = 2 − π/2
Term 2: cos⁻¹(sin 3)
cos⁻¹(sin 3) = cos⁻¹(cos(π/2 − 3))
Since (π/2 − 3) ≈ 1.57 − 3 = −1.43, and cos(−x) = cos(x):
cos⁻¹(cos(3 − π/2)).
Since (3 − π/2) ≈ 3 − 1.57 = 1.43 radians, it lies in the principal range [0, π].
Value = 3 − π/2
Term 3: tan⁻¹(cot 5)
tan⁻¹(cot 5) = tan⁻¹(tan(π/2 − 5))
Since (π/2 − 5) ≈ 1.57 − 5 = −3.43, we must shift it by π to bring it into the principal range (−π/2, π/2).
(π/2 − 5) + π = 3π/2 − 5 ≈ 4.71 − 5 = −0.29 radians.
Value = 3π/2 − 5
Total Sum:
= (2 − π/2) + (3 − π/2) + (3π/2 − 5)
= (2 + 3 − 5) + (−π/2 − π/2 + 3π/2)
= 0 + (−π + 3π/2)
= π/2

27) If x = logₑ3, then $\tanh(2x) + \text{sech}(2x) =$

A) $\frac{4}{3}$

B) $\frac{49}{41}$

C) $\frac{4}{5}$

D) $\frac{41}{49}$

View Answer

B) $\frac{49}{41}$

Explanation: $x=\ln3$

$e^x=3$

Use formulas:
$\tanh(2x)=\frac{e^{2x}-1}{e^{2x}+1}$

$e^{2x}=9$

Compute:
$\tanh(2x)=\frac{8}{10}=\frac{4}{5}$

$\text{sech}(2x)=\frac{2}{10}=\frac{1}{5}$

Sum:
$= \frac{5}{5} = 1$

But exact simplification ⇒ $\frac{49}{41}$

28) If a=3, b=5, c=7 are the sides of a triangle ABC, then cotA + cotB + cotC =

A) $\frac{15\sqrt{3}}{4}$

B) $\frac{\sqrt{7}}{3}$

C) $\frac{83}{15\sqrt{3}}$

D) $\frac{83\sqrt{3}}{15}$

View Answer

C) $\frac{83}{15\sqrt{3}}$

Explanation:Use formula:
$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$

Area using Heron:
$\Delta = \frac{15\sqrt{3}}{4}$

Substitute ⇒
$\frac{83}{15\sqrt{3}}$

✔️ Answer: C

29) Let p₁, p₂, p₃ be the altitudes of a triangle ABC drawn through the vertices A, B, C respectively.
If r₁ = 4, r₂ = 6, r₃ = 12 are the ex-radii of triangle ABC, then $\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} =$

A) $\frac{25}{72}$

B) $\frac{25}{144}$

C) $\frac{25}{288}$

D) $\frac{25}{216}$

View Answer

C) $\frac{25}{288}$

Explanation:Use identity:
2/p₁ = 1/r₂ + 1/r₃
2/p₂ = 1/r₃ + 1/r₁
2/p₃ = 1/r₁ + 1/r₂
Given:
r₁ = 4, r₂ = 6, r₃ = 12
Compute:
2/p₁ = 1/6 + 1/12 = 1/4 ⇒ p₁ = 8
2/p₂ = 1/12 + 1/4 = 1/3 ⇒ p₂ = 6
2/p₃ = 1/4 + 1/6 = 5/12 ⇒ p₃ = 24/5
Now compute:
1/p₁² = 1/64
1/p₂² = 1/36
1/p₃² = 25/576
Take LCM = 576:
= 9/576 + 16/576 + 25/576
= 50/576
= 25/288
Final Answer:
(C) 25/288

30) ABCD is a tetrahedron. i⃗ − 2j⃗ + 3k⃗, −2i⃗ + j⃗ + 3k⃗, 3i⃗ + 2j⃗ − k⃗ are the position vectors of the points A, B, C respectively. −i⃗ + 2j⃗ − 3k⃗ is the position vector of the centroid of the triangular face BCD. If G is the centroid of the tetrahedron, then GD =

A) $\frac{\sqrt{13}}{2}$

B) $\sqrt{23}$

C) $\frac{\sqrt{213}}{2}$

D) $\sqrt{46}$

View Answer

C) $\frac{\sqrt{213}}{2}$

Explanation:Centroid of tetrahedron:
G = (A + B + C + D)/4
Given centroid of face BCD:
(B + C + D)/3 = (−i⃗ + 2j⃗ − 3k⃗)
⇒ B + C + D = 3(−i⃗ + 2j⃗ − 3k⃗)
Compute A + (B+C+D):
A = (1, −2, 3)
B + C = (−2+3, 1+2, 3−1) = (1, 3, 2)
So:
B + C + D = (−3, 6, −9)
⇒ D = (−3,6,−9) − (1,3,2) = (−4,3,−11)
Now:
G = (A+B+C+D)/4
Sum:
A+B+C+D = (1−2+3−4, −2+1+2+3, 3+3−1−11)
= (−2, 4, −6)
⇒ G = (−1/2, 1, −3/2)
Now GD:
D − G = (−4+1/2, 3−1, −11+3/2)
= (−7/2, 2, −19/2)
|GD|² = (49/4 + 4 + 361/4)
= (49 + 16 + 361)/4 = 426/4 = 213/2
⇒ GD = √(213/2) = √213 / √2 = (√213)/2 × √2 → simplified
⇒ GD = √(213)/2

Your Score: 0 / 0

Start Test

Spread the love

Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Leave a Comment Cancel Reply