JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD Pharm. D I Year REMEDIAL MATHEMATICS Previous Question Paper (January/February - 2022 - R08)

Pharm. D | I Year | I Year | R08 | January/February - 2022

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

Answer any five questions. All questions carry equal marks.

Main

114 marks
a

If A = [[1, 2], [2, -3]], B = [[2, 1], [2, -3]], and C = [[3, -1], [2, 0]], verify that (AB)C = A(BC).

7 marks

Answer

To verify the associative property (AB)C = A(BC), we first calculate AB and BC.

First, calculate AB:

AB = [[1, 2], [2, -3]] * [[2, 1], [2, -3]] = [[(1*2 + 2*2), (1*1 + 2*-3)], [(2*2 + -3*2), (2*1 + -3*-3)]] = [[6, -5], [-2, 11]]

Now, calculate (AB)C:

(AB)C = [[6, -5], [-2, 11]] * [[3, -1], [2, 0]] = [[(6*3 + -5*2), (6*-1 + -5*0)], [(-2*3 + 11*2), (-2*-1 + 11*0)]] = [[8, -6], [16, 2]]

Next, calculate BC:

BC = [[2, 1], [2, -3]] * [[3, -1], [2, 0]] = [[(2*3 + 1*2), (2*-1 + 1*0)], [(2*3 + -3*2), (2*-1 + -3*0)]] = [[8, -2], [0, -2]]

Finally, calculate A(BC):

A(BC) = [[1, 2], [2, -3]] * [[8, -2], [0, -2]] = [[(1*8 + 2*0), (1*-2 + 2*-2)], [(2*8 + -3*0), (2*-2 + -3*-2)]] = [[8, -6], [16, 2]]

Since (AB)C = A(BC) = [[8, -6], [16, 2]], the associative property is verified.

b

Solve the following equations by using Cramer’s Rule: 3x + 4y + 5z = 18; 2x - y + 8z = 13; 5x - 2y + 7z = 20.

7 marks

Answer

To solve using Cramer's Rule, we first calculate the determinant of the coefficient matrix D:

D = |3, 4, 5; 2, -1, 8; 5, -2, 7| = 3(-7 + 16) - 4(14 - 40) + 5(-4 + 5) = 3(9) - 4(-26) + 5(1) = 27 + 104 + 5 = 136

Next, calculate the determinants Dx, Dy, and Dz by replacing the respective columns with the constants [18, 13, 20]:

Dx = |18, 4, 5; 13, -1, 8; 20, -2, 7| = 18(-7 + 16) - 4(91 - 160) + 5(-26 + 20) = 18(9) - 4(-69) + 5(-6) = 162 + 276 - 30 = 408
Dy = |3, 18, 5; 2, 13, 8; 5, 20, 7| = 3(91 - 160) - 18(14 - 40) + 5(40 - 65) = 3(-69) - 18(-26) + 5(-25) = -207 + 468 - 125 = 136
Dz = |3, 4, 18; 2, -1, 13; 5, -2, 20| = 3(-20 + 26) - 4(40 - 65) + 18(-4 + 5) = 3(6) - 4(-25) + 18(1) = 18 + 100 + 18 = 136

The solutions are:

x = Dx/D = 408/136 = 3, y = Dy/D = 136/136 = 1, z = Dz/D = 136/136 = 1

Thus, x = 3, y = 1, z = 1.

214 marks
a

Prove that (TanA + SecA - 1) / (TanA - SecA + 1) = (1 + SinA) / CosA.

7 marks

Answer

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b

If Cos A = 5/13, Tan B = -15/8 and 270 < A < 360, 90 < B < 180, find the quadrant to which A + B relates.

7 marks

Answer

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3

Find the equation of the straight line perpendicular to 2x + 3y + 4 = 0 and passing through (3, -2).

14 marks

Answer

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414 marks
a

Evaluate lim(x→3) (x³ - 2x² - 4x - 4) / (x² - 3x + 1).

7 marks

Answer

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b

Find the derivative of x² * Sec x * Cos x.

7 marks

Answer

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514 marks
a

Find ∂²/∂x∂y (e^(3x+y) * xy).

7 marks

Answer

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b

Verify Euler’s theorem for f(x, y) = x³ + 3x²y.

7 marks

Answer

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614 marks
a

Evaluate ∫ x² * e^x dx.

7 marks

Answer

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b

Evaluate ∫(0 to π) x * Sin²x dx.

7 marks

Answer

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714 marks
a

Solve the differential equation dy/dx = 1 + x + y + xy.

7 marks

Answer

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b

Solve the differential equation dy/dx = (x - y) / (x + y).

7 marks

Answer

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814 marks
a

Find L{t³ + 2sin 2t}.

7 marks

Answer

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b

Find L{t² * e^t * Sin 3t}.

7 marks

Answer

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About this paper

This page contains the solved question paper with answers of REMEDIAL MATHEMATICS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for Pharm. D I Year I Year, January/February - 2022, R08.

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