VITEEE Exam 2016 - Download Question Paper with Solutions (Part 3)

VITEEE Exam 2016 - Question Paper with Solutions

To prepare for the VITEEE Exam, Download VITEEE EXAM 2016 previous year mathematics question paper with solutions. VITEEE EXAM previous year mathematics question paper and sample paper will help you to score good marks

To get admission in VIT UNIVERSITY for all the campuses (Vellore, Chennai, Bhopal, Amaravati), students need to clear the VITEEE EXAM, that's one of the toughest exams to enter into VIT University in associate country, VITEEE Exam 2016 previous year question papers with solutions will help students to rearrange for the examination.

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So in this post, I will give you the VITEEE EXAM 2016 mathematics question paper with solutions. There will be 40 questions out of 120 questions. Correct answer and explanation for each question are available after each question and answer. Download links for VITEEE EXAM 2016 question paper will be available at the end of the page. Prepare well and Don't leave any questions.

VITEEE Exam 2016 - Question Paper with Solutions (Part 3)

81. The function ƒ: R → R defined by ƒ(x) = (x -1) (x - 2) (x - 3) is

A) one-one but not onto
B) onto but not one-one
C) both one-one and onto
D) neither one-one nor onto

Answer: B) onto but not one-one

Explanation: Given, ƒ(x) = (x - 1) (x - 2) (x - 3)
⇒ ƒ (1) = ƒ (2) = ƒ (3) = 0
⇒ ƒ(x) is not one-one.
For each y ∈ R, there exists x ∈ R such that ƒ(x) = y.
Therefore, ƒ is onto. Note: If a continuous function has more than one roots, then a function is always a many-one.

82. If the complex numbers z₁ z₂ and z₃ are in AP, then they lie on a

A) a circle
B) a parabola
C) line
D) ellipse

Answer: C) line

Explanation: Let' z₁, z₂ and z₃ be affixes of points A, B and C, respectively.
Since, z₁, z₂ and z₃ are in AP, therefore 2z₂ = z₁ + z₃
⇒ z₂ = z₁ + z₃ / 2
⇒ B is the mid-point of the line AC.
⇒ A, B and C are collinear.
⇒ z₁, z₂ and z₃ lie on a line.

83. Let a, b and c be in AP and |a| < 1, |b| < 1, |c| < 1.
If x = 1 + a + a²+ ... to ∞,
y = 1 + b + b² +... to ∞,
z = 1 + c + c² + ... to ∞, then x, y and z are in

A) AP
B) GP
C) HP
D) None of these

Answer: C) HP

Explanation: Given, x = 1 + a + a² + ... ∞ = 1 / 1 - a y = 1 + b + b² + ... ∞ = 1 / 1 - b and z = 1 + c + c² + ...∞ = 1 / 1 - c
Since, a, b and c are in AP.
⇒ 1 - a, 1 - b and 1 - c are in AP.
⇒ 1 / 1 - a , 1 / 1 - b and 1 / 1 - c are in HP.
⇒ x, y and z are in HP.
Note If the common ratio of a GP is not less than 1, then we do not determine the sum of an infinite GP series.

84. The number of real solutions of the equation (9/10) = - 3 + x - x² is

A) 0
B) 2
C) 1
D) None of these

Answer: A) 0

85. The lines 2x - 3y - 5 = 0 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units, then the equation of the circle is

A) x² + y² + 2x - 2y - 62 = 0
B) x² + y² + 2x - 2y - 47 = 0
C) x² + y² - 2x + 2y - 47 = 0
D) x² + y² - 2x + 2y - 62 = 0

Answer: C) x² + y² - 2x + 2y - 47 = 0

Explanation: The center of the required circle lies at the intersection of 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0.
Thus, the coordinates of the center are (1, -1).
Let r be the radius of the circle.
Then, πr² = 154 ⇒ 22 / 7 r² = 154 ⇒ r = 7
Hence, the equation of required circle is (x - 1)² + (y + 1)² = 7² ⇒ x² + y² - 2x + 2y - 47 = 0

86. The angle of depressions of the top and the foot of a chimney as seen from the top of a second chimney, which is 150 m high and standing on the same level as the first is θ and ∅ respectively, then the distance between their tops when tan θ = 4/3 and tan ∅ = 5/2 is

A) 150 / √3 m
B) 100 √3 m
C) 150 m
D) 100 m

Answer: D) 100 m

87. If one root is square of the other root of the equation x² + px + q = 0, then the relations between p and q is

A) p³ - (3p - 1) q + q² = 0
B) p³ - q (3p + 1) + q² = 0
C) p³ + q (3p - 1) + q² = 0
D) p³ + q (3p + 1) + q² = 0

Answer: A) p³ - (3p - 1) q + q² = 0

Explanation: Given, equation x² + px + q = 0 has roots α and α².
⇒ α + α² = -p and α³ = q
⇒ α (α + 1) = -p
⇒ α³ [α³ + 1 + 3α (α + 1)] = -p³
⇒ q (q + 1 - 3p) = -p³
⇒ p³ - (3p - 1) q + q² = 0

88. The coefficient of x53 in the following expansions

A) A

B) B
C) C
D) D

Answer: C) C

Explanation: 

89. If (-3,2) lies on the circle x² + y² + 2gx + 2ƒy + c = 0, which is concentric with the circle x² + y² + 6x + 8y - 5 = 0, then c is equal to

A) 11
B) -11
C) 24
D) 100

Answer: B) -11

Explanation: Equation of family of concentric circles to the circle x² + y² + 6x + 8y - 5 = 0 is
x² + y² + 6x + 8y + λ = 0
Which is similar to
x² + y² + 2gx + 2fy + c = 0
Thus, the point (-3, 2) lies on the circle
x² + y² + 6x + 8y + c = 0
∴ (-3)² + (2)² + 6(-3) + 8(2) + c = 0
⇒ 9 + 4 - 18 + 16 + c = 0
⇒ c = -11

90. If a = i + j + k, b = i + 3j + 5k and c = 7i + 9j + 11k, then the area of parallelogram having diagonals a + b and b + c is

A) 4√6 sq. units
B) 1/2√21 sq. units
C) √6 / 2 sq. units
D) √6 sq. units

Answer: A) 4√6 sq. units

Explanation: 

91. Find the trace of the matrix A given below

A) 17
B) 25
C) 3
D) 12

Answer: A) 17

Explanation:

92. The value of the determinant for

A) independent of α
B) independent of β
C) independent of α and β
D) None of the above

Answer: A) independent of α

Explanation:

93. The maximum value of 4 sin² x -12sin x + 7 is

A) 25
B) 4
C) does not exist
D) none of these

Answer: D) None of these

Explanation:

94. A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is

A) 3x - 4y + 7 = 0
B) 4x + 3y = 24
C) 3x + 4y = 25
D) x + y = 7

Answer: B) 4x + 3y = 24

Explanation: 

95. The tangent at (1, 7) to the curve x² = y - 6 touches the circle x² + y² + 16x + 12y + c = 0 at

A) (6, 7)
B) (-6, 7)
C) (6, - 7)
D) (-6, -7)

Answer: D) (-6, -7)

Explanation: 

96. The equation of straight line through the intersection of the lines x - 2y = 1 and x + 3y = 2 and parallel to 3x + 4y = 0 is

A) 3x + 4y + 5 = 0
B) 3x + 4y - 10 = 0
C) 3x + 4y - 5 = 0
D) 3x + 4y + 6 = 0

Answer: C) 3x + 4y - 5 = 0

Explanation: The intersection point of lines x - 2y = 1 and x + 3y = 2 is (7/5, 1/5)
since, required line is parallel to 3x + 4y = 0.
Therefore, the slope of the required line is - (3/4).
∴ Equation of required line which passes through (7/5, 1/5). y - (1/5) = - (3/4) (x - 7/5) ⇒ 3x / 4 + y = 21 / 20 + 1 / 5 ⇒ 3x + 4y / 4 = 21 + 4 / 20 ⇒ 3x + 4y - 5 = 0

97. Find

A) - (1/√2) tan(x/2 + π / 8) + C
B) 1/2 tan(x/2 + π/8) + C
C) 1/√2 cot(x/2 + π / 8) + C
D) - (1/√2) cot(x/2 + π/8) + C

Answer: C) 1/√2 cot(x/2 + π / 8) + C

98. The value of integral

A) π / 2 + 1
B) π / 2 - 1
C) -1
D) 1

Answer: B) π / 2 - 1

Explanation: 

99. The value of

A) 1/3
B) 1/4
C) 1/8
D) None of these

Answer: C) 1/8

Explanation: 

100. The eccentricity of the ellipse, which meets the straight line x/7 + y/2 = 1 on the axis of x and the straight line x/3 - y/5 = 1 on the axis of y and whose axes lie along the axes of coordinates, is

A) 3√2 / 7
B) 2√6 / 7
C) √3 /7
D) None of the above

Answer: B) 2√6 / 7

Explanation: Let the equation of the ellipse be x² / a² + y² / b² = 1.
It is given that it passes through (7, 0) and (0,-5).
Therefore, a² = 49 and b² = 25
the eccentricity of the ellipse is e = √ 1 - b² / a² = √ 1 - 25 / 49 = √ 24 / 49 = 2√6 / 7

101. If (x² / a²) + (y² / b²) = 1 (a > b) and x² - y² = c² cut at right angles, then

A) a² + b² = 2c²
B) b² - a² = 2c²
C) a² - b² = 2c²
D) a²b² = 2c²

Answer: C) a² - b² = 2c²

102. The equation of the comic with focus at (1, -1) directory along x - y + 1 = 0 and with eccentricity √2, is

A) x² - y² = 1
B) xy = 1
C) 2xy - 4x + 4y + 1 = 0
D) 2xy + 4x - 4y - 1 = 0

Answer: C) 2xy - 4x + 4y + 1 = 0

Explanation: Let P(x, y) be any point on the conic.
Then, √(x-1)² + (y+1)² = √2(x - y + 1 / √2)
⇒ (x-1)² + (y+1)² = (x-y+1)²
⇒ 2xy - 4x + 4y + 1 = 0

103. There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in the wrong envelope is

A) 119
B) 44
C) 59
D) 40

Answer: B) 44

Explanation: Required numbers = 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!] = 44
Note If r (0 ≤ r ≤ n) objects occupy the original places and none of the remaining (n - r) objects occupies its original places, then the number of such arrangements
= nCr ⋅ (n - r)! [1 - 1/1! + 1/2! - 1/3! +...+ (-1) n-r 1 / (n - r)!]

104. The sum of the series,

A) 3e
B) 17 / 6 e
C) 13 / 6 e
D) 19 / 6 e

Answer: B) 17 / 6 e

Explanation: 

105. The coefficient of xn in the expansion of loga (l + x) is

A) (-1) n-1 / n
B) (-1) n-1 loga e
C) (-1) n-1 / n loge a
D) (-1) n / n loga e

Answer: B) (-1) n-1 loge

Explanation: 

106. If a plane meets the coordinate axes at A, B and C in such a way that the centroid of ΔABC is at the point (1, 2, 3), the equation of the plane is

A) x/1 + y/2 + z/3 = 1
B) x/3 + y/6 + z/9 = 1
C) x/1 + y/2 + z/3 = 1/3
D) None of these

Answer: B) x/3 + y/6 + z/9 = 1

Explanation: Let the equation of the required plane be x / a + y / b + z / c = 1.
This meets the coordinate axes at A, B and C, the coordinates of the centroid of ΔABC are (a/3 , b/3, c/3)
∴ a/3 = 1 , b/3 = 2 , c/3 = 3
⇒ a = 3 , b = 6 , c = 9
Hence, the equation of the plane is x / 3 + y / 6 + z / 9 = 1

107. The area lies in the first quadrant and bounded by the circle x² + y² = 4, the line x = √3y and x-axis is

A) π sq. units
B) π / 2 sq. units
C) π / 3 sq. units
D) none of these

Answer: C) π / 3 sq. units

Explanation: 

108. Find

A) 0
B) 1
C) -1
D) e

Answer: B) 1

Explanation: 

109. Find

A) m = 1, n = 0
B) m = nπ / 2 + 1
C) n = mπ / 2
D) m = n = π / 2

Answer: C) n = mπ / 2

Explanation: 

110. The domain of the function ƒ(x) = √4 - x² / sin-1 (2 - x) is

A) [0, 2]
B) [0, 2]
C) [1, 2]
D) [1, 2]

Answer: C) [1, 2]

Explanation: Given, ƒ(x) = √4 - x² / sin-1(2 - x) √4 - x² is defined for 4 - x² ≥ 0.
⇒ x² ≤ 4
⇒ -2 ≤ x ≤ 2 and sin-1(2 - x) is defined for -1 ≤ 2 - x ≤ 1
⇒ -3 ≤ - x ≤ -1
⇒ 1 ≤ x ≤ 3
Also, sin-1(2 - x) = 0 for x = 2
∴ Domain of ƒ(x) = [-2, 2] ∩ [1,3] - {2} = [1,2]

111. The general solution of the differential equation (1 + y²) dx + (1 + x²) dy = 0 is

A) x - y = C (1 - xy)
B) x - y = C (1 + xy)
C) x + y = C (1 - xy)
D) x + y = C (1 + xy)

Answer: C) x + y = C (1 - xy)

Explanation: Given , (1 + y²) dx + (1 + x²)dy = 0
⇒ dx / 1 + x² + dy / 1 + y² = 0
On integrating, we get tan -1 x + tan-1 y = tan-1 C
⇒ x + y / 1 - xy = C
⇒ x + y = C (1 - xy)

112. The order and degree of the differential equation p = [1 + (dy / dx) ²]3/2 / d²y / dx² are, respectively

A) 2, 2
B) 2, 3
C) 2, 1
D) None of these

Answer: A) 2, 2

Explanation: Given, p = [1 + (Dy / dx) ²]3/2 / d² y / dx² ⇒ p (d²y / dx²) = [1 + (dy/dx) ²]3/2
on squaring both sides, we get p² (d²y / dx²)² = [1 + (dy/dx) ²]³
clearly, it is a second-order differential equation of degree 2.
Note if the higher order derivative is in the transcendental, then we do not determined the degree that equation.

113. The relation R defined on the set of natural numbers as {(a, b): a differs from b by 3} is given

A) {(1,4),(2,5),(3,6), ...}
B) {(4,1),(5,2),(6,3), ...}
C) {(1,3),(2,6),(3,9), ...}
D) None of the above

Answer: B) {(4,1),(5,2),(6,3), ...}

Explanation: Let R = {(a,b) : a, b ∈ N, a - b = 3} = [{(n + 3),n} : n ∈ N] = {(4,1),(5,2),(6,3), ...}

114. The solution of the differential equation dy / dx + 2yx / 1 + x² = 1 / (1 + x²)² is

A) y (1 + x²) = C + tan-1 x
B) y / (1 + x²) = C + tan-1 x
C) y log (1 + x²) = C + tan-1 x
D) y (1 + x²) = C + sin-1 x

Answer: A) y (1 + x²) = C + tan-1 x

Explanation: Given, dx/dx + 2yx / 1 + x² = 1 / (1 + x²)² which is a linear differential equation.
∴ P = 2x / 1 + x² , Q = 1 / (1 + x²)²
Now, IF = e∫2x / 1+x² dx = elog(1+x⊃) = (1 + x²)
∴ Solution of differential equation is y ⋅ (1 + x²) = ∫ 1 / (1 + x²)² ⋅ (1 + x²)dx + C
⇒ y(1 + x²) = ∫1 / 1+x² dx + C
⇒ y(1 + x²) = tan-1 x + C

115. If x, y and z are all distinct and

A) -2
B) -1
C) -3
D) None of these

Answer: B) -1

Explanation: 

116. The probability that at least one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then

A) 0.4
B) 0.8
C) 1.2
D) 1.4

Answer: C) 1.2

Explanation: 

117. If 3p and 4p are resultant of a force 5p, then angle between 3p and 5p is

A) sin-1(3/5)
B) sin - (4/5)
C) π / 3
D) None of these

Answer: B) sin - (4/5)

Explanation: 

118. If 2 tan-1(cosx) = tan-1 (2 cosec x), then the value of x is

A) 3π / 4
B) π / 4
C) π / 3
D) None of these

Answer: B) π / 4

Explanation: Given, 2 tan-1 (cos x) = tan-1 (2 cosec x)
⇒ tan-1 (2 cos x / 1 - cos² x) = tan-1 (2 cosec x)
⇒ 2 cos x / 1 - cos² x = 2 cosec x
⇒ 2 cos x / sin² x = 2 cosec x
⇒ sin x = cos x
⇒ x = π / 4

119. Let a be any element in a Boolean algebra B. If a + x = 1 and ax = 0, then

A) x = 1
B) x = 0
C) x = a
D) x = a'

Answer: D) x = a'

Explanation: Given, conditions are a + x = 1 and ax = 0.
These two conditions will be true if x = a'

120. Dual of (x + y) ⋅ (x + 1) = x + x ⋅ y + y is

A) (x ⋅ y) + (x ⋅ 0) = x ⋅ (x + y) ⋅ y
B) (x + y) + (x ⋅ 1) = x ⋅ (x + Y) ⋅ y
C) (x ⋅ y) (x ⋅ 0) = x ⋅ (x + y) ⋅ y
D) None of the above

Answer: A) (x ⋅ y) + (x ⋅ 0) = x ⋅ (x + y) ⋅ y

Explanation: Given, (x + y) ⋅ (x + 1) = x + x ⋅ y + y
Replace '⋅' by '+’, '+' by '⋅', '1' by '0',
we get (x ⋅ y) + (x ⋅ 0) = x ⋅ (x + y) ⋅ y

ALL THE BEST!

Download link: VITEEE Exam 2016 Question paper with solutions

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VITEEE Exam 2016 Question paper with solutions (Part 1)

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VITEEE Exam 2019 - Previous year question paper, Sample paper, Mock test, Syllabus

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