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Exam Details

Engineering Sciences (XE) 2007

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Questions 196
Duration 180 mins
Package Engineering Sciences (XE) - Previous Year Papers

Paper pattern & analysis

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Showing all 196 questions in this paper.

Subject distribution

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Topic distribution

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Subtopic distribution

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Difficulty distribution

Easy 196 100%

Question type distribution

Multiple Choices 196 100%

Instructions

Engineering Sciences (XE) 2007 – Instructions
  • Total number of questions: 196
  • 42 questions carry one mark each
  • 154 questions carry two marks each
  • Negative marking: 1/4 of the marks allotted to the question
  • Use of calculator is allowed
  • This is a proctored examination
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  • After three warnings, the examination window will close automatically

Syllabus

Full Syllabus

Sample questions from this paper

Questions are selected across the paper subjects wherever the paper contains that variety.

1
2007 · Unclassified
Engineering Sciences (XE) 2007
Q.1 – Q.6 carry one mark each.

Let M = . Then the maximum number of linearly independent eigenvectors of M is

A
0
B
1
C
2
D
3
2
2007 · Unclassified
Engineering Sciences (XE) 2007

Let L = limx→π/2 (sin22x / (x - π/2)2). Then L is equal to

A
-4
B
0
C
2
D
4
3
2007 · Unclassified
Engineering Sciences (XE) 2007

Let f(z) = 1/(z2 - z). The coefficient of 1/(z - 1) in the Laurent expansion of f(z) about z = 1 is

A
-1
B
-1/2
C
1/2
D
1
4
2007 · Unclassified
Engineering Sciences (XE) 2007

Let u(x,t) be the solution of the initial value problem ∂2u/∂t2 = 9∂2u/∂x2, t > 0, -∞ < x < ∞, u(x,0) = x + 5, ∂u/∂t (x,0) = 0. Then u(2,2) is

A
7
B
13
C
14
D
26
5
2007 · Unclassified
Engineering Sciences (XE) 2007

Two students take a test consisting of five TRUE/FALSE questions. To pass the test the students have to answer at least three questions correctly. Both of them know the correct answers to two questions and guess the answers to the remaining three. The probability that only one student passes the test is equal to

A
6/32
B
7/32
C
1/4
D
3/4
6
2007 · Unclassified
Engineering Sciences (XE) 2007

The equation g(x) = x is solved by Newton-Raphson iteration method, starting with an initial approximation x0 near the simple root α. If xn1 is the approximation to α at the (n+1)th iteration, then

A
xₙ₁ = xₙ - g(xₙ)/g'(xₙ)
B
xₙ₁ = xₙg'(xₙ) - g(xₙ)/g'(xₙ) - 1
C
xₙ₁ = g(xₙ)
D
xₙ₁ = xₙ - xₙg(xₙ) - g(xₙ) + 2xₙ/g'(xₙ) + 1