TANGEDCO / TNEB AE 2018 Engineering Mathematics Previous Paper

1. TNEB Previous Paper - 2018
For
Competitive Exams prep
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2. Question – TNEB AE - 2018:
The Eigen values of the matrix 𝑨 =
𝟏 𝟏 𝟏
𝟏 𝟏 𝟏
𝟏 𝟏 𝟏
are.
1. (0, 0, 0) 2. (0, 0, 1)
3. (0, 0, 3) 4. (1, 1, 1)
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Answer: 3
Explanation

3. 𝐓𝐡𝐞 𝐜𝐡𝐚𝐫𝐚𝐜𝐭𝐞𝐫𝐢𝐬𝐭𝐢𝐜𝐬 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧,
𝑨 − λ𝑰 = 𝟎
𝟏 𝟏 𝟏
𝟏 𝟏 𝟏
𝟏 𝟏 𝟏
− λ
𝟏 𝟎 𝟎
𝟎 𝟏 𝟎
𝟎 𝟎 𝟏
= 𝟎
𝟏 − λ 𝟏 𝟏
𝟏 𝟏 − λ 𝟏
𝟏 𝟏 𝟏 − λ
= 𝟎
Calculating this determinant, we obtain;
𝟏 − λ 𝟑 − 𝟑 𝟏 − λ + 𝟐 = 𝟎
λ 𝟑
− 𝟑 λ 𝟐
= 0
Thus, λ 𝟏 = 𝟎, λ 𝟐 = 𝟎, λ 𝟑 = 𝟎
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4. Question – TNEB AE - 2018:
The value of 𝒙 for which the matrix
𝟕 − 𝒙 𝟑
𝟒 𝟑 − 𝒙
is
singular.
1. 1 2. 1 and 9
3. 9 4. -1 and 9
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Answer: 2
Explanation

5. 𝟕 − 𝐱 𝟑
𝟒 𝟑 − 𝐱
= 𝟎
𝟕 − 𝐱 𝟑 − 𝐱 − 𝟏𝟐 = 𝟎
𝟐𝟏 − 𝟕𝐱 − 𝟑𝐱 + 𝐱 𝟐
− 𝟏𝟐 = 𝟎
𝐱 𝟐
− 𝟏𝟎𝐱 + 𝟗 = 𝟎
𝐱 − 𝟏 𝐱 − 𝟗 = 𝟎
𝐱 = 𝟏, 𝟗
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6. Question – TNEB AE - 2018:
The values of k for which the system of equations given by; has a
non-trivial solution is
𝟑𝒌 − 𝟖 𝒙 + 𝟑𝒚 + 𝟑𝒛 = 𝟎;
𝟑𝒙 + 𝟑𝒌 − 𝟖 𝒚 + 𝟑𝒛 = 𝟎;
𝟑𝒙 + 𝟑𝒚 + 𝟑𝒌 − 𝟖 𝒛 = 𝟎
1. 2 and 11 2. 2/3 and 11
3. 2/3 and 11/3 4. 2 and 11/3
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Answer: 3
Explanation

7. For the given system of equation to have a non trivial solution,
R(A) = 3
𝑨 =
𝟑𝒌 − 𝟖 𝟑 𝟑
𝟑 𝟑𝒌 − 𝟖 𝟑
𝟑 𝟑 𝟑𝒌 − 𝟖
𝒊𝒔 𝒕𝒉𝒆 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒓𝒊𝒙
𝑨 = 𝟎
𝟑𝒌 − 𝟖 𝟑 𝟑
𝟑 𝟑𝒌 − 𝟖 𝟑
𝟑 𝟑 𝟑𝒌 − 𝟖
= 𝟎 𝑪 𝟏 + 𝑪 𝟐 + 𝑪 𝟑
𝟑𝒌 − 𝟐 𝟑 𝟑
𝟑𝒌 − 𝟐 𝟑𝒌 − 𝟖 𝟑
𝟑𝒌 − 𝟐 𝟑 𝟑𝒌 − 𝟖
= 𝟎
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8. Question – TNEB AE - 2018:
𝒍𝒊𝒎
𝒙−𝟏
𝒚−𝟐
𝒙 𝟐
+ 𝟐𝒚 is equal to.
1. 0
2. 3
3. 5
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Answer: 3

9. Question – TNEB AE - 2018:
Which of the following double integrals in polar coordinates is
equivalent to 𝟎
∞
𝐞− 𝐱 𝟐+𝐲 𝟐
𝐝𝐱𝐝𝐲?
1. 𝟎
𝛑/𝟐
𝟎
∞
𝐞−𝐫 𝟐
𝐝𝐫𝐝𝛉 𝟐. 𝟎
𝛑/𝟐
𝟎
∞
𝐞−𝐫 𝟐
𝐫 𝐝𝐫𝐝𝛉
3. 𝟎
𝟐𝛑
𝟎
∞
𝐞−𝐫 𝟐
𝐝𝐫𝐝𝛉 𝟒. 𝟎
𝟐𝛑
𝟎
∞
𝐞−𝐫 𝟐
𝐫 𝐝𝐫𝐝𝛉
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Answer: 2

10. Question – TNEB AE - 2018:
If C is any simple closed curve enclosing the point
𝒛 = 𝒛 𝟎, then the value of 𝒛 − 𝒛 𝟎 𝒅𝒛 is.
1. 𝟎
2. 𝟐𝛑𝐢
3. 𝛑𝐢
4. 𝟒𝛑𝐢Test Shopping
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Answer: 1

11. Question – TNEB AE - 2018:
The Kernel of the Laplace transform is given by.
1. 𝒆−𝒔𝒕
2.
𝒆 𝒊𝒔𝒕
𝟐𝝅
3.
𝟐
𝝅
𝒔𝒊𝒏 𝒔𝒕 4.
𝟐
𝝅
𝒄𝒐𝒔 𝒔𝒕
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Answer: 1

12. Question – TNEB AE - 2018:
Let Laplace transform of 𝒇 𝒕 is 𝒇 𝒔 , then
1. 𝑳 𝒇 𝒕𝒂 𝒖 𝒕𝒂 = 𝒆−𝒂𝒔 𝒇(𝒔)
2. 𝑳 𝒇 𝒕 + 𝒂 𝒖 𝒕 + 𝒂 = 𝒆−𝒂𝒔
𝒇(𝒔)
3. 𝑳 𝒇 𝒕 − 𝒂 𝒖 𝒕 − 𝒂 = 𝒆−𝒂𝒔
𝒇(𝒔) 𝒘𝒉𝒆𝒓𝒆 𝒖 𝒕 − 𝒂 =
𝟎, 𝒕 < 𝒂
𝟏, 𝒕 > 𝒂
4. 𝑳 𝒇 𝒕 − 𝒂 / 𝒖 𝒕 + 𝒂 = 𝒆−𝒂𝒔 𝒇(𝒔) 𝒘𝒉𝒆𝒓𝒆 𝒖 𝒕 − 𝒂 =
𝟎, 𝒕 < 𝒂
𝟏, 𝒕 > 𝒂
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Answer: 3

13. Question – TNEB AE - 2018:
The z- transform of
𝟏
𝒏
is.
1.
𝒛
𝒛−𝟏
𝒊𝒇 𝒛 > 𝟏 2. 𝒍𝒐𝒈
𝟏
𝒛
3. 𝒍𝒐𝒈
𝒁
𝒛−𝟏
𝒊𝒇 𝒛 > 𝟏 4. 𝒛 𝟏 − 𝒏
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Answer: 3
Explanation

14. 𝒁
𝟏
𝒏
=
𝒏=𝟏
∞
𝟏
𝒏
𝒁−𝒏
= 𝒁−𝟏
+
𝒁−𝟐
𝟐
+
𝒁−𝟑
𝟑
+ ⋯
=
𝟏
𝒁
+
𝟏
𝟐
𝟏
𝒁
𝟐
+
𝟏
𝟑
𝟏
𝒁
𝟑
+ ⋯
= − −
𝟏
𝒁
−
𝟏
𝟐
𝟏
𝒁
𝟐
−
𝟏
𝟑
𝟏
𝒁
𝟑
+ ⋯
= −𝒍𝒐𝒈 𝟏 −
𝟏
𝒁
= 𝒍𝒐𝒈 𝟏 −
𝟏
𝒁
= 𝒍𝒐𝒈 𝒁 −
𝟏
𝒁
−𝟏
= 𝒍𝒐𝒈
𝒁
𝒁−𝟏
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15. Question – TNEB AE - 2018:
If 𝒇(𝒙) represented by Fourier Integral
𝒇 𝒙 = 𝟎
∞
𝑨 𝝎 𝒄𝒐𝒔𝝎𝒙 + 𝑩 𝝎 𝒔𝒊𝒏𝝎𝒙 𝒅𝝎, then 𝑨(𝝎) is defined as.
1.
𝟏
𝝅 −∞
∞
𝒇 𝒗 𝒄𝒐𝒔𝝎𝒗 𝒅𝒗
2.
𝟏
𝝅 −∞
∞
𝒇 𝒗 𝒔𝒊𝒏𝝎𝒗 𝒅𝒗
3. −∞
∞
𝒇 ꞷ 𝒄𝒐𝒔𝝎𝒗 𝒅𝒗
4. −∞
∞
𝒇 ꞷ 𝒔𝒊𝒏𝝎𝒗 𝒅𝒗Test Shopping
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Answer: 1

16. Question – TNEB AE - 2018:
In Newton- cotes formulas, if 𝒇(𝒙) is interpolated at equally spaced
nodes by a polynomial of degree four then it represents.
1. Trapezoidal rule
2. Simpson rule
3. Three-eight rule
4. Booles rule
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17. Question – TNEB AE - 2018:
If 𝒇(𝒙) is a polynomial of degree 𝒏 in 𝒙. Then 𝒏 𝒕𝒉 difference of this
polynomial is.
1. Constant
2. Variable
3. Zero
4. Ones
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18. Question – TNEB AE - 2018:
𝒇 𝒙 = 𝒇 𝟎 + 𝒙𝜵𝒇 𝟎 +
𝒙 𝒙+𝟏
𝟐!
𝜵 𝟐 𝒇 𝟎 + ⋯ +
𝒙 𝒙+𝟏 … 𝒙+𝒏−𝟏
𝒏!
𝜵 𝒏 𝒇(𝟎)
represents.
1. Newton Backward difference formula
2. Newton forward difference formula
3. Gauss forward formula
4. Newton divided difference formula
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Answer: 1

19. Question – TNEB AE - 2018:
The coefficient matrix transformed into when AX=B is solved by
Gauss – Jordan method is.
1. Lower triangular
2. Upper triangular
3. Diagonal matrix
4. Triangular
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Answer: 3

20. Question – TNEB AE - 2018:
Gauss elimination method fails if .
1. Any one of the pivots is zero or very small
2. Any one of the pivots is non zero or very large
3. Any two of the pivots are zero and one pivot are large
4. Any three of the pivots are non-zero and others are non-zero.
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Answer: 1

21. Question – TNEB AE - 2018:
A continues random variable X has a density function given by
𝒇 𝒙 = 𝒌𝒙 𝟏 − 𝒙 , 𝟎 ≤ 𝒙 ≤ 𝟏. The value of k is.
1. 2
2. 3
3. 5
4. 6
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Answer: 4
Explanation

22. 𝒇 𝒙 =
𝒌𝒙 𝟏 − 𝒙 , 𝟎 ≤ 𝒙 ≤ 𝟏
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
−∞
∞
𝒇 𝒙 𝒅𝒙 = 𝟏
−∞
𝟎
𝒇 𝒙 𝒅𝒙 + 𝟎
𝟏
𝒇 𝒙 𝒅𝒙 + 𝟏
∞
𝒇 𝒙 𝒅𝒙 = 𝟏
𝟎 + 𝟎
𝟏
𝒌𝒙 𝟏 − 𝒙 𝒅𝒙 + 𝟎 = 𝟏
𝟎
𝟏
𝒌𝒙 𝟏 − 𝒙 𝒅𝒙 = 𝟏
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23. 𝟎
𝟏
𝒌𝒙 𝟏 − 𝒙 𝒅𝒙 = 𝟏
𝒌
𝟎
𝟏
𝒙 − 𝒙 𝟐
𝒅𝒙 = 𝟏
𝒌
𝒙 𝟐
𝟐
−
𝒙 𝟑
𝟑 𝟎
𝟏
= 𝟏
𝒌
𝟏
𝟔
= 𝟏
𝒌 = 𝟔
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24. Question – TNEB AE - 2018:
Let X and Y be a bivariate random variable with correlation
coefficient 1/2, and standard deviation 2 and 3 respectively, then
𝑪𝒐𝒗(𝑿, 𝒀) is.
1. 1/3
2. 3
3. 6
4. 1/6
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Answer: 2

25. 𝝆 𝑿𝒀 =
𝑪𝒐𝒗(𝑿, 𝒀)
𝝈 𝑿 𝝈 𝒀
Where,
𝝆 𝑿𝒀 → 𝐂𝐨𝐫𝐫𝐞𝐥𝐚𝐭𝐢𝐨𝐧 𝐂𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭
𝝈 𝑿 𝝈 𝒀 → 𝑺𝑻𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏𝒔
𝑪𝒐𝒗 𝑿, 𝒀 → 𝑪𝒐𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆
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26. Question – TNEB AE - 2018:
The only discrete distribution that follows memoryless property
is.
1. Binomial
2. Poisson
3. Exponential
4. Geometric
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Answer: 4

27. Question – TNEB AE - 2018:
The random variable X takes values -1, 0 and 1 with probabilities
0.2, 0.5 and 0.3 respectively. The value of 𝑬 𝑿 𝟐 is.
1. 0.2
2. 0.3
3. 0.4
4. 0.5
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28. Question – TNEB AE - 2018:
A random variable X has the probability density function given
by 𝒇 𝒙 =
𝟏
𝟒
, −𝟐 < 𝒙 < 𝟐. The moment generating function of X is
given by.
1.
𝒆 𝟐𝒕+𝒆−𝟐𝒕
𝟒𝒕
𝟐.
𝒆 𝟐𝒕−𝒆−𝟐𝒕
𝟒𝒕
3.
𝒆 𝟐𝒕+𝒆−𝟐𝒕
𝒕
𝟒.
𝒆 𝟐𝒕−𝒆−𝟐𝒕
𝒕
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Answer: 2

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