Calculus | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) PDF Download

Question 1: What is curl of the vector field 2x²yi + 5z²j - 4yzk?    [2019 : 1 Mark, Set-ll]
(a) -14zi-2x²k
(b) 6zi + 4x²j - 2x²k
(c) - 14zi + 6yj + 2x²k
(d) 6zi - 8xyj + 2x²yk
Answer: (a)
Solution: 


Question 2: Euclidean norm (length) of the vector [4 -2 -6]^r is    [2019 : 1 Mark, Set-ll]

Answer: (b)
Solution: 

Euclidean norm length


Question 3: The following inequality is true for all x close to 0.

What is tha value of      [2019 : 1 Mark, Set-ll]
(a) 1
(b) 0
(c) 1/2
(d) 2
Answer: (d)
Solution: 



Question 4: Consider the functions: x = y In φ and y = φ In y. which one of the following the correct expression for      [2019 : 2 Mark, Set-l]




Answer: (a)
Solution: 

 ......(i)


Question 5: Which one of the following is NOT a correct statement?    [2019 : 2 Marks, Set-I]
(a) The functionhas the global minima at x = e
(b) The function has the global maxima at x = a 
(c) The function x³ has neither global minima nor global maxima 
(d) The function |x| has the global minima at x = 0
Answer: (a) 
Solution: Let y = x^1/x
log y = logx/x

y maximum (or) minimum when,
 is maximum (or) minimum



Question 6: For a small value of h, the Taylor series expansion for f(x +h) is [2019 : 1 Mark, Set-I]

Answer: (c)
Solution: Taylor series of f (x + h) at x.
f(x + h) = f(x) + (x + h - x)



Question 7: Which of the following is correct?    [2019 : 1 Mark, Set-I]

Answer: (d)
Solution: 


Question 8: The value (up to two decimal places) of a line  along C which is a straight line joining (0, 0) to (1, 1) is _____.    [2018 : 2 Marks, Set-II]
Solution:

(0, 0) to (1, 1) line is y = x


Question 9: The value of the integral     [2018 : 2 Marks, Set-I]



(d) ∏²
Answer: (b)
Solution: 



Question 10: At the point x = 0, the function f(x) = x³ has    [2018 : 1 Marks, Set-I] 
(a) local maximum 
(b) local minimum 
(c) both local maximum and minimum
(d) neither local maximum nor local minimum
Answer: (d)
Solution: 


Question 11: Consider the following definite integral:

The value of the integral is    [2017 : 2 Marks, Set-II]




Answer: (a)
Solution:



Question 12: The tangent to the curve represented by y = x In x is required to have 45° inclination with the x-axis. The coordinates of the tangent point would be    [2017 : 2 Marks, Set-II]
(a) (1,0) 
(b) (0,1) 
(c) (1,1) 
(d)
Answer: (a)
Solution:
tan 45° = In x + 1
1 = lnx + 1
⇒ Inx = 0
∴ x = 1

Putting x = 1 in the eq. of curve, we get y = 0.

Question 13:  The divergence of the vector field V = x²i + 2y³j + z⁴k at x = 1, y = 2, z = 3 is _________.    [2017 : 1 Mark, Set-II]
Solution: 


Question 14: Let w= f(x, y), where x and yare functions of t. Then, according to the chain rule dw/dt    [2017 : 1 Mark, Set-II]




Answer: (c)
Solution: W = f(x, y)
By Chain rule,


Question 15: Let x be a continuous variable defined over the interval (-∞, ∞), and  The integral  is equal to    [2017 : 1 Mark, Set-I]




Answer: (b)
Solution: 

Let  e^-x = t

Question 16:    [2017 : 1 Mark, Set-I]
Solution:  (Applying L'Hospital rule)

                               =

Question 17: The quadratic approximation of 
f(x) = x³ - 3x² - 5 a the point x = 0 is    [2016 : 2 Marks, Set-II]
(a) 3x² - 6x - 5 
(b) -3x² - 5 
(c) -3x² + 6x - 5 
(d) 3x² - 5
Answer: (b)
Solution: The quadratic approximation of f{x) at the point x = 0 is,



Question 18: The area between the parabola x² = 8y and the straight line y = 8 is______.    [2016 : 2 Marks, Set-II]
Solution: Parabola is x² = 8y
 and straight is y = 0
At the point of intersection, we have,

⇒


Question 19: The angle of intersection of the curves x² = 4y and y² = 4x at point (0, 0) is     [2016 : 2 Marks, Set-II]
(a) 0° 
(b) 30° 
(c) 45° 
(d) 90° 
Answer: (d)
Solution: Given curve,
x² = 4y     .......(i)
and y² = 4x    ........(ii)






Question 20: The area of the region bounded by the parabola y = x² + 1 and the straight line x + y = 3 is
(a) 59/6
(b) 9/2
(c) 10/3
(d) 7/6
Answer: (b)
Solution: At the point of intersection of the curves,
y = x² + 1 and x + y = 3 i.e., y = 3 - x , we have,
x² + 1 - 3 - x ⇒ x² + x - 2 = 0
⇒ x = -2, 1 and 3 - x > x² + 1



Question 21: The value of     [2016 : 2 Marks, Set-I]
(a) π/2
(b) π
(c) 3π/2 
(d) 1
Answer: (b)
Solution: 



(Using "division by x")


(Using definition of Laplace transform)
Put s - 0, we get


Question 22: What is the value of      [2016 : 1 Mark, Set-II]
(a) 1
(b) -1
(c) 0
(d) Limit does not exit
Answer: (d)
Solution: 

(i.e., put x = 0 and then y = 0)

which depends on m.

Question 23: The optimum value of the function f(x) = x² - 4x + 2 is    [2016 : 1 Mark, Set-II]
(a) 2 (maximum) 
(b) 2 (minimum) 
(c) -2 (maximum) 
(d) -2 (minimum)
Answer: (d)
Solution: 
f'(x) = 0
⇒ 2x — 4 = 0
⇒ x = 2 (stationary point)
f"(x) = 2 > 0

⇒ f(x) is minimum at x = ?
i.e., (2)² - 4(2) + 2 = -2
∴ The optimum value of f(x) is -2 (minimum)

Question 24: While minimizing the function f(x), necessary and sufficient conditions for a point x₀ to be a minima are    [2015 : 1 Mark, Set-II]
(a) f' (x₀) > 0 and f" (x₀) = 0 
(b) f'(x₀)< 0 an d f"(x₀) = 0 
(c) f' (x₀) = 0 and f" (x₀) < 0 
(d) f' (x₀) = 0 and f" (x₀) > 0
Answer: (d)
Solution: f(x) has a local minimum at x = x₀ 
if f'(x₀) = 0 and f''(x₀) > 0

Question 25: is equal to    [2015 : 1 Mark, Set-II]
(a) e^-2 
(b) e 
(c) 1 
(d) e²
Answer: (d)
Solution: 

⇒
Which is in the form of
To convert this into 0/0  form, we rewrite as,
⇒
Now it is in 0/0 form.
Using L’Hospital’s rule,


∴ y = e²

Question 26: The directional derivative of the field u(x, y, z) = x² - 3yz in the direction of the vector  at point (2, - 1, 4) is _________.    [2015 : 2 Marks, Set-I]
Solution:


Directional derivative


Question 27: The expression  is equal to    [2014 : 2 Marks, Set-II]
(a) ln x 
(b) 0 
(c) x ln x 
(d) ∞
Answer: (a)
Solution: 



Question 28:    [2014 : 1 Marks, Set-I]
(a) -∞
(b) 0 
(c) 1
(d) ∞
Answer: (c)
Solution: Put



Question 29: The value of    [2013 : 2 Marks]
(a) 0
(b) 1/15
(c) 1
(d) 8/3
Answer: (b)
Solution: 

⇒


Alternative Method:




Question 30: For the parallelogram OPQR shown in the sketch,   The area of the parallelogram is    [2011 : 2 Marks]

(a) ad - bc 
(b) ac + bd 
(c) ad + bc 
(d) ab - cd
Answer: (a)
Solution: 

The area of parallelogram OPQR in figure shown above, is the magnitude of the vector product