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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

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