JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD Pharm. D I Year REMEDIAL MATHEMATICS Previous Question Paper (October/November - 2019 - R08)

Pharm. D | I Year | I Year | R08 | October/November - 2019

Answer PaperPaper Code: PH106Maximum Marks: 70Time: 3hours

Answer any five questions. All questions carry equal marks.

Main

114 marks

Find AB and BA where A =

[[1, 2, -1], [2, 4, -1], [1, 1, 1]]

and B =

[[1, 1, 1], [3, -2, 3], [1, 4, -6]]

7 marks

Answer

To find the product AB:

AB = [[1, 2, -1], [2, 4, -1], [1, 1, 1]] * [[1, 1, 1], [3, -2, 3], [1, 4, -6]]

Calculating the elements:

Row 1: (1*1+2*3-1*1)=6, (1*1+2*-2-1*4)=-7, (1*1+2*3-1*-6)=13
Row 2: (2*1+4*3-1*1)=13, (2*1+4*-2-1*4)=-10, (2*1+4*3-1*-6)=20
Row 3: (1*1+1*3+1*1)=5, (1*1+1*-2+1*4)=3, (1*1+1*3+1*-6)=-2

Thus, AB = [[6, -7, 13], [13, -10, 20], [5, 3, -2]]

To find the product BA:

BA = [[1, 1, 1], [3, -2, 3], [1, 4, -6]] * [[1, 2, -1], [2, 4, -1], [1, 1, 1]]

Calculating the elements:

Row 1: (1*1+1*2+1*1)=4, (1*2+1*4+1*1)=7, (1*-1+1*-1+1*1)=-1
Row 2: (3*1-2*2+3*1)=2, (3*2-2*4+3*1)=1, (3*-1-2*-1+3*1)=2
Row 3: (1*1+4*2-6*1)=3, (1*2+4*4-6*1)=12, (1*-1+4*-1-6*1)=-11

Thus, BA = [[4, 7, -1], [2, 1, 2], [3, 12, -11]]

Solve the following equations by using cramer’s Rule:

x - 2y + 3z = 2; 3x + 2y - z = 12; 2x - y + 3z = 7

7 marks

Answer

The system is represented by the coefficient matrix D and constants matrix B:

D = [[1, -2, 3], [3, 2, -1], [2, -1, 3]]

Calculating the determinant |D|:

|D| = 1(6-1) - (-2)(9+2) + 3(-3-4) = 5 + 22 - 21 = 6

Since |D| ≠ 0, the system has a unique solution. Now find determinants for variables:

Dx = [[2, -2, 3], [12, 2, -1], [7, -1, 3]] = 2(6-1) + 2(36+7) + 3(-12-14) = 10 + 86 - 78 = 18

Dy = [[1, 2, 3], [3, 12, -1], [2, 7, 3]] = 1(36+7) - 2(9+2) + 3(21-24) = 43 - 22 - 9 = 12

Dz = [[1, -2, 2], [3, 2, 12], [2, -1, 7]] = 1(14+12) + 2(21-24) + 2(-3-4) = 26 - 6 - 14 = 6

Applying Cramer's Rule:

x = Dx/|D| = 18/6 = 3

y = Dy/|D| = 12/6 = 2

z = Dz/|D| = 6/6 = 1

The solution is x=3, y=2, z=1.

214 marks

Prove that

(4cos(A/2) - cot(A-15) - tan(A-15)) / (1 + 2sin(A/2)) = 0

7 marks

Answer

Prove that

Cos(π/9) Cos(2π/9) Cos(3π/9) Cos(4π/9) = 1/2^4

7 marks

Answer

314 marks

Find the equation of the line passing through the point (1, 1) and perpendicular to the line passing through the points (3, 5) and (-6, -2).

7 marks

Answer

Find the equation of the circle whose centre is (5, 7) and radius 4.

7 marks

Answer

414 marks

Find the differential coefficient of

y = xe^x + sin(log x)

7 marks

Answer

Find the derivation of

(log x) / (x^2 + e^x)

7 marks

Answer

514 marks

u = (x^2 + y^2 + z^2)^(-1/2)

, then show that

∂^2u/∂x^2 + ∂^2u/∂y^2 + ∂^2u/∂z^2 = 0

7 marks

Answer

u = sin^(-1)(x^2 + y^2)/(x + y)

, then prove that

x(∂u/∂x) + y(∂u/∂y) = 3tan u

7 marks

Answer

614 marks

Evaluate

∫ x^2 e^x dx

7 marks

Answer

Evaluate

∫[0 to π/2] (sin x) / (sin x + cos x) dx

7 marks

Answer

714 marks

Solve:

(xy + x)dx + (yx + y)dy = 0

7 marks

Answer

Solve:

dy/dx = (x + y) / x

7 marks

Answer

814 marks

Find

L[5sin(2t) + 2sin(3t)]

7 marks

Answer

Find

L[e^(-t) sin^2(t)]

7 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD Pharm. D I Year REMEDIAL MATHEMATICS Previous Question Paper (October/November - 2019 - R08)

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