Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Cayley Hamilton Theorem and Answers

Linear Algebra Multiple Choice Questions on “Cayley Hamilton Theorem”.

1. Find the inverse of the given Matrix, using Cayley Hamilton’s Theorem.
A=(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix})
a) A-1=(frac{1}{16} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
b) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-3\-6&9&11end{bmatrix})
c) A-1=(frac{1}{16} begin{bmatrix}2&-1&-1\4&-2&-6\-6&9&11end{bmatrix})
d) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
Answer: d
Explanation: For the given Matrix,
A=(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix})
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0
α3-7α2+11α-8=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.
Thus,
A3-7A2+11A-8I=0
To find A-1, multiply both the sides of the equation by A-1
A2 A A-1-7AA A-1+11A A-1-8I A-1=0
We know that A A-1=I
A2I-7AI+11I-8IA-1=0
A2-7A+11-8 A-1=0
A2-7A+11=8 A-1

8A-1=(begin{bmatrix}19&18&13\-3&1&1\15&9&7end{bmatrix}-7begin{bmatrix}4&3&2\-1&2&1\3&0&1end{bmatrix}+11begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

8A-1=(begin{bmatrix}19-28+11&18-21&13-14\-3+7&1-14+11&1-7\15-21&9&7-7+11end{bmatrix})

8A-1=(begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})

A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix}).

2. Find the value of A3 where A=(begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix}).
a) (begin{bmatrix}3&5&-1\-2&-9&2\-2&-4&-5end{bmatrix})
b) (begin{bmatrix}3&5&-1\1&-9&1\-2&-4&-5end{bmatrix})
c) (begin{bmatrix}3&5&-1\-2&-9&1\-2&-4&-5end{bmatrix})
d) (begin{bmatrix}3&5&-1\-1&-9&1\-2&-4&-5end{bmatrix})
Answer: c
Explanation: For the given Matrix,
A=(begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix})
The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-α2+3α+5=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-A2+3A+5I=0
A3=A2-3A-5I
A3=(begin{bmatrix}5&2&5\-2&-1&-2\4&2&3end{bmatrix}-3begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix}-5begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

A3=(begin{bmatrix}5+3-5&2+3&5-6\-2+0&-1-3-5&-2+3\4-6&2-6&3-3-5end{bmatrix})

A3=(begin{bmatrix}3&5&-1\-2&-9&1\-2&-4&-5end{bmatrix}).

3. Find the value of A3+19A, A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{bmatrix}).
a) (begin{bmatrix}42&-14&70\21&+21&-21\105&119&203end{bmatrix})
b) (begin{bmatrix}42&-7&70\21&-21&-21\105&119&203end{bmatrix})
c) (begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix})
d) (begin{bmatrix}42&-7&70\21&+21&-21\105&119&203end{bmatrix})
Answer: c
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{bmatrix})
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-7α2+19α-49=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-7A2+19A-49I=0
A3+19A=7A2+49I
A3+19A=7(begin{bmatrix}-1&-2&10\3&-10&-3\15&17&22end{bmatrix}+49begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
A3+19A=(begin{bmatrix}-7+49&-17&70\21&-70+49&-21\105&119&154+49end{bmatrix})
A3+19A=(begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix}).

4. Find the value of 2A3+4A2, where = (begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}).
a) (begin{bmatrix}-200&0&-24\24&-32&-24\-72&96&-56end{bmatrix})
b) (begin{bmatrix}-200&0&-24\24&-32&-12\-72&96&-56end{bmatrix})
c) (begin{bmatrix}-200&0&-24\12&-32&-24\-72&96&-56end{bmatrix})
d) (begin{bmatrix}-100&0&-12\12&-16&-12\-36&48&-28end{bmatrix})
Answer: a
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix})

The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3+2α2-12α-40=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3+2A2-12A+40I=0
A3+2A2=12A-40I
A3+2A2=(12begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}-40begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

A3+2A2=(begin{bmatrix}-60-40&0&-12\12&24-40&-12\-36&48&12-40end{bmatrix})

A3+2A2=(begin{bmatrix}-100&0&-12\12&-16&-12\-36&48&-28end{bmatrix})

2A3+4A2=(begin{bmatrix}-200&0&-24\24&-32&-24\-72&96&-56end{bmatrix}).

5. Find the value of A3-3A2-28A, A = (begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}).
a) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
b) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
c) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
d) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
Answer: b
Explanation: For the given Matrix,
A=(begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix})

The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-3α2+35α-17=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-3A2+35A-17I=0
On performing long division (α3-3α2+35α-17)/(α2-7α)
Q=α+4 and R=63α-17
Using division properties,
α3-3α2+35α-17=(α2-7α)×(α+4)+(63α-17)
α3-3α2+35α-17=(α3-3α2-28α)+( 63α-17)
0=(α3-3α2-28α)+(63α-17) ————— (From Characteristic Polynomial)
(α3-3α2-28α) = -63α+17
(A3-3A2-28A) = -63A+17I
(A3-3A2-28A) = (-63begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}+17begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}63+17&-126&-504\126&17-189&-63\252&-315&17-63end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix}).

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250+ TOP MCQs on First Order Non-Linear PDE and Answers

Fourier Analysis and Partial Differential Equations Multiple Choice Questions on “First Order Non-Linear PDE”.

1. Which of the following is an example of non-linear differential equation?
a) y=mx+c
b) x+x’=0
c) x+x2=0
d) x”+2x=0
View Answer

Answer: c
Explanation: For a differential equation to be linear the dependent variable should be of first degree. Since in equation x+x2=0, x2 is not a first power, it is not an example of linear differential equation.

2. Which of the following is not a standard method for finding the solutions for differential equations?
a) Variable Separable
b) Homogenous Equation
c) Orthogonal Method
d) Bernoulli’s Equation
View Answer

Answer: c
Explanation: The following are the different standard methods used in finding the solution of a differential equation:

  • Variable Separable
  • Homogenous Equation
  • Non-homogenous Equation reducible to Homogenous Equation
  • Exact Differential Equation
  • Non-exact Differential Equation that can be made exact with the help of integrating factors
  • Linear First Order Equation
  • Bernoulli’s Equation

3. Solution of a differential equation is any function which satisfies the equation.
a) True
b) False
View Answer

Answer: a
Explanation: A solution of a differential equation is any function which satisfies the equation, i.e., reduces it to an identity. A solution is also known as integral or primitive.

4. A solution which does not contain any arbitrary constants is called a general solution.
a) True
b) False
View Answer

Answer: a
Explanation: The solution of a partial differential equation obtained by eliminating the arbitrary constants is called a general solution.

5. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
View Answer

Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

6. A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution.
a) True
b) False
View Answer

Answer: a
Explanation: A solution which does not contain any arbitrary constants is called a general solution whereas a particular solution is derived by substituting particular values to the arbitrary constants in this solution.

7. Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.
a) True
b) False
View Answer

Answer: a
Explanation: A differential equation is said to have a singular solution if in all points in the domain of the equation the uniqueness of the solution is violated. Hence, this solution cannot be obtained from the general solution.

8. Which of the following equations represents Clairaut’s partial differential equation?
a) z=px+f(p,q)
b) z=f(p,q)
c) z=p+q+f(p,q)
d) z=px+qy+f(p,q)
View Answer

Answer: d
Explanation: Equations of the form, z=px+qy+f(p,q) are known as Clairaut’s partial differential equations, named after the Swiss mathematician, A. C. Clairaut (1713-1765).

9. Which of the following represents Lagrange’s linear equation?
a) P+Q=R
b) Pp+Qq=R
c) p+q=R
d) Pp+Qq=P+Q
View Answer

Answer: b
Explanation: Equations of the form, Pp+Qq=R are known as Lagrange’s linear equations, named after Franco-Italian mathematician, Joseph-Louis Lagrange (1736-1813).

10. A partial differential equation is one in which a dependent variable (say ‘x’) depends on an independent variable (say ’y’).
a) False
b) True
View Answer

Answer: a
Explanation: An ordinary differential equation is divided into two types, ordinary and partial differential equations.
A partial differential equation is one in which a dependent variable depends on one or more independent variables.
Example: (F(x,t,y,frac{∂y}{∂x},frac{∂y}{∂t},……)= 0. )

11. What is the complete solution of the equation, (q= e^frac{-p}{α})?
a) (z=ae^frac{-a}{α}y)
b) (z=x+e^frac{-a}{α}y)
c) (z=ax+e^frac{-a}{α} y+c)
d) (z=e^frac{-a}{α}y)
View Answer

Answer: c
Explanation: Given: (q= e^frac{-p}{α})
The given equation does not contain x, y and z explicitly.
Setting p = a and q = b in the equation, we get (b= e^frac{-a}{α}.)
Hence, a complete solution of the given equation is,
(z=ax+by+c,,with , b= e^frac{-a}{α})
(z=ax+e^frac{-a}{α} y+c.)

12. A particular solution for an equation is derived by eliminating arbitrary constants.
a) True
b) False
View Answer

Answer: b
Explanation: A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution thereby eliminating any arbitrary constants present in the solution. Such solution represents a particular member of the family of surfaces given by the complete solution.

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250+ TOP MCQs on Regions in the Complex Plane and Answers

Complex Analysis Multiple Choice Questions on “Regions in the Complex Plane”.

1. What is the shape of the region formed by the set of complex numbers z satisfying |z-ω|≤ α?
a) circle of radius ω
b) circle with center ω
c) disk of radius α
d) disk with center α
View Answer

Answer: d
Explanation: The equation |z-ω|≤ α implies that the distance of z from ω is less than or equal to α. This means that a disk is formed with center ω and radius α.

2. The complex number given by [(√3/2)+i/2]5+[(√3/2)-i/2]5 lies, on which of the following regions?
a) imaginary axis
b) real axis
c) first quadrant
d) fourth quadrant
View Answer

Answer: b
Explanation: Let z=[(√3/2)+i/2]5+[(√3/2)-i/2]5 and ω=[(√3/2)+i/2]5 ⇒ (overline{omega})=[(√3/2)-i/2]5
⇒ z=ω+(overline{omega}) ⇒ z is real (sum of conjugates is real) ⇒ z lies on real axis.

3. Find the area of the region given by 11≤|z| ≤ 19.
a) 120π sq. units
b) 180π sq. units
c) 240π sq. units
d) 320π sq. units
View Answer

Answer: c
Explanation: The region formed is an annulus of inner radius 11 units and outer radius 19 units. Therefore, the required area=π(192–112)=240π.

4. Find the largest angle of the triangle formed by thevertices z1=8(1-i), z2=8(i-1) and
Z3=10+2√7i.
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a) π/3 radians
b) 2π/3 radians
c) π/2 radians
d) 3π/4 radians
View Answer

Answer: c
Explanation: Note that z1 and z2 are the opposite ends of a diameter of a circle of radius 8√2 units, centered at the origin. Also note that z3 lies on this circle (distance of z3 from origin = 8√2). Hence, angle corresponding to z3=π/2 radians.

5. Find the equation of the circle passing through the origin and having intercepts a and b on real and imaginary axes, respectively, on the arg and plane.
” alt=”” width=”160″ height=”153″ data-src=”2020/09/complex-analysis-questions-answers-regions-complex-plane-q5″ />
a) zz̅=a(Im z)–b(Re z)
b) zz̅=a(Im z)+b(Re z)
c) zz̅=a(Re z)–b(Im z)
d) zz̅=a(Re z)+b(Im z)
View Answer

Answer: d
Explanation: Consider a point z on the circle. Therefore, arg [(z-a)/(z-ib)]=±π/2
⇒ (z-a)/(z-ib)+(z̅-a)/(z̅+ib)=0 ⇒ zz̅-a(z+z̅)/2–b(z-z̅)/2i=0
⇒ zz̅=a(Re z)+b(Im z).

6. On the arg and plane, the complex numbers z1, z2, z3, z4 are the vertices of a parallelogram. Evaluate (z4–z1+z2)/z3 .
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: For a parallelogram, the diagonals bisect each other. Hence, their midpoints coincide.
This implies that (z1+z3)/2=(z2+z4)/2 ⇒ z1+z3=z2+ z4 ⇒ ( z4– z1+ z2)/ z3=1.

7. Consider the shape formed by the set of points z=ω-1/ω, where |ω|=2. Which of the following is incorrect?
a) eccentricity=4/5
b) |z|≤3
c) shape is an ellipse
d) major axis is of length=5/2
View Answer

Answer: d
Explanation: | ω|=2 ⇒ ω=2(cosθ+isinθ) ⇒ z=x+iy=2(cosθ+isinθ)-1/2(cosθ-isinθ)(⇒ |z|≤3)
=3/2cosθ+i5/2sinθ⇒x2/(3/2)2+y2/(5/2)2=1 ⇒ ellipse ⇒ e2=1-(9/4)/(25/4)=16/25
⇒ e=4/5.

8. Find the area enclosed by the curve formed by iz3+z2–z+i=0.
a) π/2
b) π
c) 3π/4
d) 2π
View Answer

Answer: b
Explanation: Dividing the equation by i on both sides, z3-iz2+iz+1=0
⇒ z2(z-i)+i(z-i)=0 ⇒ (z-i)(z2+i)=0 ⇒ z=i or z2=-i ⇒ |z|=|i|=1 or |z2|=|z|2=|-i|=1
⇒ |z|=1 ⇒ circle of radius 1 is formed. Hence, area=π(12)=π.

9. Given a vertex of the square circumscribing the circle |z-1|=√2 as 2+√3i, which of the following is not a vertex of this square ?
a) (1-√3)+i
b) –i√3
c) (√3+i)-i
d) i√3
View Answer

Answer: d
Explanation: The given circle has z0=1 as its center and √2 as radius. Let z1=2+i√3. Now, obtain z2 by rotating z1 anticlockwise by 900 about z0 ⇒ z2=(1-√3)+i. Now, z0 is midpoint of z1 and z3 and z2 and z4.
؞(z1+z3)/2 ⇒ (2+i√3+z3)/2=1 ⇒ z3=-i√3 and(z2+z4)/2=z0 ⇒ z4=(√3+i)-i.

10. Find the area of the region bounded by arg|z|≤π/4 and |z-1|<|z-3|.
a) 1 sq. units
b) 2 sq. units
c) 3 sq. units
d) 4 sq. units
View Answer

Answer: d
Explanation: |z-1|<|z-3| ⇒ (x-1)2+y2<(x-3)2+y2 ⇒ x<2.
Therefore, a triangle is formed with base length 4 and height 2 (along x-axis). Hence, the required area=1/2×4×2=4.

11. Find the locus of z/(1-z2), where z lies on the circle of radius 1 centered at origin and z≠±1.
a) line not passing through origin
b) |z|=√2
c) real axis
d) imaginary axis
View Answer

Answer: d
Explanation: Given |z|=1 and z≠±1, write ω=z/(1-z2)=z/(zz̅-z2)=1/(z̅-z).
Hence ω is a purely imaginary number and lies on imaginary axis.

12. Describe the region given by |z-i|z||-|z+i|z||=0.
a) real axis
b) imaginary axis
c) circle centered at origin
d) quadrant 2
View Answer

Answer: a
Explanation: |z/|z|-i|=|z/|z|+i|, z≠0 ⇒ z/|z| is unimodular complex number and lies on the perpendicular bisector of i and –i ⇒ z/|z|=±1 ⇒ z=±|z| ⇒ z is real.

13. The area of the region enclosed by the curve zz̅+a(z̅+z)+a=0 is 2π. If a2–7a+10=0, find the area of the region enclosed by the curve zz̅+2a(z̅+z)+a=0.
a) 4π sq. units
b) 10π sq. units
c) 14π sq. units
d) 22π sq. units
View Answer

Answer: c
Explanation: The curve represents a circle with center–a and radius (a2–a)1/2. Therefore, Area=π(a2–a)=2π ⇒ a=-1,2. Also, a2–7a+10=0 ⇒ a=2,5. Hence, a=2.
Hence required area=π(4a2–a)=14π.

14. Find the area of the region common to the sets S1={z∈C: |z|<4}, S2={z∈C:Im[(z-1+√3i)/(1-√3i)]>0} and S3={z∈C: Re z>0}.
a) 10π/3
b) 20π/3
c) 16π/3
d) 32π/3
View Answer

Answer: b
Explanation: S1: |z|2: √3x+y>0, z lies above the line √3x+y=0; S3: Re(z)>0, z lies to the right of the imaginary circle. Therefore, required area=π×42/4+π×42/6=20π/3.
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15. In the triangle shown, if the angle corresponding to z3 is said to be π/2,
” alt=”” width=”347″ height=”115″ data-src=”2020/09/complex-analysis-questions-answers-regions-complex-plane-q15″ data-srcset=”2020/09/complex-analysis-questions-answers-regions-complex-plane-q15 347w, 2020/09/complex-analysis-questions-answers-regions-complex-plane-q15-300×99 300w” data-sizes=”(max-width: 347px) 100vw, 347px” />
Find a possible value of z3 in terms of z1, z2 and z4.
a) z4+3/5(z2– z1)eiπ/2
b) z4-3/5(z2– z1)eiπ/2
c) z4+3/5(z2– z1)e-iπ/2
d) no such z3 is possible
View Answer

Answer: d
Explanation: If we draw a circle with z2-z1 as a diameter, we see that z3 would be outside the circle, since the radius of the circle is 5(3 is possible.
” alt=”” width=”175″ height=”186″ data-src=”2020/09/complex-analysis-questions-answers-regions-complex-plane-q15a” />

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250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Questions and Answers for Freshers focuses on “Leibniz Rule – 2”.

1. Let f(x) = ex sin(x2) ⁄ x Then the value of the fifth derivative at x = 0 is given by
a) 25
b) 21
c) 0
d) 5
Answer: b
Explanation: First expanding sin(x2) ⁄x into a Taylor series we have
sin(x2)=(frac{x^2}{1!}-frac{x^6}{3!}+frac{x^{10}}{5!}….infty)
(frac{sin(x^2)}{x}=frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}….infty)
Now applying the Leibniz rule up to the fifth derivative we have
(((e^x)(frac{sin(x^2)}{x})^{(5)} = c_0^5e^x(frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}…infty))
(+c_1^5e^x(frac{1}{1!}-frac{5x^4}{3!}+frac{9x^8}{5!}…infty)+…..+c_5^5e^x(frac{5!}{3!}+frac{(9.8.7.6.5)x^4}{5!}…infty))
Now substituting x=0 we get
(((e^x)(frac{sin(x^2)}{x}))^{(5)}=c_1^5(1)+frac{5!}{3!})
= 1 + 20 = 21.

2. Let f(x) = eex assuming all the nth derivatives at x =0 to be 1 the value of the (n + 1)th derivative can be written as
a) e – 1 + 2n
b) 0
c) 1
d) None
Answer: a
Explanation: Assume y = f(x)
Taking ln(x) on both sides The function has to be written in the form ln(y) = ex
Now computing the first derivative yields
y(1) = y * ex
Now applying the Leibniz rule up to nth derivative we have
y(n+1)=(c_0^ne^xy+c_1^ne^xy^{(1)}+….+c_n^ne^xy^{(n)})
We know that in the problem it is assumed that [y(1)=y(2)=…=y(n)=1]x=0
Now, substituting x=0 we get
y(n+1)=(c_0^ne+c_1^n+….+c_n^n)
From combinatorial results we know that 2n=(c_0^ne+c_1^n+….+c_n^n)
This gives us
y(n+1)=(e+(c_0^ne+c_1^n+….+c_n^n)-c_0^n)
y(n+1)=e-1+2n

3. Let f(x) = (sqrt{sin(x)}) and let yn denote the nth derivative of f(x) at x = 0 then the value of the expression 12y(5) y(1) + 30 y(4) y(2) + 20 (y(3))2 is given by
a) 0
b) 655
c) 999
d) 1729
Answer: a
Explanation: Assume y = f(x)
Rewriting the function as y2 = sin(x)
Now applying Leibniz rule up to the sixth derivative we have
(y2)(6) = c06 y(6) y + c16 y(5) y(1) + ………+ c66 y(6) y

(y2)(6) = 2 y(6) y + 12 y(6) y(1) + 30 y(4) y(2) + 20 (y(3))2

(sin(x))(6) = -sin(x)
Now substituting x = 0 and observing that y(0) = 0 we have
sin(0) = 0 = 12 y(6) y(1) + 30 y(4) y(2) + 20 (y(3))2.

4. The fourth derivative of f(x) = sin(x)sinh(x) ⁄ x at x = 0 is given by
a) 0
b) π⁄2
c) 45
d) 4
Answer: a
Explanation: First convert the function sinh(x)⁄x into its Taylor series expansion
(frac{sinh(x)}{x}=frac{frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}….infty}{x})
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now pick up the whole function (((sin(x))(frac{sinh(x)}{x}))) and apply Leibniz rule up to the fourth derivative we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}=c_0^4sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}…..infty))
(-c_1^4cos(x)(frac{2x}{3!}+frac{4x^3}{5!}…..infty)+…..+c_4^4sin(x)(frac{4!}{5!}+frac{(7.6.5.4)x^3}{7!}…..infty))
Substituting x=0 we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}) = 0

5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is
a) 0
b) π⁄32
c) (π)2
d) cos(1)sinh(1)
Answer: a
Explanation: Assume y = f(x)
Rewriting the part sinh(x)⁄x as infinite series we have
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now the function f(x) becomes
y=(cos(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty))
Taking the third derivative of the above function using Leibniz rule we have
y(3)=(c_0^3sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)-c_1^3cos(x)(frac{2x}{3!}+frac{4x^3}{5!}….infty))
(-c_2^3sin(x)(frac{2}{3!}+frac{12x^2}{5!}….infty)+c_3^3cos(x)(frac{24x}{5!}…infty))
Now substituting x = 0 we have
y(3) = 0.

6. Let f(x) = (x2 + x + 1)sinh(x) the (1097)th derivative at x = 0 is
a) 1097
b) 1096
c) 0
d) 1202313
Answer: d
Explanation: Expanding sinh(x) into a taylor series we have
sinh(x)=(x+frac{x^3}{3!}+frac{x^5}{5!}…infty)
f(x)=(x2+x+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty))
On multiplication we get two series with odd exponents and one series with even exponent. The series with odd exponents are the only ones to contribute to the derivative at x=0
Hence it is enough to compute the derivative at for the following function
(x2+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty)=frac{x}{1!}+x^3(frac{1}{3!}+1)+x^5(frac{1}{5!}+frac{1}{3!})….infty)
Taking the 1097thderivative of this function, we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!})+(1099times 1098…4times 3)x^2(frac{1}{(1097)!}+frac{1}{(1097)!})+…infty)
Substituting x=0 we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!}))
=(1+1097*1096)=(1+1202312)=1202313

7. The 7th derivative of f(x) = (x3 + x2 + x + 1) sinh(x) at x = 0 is given by
a) 43
b) 7
c) 0
d) 34
Answer: a
Explanation: Expanding sinh(x) into a Taylor series we have
sinh(x)=(frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty)
Now rewriting the function we have
f(x)=(x3+x2+x+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
For the 7th derivative observe that, only the odd termed powers contribute to the derivative at x=0
Hence it is enough for us to find seventh derivative for
(x2+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
(frac{x}{1!}+x^3(frac{1}{3!}+frac{1}{1!})+x^5(frac{1}{5!}+frac{1}{3!})+…infty)
Taking the 7th derivative of this function we have
f(7)(x)=((7!)(frac{1}{7!}+frac{1}{5!}) + (9*8…4*3)x^2(frac{1}{7!}+frac{1}{9!}))
Now substituting x=0 yields
f(7)(0)=((7!)(frac{1}{5!}+frac{1}{7!}))=(1+7*6)=43.

8. The (1071729)th derivative of f(x) = (x6 + x4 + x2) cosh(x) at x = 0 is given by
a) 0
b) 1071
c) 1729
d) ∞
Answer: a
Explanation: Expanding cosh(x) into a Taylor series we have
cosh(x)=(frac{1}{1!}+frac{x^2}{2!}+frac{x^4}{4!}…infty)
Observe again, that the derivative in question is odd, and hence, only the odd powered terms contribute to the derivative at x = 0
Also note that, there are no odd powered terms and hence we can conclude that
The (1071729)th derivative must be 0.

9. The (17291728)th derivative of f(x) = (x2 + 1)tan-1 (x) at x = 0 is
a) 0
b) 1729
c) 1728
d) ∞
Answer: a
Explanation: Expanding the tan-1 (x) function into Taylor series we have
tan-1(x)=(frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty)
Rewrite the function as
f(x)=(x2+1)((frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty))
Now observe that the derivative in question is even. Hence, even terms are the only ones to contribute to the derivative at x = 0
Also note that there are no even powered terms in the function. One can conclude that the (17291728)th derivative at x = 0 is 0.

To practice all areas of Engineering Mathematics for Freshers,

250+ TOP MCQs on Polar Curves and Answers

Differential Calculus Multiple Choice Questions & Answers focuses on “Polar Curves”.

1. Polar equations of the circle for the given coordinate (x,y) which satisfies the equation given by (x-a)2+(y-b)2=r2 where (a,b) is the coordinates of the centre of the circle &r is the radius.
a) x = r cos⁡θ, y = r sin⁡θ
b) x = a+ r cos⁡θ, y = b + r sin⁡θ
c) y = a+r cos⁡θ, x = b + r sin⁡θ
d) x = r sin⁡θ, y = r cos⁡θ
View Answer

Answer: b
Explanation: option x = a+ r cos⁡θ, y = b + r sin⁡θ satisfies the equation (x-a)2+(y-b)2=r2
because LHS=(a+r cos⁡θ-a)2 + (b+ r sin⁡θ-b)2 = r2(cos2 θ + sin2 θ)= r2=RHS.

2. In an polar curve r=f (θ) what is the relation between θ & the coordinates (x,y)?
a) tan⁡θ = (frac{x}{y} )
b) (1+sin⁡θ) = (frac{y}{x} )
c) (1+sec2 θ) = (frac{y^2}{x^2} )
d) (1+cos⁡θ) = (frac{x}{y} )
View Answer

Answer: c
Explanation: w.k.t for the polar curve r=f (θ) x=rcos⁡θ, y=rsin⁡θ
dividing them we get (frac{y}{x} = frac{r sin⁡θ}{r cos⁡θ} = tan⁡θ)
squaring on both side (frac{y^2}{x^2} = tan^2 θ = (1+sec^2 θ)).

3. The angle between Radius vector r=a(1-cos⁡θ)and tangent to the curve is ∅ given by _______
a) ∅=(frac{π}{2})
b) ∅=π
c) ∅=(-frac{π}{2})
d) ∅=0
View Answer

Answer: a
Explanation: r= a(1-cos⁡θ)
taking logarithms on both sides we get,
log⁡r = log⁡a + log⁡(1-cos⁡θ)
differentiating w.r.t θ we get,
( frac{1}{r} frac{dr}{dθ} = 0 + frac{sin⁡θ}{1-cos⁡θ})
(frac{1}{r} frac{dr}{dθ} = frac{2 sin ⁡frac{θ}{2} cos⁡frac{θ}{2}}{2sin^2 frac{θ}{2}} = cot⁡frac{θ}{2}) ..(1),
but (cot⁡∅ = frac{1}{r} frac{dr}{dθ})….(2)
From (1)&(2)
∅=(frac{π}{2}).

4. Angle of intersection between two polar curves given by r=a(1+sin⁡θ) & r=a(1-sin⁡θ) is given by ________
a) (frac{π}{4})
b) (frac{π}{2})
c) Π
d) 0
View Answer

Answer: b
Explanation: r=a(1+sin⁡θ) : r=a(1-sin⁡θ)
taking logarithm on both the equations
log⁡r = log⁡a + log⁡(1+sin⁡θ) : log⁡r = log⁡a + log⁡(1-sin⁡θ)
differentiating on both side we get
( frac{1}{r} frac{dr}{dθ} = frac{cos⁡θ}{1+sin⁡θ} : frac{1}{r} frac{dr}{dθ} = frac{-cos⁡θ}{1-sin⁡θ})
(cot⁡∅1 = frac{cos⁡θ}{1+sin⁡θ} : cot⁡∅2 = frac{-cos⁡θ}{1-sin⁡θ})
where ∅1&∅2 are the angle between tangent & the vector respectively
(tan⁡∅1 = frac{1+sin⁡θ}{cos⁡θ} : tan⁡∅2 = frac{1-sin⁡θ}{cos⁡θ})
(tan⁡∅1 . tan⁡∅2 = frac{1+sin⁡θ}{cos⁡θ} . frac{1-sin⁡θ}{-cos⁡θ} = frac{1-sin^2 θ}{-cos^2 θ} = -frac{cos^2 θ}{cos^2 θ} = -1)
above is the condition of orthogonality of two polar curves thus
|∅1-∅2|=(frac{π}{2}).

5. One among the following is the correct explanation of pedal equation of an polar curve, r=f (θ), p=r sin(∅) (where p is the length of the perpendicular from the pole to the tangent & ∅ is the angle made by tangent to the curve with vector drawn to curve from pole)is _______
a) It is expressed in terms of p & θ only
b) It is expressed in terms of p & ∅ only
c) It is expressed in terms of r & θ only
d) It is expressed in terms of p& r only
View Answer

Answer: d
Explanation: It is expressed in terms of p& r only
where p=r sin(∅) & (tan⁡∅ = frac{r}{frac{dr}{dθ}} = r (frac{dr}{dθ}))
& r=f (θ) or after solving we get direct relationship between p & r as
(frac{1}{p^2} = frac{1}{r^2} cosec^2∅.)

6. The pedal Equation of the polar curve rn=an cos⁡nθ is given by ______
a) rn=pan
b) rn-1=pan
c) rn+1=pan+1
d) rn+1=pan
View Answer

Answer: d
Explanation: Taking logarithm for the given curve we get
n log⁡r = n log⁡a + log⁡(cos⁡nθ)
differentiating w.r.t θ, we get
(frac{n}{r} frac{dr}{dθ} = frac{-n sin⁡nθ}{cos⁡nθ} rightarrow frac{1}{r} frac{dr}{dθ} = -tan⁡θ)
thus (cot⁡∅ = cot⁡(frac{π}{2} + nθ)rightarrow ∅ = frac{π}{2} + nθ……(1))
from the eqn w.k.t p=r sin ∅
substituting from (1)
p = r sin ((frac{π}{2}) + nθ) = r cos (nθ), but we have rn = an cos⁡nθ
hence dividing them we get (frac{p}{r^n} = frac{r cos (nθ)}{a^n cos⁡nθ})
rn+1=pan.

7. The length of the perpendicular from the pole to the tangent at the point θ=(frac{π}{2}) on the curve. r=a sec2((frac{π}{2})) is _____
a) (p = frac{2a}{sqrt{3}})
b) (p = frac{4a}{sqrt{3}})
c) (p = 2asqrt{2})
d) p = 4a
View Answer

Answer: c
Explanation: Taking Logarithm on both side of the polar curve
we get log⁡r = log⁡a + 2 log⁡sec⁡((frac{θ}{2}))
differentiating w.r.t θ we get
(frac{1}{r} frac{dr}{dθ} = frac{2 sec⁡(frac{θ}{2}).tan⁡(frac{θ}{2}) }{2 sec⁡(frac{θ}{2})} = tan⁡(frac{θ}{2}))
(cot⁡∅ = cot⁡(frac{π}{2} – frac{θ}{2}) rightarrow ∅ = frac{π}{2} -frac{θ}{2})
w.k.t length of the perpendicular is given by p=r sin ∅
thus substituting ∅ value we get (p = r ,sin(frac{π}{2} – frac{θ}{2}) = r ,cos(frac{θ}{2}))
at ( θ=frac{π}{4}, p = r ,cos frac{π}{4} = frac{r}{sqrt{2}}….(1))
but ( r=a ,sec^2 (frac{θ}{2}) ,at, θ = frac{π}{4}, r=a ,sec^2 (frac{π}{4}) = 4a…(2))
from (1) & (2) (p=frac{4a}{sqrt{2}} = 2asqrt{2}).

250+ TOP MCQs on Euler’s Theorem and Answers

Engineering Mathematics Question Bank focuses on “Euler’s Theorem – 2”.

1. In euler theorem x ∂z⁄∂x + y ∂z⁄∂y = nz, here ‘n’ indicates?
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z
Answer: a
Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”.

2. If z = xn f(y⁄x) then?
a) y ∂z⁄∂x + x ∂z⁄∂y = nz
b) 1/y ∂z⁄∂x + 1/x ∂z⁄∂y = nz
c) x ∂z⁄∂x + y ∂z⁄∂y = nz
d) 1/x ∂z⁄∂x + 1/y ∂z⁄∂y = nz
Answer: c
Explanation: Since the given function is homogeneous of order n, hence by euler’s theorem
x ∂z⁄∂x + y ∂z⁄∂y = nz.

3. Necessary condition of euler’s theorem is?
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation:
Answer ‘z should be homogeneous and of order n’ is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”
Answer ‘z should not be homogeneous but of order n’ is incorrect as z should be homogeneous.
Answer ‘z should be implicit’ is incorrect as z should not be implicit.
Answer ‘z should be the function of x and y only’ is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.

4. If (z=e^{frac{x^2+y^2}{x+y}}) then, (x frac{∂z}{∂x} + y frac{∂z}{∂y}) is?
a) 0
b) zln(z)
c) z2 ln⁡(z)
d) z
Answer: b
Explanation:
Given (z=e^{frac{x^2+y^2}{x+y}}),let u=ln⁡(z)=(frac{x^2+y^2}{x+y}=frac{x(1+(frac{y}{x})^2)}{(1+frac{y}{x})}) = x f(y/x)
Hence u is homogeneous of order 1,
Hence,
(x frac{∂u}{∂x}+y frac{∂u}{∂y})=u
Putting, u = ln(z) we get,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}) = zln(z)

5. If (z=sin^{-1}frac{x^3+y^3+z^3}{x+y+z}) then, (xfrac{∂z}{∂x}+yfrac{∂z}{∂y}).
a) 2 tan(z)
b) 2 cot(z)
c) tan(z)
d) cot(z)
Answer: a
Explanation:
Given (z=sin^{-1}⁡frac{x^3+y^3+z^3}{x+y+z}), put u=sin⁡(z)=(frac{x^3+y^3+z^3}{x+y+z}=x^2 f(frac{y}{x},frac{z}{x}))
Hence, (x frac{∂u}{∂x}+y frac{∂u}{∂y}=2u)
Putting u = sin(z), we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{2Sin(z)}{Cos(z)}=2Tan(z))

6. Value of (x frac{∂u}{∂x}+y frac{∂u}{∂y}) if (u=frac{Sin^{-1} (frac{y}{x})(sqrt{x}+sqrt{y})}{x^3+y^3}) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
Answer: a
Explanation: Since the function can be written as,
u=(x^{frac{-5}{2}} frac{Sin^{-1} (frac{y}{x})(1+sqrt{frac{y}{x}})}{1+(frac{y}{x})^3}=x^n f(frac{y}{x})), by euler’s theorem,
(x frac{∂u}{∂x}+y frac{∂u}{∂y} = -frac{5}{2} u)

7. If f1(x,y) and f2(x,y) are homogeneous and of order ‘n’then the function f3(x,y) = f1(x,y) + f2(x,y) satisfies euler’s theorem.
a) True
b) False
Answer: a
Explanation: Since f1(x,y) and f2(x,y) are homogeneous and of order n hence,
(x frac{∂f_1}{∂x}+y frac{∂f_1}{∂y} = nf_1 (x,y))
(x frac{∂f_2}{∂x}+y frac{∂f_2}{∂y} = nf_2 (x,y))
Hence adding these two equations,
We get
(x frac{∂f_1+f_2}{∂x}+y frac{∂f_1+f_2}{∂y} = nf_2 (x,y)+nf_1 (x,y))
(x frac{∂f_3}{∂x}+y frac{∂f_3}{∂y} = nf_3 (x,y))
Hence f3 satisfies euler’s theorem.

8. If (z=ln⁡(frac{x^2+y^2}{x+y})-e^{frac{x^2+y^2}{x+y}}) then find (x frac{∂z}{∂x}+y frac{∂z}{∂y}).
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation:
Given (z=ln⁡(frac{x^2+y^2}{x+y})-e^frac{x^2+y^2}{x+y})
Let, (u = ln⁡(frac{x^2+y^2}{x+y})) and (v=e^(frac{x^2+y^2}{x+y})) hence z=u-v
Now, let (u’ = e^u = frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=u’)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{e^u}{e^u} = 1), ……(1)
Now, let v’ = ln(v)= (frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=v’)
Hence, by putting v’=ln(v),we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=vln(v)), …… (2)
By subtracting eq(1) and eq(2), we get
(x frac{∂z}{∂x}-y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})

9. If z = Sin-1 (x⁄y) + Tan-1 (y⁄x) then x ∂z⁄∂x + y ∂z⁄∂y is?
a) 0
b) y
c) 1 + x⁄y Sin-1 (x⁄y)
d) 1 + y⁄x Tan-1 (y⁄x)
Answer: a
Explanation: Given z = Sin-1 (x⁄y) + Tan-1 (y⁄x)
Let, u = Sin-1 (x⁄y) and v = Tan-1 (y⁄x) hence z = u + v
Now, let u’ = Sin(u) = x⁄y = f(x⁄y) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=0)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=0/e^u = 0) ,……(1)
Now, let v’= Tan(v)=y/x=f(y/x) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=0)
Hence, by putting v’=ln(v), we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=0 ),……(2)
By adding eq(1) and eq(2), we get
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})

10. If f(x,y)is a function satisfying euler’ s theorem then?
a) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
b) (frac{1}{x}^2 frac{∂^2 f}{∂x^2}+2/xy frac{∂^2 f}{∂x∂y}+frac{1}{y}^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
c) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=nf)
d) (y^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+x^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
Answer: a
Explanation: Since f satisfies euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=nz)
Differentiating it w.r.t x and y respectively we get,
(x frac{∂^2 u}{∂x^2}+frac{∂u}{∂x}+y frac{∂^2 u}{∂x∂y}=n frac{∂u}{∂x}),
and
(x frac{∂^2 u}{∂y}∂x+frac{∂u}{∂y}+y frac{∂^2 u}{∂y^2}=n frac{∂u}{∂y})
Multiplying with x and y respectively,
(x^2 frac{∂^2 u}{∂x^2}+x frac{∂u}{∂x}+xy frac{∂^2 u}{∂x∂y}=nx frac{∂u}{∂x}),
and
(xy frac{∂^2 u}{∂y}∂x+y frac{∂u}{∂y}+y^2 frac{∂^2 u}{∂y^2}=ny frac{∂u}{∂y})
Adding above equations we get
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}=n(n-1)u)

11. If (u = Tan^{-1} (frac{x^3+y^3}{x+y})) then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) is?
a) Sin(4u) – Cos(2u)
b) Sin(4u) – Sin(2u)
c) Cos(4u) – Sin(2u)
d) Cos(4u) – Cos(2u)
Answer: b
Explanation:
Let, v = Tan(u) = x2 f(y/x)
By euler’s theorem,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y} = 2frac{Tan(u)}{Sec^2 (u)} = Sin(2u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y} = g(u)[g’(u)-1])
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = Sin(2u)[2Cos(2u)-1] = Sin(4u)-Sin(2u)

12. If (u = e^{frac{(x^2+y^2)}{x+y}}) Then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y})=?
a) u ln⁡(u)
b) u ln⁡(u)2
c) u [1+ln⁡(u)]
d) 0
Answer: b
Explanation: Let, v = ln(u) = (frac{x^2+y^2}{x+y} = x f(frac{y}{x}))
Hence by applying euler theorem,
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=v)
Hence,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y}=u ln⁡(u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = g(u)[g’(u)-1]
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = u ln(u)[1+ln⁡(u)-1] = u ln⁡(u)2

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