Engineering Entrance Exam Preparation Resources

Engineering Entrance

MHT-CET

Q. Find the integral of e^x dx. (2022)

A. e^x + C
B. e^x
C. x e^x + C
D. ln(e^x) + C

Solution

The integral of e^x is e^x + C.

Correct Answer: A — e^x + C

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Q. Find the integral of sin(x) with respect to x. (2020)

A. -cos(x) + C
B. cos(x) + C
C. sin(x) + C
D. -sin(x) + C

Solution

The integral of sin(x) is -cos(x) + C.

Correct Answer: A — -cos(x) + C

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Q. Find the integral of sin(x). (2020)

A. -cos(x) + C
B. cos(x) + C
C. sin(x) + C
D. -sin(x) + C

Solution

The integral of sin(x) is -cos(x) + C.

Correct Answer: A — -cos(x) + C

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Q. Find the integral of sin(x)dx. (2020)

A. -cos(x) + C
B. cos(x) + C
C. sin(x) + C
D. -sin(x) + C

Solution

The integral of sin(x) is -cos(x) + C.

Correct Answer: A — -cos(x) + C

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Q. Find the integral of x^5 dx. (2020)

A. (1/6)x^6 + C
B. (1/5)x^6 + C
C. (1/4)x^6 + C
D. (1/7)x^6 + C

Solution

The integral is (1/6)x^6 + C.

Correct Answer: B — (1/5)x^6 + C

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Q. Find the length of the diagonal of a cuboid with dimensions 3, 4, and 12 units. (2020)

A. √169
B. √145
C. √153
D. √157

Solution

Diagonal = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13.

Correct Answer: C — √153

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Q. Find the length of the diagonal of a rectangular box with dimensions 2, 3, and 6. (2023)

A. √49
B. √36
C. √45
D. √50

Solution

Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7.

Correct Answer: A — √49

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Q. Find the length of the diagonal of a rectangular box with dimensions 2, 3, and 6 units. (2022)

A. √49
B. √45
C. √36
D. √50

Solution

Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7 units.

Correct Answer: A — √49

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Q. Find the limit: lim (x -> 0) (x^2)/(sin(x)) (2023)

A. 0
B. 1
C. 2
D. Undefined

Solution

As x approaches 0, sin(x) approaches x, thus lim (x -> 0) (x^2/sin(x)) = lim (x -> 0) (x^2/x) = lim (x -> 0) x = 0.

Correct Answer: A — 0

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Q. Find the limit: lim (x -> 1) (x^4 - 1)/(x - 1) (2023)

A. 0
B. 1
C. 4
D. Undefined

Solution

Factoring gives ((x - 1)(x^3 + x^2 + x + 1))/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Thus, lim (x -> 1) = 4.

Correct Answer: A — 0

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Q. Find the limit: lim (x -> 2) (x^2 + 3x - 10)/(x - 2) (2021)

A. 0
B. 1
C. 5
D. 7

Solution

Factoring gives (x - 2)(x + 5)/(x - 2). For x ≠ 2, this simplifies to x + 5. Evaluating at x = 2 gives 7.

Correct Answer: D — 7

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Q. Find the limit: lim (x -> 2) (x^2 - 3x + 2)/(x - 2) (2021)

A. 1
B. 2
C. 0
D. Undefined

Solution

The expression is undefined at x=2. The limit does not exist as the function approaches infinity.

Correct Answer: D — Undefined

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Q. Find the limit: lim (x -> 3) (x^2 - 9)/(x - 3) (2023)

A. 0
B. 3
C. 6
D. 9

Solution

The expression can be factored as ((x - 3)(x + 3))/(x - 3). For x ≠ 3, this simplifies to x + 3. Thus, lim (x -> 3) (x + 3) = 6.

Correct Answer: A — 0

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Q. Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

A. 1
B. 5
C. 9
D. 7

Solution

The maximum occurs at x = -b/(2a) = -4/(2*-1) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

Correct Answer: B — 5

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Q. Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

A. (0, 1)
B. (1, 3)
C. (2, 5)
D. (3, 1)

Solution

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.

Correct Answer: B — (1, 3)

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Q. Find the local maximum of f(x) = -x^3 + 3x^2 + 4. (2020)

A. 4
B. 5
C. 6
D. 3

Solution

Set f'(x) = 0 to find critical points. The local maximum occurs at x = 2. f(2) = 5.

Correct Answer: B — 5

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Q. Find the magnitude of the vector A = 3i - 4j. (2020)

A. 5
B. 7
C. 10
D. 12

Solution

|A| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.

Correct Answer: A — 5

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Q. Find the maximum area of a triangle with a base of 10 m and height varying. (2020)

A. 25
B. 50
C. 75
D. 100

Solution

Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.

Correct Answer: B — 50

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Q. Find the maximum area of a triangle with a base of 10 units and height as a function of x. (2022)

A. 25
B. 50
C. 75
D. 100

Solution

Area = 1/2 * base * height. Max area occurs when height is maximized, which is 10 units, giving Area = 50.

Correct Answer: B — 50

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Q. Find the maximum area of a triangle with a base of 10 units and height as a function of the base. (2021)

A. 25
B. 50
C. 30
D. 40

Solution

Area = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.

Correct Answer: B — 50

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Q. Find the maximum area of a triangle with a fixed perimeter of 30 cm. (2022)

A. 75 cm²
B. 100 cm²
C. 50 cm²
D. 60 cm²

Solution

For maximum area, the triangle should be equilateral. Area = (sqrt(3)/4) * (10)^2 = 75 cm².

Correct Answer: A — 75 cm²

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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 32t + 48. (2020)

A. 48
B. 64
C. 80
D. 32

Solution

The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.

Correct Answer: A — 48

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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48. (2020)

A. 48
B. 64
C. 80
D. 32

Solution

The maximum occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

Correct Answer: B — 64

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Q. Find the maximum value of the function f(x) = -2x^2 + 8x - 3. (2021) 2021

A. 3
B. 8
C. 12
D. 6

Solution

The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = -8/(2*-2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.

Correct Answer: B — 8

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Q. Find the midpoint of the line segment joining the points (2, 3) and (4, 7). (2022) 2022

A. (3, 5)
B. (2, 5)
C. (4, 5)
D. (3, 4)

Solution

Midpoint = ((2+4)/2, (3+7)/2) = (3, 5).

Correct Answer: A — (3, 5)

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Q. Find the minimum value of f(x) = 4x^2 - 16x + 20. (2022)

A. 4
B. 5
C. 6
D. 7

Solution

The vertex gives the minimum at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

Correct Answer: A — 4

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Q. Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

A. 2
B. 3
C. 4
D. 5

Solution

The vertex form gives the minimum at x = 2. f(2) = 2.

Correct Answer: A — 2

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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021)

A. 3
B. 5
C. 4
D. 6

Solution

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

Correct Answer: A — 3

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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021) 2021

A. 3
B. 5
C. 4
D. 6

Solution

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

Correct Answer: A — 3

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Q. Find the minimum value of the function f(x) = 2x^2 - 8x + 10. (2022)

A. 2
B. 4
C. 6
D. 8

Solution

The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.

Correct Answer: B — 4

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Engineering Entrance

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