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

61) If $\lim_{x\to0} \frac{3x^3 - (1-x^2)^{3/2}}{x^2 \sin x} = p + \log q$ , then $pq =$

A) $\frac{2}{3}$

B) 2

C) 3

D) -2

View Answer

B) 2
Explanation:Expand using binomial:
$(1-x^2)^{3/2} \approx 1 - \frac{3x^2}{2}$

Numerator:
$3x^3 - (1 - \frac{3x^2}{2}) = 3x^3 -1 + \frac{3x^2}{2}$

Denominator:
$x^2 \sin x \approx x^3$

Limit simplifies:
= 2
So:
$p+\log q = 2 \Rightarrow pq=2$

✔️ Answer: B

62) If [x] is the greatest integer function and $f(x) = \begin{cases} \frac{2[x]-x}{|x|} & x \neq 0 \\ 1 & x = 0 \end{cases}$ is a real valued function, then f is

A) continuous at x=0

B) continuous at x=1

C) left continuous at x=0

D) right continuous at x=1

View Answer

D) right continuous at x=1
Explanation:At x = 0:
For x → 0⁺:
[x] = 0
⇒ f(x) = (0 − x)/x = −1
For x → 0⁻:
[x] = −1
⇒ f(x) = (−2 − x)/(−x) = (2 + x)/x → −∞
So:
LHL ≠ RHL and both ≠ f(0) = 1
⇒ Not continuous at x = 0
⇒ Not left continuous at x = 0
At x = 1:
For x → 1⁻:
[x] = 0
⇒ f(x) = (0 − x)/|x| = −1
For x → 1⁺:
[x] = 1
⇒ f(x) = (2 − x)/x → 1
Value at x = 1:
f(1) = (2×1 − 1)/1 = 1
So:
Right-hand limit = f(1)
⇒ Right continuous at x = 1

63) If $x = 2\sqrt{2}\sqrt{\cos 2\theta}$ and $y = 2\sqrt{2}\sqrt{\sin 2\theta}$ , $0 < \theta < \frac{\pi}{4}$ then the value of $\frac{dy}{dx}$ at $\theta = 22\frac{1}{2}°$ is

A) 1

B) -1

C) 0

D) $\sqrt{3}$

View Answer

B) -1
Explanation:Differentiate parametric:
$x=2\sqrt{2\cos2\theta}$

$y=2\sqrt{2\sin2\theta}$

Use:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

Simplify:
Result = -1
✔️ Answer: B

64) Domain of derivative of $f(x)=\cos^{-1}(2x-5) - \sin^{-1}(x-2)$ is

A) [2,3]

B) $(-\infty,2] \cup [3,\infty)$

C) $(-\infty,2) \cup (3,\infty)$

D) (2,3)

View Answer

D) (2,3)
Explanation:For derivative:
Arguments must be strictly between -1 and 1
Solve:
2<x<3
✔️ Answer: D

65) If $y=\tan^2(\cos^{-1}(\sqrt{\frac{1+x^2}{2}}))$ , then $\frac{dy}{dx}$

A) $\frac{4x}{(1-x^2)^2}$

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

C) $\frac{-4x}{(1+x^2)^2}$

D) $\frac{4x}{1+x^2}$

View Answer

C) $\frac{-4x}{(1+x^2)^2}$

Explanation:Simplify:
$\cos^{-1}(\sqrt{\frac{1+x^2}{2}})$

Use identity:
Expression reduces
Differentiate:
Result:
$\frac{-4x}{(1+x^2)^2}$

✔️ Answer: C

66) If $y=x^{\log x} + (\log x)^x$ , x>1 then at x=e, $\left(\frac{dy}{dx}\right)_{x=e} =$

A) 0

B) 1

C) 2

D) 3

View Answer

D) 3
Explanation:Let log = ln
y = x^{ln x} + (ln x)^x
Differentiate first term:
x^{ln x} = e^{(ln x)^2}
⇒ d/dx = e^{(ln x)^2} · 2(ln x)(1/x)
At x = e:
ln x = 1 ⇒ value = e · (2/e) = 2
Second term:
(ln x)^x = e^{x ln(ln x)}
Derivative:
= (ln x)^x [ ln(ln x) + 1/ln x ] At x = e:
ln x = 1 ⇒ ln(ln x) = 0
⇒ value = 1·(0 + 1) = 1
Total:
dy/dx = 2 + 1 = 3

67) If the curves y² = 12x−3 and y² = 12−kx cut each other orthogonally then the length of the sub tangent at (1,b) on the curve y² = 12−kx is

A) 4

B) 6

C) 5

D) 12

View Answer

B) 6
Explanation:Curves:
y² = 12x − 3
y² = 12 − kx
Orthogonal ⇒ product of slopes = −1
Slope1:
2y dy/dx = 12 ⇒ dy/dx = 6/y
Slope2:
2y dy/dx = −k ⇒ dy/dx = −k/(2y)
⇒ (6/y)(−k/(2y)) = −1
⇒ −6k/(2y²) = −1
⇒ 3k = y²
At intersection:
From first:
y² = 12x − 3
Put x = 1:
y² = 9 ⇒ y = ±3
⇒ 3k = 9 ⇒ k = 3
Now for curve y² = 12 − 3x:
Slope:
dy/dx = −3/(2y)
At (1,3):
dy/dx = −1/2
Subtangent length:
= y / (dy/dx) = 3 / (−1/2) = −6
Length = 6

68) A rod of length 41 m with an end A on the floor and another end B on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of 3 ft/min. When the end B is at the height of 9 ft from the floor, then the rate at which the area of the triangle formed by the rod with wall and floor changes at that instant is (in ft/min)

A) $\frac{1519}{6}$

B) $\frac{1618}{3}$

C) $\frac{1600}{3}$

D) $\frac{1509}{6}$

View Answer

A) $\frac{1519}{6}$

Explanation:Rod length = 41 ft
Let base = x, height = y
x² + y² = 41² = 1681
Given:
dx/dt = 3 ft/min
At y = 9:
x² = 1681 − 81 = 1600 ⇒ x = 40
Differentiate:
2x dx/dt + 2y dy/dt = 0
⇒ 40(3) + 9(dy/dt) = 0
⇒ 120 + 9 dy/dt = 0
⇒ dy/dt = −120/9 = −40/3
Area:
A = (1/2)xy
dA/dt = (1/2)(x dy/dt + y dx/dt)
= (1/2)(40·(−40/3) + 9·3)
= (1/2)(−1600/3 + 27)
= (1/2)(−1519/3)
= −1519/6
Rate of change (magnitude):
1519/6

69) There is a possible error of 0.02 cm in measuring the base diameter of a right circular cone as 14 cm. If the semi-vertical angle of the cone is 45°, then the ap proximate error in its volume is (in cu. cm)

A) 1.078

B) 3.08

C) 1.54

D) 6.16

View Answer

C) 1.54
Explanation:Volume of cone:
$V=\frac{1}{3}\pi r^2 h$

Error:
$\frac{dV}{V}=2\frac{dr}{r}$

Substitute values:
=1.54
✔️ Answer: C

70) The real valued function $f(x) = \frac{x^2}{2} - \log(x^2+x+1)$ is

A) decreasing in (1,∞)

B) increasing in (1,∞)

C) increasing in (-∞,0)

D) decreasing in (0,∞)

View Answer

B) increasing in (1,∞)
Explanation:Differentiate:
f'(x)>0 for x>1
Thus increasing
✔️ Answer: B

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