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Detailed Solutions Mathematics-I A-01/C-01/T-01 December 2004

1. (a) D Let y= mx² be equation of curve. As x→0, y also tends to zero.

, which depends on m.

Thus it does not exist.

(b) A

Eliminating we get

f_x= - 3x² = 0 , f_y= 2y = 0

gives (0,0) is a critical point.

∆f (x, y)= f(∆x,∆y)= (∆y)² –(∆ x)³

> 0 , if (∆y)² >(∆ x)³

< 0 , if (∆y)² < (∆ x)³

This means in the neighborhood of (0,0) f changes sign. Thus (0,0) is neither a point of maximum nor minimum.

(d) A y = (x – k)² Diff. w.r.t. x

y₁ = 2(x – k) => y₁ = 2

For orthogonal trajectories y₁ is replaced by -1/y₁.

Therefore, -1/y₁ = 2

=> 2dy + dx = 0

Integrating, we get y^3/2 = ¾ (c-x)

(e) B yy” – (y’)² = 0

Because, y₁, y₂ are solutions

Therefore, y₁y₁” – (y₁’)² = 0

y₂y₂” – (y₂’)² = 0

Now (c₁y₁ + c₂y₂) (c₁y₁ + c₂y₂)” – ((c₁y₁ + c₂y₂)’)²

= (c₁y₁ + c₂y₂) (c₁y₁” + c₂y₂”) – (c₁y₁’² + c₂y₂’²) - 2 c₁y₁’c₂y₂’

= c₁²(y₁y₁” – (y₁’)²) + c₂² (y₂y₂” – (y₂’)²) + c₁ c₂ (y₁ y₂”+y₂y₁” - 2y₁’y₂’)

= 0, if c₁c₂ = 0.

(f) B Since A, B are square matrix of order n s.t. AB = 0, then rank of both A and B is less than n.

(g) D Because i is one eigen value so another eigen value must be – i.

(h) C Let I =

Let cosq = t. –sinqdq = dt

= 0

2. (b) Because x + y = 2e^qcosj, x – y = 2ie^qsinj

Adding & Subtracting, we get

2x = 2e^q(cosj + isinj) = 2e^{q + i}^j

=> x = e^{q + i}^j

Similarly y = e^{q - i}^j

Let v = f (x, y), x = g (q, j), y = h (q, j)

3. (a) f(x, y) = x^y , f(1, 1) = 1

f_x (x, y) = yx^y-1 , f_x(1, 1) = 1

f_y (x, y) = x^ylog x , f_y(1, 1) = 0

(x, y) = y (y-1) x^y-2 , (1, 1) = 0

(x, y) = x^y(log x)² , (1, 1) = 0

f_xy (x, y) = x^y-1+ yx^y-1 log x , f_xy(1, 1) = 1

(x, y) = y (y-1) (y-2) x^y-3 , (1, 1) = 0

(x, y) = (2y-1)x^y-2 + y (y-1) x^y-2 log x , (1, 1) = 1

(x, y) = yx^y-1 (log x)² + 2 log x x^y-1 , (1, 1) = 0

By Taylor’s Theorem

3. (b) Let f = x² + y²+ z² + xy + xz + zy

g = x + y + z – 1 = 0

h = x + 2y + 3z – 3 = 0

Let l₁, l₂ be two constants. Using Lagrange’s multiplier method, we get

F = f + l₁g + l₂h OR

F = x² + y²+ z² + xy + xz + zy + l₁(x + y + z – 1) + l₂(x + 2y + 3z – 3)

For extreme values,

, x + y + z = 1, x + 2y + 3z = 3.

=> 2x + y + z + l₁ + l₂ = 0 => x + l₁ + l₂+ 1 = 0

2y + x + z + l₁ + 2l₂ = 0 y + l₁ + 2l₂+ 1 = 0 (A)

2z + x + y + l₁ + 3l₂ = 0 z + l₁ + 3l₂+ 1 = 0

Adding (A) and using x + y + z = 1, we get

3l₁ + 6l₂ + 4 = 0

Multiplying equation (ii) of ‘A’ by 2 and (iii) by 3 and adding all and using

x +2 y +3 z = 1, we get 6l₁ + 14l₂ + 9 = 0

Solving, 3l₁ + 6l₂ + 4 = 0

6l₁ + 14l₂ + 9 = 0, we get

l₁ = –1 /3, l₂ = –1 /2

From (A), we get

x = –1/6, y = 1/3, z = 5/6

Therefore, (–1/6, 1/3, 5/6) is a point of extremum, with extreme value

F(–1/6, 1/3, 5/6)= (-1/6)² + (1/3)² + (5/6)² – 1/6*1/3 – 1/6 * 5/6 + 1/3 * 5/6 =11/12

4. (a) Applying R₁ R₃, R₂ R₄

R₃ R₄

R₃àR₃ – 3R₁, R₄ à R₄ – 9R₁

R₄àR₄ – R₂, R₃ à R₂ – 2R₃

R₄à 9R₃ + 5R₄

Thus |A| 0 Hence, rank of A = 4.

(b) Let

y₁ = 1.25 z₁ + 3 z₂ – 2.3 z₃

y₂ = 0.75 z₁ + 2 z₂ – 2.3 z₃

y₃ = 0.5 z₁ - z₂+ z₃

5. (a) Let A be a square matrix of order n.

Then |A – lI | = (-1)ⁿlⁿ + k₁l^n-2 + ----- + k_n = 0

where k’s are expressible in terms of elements a_ij of matrix A. The roots of this equation are eigen values of matrix A. Since this is n^th polynomial in l which has n distinct roots which are either real or complex conjugates.

Hence, eigen values of matrix are either real or complex conjugates.

If l is an eigen value of orthogonal matrix then 1/l is an eigen value of A^-1. Because A is an orthogonal matrix. Therefore A^-1 is same as A’.

Therefore 1/l is eigen value of A’. But A and A’ have same eigen values.

Hence, 1/l is also an eigen value of A. The product of eigen value of orthogonal matrix = 1 and hence if the order of A is odd it must have 1 as eigen value. Since product of eigen value of matrix A is equal to its determinant. Therefore |A| = ±1.

(b) |A – lI | = 0

=> l³ + l² – 21l - 45 = 0

=> l = 5, -3, - 3

Eigen values are 5, -3, -3

For l = 5, eigen vectors are obtained from

=> -7x + 2y - 3z = 0

2x – 4y – 6z = 0

-x –2y – 5z = 0

Solving we get,

i.e. (1, 2, -1)’ is an eigen vector

For l = -3, eigen vectors are obtained from

i.e. x +2y – 3z = 0

There are two linearly independent eigen vectors for l = -3. These are obtained by putting x = 0 and y = 0 respectively in the equation.

Let x = 0 then 2y – 3z = 0

i.e. is an eigen vector

Let y = 0, then x – 3z = 0

is an eigen vector.

Eigen vectors corresponding to 5, -3 , -3 are

[1, 2, -1]’ , [0, 3, 2]’ and [3, 0, 1]’.

6. (a) A =

| A - lI | = 0

=> l² - 7l + 6 = 0

=> l = 1, 6

For l = 1,

Therefore, x + y = 0

Eigen vector is (1, -1)’

For l = 6,

Therefore, x - 4y = 0

Eigen vector is (4, 1)’

Therefore, modal matrix is X = and

X^-1 =

D = X^-1AX = which is diagonal matrix

Also A = XDX^-1

A⁵⁰ = XD⁵⁰X^-1

X^-1

(b) The system of equation can be written as AX = B

Rank(A) = Rank(A,B) = 3. Thus system is consistent. Homogeneous system AX=0, has 5 – 3 = 2 linearly independent solutions. Clearly (, -1, 3, 1, 0),

(, 1, -2, 0, 1) are linearly independent and satisfy the homogeneous system

AX = 0. Also (, 2, 0, 0, 0) is a particular solution of non-homogeneous system AX = B. Thus general solution of non homogeneous system is

(x₁, x₂, x₃, x₄, x₅) = (, 2, 0, 0, 0) + a (, -1, 3, 1, 0) + b (, 1, -2, 0, 1), where a, b are arbitrary.

For I₁, region of integration is bonded by the lines x = 1, x = 2, y=0 y = 1 i.e. region R₁ in figure. For I₂, region of integration is bonded by the lines x = y,

x = 2, y = 1, y = 2 i.e. region R₂ in figure.

Now the region R of integration i.e. union of R₁ and R₂ is bonded by the lines

y = 0, y = x, x = 1, x = 2

(b) Let V be the volume of solid. The two surfaces intersect at z = 1. Therefore

Let . Then dydx = rdrdθ, r varies from 0 to and θ varies from 0 to 2π. Then

8. (a) Since m be an integrating factor for differential equation Mdx + Ndy = 0. Thus m(Mdx + Ndy) = 0 is an exact differential equation.

Also dj = m (Mdx + Ndy) (given)

Because, j = constant, is a solution.

Let G(j) be any function of j

Therefore G(j)dj = m G(j) (Mdx + Ndy).

Let

Since terms on left is an exact differential, the terms on right must be an exact differential. Therefore, m G(j) is an integrating factor of differential equation.

(b)