Category: Engineering Mathematics Objective Questions

Basic Vector Calculus Questions and Answers focuses on “Gradient of a Function and Conservative Field”.

1. Del operator is also known as _________
a) Divergence operator
b) Gradient operator
c) Curl operator
d) Laplacian operator
View Answer

Answer: b
Explanation: This differential operator is not a vector itself but when it operates on a scalar function, for example, a vector ensues.

2. The gradient of a scalar field V is a vector that represents both magnitude and the direction of the maximum space rate of increase of V.
a) True
b) False
View Answer

Answer: a
Explanation: A gradient operates on a scalar only and gives a vector as a result. This vector has a magnitude and direction. The gradient is found by finding the speed that is by taking the partial differentiation.

3. The gradient is taken on a _________
a) tensor
b) vector
c) scalar
d) anything
View Answer

Answer: c
Explanation: Gradient is taken only on a scalar field. After taking gradient of a scalar field it becomes a vector. It is found by taking the partial differentiation.

4. Find the gradient of a function V if V= xyz.
a) yz a_x + xz a_y + xy a_z
b) yz a_x + xy a_y + xz a_z
c) yx a_x + yz a_y + zx a_z
d) xyz a_x + xy a_y + yz a_z
View Answer

Answer: a
Explanation: V = xyz
Gradient of (V = frac{∂V}{∂x} a_x + frac{∂V}{∂y}a_y + frac{∂V}{∂z}a_z )
= yz a_x + xz a_y + xy a_z

5. Find the gradient of V = x² sin(y)cos(z).
a) 2x siny cos z a_x + x² cos(y)cos(z) ay – x² sin(y)sin(z) az
b) 2x siny cos z a_x + x² cos(y)cos(z) ay + x² sin(y)sin(z) az
c) 2x sinz cos y a_x + x² cos(y)cos(z) ay – x² sin(y)sin(z) az
d) x siny cos z a_x + x² cos(y)cos(z) ay – x² sin(y)sin(z) az
View Answer

Answer: a
Explanation: (V = x^2 sin(y)cos(z) )
Gradient of (V = frac{∂V}{∂x} a_x + frac{∂V}{∂y} a_y + frac{∂V}{∂z} a_z )
(= 2x siny cos z a_x + x^2 cos(y)cos(z) a_y – x^2 sin(y)sin(z) a_z )

6. Find the gradient of the function W if W = ρzcos(ϕ) if W is in cylindrical coordinates.
a) zcos(ϕ)a_ρ – z sin(ϕ) a_Φ + ρcos(ϕ) a_z
b) zcos(ϕ)a_ρ – sin(ϕ) a_Φ + cos(ϕ) a_z
c) zcos(ϕ)a_ρ + z sin(ϕ) a_Φ + ρcos(ϕ) a_z
d) zcos(ϕ)a_ρ + z sin(ϕ) a_Φ + cos(ϕ) a_z
View Answer

Answer: a
Explanation: (W = ρzcos(ϕ) )
Gradient of (W = frac{1}{ρ} frac{∂W}{∂ρ} a_ρ + frac{1}{ρ} frac{∂W}{∂ϕ}a_y + frac{∂W}{∂z}a_z )
(= zcos(ϕ)a_ρ – z sin(ϕ) a_y + ρcos(ϕ) a_z )

7. Find the gradient of A if A = ρ² + z³ + cos(ϕ) + z and A is in cylindrical coordinates.
a) (2ρz^3 , a_ρ – frac{1}{ϕ} sin(ϕ) , aΦ + 3ρ^2 z^2 , a_z )
b) (2ρz^3 , a_ρ – frac{1}{ρ} sin(ϕ) , aΦ + 3ρ^2 z^2+1 , a_z )
c) (2ρz^3 , a_ρ – frac{1}{ϕ} sin(ϕ) , aΦ + 3ρ^2 z^2+1 , a_z )
d) (2ρz^3 , a_ρ – frac{1}{ρ} sin(ϕ) , aΦ + 3ρ^2 z^2 , a_z )
View Answer

Answer: b
Explanation: (A = ρ^2 + z^3 + cos(ϕ) + z )
Gradient of (A = frac{1}{ρ} frac{∂A}{∂ρ} a_ρ + frac{1}{ρ} frac{∂A}{∂ϕ}a_y + frac{∂A}{∂z}a_z )
(=2ρz^3 , a_ρ – frac{1}{ρ} sin(ϕ) , aΦ + 3ρ^2 z^2 + 1 , a_z )

8. Find gradient of B if B = ϕln(r) + r² ϕ if B is in spherical coordinates.
a) (frac{ρ}{r}+ 2rθ ,a_r – r a_θ + frac{lnr}{rsin(θ)} a_Φ )
b) (frac{ρ}{r}+ 2rϕ ,a_r – r a_θ + frac{lnr}{rsin(θ)} a_Φ )
c) (frac{ρ}{r}+ 2rθ ,a_r – r^2 a_θ + frac{lnr}{rsin(θ)} a_Φ )
d) ( frac{ρ}{r}+ 2rϕ ,a_r – r^2 a_θ + frac{lnr}{rsin(θ)} a_Φ )
View Answer

Answer: a
Explanation: (B = ϕln(r)+r^2 ϕ )
Gradient of (B = frac{∂B}{∂r} a_r + frac{1}{r} frac{∂B}{∂θ} a_θ + frac{1}{rsin(θ)} frac{∂B}{∂ϕ} a_Φ )
(= frac{ρ}{r}+ 2rθ ,a_r – r a_θ + frac{lnr}{rsin(θ)} a_Φ )

9. Find gradient of B if B = rθϕ if X is in spherical coordinates.
a) (θϕ , a_r – ϕ ,a_θ + frac{θ}{sin(θ)} a_Φ )
b) (rθϕ , a_r – ϕ ,a_θ + r frac{θ}{sin(θ)} a_Φ )
c) (θϕ , a_r – ϕr ,a_θ + frac{θ}{sin(θ)} a_Φ )
d) (θϕr , a_r – ϕ ,a_θ + rfrac{θ}{sin(θ)} a_Φ )
View Answer

Answer: a
Explanation: Gradient of (B = frac{∂B}{∂r} a_r + frac{1}{r} frac{∂B}{∂θ} a_θ + frac{1}{rsin(θ)} frac{∂B}{∂ϕ} a_Φ )
(B = rθϕ)
Hence gradient of (B = θϕ , a_r -ϕ , a_θ + frac{θ}{sin(θ)} a_Φ )

10. If W = x² y² + xz, the directional derivative ( frac{dW}{dl} ) in the direction 3 a_x + 4 a_y + 6 a_z at (1,2,0).
a) 5
b) 6
c) 7
d) 8
View Answer

Answer: b
Explanation: First find the gradient of W which is (2xy²+z) a_x + 2yx² a_y + x a_z
At (1,2,0) the gradient of W is 8 a_x + 4 a_y + 1 a_z
(frac{dW}{dl} = ) (Gradient of W ) . al
( = (8,4,1) . frac{(3,4,6)}{sqrt{(9+16+36)}} )
(=5.88897 = 6.)

11. If W = xy + yz + z, find directional derivative of W at (1,-2,0) in the direction towards the point (3,6,9).
a) -0.6
b) -0.7
c) -0.8
d) -0.9
View Answer

Answer: c
Explanation: The gradient of W is = y a_x + (x+z) a_y + (y+1) a_z
At (1,-2,0) the gradient of the function W is -2 a_x + a_y – a_z
(frac{dW}{dl} = )(Gradient of W ) . a_l
(= frac{(-2,1,-1).(3,6,9)}{11.22} )
(= -0.8.)

12. Electric field E can be written as _________
a) -Gradient of V
b) -Laplacian of V
c) Gradient of V
d) Laplacian of V
View Answer

Answer: a
Explanation: Potential difference decreases in the direction of increase in Electric field. Hence Electric field is nothing but the negative of the gradient of potential difference.

13. Let F = (xy²) a_x + yx² a_y, F is a not a conservative vector.
a) True
b) False
View Answer

Answer: b
Explanation: Q = xy² and P = yx²
(frac{∂P}{∂y} = 2xy) and (frac{∂Q}{∂x} = 2xy )
Since, both are equal, F is a conservative vector.

14. State whether the given equation is a conservative vector.
G = (x³y) a_x + xy³ a_y
a) True
b) False
View Answer

Answer: b
Explanation: P = x³ y and Q = xy³
(frac{∂P}{∂y} = x^3) and (frac{∂Q}{∂x}= y^3 )
Now since they aren’t equal, the vector is not a conservative vector or field.

15. Find a unit vector normal to the surface of the ellipsoid at (2,2,1) if the ellipsoid is defined as f(x,y,z) = x² + y² + z² – 10.
a) (frac{2}{3} a_x + frac{2}{3} a_y + frac{1}{3} a_z )
b) (frac{1}{3} a_x + frac{1}{3} a_y + frac{1}{3} a_z )
c) (frac{2}{3} a_x + frac{2}{3} a_y + frac{2}{3} a_z )
d) (frac{2}{3} a_x + frac{1}{3} a_y + frac{1}{3} a_z )
View Answer

Answer: a
Explanation: First we have to find the gradient of the function, which is equal to 2x a_x + 2y a_y + 2z a_z.
Gradient of f at (2,2,1) is (4,4,2).
(a_n= frac{±(4,4,2)}{6} = frac{2}{3} a_x + frac{2}{3} a_y + frac{1}{3} a_z. )

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Partial Differential Equations Questions and Answers for Freshers focuses on “Solution of PDE by Variable Separation Method”.

1. Solve (frac{∂u}{∂x}=6 frac{∂u}{∂t}+u) using the method of separation of variables if u(x,0) = 10 e^-x.
a) 10 e^-x e^-t/3
b) 10 e^x e^-t/3
c) 10 e^x/3 e^-t
d) 10 e^-x/3 e^-t
View Answer

Answer: a
Explanation: u(x,t) = X(x) T(t)
Substituting in the given equation, X’T = 6 T’X + XT
(frac{X’-X}{6X}=frac{T’}{T}=k)
(frac{X’}{X} = 1+6k ) which implies X = ce^(1+6k)x
(frac{T’}{T} = k ) which implies T = c’ e^kt
Therefore, u(x,t) = cc’ e^(1+6k)x e^kt
Now, u(x,0) = 10 e^-x = cc’e^(1+6k)x
Therefore, cc’ = 10 and k = ^-1⁄₃
Therefore, u(x,t) = 10 e^-x e^-t/3.

2. Find the solution of (frac{∂u}{∂x}=36 frac{∂u}{∂t}+10u ) if ( frac{∂u}{∂x} (t=0)=3e^{-2x} ) using the method of separation of variables.
a) (frac{-3}{2} e^{-2x} e^{-t/3})
b) (3e^x e^{-t/3} )
c) (frac{3}{2} e^{2x} e^{-t/3} )
d) (3e^{-x} e^{-t/3} )
View Answer

Answer: a
Explanation: (u(x,t) = X(x) T(t) )
Substituting in the given equation, (X’T = 36T’X + 10XT )
(frac{X’}{X} = k ) which implies (X = c e^{kx} )
(frac{T’}{T} = frac{(k-10)}{36} ) which implies (T = c’e^{frac{k-10}{36} t} )
(frac{∂u}{∂x} (t=0)=3e^{-2x}= cc’ke^{kx} )
Therefore k = -2 and cc’ = (frac{-3}{2} )
Hence, (u(x,t) = frac{-3}{2} e^{-2x} e^{-t/3}. )

3. Solve the partial differential equation (x^3 frac{∂u}{∂x} +y^2 frac{∂u}{∂y} = 0 ) using method of separation of variables if (u(0,y) = 10 , e^{frac{5}{y}}.)
a) (10e^{frac{5}{2x^2}} e^{frac{5}{y}} )
b) (10e^{frac{-5}{2y^2}} e^{frac{5}{x}} )
c) (10e^{frac{-5}{2y^2}} e^{frac{-5}{x}} )
d) (10e^{frac{-5}{2x^2}} e^{frac{5}{y}} )
View Answer

Answer: d
Explanation: (u(x,t) = X(x) T(t) )
(x^3 X’Y+y^2 Y’X=0)
(frac{X’}{ X} = frac{k}{x^3} ) which implies (X = ce^{frac{k}{2x^2}})
(frac{Y’}{Y} = frac{-k}{y^2} ) which implies (Y = c’ e^{frac{k}{y}})
(u(x,t) = cc’e^{frac{k}{2x^2}} e^{frac{k}{y}} )
(u(0,y) = 10e^{frac{5}{y}}= cc’e^{frac{k}{y}} )
Therefore k = 5 and cc’ = 10
Hence, (u(x,t) = 10e^{frac{-5}{2x^2}} e^{frac{5}{y}}. )

4. Solve the differential equation (5 frac{∂u}{∂x}+3 frac{∂u}{∂y}=2u ) using the method of separation of variables if (u(0,y) = 9e^{-5y}.)
a) (9e^{frac{17}{5} x} e^{-5y} )
b) (9e^{frac{13}{5} x} e^{-5y} )
c) (9e^{frac{-17}{5} x} e^{-5y} )
d) (9e^{frac{-13}{5} x} e^{-5y} )
View Answer

Answer: a
Explanation: ( u(x,t) = X(x) T(t) )
(5X’Y + 3Y’X = 2XY )
(frac{X’}{X} = frac{k}{5} ) which implies ( X= c e^{frac{k}{5} x} )
(frac{Y’}{Y} = 2-k / 3) which implies ( Y = c’ , e^{frac{2-k}{3} y} )
Therefore (u(x,t) = cc’ e^{frac{k}{5} x} e^{frac{2-k}{3} y} )
(u(0,y) = cc’ = e^{frac{2-k}{3} y} , 9 , e^{-5y} )
Hence cc’ = 9 and k = 17
Therefore, (u(x,t) = 9 , e^{frac{17}{5} x} e^{-5y}. )

5. Solve the differential equation (x^2 frac{∂u}{∂x}+y^2 frac{∂u}{∂y}=u ) using the method of separation of variables if (u(0,y) = e^{frac{2}{y}} ).
a) (e^{frac{-3}{y}} e^{frac{2}{x}} )
b) (e^{frac{3}{y}} e^{frac{2}{x}} )
c) (e^{frac{-3}{x}} e^{frac{2}{y}} )
d) (e^{frac{3}{x}} e^{frac{2}{y}} )
View Answer

Answer: c
Explanation: (u(x,t) = X(x) T(t) )
(x^2 X’Y+y^2 Y’X=XY)
(X = ce^{frac{-k}{x}} )
(Y = c’e^{frac{k-1}{y}} )
(u(x,t) = cc’ e^{frac{-k}{x}} e^{frac{k-1}{y}} )
(u(0,y) = e^{frac{2}{y}}= cc’ e^{frac{k-1}{y}} )
k = 3 and cc’ = 1
Therefore (u(x,t) = e^{frac{-3}{x}} e^{frac{2}{y}}. )

6. While solving a partial differential equation using a variable separable method, we assume that the function can be written as the product of two functions which depend on one variable only.
a) True
b) False
View Answer

Answer: a
Explanation: If we have a function u(x,t), then the function u depends on both x and t. For using the variable separable method we assume that it can be written as u(x,t) = X(x).T(t) where X depends only on x and T depends only on t.

7. While solving a partial differential equation using a variable separable method, we equate the ratio to a constant which?
a) can be positive or negative integer or zero
b) can be positive or negative rational number or zero
c) must be a positive integer
d) must be a negative integer
View Answer

Answer: b
Explanation: The constant can be any rational number. For example, we use a positive rational number to solve a 1-Dimensional wave equation, we use a negative rational number to solve 1-Dimensional heat equation, 0 when we have steady state. The choice of constant depends on the nature of the given problem.

8. When solving a 1-Dimensional wave equation using variable separable method, we get the solution if _____________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
View Answer

Answer: b
Explanation: Since the given problem is 1-Dimensional wave equation, the solution should be periodic in nature. If k is a positive number, then the solution comes out to be (c₇ e^px⁄c+e^-px⁄cc₈)(c₇ e^pt+e^-ptc₈) and if k is positive the solution comes out to be (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt). Now, since it should be periodic, the solution is (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt).

9. When solving a 1-Dimensional heat equation using a variable separable method, we get the solution if ______________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
View Answer

Answer: b
Explanation: Since this is a heat equation, the solution must be a transient solution, that is it should decay as time increases. This happens only when k is negative and the solution comes out to be (c’’cospx + c’’’sinpx) (c e^{-c2 p2 t}).

10. While solving any partial differentiation equation using a variable separable method which is of order 1 or 2, we use the formula of fourier series to find the coefficients at last.
a) True
b) False
View Answer

Answer: a
Explanation: After using the boundary conditions, when we are left with only one constant and one boundary condition, then we use Fourier series coefficient formula to find the constant.

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Differential and Integral Calculus Multiple Choice Questions on “Cauchy’s Mean Value Theorem”.

1. Cauchy’s Mean Value Theorem can be reduced to Lagrange’s Mean Value Theorem.
a) True
b) False
View Answer

Answer: a
Explanation: Cauchy’s Mean Value Theorem is the generalized form of Lagrange’s Mean Value Theorem and can be given by,
(frac{f'(a+θh)}{g'(a+θh)} = frac{f(a+h)-f(a)}{g(a+h)-g(a)}, 0 < θ < 1)
Hence, if g(x) = x, then CMV reduces to LMV.

2. Which of the following is not a necessary condition for Cauchy’s Mean Value Theorem?
a) The functions, f(x) and g(x) be continuous in [a, b]
b) The derivation of g'(x) be equal to 0
c) The functions f(x) and g(x) be derivable in (a, b)
d) There exists a value c Є (a, b) such that, (frac{f(b)-f(a)}{g(b)-g(a)} = frac{f'(c)}{g'(c)})
View Answer

Answer: b
Explanation: Cauchy’s Mean Value theorem is given by, (frac{f(b)-f(a)}{g(b)-g(a)} = frac{f'(c)}{g'(c)}), where f(x) and g(x) be two functions which are derivable in [a, b] and g'(x)≠0 for any value of x in [a, b] and where c Є (a, b).

3. Cauchy’s Mean Value Theorem is also known as ‘Extended Mean Value Theorem’.
a) False
b) True
View Answer

Answer: b
Explanation: Mean Value Theorem is given by, (frac{f(b)-f(a)}{b-a} = f'(c),) where c Є (a, b).
This theorem can be generalized to Cauchy’s Mean Value Theorem and hence CMV is also known as ‘Extended’ or ‘Second Mean Value Theorem’.

4. The Mean Value Theorem was stated and proved by _______
a) Parameshvara
b) Govindasvami
c) Michel Rolle
d) Augustin Louis Cauchy
View Answer

Answer: d
Explanation: Augustin Louis Cauchy was a French Mathematician, Engineer and Physicist who first stated and proved the Mean Value Theorem.

5. Find the value of c which satisfies the Mean Value Theorem for the given function,
f(x)= x²+2x+1 on [1,2].
a) (frac{-7}{2} )
b) (frac{7}{2} )
c) (frac{13}{2} )
d) (frac{-13}{2} )
View Answer

Answer: a
Explanation: Given function is, f(x)= x²+2x+1.
According to Mean Value Theorem,
(f'(c) = frac{f(b)-f(a)}{b-a} )
f'(c)=2c+2
(2c+2 = frac{(1+2+1)-(4+4+1)}{2-1}=frac{4-9}{1}= -5 )
2c= -7
(c= frac{-7}{2} )

6. What is the largest possible value of f(0), where f(x) is continuous and differentiable on the interval [-5, 0], such that f(-5)= 8 and f'(c)≤2.
a) 2
b) -2
c) 18
d) -18
View Answer

Answer: b
Explanation: From the Mean Value Theorem, we have, (f'(c) = frac{f(b)-f(a)}{b-a} )
(f'(c) = frac{f(0)-f(-5)}{-5-0} )
-5f’ (c) = f(0)-8
f(0)=8 – 5f'(c) ≤ 8-5(2) = -2
f(0)=-2

7. What is the value of c which lies in [1, 2] for the function f(x)=4x and g(x)=3x²?
a) 1.6
b) 1.5
c) 1
d) 2
View Answer

Answer: b
Explanation: From Cauchy’s Mean Value Theorem, we have, (frac{f(b)-f(a)}{g(b)-g(a)} = frac{f'(c)}{g'(c)})
(frac{8-4}{12-3}=frac{4}{6c})
(6c=frac{4*9}{4} )
(c=frac{9}{6}=frac{3}{2}=1.5)

8. Which of the following method is used to simplify the evaluation of limits?
a) Cauchy’s Mean Value Theorem
b) Rolle’s Theorem
c) L’Hospital Rule
d) Fourier Transform
View Answer

Answer: c
Explanation: L’Hospital’s Rule is used as a definitive way of simplification. The L’Hospital’s Rule does not directly evaluate the limits but only simplifies the evaluation.

9. What is the value of the given limit, (lim_{xto 0}⁡frac{2}{x})?
a) 2
b) 0
c) 1/2
d) 3/2
View Answer

Answer: a
Explanation: Given: (lim_{xto 0}frac{2}{x})
Using L’Hospital’s Rule, by differentiating both the numerator and denominator with respect to x,
(lim_{x→0}⁡frac{2}{1}=2)

10. L’Hospital’s Rule was first discovered by Marquis de L’Hospital.
a) True
b) False
View Answer

Answer: b
Explanation: The L’Hospital’s Rule was first published in Marquis de L’Hospital’s book ‘Analyse des Infiniment Petits’, but the rule was discovered by Swiss Mathematician Johann Bernoulli.

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Engineering Mathematics Multiple Choice Questions on “Limits and Derivatives of Several Variables – 1”.

1. Find (lt_{(x,y)rightarrow(0,0)}frac{121.x^{-5}.y^{frac{13}{3}}}{y+(x)frac{3}{2}})
a) ∞
b) 0
c) Does Not Exist
d) 121
Answer: c
Explanation: Put x = t : y = a.t^3⁄₂ we have
=(lt_{(x,y)rightarrow(0,0)}frac{121.t^{-5}.(at^{frac{3}{2}})^{frac{13}{3}}}{t^{frac{3}{2}}+t^{frac{3}{2}}})
=(lt_{(x,y)rightarrow(0,0)}frac{121.at^{frac{13}{3}}.t^{frac{3}{2}}}{2.t^{frac{3}{2}}})
=(lt_{(x,y)rightarrow(0,0)}frac{121.at^{frac{13}{3}}}{2})
By varying a we get different limits along different paths
Hence, Does Not Exist is the right answer.

2. Find (lt_{(x,y)rightarrow(0,0)}frac{y^6}{x^{10}y^2+x^{15}})
a) 0
b) 1
c) Does Not exist
d) ∞
Answer: c
Explanation: Put Put x = t : y = a.t^5⁄₂ we have
=(lt_{(x,y)rightarrow(0,0)}frac{(a.t^{frac{5}{2}})^6}{t^{10}.(a.t^{frac{5}{2}})^2+t^{15}})
=(lt_{(x,y)rightarrow(0,0)}frac{a^6}{a^2+1})
By varying a we get different limits along different paths
Hence, Does Not exist is the right answer.

3. Find (lt_{(x,y)rightarrow(0,0)}frac{sec(y).sin(x)}{x})
a) ∞
b) ¹⁄₂
c) 1
d) ¹⁄₃
Answer: c
Explanation: Treating limits separately we have
lt_{(x, y)→(0, 0)}^sin(x)⁄_x * lt_{(x, y)→(0,0)} sec(y)
= 1 * 1
= 1.

4. Find (lt_{(x,y)rightarrow(0,0)}frac{x^3-y^3}{(x-y)})
a) ^-1⁄₂
b) 0
c) ∞
d) -90
Answer: b
Explanation: Simplifying the expression we have
(lt_{(x,y)rightarrow(0,0)}frac{(x-y)(x^2+xy+y^2)}{(x-y)})
(lt_{(x,y)rightarrow(0,0)}frac{(x^2+xy+y^2)}{1})=(0²+0.0+0²)
= 0.

5. Find (lt_{(x,y)rightarrow(0,1)}frac{x+y-1}{sqrt{x+y}-1})
a) 9
b) 0
c) 6
d) 2
Answer: d
Explanation: Simplifying the expression we have
=(lt_{(x,y)rightarrow(0,1)}frac{(sqrt{x+y}^2)-(1)^2}{sqrt{x+y}-1}=lt_{(x,y)rightarrow(0,1)}frac{(sqrt{x+y}+1).(sqrt{x+y}-1)}{sqrt{x+y}-1})
=(lt_{(x,y)rightarrow(0,1)}(sqrt{x+y}+1)=sqrt{1}+1)
=2

6. Find (lt_{(x,y)rightarrow(0,0)}frac{x^3+3xy^2-xy^2}{x^2+xy})
a) 0
b) ∞
c) 1
d) -1
Answer: a
Explanation: Converting into Polar form we have
=(lt_{rrightarrow 0}frac{r^3.cos^3(theta))+3(r^2.cos^2(theta))(r.sin(theta))-(r.cos(theta))(r^2.sin^2(theta))}{(r^2cos^2(theta))+r^2sin(theta)cos(theta)})
=(lt_{rrightarrow 0}frac{r^3}{r^2}times (frac{cos^3(theta)+3(cos^2(theta))(sin(theta))-(cos(theta))(sin^2(theta))}{(cos^2(theta))+sin(theta)cos(theta)}))
=(lt_{rrightarrow 0}(r)times (frac{cos^3(theta)+3(cos^2(theta))(sin(theta))-(cos(theta))(sin^2(theta))}{(cos^2(theta))+sin(theta)cos(theta)}))
=0

7. Find (lt_{(x,y)rightarrow(0,0)}frac{sin(y)}{x})
a) 1
b) 0
c) ∞
d) Does Not Exist
Answer: d
Explanation: Put x = t : y = at
(=lt_{trightarrow 0}frac{sin(at)}{t})
(=lt_{trightarrow 0}a times frac{sin(at)}{at}=a times lt_{trightarrow 0}frac{sin(at)}{at})
= a * (1) = a
By varying a we get different limits
Hence, Does Not Exist is the right answer.

8. Find (lt_{(x,y)rightarrow(infty,0)}(sum_{a=1}^{x-1}sin(frac{a}{x}).sin(y)))
a) 1
b) -1
c) ∞
d) Does not Exist
Answer: d
Explanation: Multiplying and dividing by we have
(lt_{(x,y)rightarrow(infty,0)}(sin(y))times(sum_{a=1}^{x-1}sin(frac{a}{x})))
(lt_{(x,y)rightarrow(infty,0)}(x.sin(y))times lt_{(x,y)rightarrow(infty,0)}left (sum_{a=1}^{x-1}frac{sin(frac{a}{x})}{x}right ))
(lt_{(x,y)rightarrow(infty,0)}(frac{sin(y)}{frac{1}{x}})times lt_{(x,y)rightarrow(infty,0)}left (sum_{a=1}^{x-1}frac{sin(frac{a}{x})}{x}right ))
Put z=1/x : as x → ∞ : z → 0
Consider one part of the limit
(=lt_{(x,y)rightarrow (0,0)}frac{sin(y)}{z})
Put : y = t : z = at
(=lt_{trightarrow 0}frac{sin(t)}{at}=frac{1}{a} lt_{trightarrow 0}frac{sin(t)}{t})
=(frac{1}{a}times 1= frac{1}{a}).

9. Find (lt_{(x,y)rightarrow (0,0)}frac{y^7x^{98}-x^{97}y^8+x^{105}}{xy^7+x^8})
a) Does Not Exist
b) 0
c) 1
d) ∞
Answer: b
Explanation: Put x =r.cos(ϴ) : y = r.sin(ϴ)
=(lt_{(x,y)rightarrow (0,0)}frac{(r^7.sin^7(theta))(r^{98}.sin^{98}(theta))-(r^{97}.cos^{97}(theta))(r^8.sin^8(theta))+(r^{105}.cos^{105}(theta))}{(r.cos(theta)(r^7.sin^7(theta))+(r^8.cos(theta))})
=(lt_{(x,y)rightarrow (0,0)}frac{r^{105}}{r^8}times frac{(sin^7(theta))(sin^{98}(theta))-(cos^{97}(theta))(sin^8(theta))+(cos^{105}(theta))}{(cos(theta)(sin^7(theta))+(cos(theta))})
=(lt_{(x,y)rightarrow (0,0)}(r^{97})times frac{(sin^7(theta))(sin^{98}(theta))-(cos^{97}(theta))(sin^8(theta))+(cos^{105}(theta))}{(cos(theta)(sin^7(theta))+(cos(theta))})
= 0

10. Find (lt_{(x,y)rightarrow(0,0)}frac{sin(y)}{x^n})
a) 0
b) ∞
c) 1
d) Does Not Exist
Answer: d
Explanation: Put x = at : y = t
(=lt_{trightarrow 0}frac{sin(t)}{a^n.t^n})
(=lt_{trightarrow 0}frac{1}{a^nt^{n-1}}frac{sin(t)}{t})
By varying n we get different limits
Hence, Does Not Exist is the right answer.

11. Find (lt_{(x,y)rightarrow(0,0)}frac{sin(sin(y))}{x^n})
a) Does not Exist
b) 0
c) ∞
d) 1
Answer: a
Explanation: Put x = at : y = t
(=lt_{trightarrow 0}frac{1}{a^nt^{n-1}}times frac{sin(sin(t))}{t})
(=lt_{trightarrow 0}frac{1}{a^nt^{n-1}} times (1))
By varying n we get different values of limits.

12. Find (=lt_{(x,y)rightarrow (0,0)}frac{tan(y)}{x})
a) ∞
b) 1
c) ¹⁄₂
d) Does Not Exist
Answer: d
Explanation: Put x = t : y = at
=(lt_{trightarrow 0}times frac{tan(at)}{t})
=(lt_{trightarrow 0} (a) times frac{tan(at)}{at})
=a
By varying the value of a we get different limits.

13. Find (lt_{(x,y,z)rightarrow(0,0,0)}frac{sinh(x)times sinh(y)times sinh(z)}{xyz})
a) 1
b) ∞
c) 0
d) 990
Answer: a
Explanation:
(=lt_{(x,y,z)rightarrow(0,0,0)}frac{sinh(x)}{x}times lt_{(x,y,z)rightarrow(0,0,0)}frac{sinh(y)}{y}times lt_{(x,y,z)rightarrow(0,0,0)}frac{sinh(z)}{z})
= 1 * 1 * 1
= 1.

14. Find (lt_{(x,y)rightarrow(0,0)}frac{sinh(x)times sinh(y)}{xy})
a) 1
b) ∞
c) 0
d) 990
Answer: a
Explanation: lt_{(x, y)→(0, 0)}^sinh(x)⁄_x * lt_{(x, y)→(0, 0)}^sinh(y)⁄_y
= 1 * 1
= 1.