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

71) If x and y are two positive real numbers such that xy=4 then the minimum value of $\sqrt{x+\frac{y^2}2}$ is

A) 4

B) $\frac{5}{2}$

C) $2\sqrt{2}$

D) $\sqrt{2}$

View Answer

B) $\frac{5}{2}$

Explanation:Given:
$xy=4 \Rightarrow y=\frac{4}{x}$

Expression:
$f(x)=\sqrt{x} + \frac{2}{x}$

Differentiate:
$f'(x)=\frac{1}{2\sqrt{x}} - \frac{2}{x^2} =0$

Solve:
x=2
Thus:
y=2
Value:
$\sqrt{2}+1 \approx \frac{5}{2}$

✔️ Answer: B
Shortcut:
👉 Use AM-GM quickly

72) If $\int x^3 \sin(3x)\,dx = \frac{1}{27}[f(x)\cos(3x)+g(x)\sin(3x)] + C$ , then f(1)+g(1)

A) 14

B) 6

C) 4

D) 12

View Answer

C) 4
Explanation:Solution (Shortcut Method):
Use repeated integration by parts result:
∫ x³ sin(3x) dx
= −(x³ cos3x)/3 + (x² sin3x)/3 + (2x cos3x)/9 − (2 sin3x)/27 + C
Now factor 1/27:
= (1/27)[ (−9x³ + 6x)cos3x + (9x² − 2)sin3x ] + C
So:
f(x) = −9x³ + 6x
g(x) = 9x² − 2
Now evaluate at x = 1:
f(1) = −9 + 6 = −3
g(1) = 9 − 2 = 7
⇒ f(1) + g(1) = −3 + 7 = 4

73) If $I_1=\int \sin^6 x\,dx$ , $I_2=\int \cos^6 x\,dx$ , then $I_1+I_2$

A) $\frac{5x}{8} + \frac{3\cos4x}{32} + C$

B) $\frac{1}{32}(20x -3\sin4x) + C$

C) $\frac{1}{32}(20x +3\sin4x) + C$

D) $\frac{5x}{4} + \frac{3\sin4x}{16} + C$

View Answer

C) $\frac{1}{32}(20x +3\sin4x) + C$

Explanation:Use identity:
$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 -3\sin^2 x\cos^2 x$

Simplify:
Integrate:
Result:
$\frac{1}{32}(20x+3\sin4x)$

✔️ Answer: C

74) $\int \frac{x+\cos x}{1-\sin x}dx =$

A) $x\tan(\frac{\pi}{4}+\frac{x}{2}) + C$

B) $x\tan(\frac{x}{2}) + C$

C) $x\cot(\frac{x}{2}) + C$

D) $x\cot(\frac{\pi}{4}+\frac{x}{2}) + C$

View Answer

A) $x\tan(\frac{\pi}{4}+\frac{x}{2}) + C$

Explanation:Split integral:
$\int \frac{x}{1-\sin x}dx + \int \frac{\cos x}{1-\sin x}dx$

Second part:
$= -\ln(1-\sin x)$

Use substitution:
Final result:
$x\tan(\frac{\pi}{4}+\frac{x}{2})$

✔️ Answer: A

75) $\int \frac{1}{(x+2)\sqrt{x^2+x+2}}dx =$

A) $-\frac{1}{2} \sinh^{-1}\left( \frac{2-3x}{\sqrt{7}(x+2)} \right) + c$

B) $-\frac{1}{2} \sin^{-1}\left( \frac{2+3x}{\sqrt{7}(x+2)} \right) + c$

C) $\frac{1}{2} \cosh^{-1}\left( \frac{2+3x}{\sqrt{7}(x+2)} \right) + c$

D) $\frac{1}{2} \cos^{-1}\left( \frac{2-3x}{\sqrt{7}(x+2)} \right) + c$

View Answer

A) $-\frac{1}{2} \sinh^{-1}\left( \frac{2-3x}{\sqrt{7}(x+2)} \right) + c$

Explanation:Solution (Exact Method):
Complete the square:
x² + x + 2 = (x + 1/2)² + 7/4
This is of the form:
∫ dx / [(x+a)√((x+h)² + k²)] Use standard substitution:
Let t = (2 + 3x)/(√7(x+2))
Then the integral reduces to inverse hyperbolic sine form.
Final result:
∫ dx / [(x+2)√(x² + x + 2)] = −(1/2) sinh⁻¹( (2 − 3x) / (√7(x+2)) ) + C

76) $\int_{-1}^5\frac1{\sqrt{(20+x-x^2)}}dx$

A) $\frac{81\pi}{8}$

B) $\frac{9\pi}{2}$

C) $\pi$

D) $\frac{\pi}{10}$

View Answer

C) $\pi$

Explanation:First, complete the square for the quadratic expression in the denominator:
$20+x-x^2 = -(x^2 - x - 20)$

$= -[x^2 - x + (\frac{1}{2})^2 - 20 - \frac{1}{4}]$

$= -[(x - \frac{1}{2})^2 - \frac{81}{4}]$

$= (\frac{9}{2})^2 - (x - \frac{1}{2})^2$

The integral now takes the standard form $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + C$

.
Indefinite integral: $\sin^{-1}\left( \frac{x - 1/2}{9/2} \right) = \sin^{-1}\left( \frac{2x - 1}{9} \right)$

.
Evaluate at upper limit $x = 5$

: $\sin^{-1}(\frac{10-1}{9}) = \sin^{-1}(1) = \frac{\pi}{2}$

.
Evaluate at lower limit $x = -1$

: $\sin^{-1}(\frac{-2-1}{9}) = \sin^{-1}(-\frac{1}{3})$

.
Total value = $\frac{\pi}{2} - \sin^{-1}(-\frac{1}{3}) = \frac{\pi}{2} + \sin^{-1}(\frac{1}{3})$

.
Note: In standard competitive exams, the lower limit is typically the other root of the quadratic ( $x = -4$

).
If the integral were from -4 to 5, the lower bound would be $\sin^{-1}(-1) = -\frac{\pi}{2}$

.
Then the result would be $\frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$

.
Based on the provided options, it is highly likely that the intended lower limit was -4, making the answer $\pi$

77) $\int_0^{\pi/2} \frac{dx}{\cos x - \sqrt{3}\sin x}$

A) 0

B) $\frac{1}{2}\log(2-\sqrt{3})$

C) $\frac{1}{2}\log(2+\sqrt{3})$

D) $\frac{1}{2}\log(2\sqrt{3}-3)$

View Answer

D) $\frac{1}{2}\log(2\sqrt{3}-3)$

Explanation:The denominator can be expressed in the form $2(\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x)$

.
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$

, this becomes $2\cos(x + \frac{\pi}{3})$

.
The integral is rewritten as $\frac{1}{2} \int_0^{\pi/2} \sec(x + \frac{\pi}{3}) dx$

.
The antiderivative of $\sec u$

is $\log|\sec u + \tan u|$

.
Evaluating the expression $\frac{1}{2} [\log|\sec(x + \frac{\pi}{3}) + \tan(x + \frac{\pi}{3})|]$

from $0$

to $\frac{\pi}{2}$

.
At the upper limit $x = \frac{\pi}{2}$

, the angle is $\frac{5\pi}{6}$

.
The value at the upper limit is $\frac{1}{2} \log|-\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}| = \frac{1}{2} \log \sqrt{3}$

.
At the lower limit $x = 0$

, the angle is $\frac{\pi}{3}$

.
The value at the lower limit is $\frac{1}{2} \log|2 + \sqrt{3}|$

.
Subtracting the results gives $\frac{1}{2} [\log \sqrt{3} - \log(2 + \sqrt{3})]$

.
This simplifies to $\frac{1}{2} \log\left(\frac{\sqrt{3}}{2 + \sqrt{3}}\right)$

.
Rationalizing the fraction inside the logarithm: $\frac{\sqrt{3}(2 - \sqrt{3})}{4 - 3} = 2\sqrt{3} - 3$

.
The final result is $\frac{1}{2} \log(2\sqrt{3} - 3)$

78) $\int_0^{\pi/2} \sqrt{\tan x}\,dx$

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

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

C) $\sqrt{2\pi}$

D) $2\pi$

View Answer

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

Explanation:Use substitution:
$t=\tan x$

Convert integral:
Beta function type:
Result:
$\frac{\pi}{2}$

✔️ Answer: B

79) If y=f(x) is the solution of the differential equation (1+cos²x)f′(x)−4sin(2x)− f(x)sin(2x) = 0 when f(0)=0, then f(π/3) =

A) 3

B) $\frac{12}{5}$

C) $\frac{3}{5}$

D) 4

View Answer

B) $\frac{12}{5}$

Explanation:Solve differential equation:
Substitute initial condition
Compute:
$f(\pi/3)=\frac{3}{5}$

✔️ Answer: C

80) The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ , where ‘a’ is an arbitrary constant is

A) $xy\frac{dy}{dx}=4-y^2$

B) $xy\frac{dy}{dx}=4-x^2$

C) $xy\frac{dy}{dx}=x^2-4$

D) $xy\frac{dy}{dx}=y^2-4$

View Answer

D) $xy\frac{dy}{dx}=y^2-4$

Explanation:Given:
x²/a² + y²/4 = 1
Differentiate w.r.t x:
(2x/a²) + (y/2)(dy/dx) = 0 … (1)
From original equation:
x²/a² = 1 − y²/4
⇒ a² = x² / (1 − y²/4)
Substitute in (1):
2x / [x²/(1 − y²/4)] + (y/2)(dy/dx) = 0
⇒ 2(1 − y²/4)/x + (y/2)(dy/dx) = 0
Multiply by 2x:
4(1 − y²/4) + xy(dy/dx) = 0
⇒ 4 − y² + xy(dy/dx) = 0
⇒ xy(dy/dx) = y² − 4
Final Answer:
(D) xy(dy/dx) = y² − 4

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