GATE Engineering Mathematics: Calculus questions with solutions

Q1. The sum of the following infinite series is: 1/1! + 1/2! + 1/3! + 1/4! + 1/5! +...

1. π
2. 1 + e
3. e − 1
4. e

Answer: e − 1

The series represents the Taylor expansion of the exponential function e^x evaluated at x=1, which sums to e. However, since the series starts from 1/1! instead of 1/0!, we subtract 1 from e, resulting in e - 1.

Q2. Which one of the following options is the correct Fourier series of the periodic function f(x) described below: f(x) = { 0 if −2 < x < −1 { 2k if −1 < x < 1; period = 4 { 0 if 1 < x < 2

1. f(x) = k/2 + (2k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)
2. f(x) = k/2 + (2k/π) (sin(πx/2) − (1/3) sin(3πx/2) + (1/5) sin(5πx/2) + …)
3. f(x) = k + (4k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)
4. f(x) = k + (4k/π) (sin(πx/2) − (1/3) sin(3πx/2) + (1/5) sin(5πx/2) + …)

Answer: f(x) = k + (4k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)

The pulse has height 2k on -1<x<1 with period 4, so the DC term is (1/4)*2k*2 = k, and a_n = (4k/(n*pi))*sin(n*pi/2) giving 4k/pi for n=1. Since f is even the series uses cosines: f = k + (4k/pi)(cos(pi x/2) - (1/3)cos(3pi x/2) + ...). That is option C, not the stored choice.

Q3. The value of lim_(x→∞) (x − √(x² + x)) is equal to

1. -1
2. -0.5
3. -2
4. 0

Answer: -0.5

As x approaches infinity, the expression can be simplified by factoring out x from the square root, leading to a limit that evaluates to -0.5.

Q4. The following inequality is true for all x close to 0. 2 − x²/3 < x sin x/(1 − cos x) < 2 What is the value of lim(x→0) x sin x/(1 − cos x) ?

1. 0
2. 1/2
3. 1
4. 2

Answer: 2

As x approaches 0, the expression x sin x/(1 - cos x) simplifies to 2, which can be confirmed using L'Hôpital's rule or Taylor series expansion, showing that the limit converges to 2.

Q5. A closed thin-walled tube has thickness, t, mean enclosed area within the boundary of the centreline of tube’s thickness, Aₘ, and shear stress, τ. Torsional moment of resistance, T, of the section would be

1. 0.5 τ Aₘ t
2. τ Aₘ t
3. 2 τ Aₘ t
4. 4 τ Aₘ t

Answer: 2 τ Aₘ t

The torsional moment of resistance for a closed thin-walled tube is derived from the shear stress acting over the mean enclosed area and the thickness of the tube. The factor of 2 accounts for the contribution of shear stress across both walls of the tube, leading to the formula T = 2 τ Aₘ t.

Q6. Consider the hemi-spherical tank of radius 13 m as shown in the figure (not drawn to scale). What is the volume of water (in m³) when the depth of water at the centre of the tank is 6 m?

1. 78π
2. 156π
3. 396π
4. 468π

Answer: 396π

For a hemispherical bowl R=13 with water depth h=6, cap volume = pi*h^2*(3R-h)/3 = pi*36*(39-6)/3 = pi*36*11 = 396*pi. Stored 156*pi is wrong; correct is 396*pi.

Q7. Q.10 In an equilateral triangle PQR, side QR is divided into six equal parts. The length of each subdivided part is 4. The minimum area of the triangle is

1. 18
2. 24
3. 48√3
4. 144√3

Answer: 144√3

The area of an equilateral triangle can be calculated using the formula A = (sqrt(3)/4) * a², where 'a' is the length of a side. Since each subdivided part of side QR is 4, the total length of side QR is 6 * 4 = 24, leading to an area of (sqrt(3)/4) * 24² = 144√3.

Q8. In a certain month, the reference crop evapotranspiration at a location is 6 mm/day. If the crop coefficient and soil coefficient are 1.2 and 0.8, respectively, the actual evapotranspiration in mm/day is

1. 5.76
2. 7.20
3. 6.80
4. 8.00

Answer: 5.76

The actual evapotranspiration can be calculated by multiplying the reference crop evapotranspiration by the crop coefficient and the soil coefficient. In this case, 6 mm/day multiplied by 1.2 (crop coefficient) and 0.8 (soil coefficient) results in 5.76 mm/day.

Q9. Consider the polynomial f(x) = x³ - 6x² + 11x - 6 on the domain S given by 1 ≤ x ≤ 3. The first and second derivatives are f'(x) and f''(x). Consider the following statements: I. The given polynomial is zero at the boundary points x = 1 and x = 3. II. There exists one local maxima of f(x) within the domain S. III. The second derivative f''(x) > 0 throughout the domain S. IV. There exists one local minima of f(x) within the domain S. The correct option is:

1. Only statements I, II and III are correct.
2. Only statements I, II and IV are correct.
3. Only statements I and IV are correct.
4. Only statements II and IV are correct.

Answer: Only statements I, II and IV are correct.

Statements I, II, and IV are correct because the polynomial evaluates to zero at the endpoints, has one local maximum, and one local minimum within the given domain. Statement III is incorrect as the second derivative does not remain positive throughout the entire interval.

Q10. If x satisfies the equation 4^(8x) = 256, then x is equal to ______.

1. 1/2
2. log₁₆ 8
3. 2/3
4. log₄ 8

Answer: 1/2

To solve for x, we can rewrite 256 as a power of 4: 256 = 4⁴. Thus, the equation becomes 4^(8x) = 4⁴, leading to 8x = 4. Dividing both sides by 8 gives x = 1/2.

Q11. The area of the region bounded by the parabola y = x² + 1 and the straight line x + y = 3 is

1. 59/6
2. 9/2
3. 10/3
4. 7/6

Answer: 9/2

Intersections of y=x^2+1 and y=3-x: x^2+x-2=0 gives x=-2 and x=1. Area = integral from -2 to 1 of (2 - x - x^2) dx = 27/6 = 9/2. Stored answer 7/6 is wrong.

Q12. The optimum value of the function f(x) = x² - 4x + 2 is

1. 2 (maximum)
2. 2 (minimum)
3. -2 (maximum)
4. -2 (minimum)

Answer: -2 (minimum)

f'(x)=2x-4=0 gives x=2; f''(x)=2>0 so it is a minimum. f(2)=4-8+2=-2. The optimum is -2 (minimum), option 3, not 2 (minimum).

Q13. What is the value of lim(x→0, y→0) xy / (x² + y²) ?

1. 1
2. -1
3. 0
4. Limit does not exist

Answer: Limit does not exist

The limit does not exist because the value of the expression xy / (x² + y²) approaches different values depending on the path taken towards (0,0). For example, approaching along the line y = x gives a different limit than approaching along the line y = -x.

Q14. The angle of intersection of the curves x² = 4y and y² = 4x at point (0, 0) is

1. 0°
2. 30°
3. 45°
4. 90°

Answer: 90°

The curves x² = 4y and y² = 4x are perpendicular at the origin (0, 0) because their slopes at that point are negative reciprocals of each other, indicating an angle of intersection of 90°.

Q15. The quadratic approximation of f(x) = x³ - 3x² - 5 at the point x = 0 is

1. 3x² - 6x - 5
2. -3x² - 5
3. -3x² + 6x - 5
4. 3x² - 5

Answer: -3x² - 5

The quadratic approximation at a point is derived from the function's value and its first two derivatives at that point. For f(x) = x³ - 3x² - 5 at x = 0, the function value is -5 and the second derivative is -6, leading to the approximation -3x² - 5.

Q16. Seven identical cylindrical chalk-sticks are fitted tightly in a cylindrical container. The figure below shows the arrangement of the chalk-sticks inside the cylinder. The length of the container is equal to the length of the chalk-sticks. The ratio of the occupied space to the empty space of the container is

1. 5/2
2. 7/2
3. 9/2
4. 3

Answer: 7/2

Seven equal cylinders of radius r pack so the container radius is 3r. Occupied area = 7*pi*r^2, total = pi*(3r)^2 = 9*pi*r^2, empty = 2*pi*r^2. Ratio occupied:empty = 7/2 (option 1), not 9/2.

Q17. The function f(x) = x³ − 27x + 4, 1 ≤ x ≤ 6 has

1. Maxima point
2. Minima point
3. Saddle point
4. Inflection point

Answer: Minima point

f'(x) = 3x^2 - 27 = 0 gives x = 3 (within 1<=x<=6), and f''(x) = 6x = 18 > 0, so x = 3 is a local minimum. The function therefore has a minima point, option index 1, not an inflection point.

Q18. Integration of ln(x) with x i.e. ∫ ln(x) dx = _______.

1. x·ln(x) − x + Constant
2. x − ln(x) + Constant
3. x·ln(x) + x + Constant
4. ln(x) − x + Constant

Answer: x·ln(x) − x + Constant

The correct option is derived from integration by parts, where we let u = ln(x) and dv = dx. This results in the formula x·ln(x) - x, plus a constant of integration, which accurately represents the indefinite integral of ln(x).

Q19. Consider the function given below and pick one or more CORRECT statement(s) from the following choices. f(x) = x³ - 15/2 x² + 18x + 20

1. f(x) has a local minimum at x = 3.
2. f(x) has a local maximum at x = 3.
3. f(x) has a local minimum at x = 2.
4. f(x) has a local maximum at x = 2.

Answer: f(x) has a local minimum at x = 3.

f'(x) = 3x^2 - 15x + 18 = 3(x-2)(x-3) gives critical points x=2 and x=3, and f''(x)=6x-15 gives f''(2)=-3<0 (local max at 2) and f''(3)=3>0 (local min at 3). So the correct statements are local min at x=3 and local max at x=2; the stored 'local maximum at x=3' is wrong.

Q20. Consider the following three functions. f1 = 10ⁿ, f2 = n^(log n), f3 = n^(sqrt(n)) Which one of the following options arranges the functions in the increasing order of their growth rate as n tends to infinity?

1. f1, f2, f3
2. f2, f1, f3
3. f3, f2, f1
4. f2, f3, f1

Answer: f2, f3, f1

As n approaches infinity, f2 grows at a rate of n raised to the logarithm of n, which is slower than f3 (n raised to the square root of n) but faster than f1 (10 raised to the power of n). Therefore, the correct order of growth rates is f2, f3, and then f1.

Q21. For positive non-zero real variables p and q, if log(p² + q²) = log p + log q + 2 log 3, then, the value of (p⁴ + q⁴)/(p² q²) is

1. 79
2. 81
3. 9
4. 83

Answer: 79

The condition log(p^2+q^2) = log p + log q + 2log 3 gives p^2+q^2 = 9pq. Then (p^4+q^4)/(p^2q^2) = ((p^2+q^2)^2 - 2p^2q^2)/(p^2q^2) = (81p^2q^2 - 2p^2q^2)/(p^2q^2) = 79, option 0.

Q22. Let f: R → R be a function such that f(x) = max{x, x³}, x ∈ R, where R is the set of all real numbers. The set of all points where f(x) is NOT differentiable is

1. {-1,1,2}
2. {-2,-1,1}
3. {0,1}
4. {-1,0,1}

Answer: {-1,0,1}

The function f(x) = max{x, x³} is not differentiable at points where the two expressions x and x³ intersect or change dominance, which occurs at x = -1, 0, and 1. At these points, the left-hand and right-hand derivatives do not match, indicating non-differentiability.

Q23. lim_(x→∞) (x − sin x)/(x + cos x) equals

1. 1
2. -1
3. ∞
4. -∞

Answer: 1

As x approaches infinity, the terms involving sine and cosine become negligible compared to the linear terms x in both the numerator and denominator. Thus, the limit simplifies to the ratio of the leading coefficients, which is 1.

Q24. A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve 3x⁴ - 16x³ + 24x² + 37 is

1. 0
2. 1
3. 2
4. 3

Answer: 1

The function is a polynomial of degree four, and by analyzing its first derivative, we can determine the critical points. In this case, there is only one local extremum, indicating that the curve has one distinct extremum.

Q25. ∫ from 0 to π/4 of (1 − tan x)/(1 + tan x) dx evaluates to

1. 0
2. 1
3. ln 2
4. ½ ln 2

Answer: ½ ln 2

(1-tanx)/(1+tanx) = (cosx-sinx)/(cosx+sinx), whose integral is ln|sinx+cosx|. Evaluated from 0 to pi/4: ln(sqrt2) - ln(1) = (1/2)ln2.

Q26. What is the value of lim[n→∞] (1 - 1/n)ⁿ ?

1. 0
2. e⁻¹
3. e^-n
4. 1

Answer: e⁻¹

As n approaches infinity, the expression (1 - 1/n)ⁿ converges to e⁻¹, which is derived from the limit definition of the exponential function. This limit represents the continuous compounding of a decreasing quantity.

Q27. Consider the function f(x) = sin(x) in the interval x ∈ [π/4, 7π/4]. The number and location(s) of the local minima of this function are

1. One, at π/2
2. One, at 3π/2
3. Two, at π/2 and 3π/2
4. Two, at π/4 and 3π/2

Answer: One, at 3π/2

The function f(x) = sin(x) has a local minimum at 3π/2 within the given interval, as this is the point where the sine function reaches its lowest value of -1, while there are no other local minima in the specified range.

Q28. A function f(x) is continuous in the interval [0,2]. It is known that f(0) = f(2) = -1 and f(1) = 1. Which one of the following statements must be true?

1. There exists a y in the interval (0,1) such that f(y) = f(y + 1)
2. For every y in the interval (0,1), f(y) = f(2 - y)
3. The maximum value of the function in the interval (0,2) is 1
4. There exists a y in the interval (0,1) such that f(y) = -f(2 - y)

Answer: There exists a y in the interval (0,1) such that f(y) = -f(2 - y)

This statement is true due to the Intermediate Value Theorem and the properties of continuous functions. Since f(0) = -1 and f(2) = -1, and f(1) = 1, there must be some point in (0,1) where the function takes on the value that is the negative of its value at the corresponding point in (1,2), ensuring that f(y) equals -f(2 - y) for some y in that interval.

Q29. A non-zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?

1. f(0)f(4) < 0
2. f(0)f(4) > 0
3. f(0) + f(4) > 0
4. f(0) + f(4) < 0

Answer: f(0)f(4) < 0

Write f(x)=a(x-1)(x-2)(x-3) with a nonzero. Then f(0)=a(-1)(-2)(-3)=-6a and f(4)=a(3)(2)(1)=6a, so f(0)f(4)=-36a^2, which is strictly negative for every nonzero a. Hence the statement that must be true is f(0)f(4)<0 (index 0), not f(0)f(4)>0.

Q30. The value of the integral given below is ∫₀^π x² cos x dx

1. −2π
2. π
3. −π
4. 2π

Answer: −2π

The integral evaluates to -2π because the integration by parts method applied to the function x² cos x results in a negative value due to the oscillatory nature of the cosine function and the polynomial growth of x².

Q31. lim_(x→4) sin(x − 4) / (x − 4) = _________.

1. 0
2. 1
3. 4
4. ∞

Answer: 1

As x approaches 4, the expression sin(x - 4) behaves like its argument (x - 4) due to the small angle approximation, leading to the limit being equal to 1.

Q32. Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(-x)) is 10, then the degree of (g(x) - g(-x)) is

1. 8
2. 9
3. 10
4. 11

Answer: 9

deg(f(x)+f(-x))=10 means the highest even-degree term of f is x^10. Then g=f' has an x^9 term, and g(x)-g(-x) retains only the odd-degree terms of g; the highest such term is x^9, so the degree is 9.

Q33. Which one of the following is a closed form expression for the generating function of the sequence {aₙ}, where aₙ = 2n + 3 for all n = 0, 1, 2,... ?

1. 3/(1 - x)²
2. 3x/(1 - x)²
3. (2 - x)/(1 - x)²
4. (3 - x)/(1 - x)²

Answer: (3 - x)/(1 - x)²

The correct option represents the generating function for the sequence defined by aₙ = 2n + 3. It combines the contributions from the linear term (2n) and the constant term (3) in a way that aligns with the properties of generating functions, specifically using the formula for the generating function of a linear sequence.

Q34. Compute lim x→3 (x⁴ - 81) / (2x² - 5x - 3)

1. 1
2. 53/12
3. 108/7
4. Limit does not exist

Answer: 108/7

x^4-81 = (x-3)(x+3)(x^2+9) and 2x^2-5x-3 = (2x+1)(x-3). Cancelling (x-3) gives (x+3)(x^2+9)/(2x+1); at x=3 this is 6*18/7 = 108/7.

Q35. A function y(x) is defined in the interval [0, 1] on the x-axis as y(x) = { 2 if 0 ≤ x < 1/3 3 if 1/3 ≤ x < 3/4 1 if 3/4 ≤ x ≤ 1 } Which one of the following is the area under the curve for the interval [0, 1] on the x-axis?

1. 5/6
2. 6/5
3. 13/6
4. 6/13

Answer: 13/6

Area = 2*(1/3) + 3*(3/4 - 1/3) + 1*(1 - 3/4) = 2/3 + 3*(5/12) + 1/4 = 8/12 + 15/12 + 3/12 = 26/12 = 13/6. The correct option is index 2, not the stored 5/6.

Q36. Which one of the following is the closed form for the generating function of the sequence {aₙ}_(n≥0) defined below? aₙ = { n+1, n is odd { 1, otherwise

1. x(1+x²)/(1-x²)² + 1/(1-x)
2. x(3-x²)/(1-x²)² + 1/(1-x)
3. 2x/(1-x²)² + 1/(1-x)
4. x/(1-x²)² + 1/(1-x)

Answer: x(1+x²)/(1-x²)² + 1/(1-x)

The correct option accurately represents the generating function by combining the contributions from the odd-indexed terms, which follow the pattern of n+1, and the even-indexed terms, which are constant at 1. The use of the generating function formula for sequences allows for the correct summation of these two distinct behaviors.

Q37. Consider two functions of time (t), f(t) = 0.01 t² and g(t) = 4 t, where 0 < t < ∞. Now consider the following two statements: (i) For some t > 0, g(t) > f(t). (ii) There exists a T, such that f(t) > g(t) for all t > T. Which one of the following options is TRUE?

1. only (i) is correct
2. only (ii) is correct
3. both (i) and (ii) are correct
4. neither (i) nor (ii) is correct

Answer: both (i) and (ii) are correct

Both statements are correct because for small values of t, g(t) can exceed f(t), but as t increases, the quadratic growth of f(t) eventually surpasses the linear growth of g(t), establishing a point T beyond which f(t) is always greater than g(t).

Q38. f(x) and g(y) are functions of x and y, respectively, and f(x) = g(y) for all real values of x and y. Which one of the following options is necessarily TRUE for all x and y?

1. f(x) = 0 and g(y) = 0
2. f(x) = g(y) = constant
3. f(x) ≠ constant and g(y) ≠ constant
4. f(x) + g(y) = f(x) − g(y)

Answer: f(x) = g(y) = constant

The correct option is true because if f(x) equals g(y) for all real values of x and y, then both functions must take on the same constant value across their entire domains to maintain equality.

Q39. Let f(x) = x³ + 15x² - 33x - 36 be a real-valued function. Which of the following statements is/are TRUE?

1. f(x) does not have a local maximum.
2. f(x) has a local maximum.
3. f(x) does not have a local minimum.
4. f(x) has a local minimum.

Answer: f(x) has a local maximum.

The function f(x) has a local maximum because its first derivative changes from positive to negative at a certain point, indicating that the function reaches a peak there.

Q40. Let f and g be functions of natural numbers given by f(n) = n and g(n) = n². Which of the following statements is/are TRUE?

1. f ∈ O(g)
2. f ∈ Ω(g)
3. f ∈ o(g)
4. f ∈ Θ(g)

Answer: f ∈ O(g)

The statement f ∈ O(g) is true because the function f(n) = n grows at a rate that is asymptotically less than or equal to the growth rate of g(n) = n², meaning there exists a constant C such that f(n) is bounded above by C * g(n) for sufficiently large n.

Q41. Let f(x) be a continuous function from R to R such that f(x) = 1 − f(2 − x) Which one of the following options is the CORRECT value of ∫₀² f(x) dx ?

1. 0
2. 1
3. 2
4. −1

Answer: 1

The equation f(x) = 1 - f(2 - x) implies that the function has a symmetry around x = 1. When integrating from 0 to 2, the contributions from f(x) and f(2 - x) balance out, leading to the integral equating to 1.

Q42. In the given figure, PQRS is a square of side 2 cm and PLMN is a rectangle. The corner L of the rectangle is on the side QR. Side MN of the rectangle passes through the corner S of the square. What is the area (in cm²) of the rectangle PLMN? Note: The figure shown is representative.

1. 2√2
2. 2
3. 8
4. 4

Answer: 4

The area of rectangle PLMN is calculated by multiplying its length and width. Given that one side of the rectangle is aligned with the side of the square and the other side extends to the corner S, the dimensions of the rectangle can be determined to be 2 cm by 2 cm, resulting in an area of 4 cm².

Q43. The value of x such that x > 1, satisfying the equation ∫₁^x t ln t dt = 1/4 is

1. √e
2. e
3. e²
4. e − 1

Answer: √e

The correct option, √e, satisfies the integral equation because when evaluated, it results in the area under the curve from 1 to √e equaling 1/4, confirming that it meets the condition of the problem.

Q44. For |x| << 1, coth(x) can be approximated as

1. x
2. x²
3. 1/x
4. 1/x²

Answer: 1/x

For small values of x, the hyperbolic cotangent function, coth(x), behaves similarly to the reciprocal of x, which is why it can be approximated as 1/x when |x| is much less than 1.

Q45. lim_(θ→0) sin(θ/2) / θ is

1. 0.5
2. 1
3. 2
4. not defined

Answer: 0.5

As θ approaches 0, the limit can be evaluated using the small angle approximation for sine, where sin(x) is approximately equal to x. Thus, sin(θ/2) approaches θ/2, leading to the limit simplifying to (θ/2) / θ, which equals 0.5.

Q46. Which one of the following functions is strictly bounded?

1. 1/x²
2. e^x
3. x²
4. e^(-x²)

Answer: e^(-x²)

The function e^(-x²) is strictly bounded because it approaches 0 as x approaches positive or negative infinity, and it has a maximum value of 1 at x=0, ensuring it remains within a fixed range.

Q47. For the function e^(-x), the linear approximation around x = 2 is

1. (3 - x)e⁻²
2. 1 - x
3. [3 + 2√2 - (1 + √2)x]e⁻²
4. e⁻²

Answer: (3 - x)e⁻²

The linear approximation of a function at a point is given by the formula f(a) + f'(a)(x - a). For the function e^(-x), evaluating at x = 2 gives the correct linear approximation as (3 - x)e⁻², which represents the tangent line at that point.

Q48. Consider the function f(x) = x² − x − 2. The maximum value of f(x) in the closed interval [−4, 4] is

1. 18
2. 10
3. −2.25
4. indeterminate

Answer: 18

The function f(x) = x² − x − 2 is a quadratic function that opens upwards, meaning it has a minimum point. To find the maximum value on the closed interval [-4, 4], we evaluate the function at the endpoints, and since f(4) = 18 is greater than f(-4) = 10, the maximum value is 18.

Q49. A 5-point sequence x[n] is given as x[-3] = 1, x[-2] = 1, x[-1] = 0, x[0] = 5, x[1] = 1. Let X(e^(jω)) denote the discrete-time Fourier transform of x[n]. The value of ∫_(-π)^(π) X(e^(jω)) dω is

1. 5
2. 10π
3. 16π
4. 5 + j10π

Answer: 10π

By the IDTFT, x[0] = (1/2pi) integral of X(e^jw) dw, so the integral equals 2*pi*x[0] = 2*pi*5 = 10*pi. The stored answer 5 is wrong; correct is 10*pi.

Q50. For real values of x, the minimum value of the function f(x) = exp(x) + exp(−x) is

1. 2
2. 1
3. 0.5
4. 0

Answer: 2

The function f(x) = exp(x) + exp(−x) represents the sum of an exponential function and its reciprocal, which reaches its minimum value when both terms are equal, specifically at x = 0, yielding f(0) = 2.