VITEEE Exam 2015 - Question paper with Solutions
So in this post, I will give you the VITEEE EXAM 2015 previous year 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 2015 question paper will be available at the end of the page. Prepare well and Don't leave any questions.
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VITEEE Exam 2015 - Question paper with Solutions (Part 3)
81. Find
A) A
B) B
C) C
D) None of the above
B) B
C) C
D) None of the above
Answer: A) A
Explanation:
82. Which of the following options is not the asymptote of
the curve 3x³ + 2x²y - 7xy² + 2y³ - 14xy + 7y² + 4x + 5y =0?
A) y = - (1/2) x - (5/6)
B) y = x - (7/6)
C) y = 2x + (3/7)
D) y = 3x - (3/2)
A) y = - (1/2) x - (5/6)
B) y = x - (7/6)
C) y = 2x + (3/7)
D) y = 3x - (3/2)
Answer: C) y = 2x + (3/7)
Explanation: ∅₃(m) = 3+ 2m - 7m² + 2m³ ∅₂(m)
= -14m + 7m² ∅' ₃(m) = 2 - 14m + 6m²
Now,∅₃(m) = 0 gives 3 + 2m - 7m² + 2m³ = 0 ⇒ (1-m)(3+5m-2m²) = 0 ⇒ (1-m)(1+2m)(3-m) = 0
So that, m = -(1/2),1,3
Thus, c giving equation is c∅' n(m) + ∅n-1(m) = 0, which in the present case becomes c(2-14m+6m²)+(-14m+7m²)= 0 ⇒ c = 14m - 7m² / 2 - 14m + 6m²
Thus, when m = -(1/2), c = -(5/6) When m = 1, c = -(7/6)
When m = 3, c = -(3/2)
Therefore, the asymptotes are y = -(1/2)x -(5/6) y = x -(7/6) and y = 3x - (3/2).
83. If N is a set of natural numbers, then under binary operation a ⋅ b = a + b, (N,.) is
A) quasi-group
B) semi-group
C) monoid
D) group
Now,∅₃(m) = 0 gives 3 + 2m - 7m² + 2m³ = 0 ⇒ (1-m)(3+5m-2m²) = 0 ⇒ (1-m)(1+2m)(3-m) = 0
So that, m = -(1/2),1,3
Thus, c giving equation is c∅' n(m) + ∅n-1(m) = 0, which in the present case becomes c(2-14m+6m²)+(-14m+7m²)= 0 ⇒ c = 14m - 7m² / 2 - 14m + 6m²
Thus, when m = -(1/2), c = -(5/6) When m = 1, c = -(7/6)
When m = 3, c = -(3/2)
Therefore, the asymptotes are y = -(1/2)x -(5/6) y = x -(7/6) and y = 3x - (3/2).
83. If N is a set of natural numbers, then under binary operation a ⋅ b = a + b, (N,.) is
A) quasi-group
B) semi-group
C) monoid
D) group
Answer: B) semi-group
Explanation: The structure (N, ⋅)
satisfies the closure property, associativity and commutativity. But the
identity element 0 does not belong to N. Hence, N is a semi-group.
84. ∫ dx / (cos x + √3 sin x) equals
A) 1/2 log tan (x/2 + π/12) + C
B) 1/3 log tan (x/2 - π/12) + C
C) log tan (x/2 + π/6) + C
D) 1/2 log tan (x/2 - π/6) + C
84. ∫ dx / (cos x + √3 sin x) equals
A) 1/2 log tan (x/2 + π/12) + C
B) 1/3 log tan (x/2 - π/12) + C
C) log tan (x/2 + π/6) + C
D) 1/2 log tan (x/2 - π/6) + C
Answer: A) 1/2 log tan (x/2 + π/12) + C
Explanation: ∫ dx / cosx + √3sinx = 1/2∫ dx / 1/2cosx + √3/2sinx = 1/2∫
dx / cosÏ€/3cosx + sinÏ€/3sinx = 1/2∫ dx / cos(x-(Ï€/3)) = 1/2∫sec(x-(Ï€/3))dx =
1/2log tan(x/2 - π/6 + π/4) + C = 1/2log tan(x/2 + π/12) + C
85. If (2,7,3) is one end of a diameter of the sphere x² + y² + z² - 6x - 12y -2z + 20 = 0, then the coordinates of the other end of the diameter are
A) (-2,5,-1)
B) (4,5,1)
C) (2,-5,1)
D) (4,5,-1)
85. If (2,7,3) is one end of a diameter of the sphere x² + y² + z² - 6x - 12y -2z + 20 = 0, then the coordinates of the other end of the diameter are
A) (-2,5,-1)
B) (4,5,1)
C) (2,-5,1)
D) (4,5,-1)
Answer: D) (4, 5,-1)
Explanation: Given equation of sphere is x² + y² + z² - 6x - 12y - 2z +
20 = 0
Centre ≡ (3, 6, 1)
If one of the end of diameter is (2, 7, 3).
Let the other end of the diameter be (x, y, z).
∴ (3,6,1) = (2+x/2 , 7+y/2 , 3+z/2)
⇒ 2 + x = 6
⇒ x = 4 7 + y = 12
⇒ y = 5 and 3 + z = 2 ⇒ z = -1
Therefore, (x,y,z)≡(4,5,-1)
86. The two lines x = my + n, z = py + q and x = m'y + n', z = p'y + q' are perpendicular, to each other, if
A) mm' + pp' = 1
B) m/m' + p/p' = -1
C) m/m' + p/p' = 1
D) mm' + pp' = -1
Centre ≡ (3, 6, 1)
If one of the end of diameter is (2, 7, 3).
Let the other end of the diameter be (x, y, z).
∴ (3,6,1) = (2+x/2 , 7+y/2 , 3+z/2)
⇒ 2 + x = 6
⇒ x = 4 7 + y = 12
⇒ y = 5 and 3 + z = 2 ⇒ z = -1
Therefore, (x,y,z)≡(4,5,-1)
86. The two lines x = my + n, z = py + q and x = m'y + n', z = p'y + q' are perpendicular, to each other, if
A) mm' + pp' = 1
B) m/m' + p/p' = -1
C) m/m' + p/p' = 1
D) mm' + pp' = -1
Answer: D) mm' + pp' = -1
Explanation: Given equations of lines are x = my + n, z = py + q and x =
m'y + n',z = p'y + q'
These equations can be rewritten as x - n / m = y - 0 / 1 = z - q / p and x - n' / m' = y - 0 / 1 = z - q' / p'
These lines will be perpendicular, if mm' + 1 + pp' = 0⇒ mm' + pp' = -1
87. A tetrahedron has vertices at O (0, 0, 0), A (1,-2, 1), B (- 2, 1, 1) and C (1, - 1, 2). Then, the angle between the faces OAB and ABC will be
A) cos-1 (1/2)
B) cos-1 (-1/6)
C) cos-1 (-1/3)
D) cos-1 (1/4)
These equations can be rewritten as x - n / m = y - 0 / 1 = z - q / p and x - n' / m' = y - 0 / 1 = z - q' / p'
These lines will be perpendicular, if mm' + 1 + pp' = 0⇒ mm' + pp' = -1
87. A tetrahedron has vertices at O (0, 0, 0), A (1,-2, 1), B (- 2, 1, 1) and C (1, - 1, 2). Then, the angle between the faces OAB and ABC will be
A) cos-1 (1/2)
B) cos-1 (-1/6)
C) cos-1 (-1/3)
D) cos-1 (1/4)
Answer: C) cos-1 (-1/3)
88. If a line segment OP makes angles of π/4 and π/3 with X-axis and Y-axis, respectively. Then, the direction cosines are
A) 1/√2 , √3/2 , 1/√2
B) 1/√2 , 1/2 , 1/√2
C) 1, √3 , 1
D) 1, 1/√3 , 1
88. If a line segment OP makes angles of π/4 and π/3 with X-axis and Y-axis, respectively. Then, the direction cosines are
A) 1/√2 , √3/2 , 1/√2
B) 1/√2 , 1/2 , 1/√2
C) 1, √3 , 1
D) 1, 1/√3 , 1
Answer: B) 1/√2, 1/2, 1/√2
Explanation: Let α, β and γ be the angles made by the line segment OP
with X-axis, Y-axis and Z-axis, respectively.
Given, α = π / 4 and β = π / 3
∴ cos²Î± + cos²Î² + cos²Î³ = 1
∴ cos² Ï€ / 4 + cos² Ï€ / 3 + cos² γ = 1
⇒ (1/√2)² + (1/2)² + cos²Î³ = 1
⇒ 1/2 + 1/4 + cos²Î³ = 1
⇒ 3/4 + cos²Î³ = 1
⇒ cos²Î³ = 1/4
⇒ cosγ = 1/√2
∴ = γ = Ï€ / 4
Hence, direction cosines are cosα , cosβ cosγ i.e. 1/√2 , 1/2 , 1/√2 .
89. If p, q, rare simple propositions with truth values T, F, T, then the truth value of (∼ p v q) ∧ r ⇒ p is
A) true
B) false
C) true if r is false
D) true, if q is true
Given, α = π / 4 and β = π / 3
∴ cos²Î± + cos²Î² + cos²Î³ = 1
∴ cos² Ï€ / 4 + cos² Ï€ / 3 + cos² γ = 1
⇒ (1/√2)² + (1/2)² + cos²Î³ = 1
⇒ 1/2 + 1/4 + cos²Î³ = 1
⇒ 3/4 + cos²Î³ = 1
⇒ cos²Î³ = 1/4
⇒ cosγ = 1/√2
∴ = γ = Ï€ / 4
Hence, direction cosines are cosα , cosβ cosγ i.e. 1/√2 , 1/2 , 1/√2 .
89. If p, q, rare simple propositions with truth values T, F, T, then the truth value of (∼ p v q) ∧ r ⇒ p is
A) true
B) false
C) true if r is false
D) true, if q is true
Answer: A) true
Explanation: ∼ p ∨ q
means F ∨ F = F, ∼ r
means F ∴ [(∼ p ∨ q) ∧ ∼ r] ⇒ p
means T
90. On the interval [0, 1], the function x25 (1-x) 75 takes its maximum value at the point
A) 0
B) 1/4
C) 1/2
D) 1/3
90. On the interval [0, 1], the function x25 (1-x) 75 takes its maximum value at the point
A) 0
B) 1/4
C) 1/2
D) 1/3
Answer: B) 1/4
Explanation:
91. If |z| ≥ 3, then the least value of | z + 1/4| is
A) 11/2
B) 11/4
C) 3
D) ¼
A) 11/2
B) 11/4
C) 3
D) ¼
Answer: B) 11/4
Explanation: |z + 1/4| = |z - (-1/4)| ≥ ||z| - |-1/4|| = ||z| - 1/4| ≥ |3
- 1/4| = 11/4 Hence, |z + 1/4| ≥ 11/4
92. The normal at the point (at²₁, 2at₁) on the parabola meets the parabola again in the point (at²₂, 2at₂), then
A) t₂ = -t₁ + 2/t₁
B) t₂ = -t₁ - 2/t₁
C) t₂ = t₁ - 2/t₁
D) t₂ = t₁ + 2/t₁
92. The normal at the point (at²₁, 2at₁) on the parabola meets the parabola again in the point (at²₂, 2at₂), then
A) t₂ = -t₁ + 2/t₁
B) t₂ = -t₁ - 2/t₁
C) t₂ = t₁ - 2/t₁
D) t₂ = t₁ + 2/t₁
Answer: B) t₂ = -t₁ - 2/t₁
Explanation: Equation of the normal at point (at²₁ , 2at₁) on parabola is y = t₁x + 2at₁ + at³₁It also passes through (at²₂ , 2at₂),
then 2at₂ = - t₁(at²₂) + 2at₁ + at³₁
⇒ 2t₂ - 2t₁ = -t₁(t²₂ - t²₁)
⇒ 2(t₂ - t₁) = -t₁(t₂ + t₁)(t₂ - t₁)
⇒ 2 = -t₁(t₂ + t₁)
⇒ t₁ + t₂ = -2 / t₁
⇒ t₂ = -t₁ - 2 / t₁
93. Find
A) tan-1(1)
B) sin-1(1/2)
C) sec-1(1)
D) tan-1(1/√3)
B) sin-1(1/2)
C) sec-1(1)
D) tan-1(1/√3)
Answer: C) sec-1(1)
Explanation: ∴ cosθ = a₁b₁ + a₂b₂ +
a₃b₃ / √a²₁ + a²₂ + a²₃ √b²₁ + b²₂ + b²₃
= 1 * 2 + (-1) * (-1) + 2 * (1) / √ 1 + 1 + 4 √4 + 1 + 1 = 2 + 2 + 2 / 6 = 6/6 = 1
That means, θ = 0° or θ = 2Ï€
∴ sec2Ï€ = 1
∴ 2Ï€ = sec-1(1)
⇒ θ = sec-1(1)
94. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 3Ï€ / 2 is
A) (4√2 - 2) sq. units
B) (4√2 + 2) sq. units
C) (4√2 - 1) sq. units
D) (4√2 + 1) sq. units
= 1 * 2 + (-1) * (-1) + 2 * (1) / √ 1 + 1 + 4 √4 + 1 + 1 = 2 + 2 + 2 / 6 = 6/6 = 1
That means, θ = 0° or θ = 2Ï€
∴ sec2Ï€ = 1
∴ 2Ï€ = sec-1(1)
⇒ θ = sec-1(1)
94. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 3Ï€ / 2 is
A) (4√2 - 2) sq. units
B) (4√2 + 2) sq. units
C) (4√2 - 1) sq. units
D) (4√2 + 1) sq. units
Answer: A) (4√2 - 2) sq. units
Explanation:
95. If a, b and c are three non-coplanar vectors, then (a
+ b - c) ⋅ [(a - b * [b - c)] equals
A) 0
B) a ⋅ b * c
C) a ⋅ c * b
D) 3a ⋅ b * c
Answer: B) a ⋅ b
* c
Explanation: (a + b - c) ⋅ [(a
- b) * (b - c)] = (a + b - c) ⋅ [a
* b - a * c - b * b + b * c] = a ⋅ a
* b - a ⋅ a * c + a ⋅ b
* c + b ⋅ a * b - b ⋅ a
* c + b ⋅ b * c - c ⋅ a
* b + c ⋅ a * c - c ⋅ b
* c = a ⋅ b * c - b ⋅ a
* c - c ⋅ a * b = [a b c] - [b a c] = [c a b] = [a b c] + [a b
c] - [a b c] = [a b c] = a ⋅ b
* c
96. If there is an error of m% in measuring the edge of the cube, then the per cent error in estimating its surface area is
A) 2 m
B) 3 m
C) 1 m
D) 4 m
96. If there is an error of m% in measuring the edge of the cube, then the per cent error in estimating its surface area is
A) 2 m
B) 3 m
C) 1 m
D) 4 m
Answer: A) 2 m
Explanation: We know that the surface area A of a cube of side x is given
by A = 6x².
On differentiating w.r.t. x, we get dA / dx = 12x
Let the change in x be Δx = m% of x = mx / 100
Now, change in surface area, ΔA = (dA/dx) Δx = (12x) Δx = 12x * (mx/100) = 12x²m / 100
∴ the approximate change in surface area = 2m / 100 * 6 x² = 2m% of original surface area
97. If the rectangular hyperbola is x² - y² = 64. Then, which of the following is not correct?
A) The length of latus rectum is 16
B) The eccentricity is √2
C) The asymptotes are parallel to each other
D) The directrices are x = ±4√2
On differentiating w.r.t. x, we get dA / dx = 12x
Let the change in x be Δx = m% of x = mx / 100
Now, change in surface area, ΔA = (dA/dx) Δx = (12x) Δx = 12x * (mx/100) = 12x²m / 100
∴ the approximate change in surface area = 2m / 100 * 6 x² = 2m% of original surface area
97. If the rectangular hyperbola is x² - y² = 64. Then, which of the following is not correct?
A) The length of latus rectum is 16
B) The eccentricity is √2
C) The asymptotes are parallel to each other
D) The directrices are x = ±4√2
Answer: C) The asymptotes are parallel to each other
Explanation: Given equation of rectangular hyperbola is x² - y² = 8²
the length of latusrectum = 2 * (8) = 16 and eccentricity = √2
the asymptotes are perpendicular lines. I.e. x ± y = 0
Now, directrices are x = ± 8 / √2 = ± 4 √2
98. The equation of tangents to the hyperbola 3x² - 2y² = 6, which is perpendicular to the line x - 3y = 3, are
A) y = -3x ± √15
B) y = 3x ± √6
C) y = -3x ± √6
D) y = 2x ± √15
the length of latusrectum = 2 * (8) = 16 and eccentricity = √2
the asymptotes are perpendicular lines. I.e. x ± y = 0
Now, directrices are x = ± 8 / √2 = ± 4 √2
98. The equation of tangents to the hyperbola 3x² - 2y² = 6, which is perpendicular to the line x - 3y = 3, are
A) y = -3x ± √15
B) y = 3x ± √6
C) y = -3x ± √6
D) y = 2x ± √15
Answer: A) y = -3x ± √15
Explanation: Given, equation of hyperbola is 3x² - 2y² =
6 ⇒ x² / 2 - y² / 3 = 1on comparing with x² / a² - y² / b² = 1,
we get a² = 2 and b² = 3
given, equation of line is x - 3y = 3.
∴ Slope of given line = 1/3 Slope of line perpendicular to given line, m = -3
Now, the equation of tangents are y = mx ± √a² m² - b² = -3x ± √2 * 9 - 3 = -3x± √18 - 3 = -3x± √15
99. Find
A) 1
B) 0
C) -2
D) -1
B) 0
C) -2
D) -1
Answer: D) -1
Explanation:
100. The area of the region bounded by the curves x² + y² =
9 and x + y = 3 is
A) 9Ï€/4 + 1/2
B) 9Ï€/4 - 1/2
C) 9(Ï€/4 - 1/2)
D) 9(Ï€/4 + 1/2)
Answer: C) 9(Ï€/4 - 1/2)
Explanation:
101. For any three vectors a,b and c , [ a + b , b + c , c
+ a] is
A) [ a b c ]
B) 3[ a b c]
C) 2[ a b c]
D) 0
A) [ a b c ]
B) 3[ a b c]
C) 2[ a b c]
D) 0
Answer: C) 2[a b c]
Explanation: [a + b , b + c , c + a] = (a + b) ⋅ [(b + c) * (c + a)]
= (a + b)⋅ [b * c + b * a + c * c + c * a] = (a + b) ⋅ (b * c + b * a + c * a)[∴ c * c = 0]
= a⋅ (b * c) + a ⋅ (b * a) + a ⋅ (c * a) + b ⋅ (b * c) + b ⋅ (b * a) + b ⋅ (c * a)
= a⋅ (b * c) + b ⋅ (c * a)
= [a b c] + [b c a] = [a b c] + [a b c] = 2[a b c]
102. ∫Ï€/2 0 sin 2x ⋅ log tan x dx is equal to
A) 0
B) 2
C) 4
D) 7
= (a + b)⋅ [b * c + b * a + c * c + c * a] = (a + b) ⋅ (b * c + b * a + c * a)[∴ c * c = 0]
= a⋅ (b * c) + a ⋅ (b * a) + a ⋅ (c * a) + b ⋅ (b * c) + b ⋅ (b * a) + b ⋅ (c * a)
= a⋅ (b * c) + b ⋅ (c * a)
= [a b c] + [b c a] = [a b c] + [a b c] = 2[a b c]
102. ∫Ï€/2 0 sin 2x ⋅ log tan x dx is equal to
A) 0
B) 2
C) 4
D) 7
Answer: A) 0
Explanation:
103. If the mean and variance of a binomial distribution
are 4 and 2, respectively. Then, the probability of at least 7 successes is
A) 3/214
B) 4/173
C) 9/256
D) 7/231
Answer: C) 9/256
Explanation: Given, mean = 4 and variance = 2 ⇒ np = 4 and npq = 2∴ npq / np = 2/4 ⇒ q = 1/2
Then, p = 1 - q = 1 - 1/2 = 1/2 Mean = np = 4
⇒ n * 1/2 = 4 ⇒ n = 8
∴ P(X = r) = nCr pr qn-r = 8Cr(1/2)8 [∴ p = q = 1/2]
The required probability of at least 7 successes is P(X ≥ 7) = P(X = 7) + P(X = 8) = (8C7 + 8C8)(1/2)8 = (8!/7!1! + 8! /8! 0!)(1/2)8 = (8 + 1) (1/2)8 = 9 / 256
104. The shortest distance between the lines
A) 234/7 units
B) 288/21 units
C) 221/3 units
D) 234/21 units
B) 288/21 units
C) 221/3 units
D) 234/21 units
Answer: B) 288/21 units
105. If the plane passing through the point (2, 2, 1) and is perpendicular to the planes 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0. Then, the equation of the plane is
A) 2x - y - z = 0
B) 2x + 3y + z - 1 = 0
C) 2x + y + z + 3 = 0
D) x - y + z - 1 = 0
105. If the plane passing through the point (2, 2, 1) and is perpendicular to the planes 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0. Then, the equation of the plane is
A) 2x - y - z = 0
B) 2x + 3y + z - 1 = 0
C) 2x + y + z + 3 = 0
D) x - y + z - 1 = 0
Answer: A) 2x - y - z = 0
Explanation: Equation of plane passes through (2,2,1) is a(x-2) + b(y-2)
+ c(z-1) = 0 ...(i)
Since, above plane is perpendicular to 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0
∴ 3a + 2b + 4c = 0 ...(ii) and
2a + b + 3c = 0 ...(iii) [∴ for perpendicular, a₁a₂ + b₁b₂ + c₁c₂ = 0]
On multiplying Eq. (iii) by 2, we get 4a + 2b + 6c = 0 ...(iv)
On subtracting Eq. (iv) from Eq. (ii), we get -a - 2c = 0⇒ a = -2c ⇒ c = -a / 2
On putting c = -a / 2 in Eq.(iii), we get 2a + b - 3a / 2 = 0⇒ 4a + 2b - 3a = 0 ⇒ 2b = -a ⇒ b = -a / 2
On putting b = -a / 2 and c = -a / 2 in Eq.(i), we get a(x-2)-a/2(y-2)-a/2(z-1) = 0⇒ a/2[2(x-2)-(y-2)-(z-1)] = 0 ⇒ 2x - 4 - y + 2 - z + 1 = 0 ⇒ 2x - y - z - 1 = 0
106. From a city population, the probability of selecting a male or smoker is 7/10, a male smoker is 2/5 and a male, if a smoker is already selected, is 2/3. Then, the probability of
A) selecting a male is 3/2
B) selecting a smoker is 1/5
C) selecting a non-smoker is 2/5
D) selecting a smoker, if a male is first selected, is given by 8/5
Since, above plane is perpendicular to 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0
∴ 3a + 2b + 4c = 0 ...(ii) and
2a + b + 3c = 0 ...(iii) [∴ for perpendicular, a₁a₂ + b₁b₂ + c₁c₂ = 0]
On multiplying Eq. (iii) by 2, we get 4a + 2b + 6c = 0 ...(iv)
On subtracting Eq. (iv) from Eq. (ii), we get -a - 2c = 0⇒ a = -2c ⇒ c = -a / 2
On putting c = -a / 2 in Eq.(iii), we get 2a + b - 3a / 2 = 0⇒ 4a + 2b - 3a = 0 ⇒ 2b = -a ⇒ b = -a / 2
On putting b = -a / 2 and c = -a / 2 in Eq.(i), we get a(x-2)-a/2(y-2)-a/2(z-1) = 0⇒ a/2[2(x-2)-(y-2)-(z-1)] = 0 ⇒ 2x - 4 - y + 2 - z + 1 = 0 ⇒ 2x - y - z - 1 = 0
106. From a city population, the probability of selecting a male or smoker is 7/10, a male smoker is 2/5 and a male, if a smoker is already selected, is 2/3. Then, the probability of
A) selecting a male is 3/2
B) selecting a smoker is 1/5
C) selecting a non-smoker is 2/5
D) selecting a smoker, if a male is first selected, is given by 8/5
Answer: C) selecting a non-smoker is 2/5
Explanation:
107. At t = 0, the function Æ’ (t) = sin t / t has
A) a minimum
B) a discontinuity
C) a point of inflexion
D) a maximum
A) a minimum
B) a discontinuity
C) a point of inflexion
D) a maximum
Answer: D) a maximum
108. Using Rolle's theorem, the equation a0xn + a₁xn-1 + ...+ an = 0 has at least one root between 0 and 1 , if
A) a0 / n + a₁ / n - 1 + ... + an-1 = 0
B) a0 / n -1 + a₁ / n - 2 + ... + an-2 = 0
C) na0 + (n - 1)a₁ + ... + an-1 = 0
D) a0 / n + 1 + a₁ / n + ...+ an = 0
108. Using Rolle's theorem, the equation a0xn + a₁xn-1 + ...+ an = 0 has at least one root between 0 and 1 , if
A) a0 / n + a₁ / n - 1 + ... + an-1 = 0
B) a0 / n -1 + a₁ / n - 2 + ... + an-2 = 0
C) na0 + (n - 1)a₁ + ... + an-1 = 0
D) a0 / n + 1 + a₁ / n + ...+ an = 0
Answer: D) a0 / n + 1 + a₁ /
n + ...+ an = 0
Explanation: Consider the function f defined by Æ’(x) = a0 xn+1 / n + 1 +
an xn / n +.... + an-1 x²/2 + anx
Since, Æ’(x) is a polynomial, so it is continous and differentiable for all x.
Consequently, Æ’(x) is continuous in the closed interval [0,1] and differentiable in the open interval(0,1).
Also, Æ’ (0) = 0 and Æ’ (1) = a0 / n+1 + a₁ / n + ... + an-1 / 2 + an = 0 [say]
i.e. Æ’ (0) = Æ’ (1) Thus, all the three conditions of Rolle's Theorem are satisfied.
Hence, there is at least one value of x in the open interval (0, 1), where Æ’'(x) = 0 i.e. a0xn + a₁xn-1 +.... +an = 0
109. Which of the following inequality is true for x > 0?
A) Log (1+x) < x/1+x < x
B) x / 1+x < x < log (1+x)
C) x < log (1+x) < x/1+x
D) x / 1+x < log (1+x) < x
Since, Æ’(x) is a polynomial, so it is continous and differentiable for all x.
Consequently, Æ’(x) is continuous in the closed interval [0,1] and differentiable in the open interval(0,1).
Also, Æ’ (0) = 0 and Æ’ (1) = a0 / n+1 + a₁ / n + ... + an-1 / 2 + an = 0 [say]
i.e. Æ’ (0) = Æ’ (1) Thus, all the three conditions of Rolle's Theorem are satisfied.
Hence, there is at least one value of x in the open interval (0, 1), where Æ’'(x) = 0 i.e. a0xn + a₁xn-1 +.... +an = 0
109. Which of the following inequality is true for x > 0?
A) Log (1+x) < x/1+x < x
B) x / 1+x < x < log (1+x)
C) x < log (1+x) < x/1+x
D) x / 1+x < log (1+x) < x
Answer: D) x / 1+x < log (1+x) < x
Explanation: Let Æ’(x) = log(1+x) - x / 1+x
∴ Æ’'(x) = 1 / 1 + x - (1 + x) ⋅ 1 - x ⋅ 1 / (1+ x)² = 1 / 1 + x - 1 / (1 + x)² = 1 + x - 1 / (1 + x)² = x / (1 + x)² which is positive. [∴ x > 0]
∴ Æ’(x) is monotonic increasing, when x > 0. ⇒ Æ’(x) > Æ’(0)
Now, Æ’(0) = log 1 - 0 = 0
∴ Æ’(x) > 0 ⇒ log(1 + x) - x / 1 + x > 0 ⇒ x / 1 + x < log(1 + x) ...(i)
Also, for x > 0 , x² > 0⇒ x² + x > x ⇒ x(x+1) > x ⇒ x > x / x+1 ...(ii)
From Eqs. (i) And (ii), we get x / (x+1) < log (1 + x) < x [∴ log (1 + x) < x for x > 0]
110. The solution of d²x / dy² - x = k, where k is a non-zero constant, vanishes when y = 0 and tends of finite limit as y tends to infinity, is
A) x = k (1 + e-y)
B) x = k (ey + e-y -2)
C) x = k (e-y -1)
D) x = k (ey -1)
∴ Æ’'(x) = 1 / 1 + x - (1 + x) ⋅ 1 - x ⋅ 1 / (1+ x)² = 1 / 1 + x - 1 / (1 + x)² = 1 + x - 1 / (1 + x)² = x / (1 + x)² which is positive. [∴ x > 0]
∴ Æ’(x) is monotonic increasing, when x > 0. ⇒ Æ’(x) > Æ’(0)
Now, Æ’(0) = log 1 - 0 = 0
∴ Æ’(x) > 0 ⇒ log(1 + x) - x / 1 + x > 0 ⇒ x / 1 + x < log(1 + x) ...(i)
Also, for x > 0 , x² > 0⇒ x² + x > x ⇒ x(x+1) > x ⇒ x > x / x+1 ...(ii)
From Eqs. (i) And (ii), we get x / (x+1) < log (1 + x) < x [∴ log (1 + x) < x for x > 0]
110. The solution of d²x / dy² - x = k, where k is a non-zero constant, vanishes when y = 0 and tends of finite limit as y tends to infinity, is
A) x = k (1 + e-y)
B) x = k (ey + e-y -2)
C) x = k (e-y -1)
D) x = k (ey -1)
Answer: C) x = k (e-y -1)
Explanation: Rewriting given differential equation as, (D² - 1) x = k ...
(i) where, D = d / Dy
Its auxiliary equation is m² - 1 = 0, so that m = 1, -1
Hence, CF = C₁ey + C₂e-y where, C₁, C₂ are arbitrary constants.
Now, PI = 1 / D² - 1 k = k⋅ 1 / D² - 1 e0⋅y = k ⋅ 1 / 0² - 1 e0⋅y = - k
So, solution of Eq.(i) is x = C₁ey + C₂e-y - k ...(ii)
Given that x = 0, when y = 0
Hence, Eq.(ii) gives 0 = C₁ + C₂ - k ⇒ C₁ + C₂ = k ...(iii)
Multiplying both sides of Eq. (ii) by e-y ,
we get x⋅ e-y = C₁ + C₂e-2y - ke-y ...(iv)
Given that x → m when y → ∞ , m being a finite quantity.
So, Eq. (iv) becomes x * 0 = C₁ + C₂ * 0 - (k * 0) ⇒ C₁ = 0
From Eqs. (iv) And (v), we get C₁ = 0 and C₂ = k
Hence, Eq. (ii) becomes x = ke-y - k = k (e-y - 1)
111. The differential equation (3x + 4y + 1)dx + (4x + 5y + 1)dy = 0 represents a family of
A) circles
B) parabolas
C) ellipses
D) hyperbolas
Its auxiliary equation is m² - 1 = 0, so that m = 1, -1
Hence, CF = C₁ey + C₂e-y where, C₁, C₂ are arbitrary constants.
Now, PI = 1 / D² - 1 k = k⋅ 1 / D² - 1 e0⋅y = k ⋅ 1 / 0² - 1 e0⋅y = - k
So, solution of Eq.(i) is x = C₁ey + C₂e-y - k ...(ii)
Given that x = 0, when y = 0
Hence, Eq.(ii) gives 0 = C₁ + C₂ - k ⇒ C₁ + C₂ = k ...(iii)
Multiplying both sides of Eq. (ii) by e-y ,
we get x⋅ e-y = C₁ + C₂e-2y - ke-y ...(iv)
Given that x → m when y → ∞ , m being a finite quantity.
So, Eq. (iv) becomes x * 0 = C₁ + C₂ * 0 - (k * 0) ⇒ C₁ = 0
From Eqs. (iv) And (v), we get C₁ = 0 and C₂ = k
Hence, Eq. (ii) becomes x = ke-y - k = k (e-y - 1)
111. The differential equation (3x + 4y + 1)dx + (4x + 5y + 1)dy = 0 represents a family of
A) circles
B) parabolas
C) ellipses
D) hyperbolas
Answer: D) hyperbolas
Explanation: The given differential equation is (3x + 4y + 1)
dx + (4x + 5y + 1) dy = 0 ...(i)Comparing Eq. (i) with Mdx + Ndy = 0, we get M = 3x + 4y + 1 and N = 4x + 5y + 1
Here, ∂M / ∂y = ∂N / ∂x = 4
Hence, Eq.(i) is exact and solution is given by ∫(3x + 4y +1) dx + ∫(5y + 1) dy = C
⇒ 3x² / 2 + 4xy + x + 5y² / 2 + y - C = 0
⇒ 3x² + 8xy + 2x + 5y² + 2y - 2C = 0
⇒ 3x² + 2 ⋅ 4xy + 2x + 5y² + 2y + C' = 0 ...(ii) where, C' = -2C
On comparing Eq.(ii) with standard form of conic section ax² + 2hxy + by² + 2gx + 2fy + C = 0
We get, a = 3, h = 4, b = 5 Here, h² - ab = 16 - 15 = 1 > 0
Hence, the solution of differential equation represents family of hyperbolas.
112. Find
A) A
B) B
C) C
D) D
B) B
C) C
D) D
Answer: B) B
Explanation:
113. If A, B, C are three events associated with the random
experiment, then P(A)P(B/A)P(C/A)∩B) is
A) P(A∪B∪C)
B) P(A∩B∩C)
C) P(C/A)∩B)
D) P(B/A)
A) P(A∪B∪C)
B) P(A∩B∩C)
C) P(C/A)∩B)
D) P(B/A)
Answer: B) P (A∩B∩C)
Explanation: P(A)P(B/A)P(C/A) ∩ B) = P(A ∩ B)P(C/A)∩ B) = P(A
∩ B)P[C ∩ (A ∩ B)] / P(A ∩ B) = P(A ∩ B ∩ C)114. Find
A) 4
B) 1
C) 2
D) 3
B) 1
C) 2
D) 3
Answer: D) 3
Explanation:
115. The probability of at least one double-six
being thrown in n throws with two ordinary dice is greater than 99%. Then, the
least numerical value of n is
A) 100
B) 164
C) 170
D) 184
A) 100
B) 164
C) 170
D) 184
Answer: B) 164
Explanation: The probability of getting a double six in one
throw with two dice = 1/6 * 1/6 = 1/36
∴ p = 1/36 , q = 1 - p = 1 - 1/36 = 35/36
Now, (p + q)m = qn + nC₁qn-1p + nC₂qn-2p² + ... + nCrqn-rpr + ... + pn
The probability of getting at least one double six in n throws with two dice = (q + p)n - qn = 1 - qn = 1 - (35/36)n
∴ 1 - (35 / 36)n > 0.99
⇒ (35 / 36)n < 0.01
⇒ n(log 35 - log 36) < log 0.01
⇒ n[15441 - 15563] < -2
⇒ - 0.0122n < - 2
⇒ 0.0122n > 2
⇒ n > 2 / 0.0122
⇒ n > 163.9
So, the least value of n is 164.
116. Find the value of k for which the simultaneous equations x + y + z = 3; x + 2y + 3z = 4 and x + 4y + kz = 6 will not have a unique solution.
A) 0
B) 5
C) 6
D) 7
∴ p = 1/36 , q = 1 - p = 1 - 1/36 = 35/36
Now, (p + q)m = qn + nC₁qn-1p + nC₂qn-2p² + ... + nCrqn-rpr + ... + pn
The probability of getting at least one double six in n throws with two dice = (q + p)n - qn = 1 - qn = 1 - (35/36)n
∴ 1 - (35 / 36)n > 0.99
⇒ (35 / 36)n < 0.01
⇒ n(log 35 - log 36) < log 0.01
⇒ n[15441 - 15563] < -2
⇒ - 0.0122n < - 2
⇒ 0.0122n > 2
⇒ n > 2 / 0.0122
⇒ n > 163.9
So, the least value of n is 164.
116. Find the value of k for which the simultaneous equations x + y + z = 3; x + 2y + 3z = 4 and x + 4y + kz = 6 will not have a unique solution.
A) 0
B) 5
C) 6
D) 7
Answer: D) 7
Explanation:
117. If the complex number z lies on a circle
with centre at the origin and radius 1/4, then the complex number -1 + 8z lies
on a circle with radius
A) 4
B) 1
C) 3
D) 2
A) 4
B) 1
C) 3
D) 2
Answer: D) 2
Explanation: Given, |z| = 1/4 Let z' = -1 + 8z ⇒ z = z' + 1 / 8 ⇒ |z| = |z' + 1| / 8 ⇒ 1/4 = |z' + 1| / 8 ⇒ |z' + 1| = 2 Hence, z' lies on a circle
with center (-1, 0) and radius 2.
118. If line y= 2x + c is a normal to the ellipse x² / 9 + y² / 16 = 1, then
A) c = 2/3
B) c = √73/5
C) c = 14/√73
D) c = √5/7
118. If line y= 2x + c is a normal to the ellipse x² / 9 + y² / 16 = 1, then
A) c = 2/3
B) c = √73/5
C) c = 14/√73
D) c = √5/7
Answer: C) c = 14/√73
Explanation: If the line y = mx + c is a
normal to the ellipse x² / a² + y² / b² = 1,then c² = m²(a² - b²)² / a² + b²m²
[Hence m = 2, a² = 9 and b² = 16] = (2)²(9 - 16)² / 9 + 16 * (2)² = 4 * 49 / 9 + 64 = 4 * 49 / 73 = 196 / 73
∴ c = 14 / √73
119. Find
A) 13
B) 12
C) 9
D) 14
B) 12
C) 9
D) 14
Answer: B) 12
120. If p: It rains today, q: I go to school, r: I shall meet any friends and s: I shall go for a movie, then which of the following is the proportion? If it does not rain or if I do not go to school, then I shall meet my friend and go for a movie.
A) (∼ p ∧ ∼ q) ⇒ (r ∧ s)
B) ∼ (p ∧ q) ⇒ (r ∧ s)
C) ∼ (p ∨ q) ⇒ (r ∨ s)
D) None of these
120. If p: It rains today, q: I go to school, r: I shall meet any friends and s: I shall go for a movie, then which of the following is the proportion? If it does not rain or if I do not go to school, then I shall meet my friend and go for a movie.
A) (∼ p ∧ ∼ q) ⇒ (r ∧ s)
B) ∼ (p ∧ q) ⇒ (r ∧ s)
C) ∼ (p ∨ q) ⇒ (r ∨ s)
D) None of these
Answer: B) ∼ (p ∧ q) ⇒ (r ∧ s)
Explanation: Correct result is (∼ p ∨ ∼ q) ⇒ (r ∧ s) or ∼(p ∧ q) ⇒ r ∧ s
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