JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD Pharm. D I Year REMEDIAL MATHEMATICS Previous Question Paper (February/March - 2023 - R08)

Pharm. D | I Year | I Year | R08 | February/March - 2023

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

Answer any five questions. All questions carry equal marks.

Main

1

If

A = [[1, 2, 2], [2, 1, 2], [2, 2, 1]]
then show that
A2 - 4A - 5I = 0
.

14 marks

Answer

To show that A2 - 4A - 5I = 0, we first calculate A2 by multiplying matrix A with itself:

A2 = [[1, 2, 2], [2, 1, 2], [2, 2, 1]] * [[1, 2, 2], [2, 1, 2], [2, 2, 1]]

Performing matrix multiplication:

A2 = [[(1*1+2*2+2*2), (1*2+2*1+2*2), (1*2+2*2+2*1)], [(2*1+1*2+2*2), (2*2+1*1+2*2), (2*2+1*2+2*1)], [(2*1+2*2+1*2), (2*2+2*1+1*2), (2*2+2*2+1*1)]]
A2 = [[9, 8, 8], [8, 9, 8], [8, 8, 9]]

Next, calculate 4A:

4A = 4 * [[1, 2, 2], [2, 1, 2], [2, 2, 1]] = [[4, 8, 8], [8, 4, 8], [8, 8, 4]]

Next, identify 5I where I is the identity matrix:

5I = 5 * [[1, 0, 0], [0, 1, 0], [0, 0, 1]] = [[5, 0, 0], [0, 5, 0], [0, 0, 5]]

Now, compute the expression A2 - 4A - 5I:

[[9, 8, 8], [8, 9, 8], [8, 8, 9]] - [[4, 8, 8], [8, 4, 8], [8, 8, 4]] - [[5, 0, 0], [0, 5, 0], [0, 0, 5]]

Subtracting the matrices element-wise:

[[9-4-5, 8-8-0, 8-8-0], [8-8-0, 9-4-5, 8-8-0], [8-8-0, 8-8-0, 9-4-5]] = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Since the resulting matrix is the zero matrix, it is shown that A2 - 4A - 5I = 0.

214 marks
a

If

tan(A-B) = 7/24
and
tan B = 3/4
where A and B are acute, show that
A - B = tan-1(1/2)
.

7 marks

Answer

Given tan(A - B) = 7/24. The question asks to show A - B = tan-1(1/2). However, based on the provided value tan(A - B) = 7/24, the result is simply A - B = tan-1(7/24). If the objective is to find tan A, we use the formula:

tan A = tan((A - B) + B) = (tan(A - B) + tan B) / (1 - tan(A - B)tan B)

Substituting the values tan(A - B) = 7/24 and tan B = 3/4:

tan A = (7/24 + 3/4) / (1 - (7/24)(3/4)) = (7/24 + 18/24) / (1 - 21/96) = (25/24) / (75/96) = (25/24) * (96/75) = 4/3

Thus, tan A = 4/3. The expression A - B = tan-1(1/2) does not follow from the provided tan(A - B) = 7/24.

b

Prove that

cos2A + cos2(120° + A) + cos2(120° - A) = 3/2
.

7 marks

Answer

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314 marks
a

Find the equation of the straight line passing through the point (5, 4) and parallel to the line

2x + 3y + 7 = 0
.

7 marks

Answer

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b

Find the circumcentre of the triangle formed by the points (1, 5), (0, -3), (-5, 1).

7 marks

Answer

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4

If

y = cos(m log x)
then show that
x2yn+2 + (2n+1)xyn+1 + (m2 + n2)yn = 0
.

14 marks

Answer

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5

Find the first and second order partial derivatives of

f(x, y) = log(x2 + y2)
.

14 marks

Answer

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614 marks
a

Evaluate

∫ (1 - tan x)/(1 + tan x) dx
.

7 marks

Answer

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b

Evaluate

∫0π/4 log(1 + tan x) dx
.

7 marks

Answer

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7

Solve

(ex + 1) y dy + (y + 1) ex dx = 0
.

14 marks

Answer

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8

Find the Laplace transform of (a)

eat cos t
(b)
t3e-3t
.

14 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, February/March - 2023, R08.

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