VITEEE Exam 2014 - Question paper with solutions
Download VITEEE EXAM 2014 previous year mathematics question paper with solutions. VIT conducts VITEEE Exam to get admission for BTech programmes. Every year more than one lakh of student are writing VITEEE Exam to get admission in VIT University. Student need to pass the VITEEE exam for admission in VIT
Hello guys, Welcome again! This is the last post of VITEEE EXAM 2014 question paper with solutions. My previous post is having 2 part of VITEEE EXAM 2014 question paper with solutions and this is the last part. So, in this post, I will finish the remaining 40 questions of VITEEE EXAM 2014 question paper with solutions.
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To get admission in Vellore Institute of Technology(VIT) for all the campuses (Vellore, Chennai, Bhopal, Amaravati), students need to clear the VITEEE examination, that's one in each of the toughest exams to enter into one in each of the foremost effective engineering university in associate country, VITEEE test 2014 previous year question papers with solutions will facilitate students to rearrange for the examination.
There will be 40 questions out of 120 questions. Correct answer and explanation for each question are available after each question. Download links for this question paper will be available at the end of the page.
VITEEE Exam 2014 Questions with Solution (Part 3)
81. If the vertices of a triangle are A(0,4,1), B(2,3,-1) and C(4,5,0), then the orthocentre of ΔABC, is
A) (4,5,0)
Answer: B) (2,3,-1)
A) (4,5,0)
B) (2,3,-1)
C) (-2,3,-1)
D) (2,0,2)
Answer: B) (2,3,-1)
Explanation: Given, vertices of ΔABC are A(0, 4, 1), B(2, 3, -1) and C(4, 5, 0).
Now , AB = √(2-0)² + (3-4)² + (-1-1)² = √4 + 1 + 4 = 3 BC = √(4-2)² + (5-3)² + (0+1)² = √4 + 4 + 1 = 3 and CA = √(4-0)² + (5-4)² +(0-1)² = √16 + 1 + 1 = 3√2
∴ AB² + BC² = AC² ∴ ΔABC is a right angled triangle.
We know that the orthocentre of a right-angled triangle is the vertex containing 90° angle.
∴ Orthocentre is point B(2,3,-1).
∴ AB² + BC² = AC² ∴ ΔABC is a right angled triangle.
We know that the orthocentre of a right-angled triangle is the vertex containing 90° angle.
∴ Orthocentre is point B(2,3,-1).
82. The equation of normal to the curve y = (1 + x)y + sin-1 (sin² x) at x = 0 is
A) x + y = 1
B) x - y = 1
C) x + y = -1
D) x - y = -1
Answer: A) x + y = 1
Explanation: Given curve is y = (1 + x)y + sin-1(sin²x)
On differentiating w.r.t. x, we get dy/dx = (1+x)y [y/1+x + 1log(1+x)dy/dx] + 2sin x cos x / √1 - sin4 x ⇒ (dy / dx) at (0,1) = + 1 [∴ at x = 0 , y = 1] Slope of normal at (x=0) = -1
∴ Equation of normal at x = 0 and y = 1 is y - 1 = -1 (x - 0) ⇒ y - 1 = -x ⇒ x + y = 1
83. The value of c from the Lagrange's mean value theorem for which
A) 5
B) 1
C) √15
D) None of these
Answer: C) √15
Explanation: It is clear that Æ’(x) has a definite and unique value for each x ∈ [1,5].
Thus, for every point in the interval [1,5], the value of Æ’(x) exists. So, Æ’(x) is continuous in the interval [1,5].
Also, Æ’'(x) = -x / √25 - x² , which clearly exists for all x in an open interval(1,5).
Hence , Æ’'(x) is differentiable in (1,5).
So, there must be a value c∈ [1,5] such that Æ’' (c) = Æ’(5) - Æ’(1) / 5 - 1 = 0 - √24 / 4 = 0 - 2√6 / 4 = -√6 / 2
But Æ’'(c) = -c / √25 - c²∴ -c / √25 - c² = -(√6 / 2) ⇒ 4c² = 6(25 - c²) ⇒ 4c² = 150 - 6c² ⇒ 10c² = 150 ⇒ c² = 15 ⇒ c = ± √15
∴ c = √15 ∈ [1,5]
84. Find
A) A
B) |A|
C) |A| ⋅ l
D) None of these
Answer: C) |A| ⋅ l
Explanation:
85. If there is an error of k% in measuring the edge of a cube, then the per cent error in estimating its volume is
A) k
B) 3k
C) k/3
D) None of these
Answer: B) 3k
Explanation: We know that, the volume V of a cube of side x is given by V = x³ On differentiating w.r.t. x, we get ⇒ dV / dx = 3x²
Let the change in x be Δx = k% of x = kx / 100 Now, the change in volume, ΔV = (dV/dx)Δx = 3x²(Δx) = 3x²(kx / 100) = 3x³ ⋅ k / 100 ∴ The approximate change in volume = 3kx³ / 100 = 3k / 100 ⋅ x³ = 3k% of original volume
86. If the system of equations x + ky - z = 0 , 3x - ky - z = 0 and x - 3y + z = 0, has non-zero solution, then k is equal to
A) -1
B) 0
C) 1
D) 2
Answer: C) 1
Explanation:
87. If the points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x + 3y - 2z = k, then
A) k < 1
B) k > 2
C) k < 1 or k > 2
D) 1 < k < 2
Answer:D) 1 < k < 2
Explanation: Since, the points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x + 3y - 2z - k = 0 ( 2+ 6 - 6 - k) (4 - 3 - k) < 0 ⇒ (k - 1)(k - 2) < 0 ...(i) ∴ 1 < k < 2
88. Find
A) 1/4
B) 1/2
C) 0
D) -(1/4)
Answer: D) -(1/4)
Explanation:
89. Let Æ’'(x), be differentiable ∀x. If Æ’(1) = - 2 and Æ’'(x) ≥ 2 ∀ x ∈[1,6], then
A) Æ’(6) < 8
B) Æ’(6) ≥ 8
C) Æ’(6) ≥ 5
D) Æ’(6) ≤ 5
Answer: B) Æ’(6) ≥ 8
Explanation: Since, Æ’'(x) is differentiable ∀ x ∈ [1,6]
∴ By Langrange's mean value theorem, Æ’(x) = Æ’(6) - Æ’(1) / 6 - 1 ∴ Æ’'(x) ≥ 2 ∀ x ∈ [1,6] (given) ⇒
∴ (6) + 2 / 5 ≥ 2 [∴ Æ’(1) = -2] ⇒ Æ’(6) ≥ 10 - 2 ⇒ Æ’(6) ≥ 8
90. Find
A) 1
B) 0
C) 2
D) None of these
Answer: B) 0
Explanation:
91. Two lines (x - 1) / 2 = (y + 1) / 3 = (z - 1) / 4 and (x - 3) / 1 = (y - k) / 2 = z intersect at a point, if k is equal to
A) 2/9
B) 1/2
C) 9/2
D) 1/6
Answer: C) 9/2
Explanation: x -1 / 2 = y + 1 / 3 = z - 1 / 4 = r (say) ⇒ x = 2r + 1, y = 3r - 1, z = 4r + 1
Since, the two lines intersect. So, putting above values in second line, we get
2r + 1 - 3 / 1 = 3r - 1 - k / 2 = 4r + 1 / 1
Taking 1st and 3rd terms, we get 2r - 2 = 4r + 1⇒ r = -3/2
Also, taking 2nd and 3rd terms, we get 3r - 1 - k = 8r + 2⇒ k = -5r - 3 = 15 / 2 - 3 = 9 / 2
92. The minimum value of x / log x is
A) e
B) 1/e
C) e²
D) e³
Answer: A) e
Explanation: Let Æ’(x) = x / log x On differentiating w.r.t. x,
we get Æ’'(x) = log x- 1 / (log x)² For maxima and minima, put Æ’'(x) = 0. log x - 1 = 0⇒ x = e
Now, Æ’"(x) = (log x)²⋅ 1/x -(logx - 1) ⋅ 2log x / x / (log x)4 ⇒ Æ’"(e) = 1/e - 0 / 1 = 1 / e > 0
∴ Æ’(x) is minimum at x = e.
Hence, minimum value of Æ’(x) at x = e is Æ’(e) = e / log e = e
93. The triangle formed by the tangent to the curve Æ’(x) = x² + bx - b at the point (1,1) and the coordinate axes lies in the first quadrant. If its area is 2, then the value of b is
A) -1
B) 3
C) -3
D) 1
Answer: C) -3
Explanation:
94. The statement (p ⇒ q) ⇔ (∼ p ∧ q) is a
A) tautology
B) contradiction
C) Neither (A) or (B)
D) None of these
Answer: C) Neither (A) or (B)
95. If x + iy = 3 /( 2 + cosθ + isinθ) , then x² + y² is equal to
A) 3x - 4
B) 4x - 3
C) 4x + 3
D) None of these
Answer: B) 4x - 3
96. The negation of (∼ p ∧ q) ∨ (p ∧ ∼ q) is
A) (p∨ ∼q) ∨ (∼ p∨q)
B) (p∨ ∼q) ∧ (∼ p∨q)
C) (p∧ ∼q) ∧ (∼p∨q)
D) (p∧ ∼q) ∧ (p∨ ∼q)
Answer: B) (p∨ ∼q) ∧ (∼ p∨q)
Explanation: Let S : (∼ p ∧ q) ∨ (p ∧ ∼ q) ⇒ ∼
S : ∼ [(∼ p ∧ q) ∨ (p ∧ ∼ q)] ⇒ ∼ S : ∼ (∼ p ∧ q) ∧ ∼ (p ∧ ∼ q) ⇒ ∼ S :(p ∨ ∼ q) ∧ (∼ p ∨ q)
97. The normals at three points P, Q and R of the parabola y² = 4αx meet at (h,k). The centroid of the ΔPQR lies on
A) x = 0
B) y = 0
C) x = -a
D) y = a
Answer: B) y = 0
Explanation: We knew that the sum of ordinates of feet of normals drawn from a point to the parabola, y² = 4ax is always zero.
Now, as normals at three points P, Q and R of parabola y² = 4ax meet at (h,k). ⇒ The normals from (h,k) to y² = 4ax meet the parabola at P,Q and R.
⇒ y-coordinates y₁, y₂, Y₃ of these points P, Q and R will be zero.
⇒ y - coordinate of the centroid pf ΔPQR i.e., y₁ + y₂ + y₃ / 3 = 0 / 3 = 0
Hence , centroid lies on y = 0.
98. The minimum area of the triangle formed by any target to the ellipse ( x² / a²) + (y² / b²) = 1 with the coordinate axes is
A) a² + b²
B) (a + b)² / 2
C) ab
D) (a - b)² / 2
Answer: C) ab
Explanation:
99. If the line lx + my - n = 0 will be a normal to the hyperbola, then (a² / l²) - (b² / m²) = (a² + b²)² / k , where k is equal to
A) n
B) n²
C) n³
D) none of these
Answer: B) n²
Explanation: The equation of any normal to x² / a² - y² / b² = 1 is axcos∅ + bycot∅ = a² + b² ⇒ axcos∅ + bycot∅ - (a² + b²) = 0 ...(i)
The straight line lx + my - n = 0 will be normal to the hyperbola x² / a² - y² / b² = 1, then Eq. (i) and lx + my - n = 0 represent the same line. ∴ acos∅ / l = bcot∅ / m = a² + b² / n ⇒ sec∅ = na / l(a² + b²) and tan∅ = nb / m(a² + b²)
∴ sec² ∅ - tan² ∅ = 1
∴ n²a² / l²(a² + b²)² - n²b² / m²(a² + b²)² = 1 ⇒ a² / l² - b² / m² = (a² + b²)² / n²
But given equation of normal is a² / l² - b² / m² = (a² + b²)² / k
∴ k = n²
100. If cosα + isinα, b = cosβ + isinβ, c= cos γ + isin γ and b / c + c / a + a / b = 1 , then cos(β - γ) + cos(γ - α) + cos(α - β) is equal to
A) 3 / 2
B) -(3 / 2)
C) 0
D) 1
Answer: D) 1
Explanation: We have, a = cos α + i sinα b = cosβ + i sinβ and c = cosγ + i sinγ
Now, b / c = cosβ + isinβ / cosγ + i sinγ * cos γ - i sinγ / cosγ - i sinγ = cos β ⋅ cos γ + sinβ ⋅ sinγ + i [ sinβ ⋅ cosγ - sinγ ⋅ cosβ] ⇒ b / c = cos(β - γ) + i sin(β - γ) ...(i) Similarly, c / a = cos(γ - α) + i sin(γ - α) ...
(ii) and a / b = cos(α - β) + i sin(α - β) ...
(iii) On adding Eqs. (i),(ii), and (iii),
we get cos(β - α) + cos(γ - α) + cos(α - β) + i[sin(β - γ) + sin(γ - α) + sin(α - β)] = 1 [∴ b / c + c / a + a/ b = 1 given]
On equating real parts, we get cos(β - γ) + cos(γ - α) + cos(α - β) = 1
101. If |z + 4| ≤ 3, then the greatest and the least value of |z + 1| are
A) -1 , 6
B) 6 , 0
C) 6 , 3
D) None of these
Answer: B) 6 , 0
Explanation: We have, |z + 4| ≤ 3 ⇒ -3 ≤ z + 4 ≤ 3 ⇒ -6 ≤ z + 1 ≤ 0 ⇒ 0 ≤ -(z + 1)≤ 6 ⇒ 0 ≤ |z + 1| ≤ 6 Hence, the greatest and least value are 6 and 0.
102. The angle between lines joining the origin to the point of intersection of the line √3x + y = 2 and the curve y² - x² = 4 is
A) tan-1 2 / √3
B) π / 6
C) tan-1 (√3 / 2)
D) π / 2
Answer: C) tan-1 (√3 / 2)
Explanation: On homogenising y² - x² = 4 with the help of the line √3x + y = 2, we get y² - x² = 4(√3x + y)² / 4 ⇒ y² - x² = 3x² + y² + 2√3xy ⇒ 4x² + 2√3xy = 0
On comparing with ax² + 2hxy + by² = 0, we get a = 4 , h = √3 and b = 0 ∴ The angle between the lines is tan θ = 2 ⋅ √h² - ab / a + b = 2 ⋅ √3 - 0 / 4 + 0 θ = tan-1 (√3 / 2)
103. If the area of the triangle on the complex plane formed by the points z, z + iz and iz is 200, then the value of 3|z| must be equal to
A) 20
B) 40
C) 60
D) 80
Answer: C) 60
Explanation:
104. The equation of the chord of the hyperbola 25x² - 16y² = 400 which is bisected at the point (6,2), is
A) 6x - 7y = 418
B) 75x - 16y = 418
C) 25x - 4y = 400
D) None of these
Answer: B) 75x - 16y = 418
Explanation: Given equation of hyperbola is 25x² - 16y² = 400. If (6,2) is the mid-point of the chord, then equation of chord is T = S₁ ⇒ 25(6x) - 16(2y) = 25(36) - 16(4) ⇒ 75x - 16y = 450 - 32 (divide by 2) ⇒ 75x - 16y = 418
105. If a plane meets the coordinate axes at A, B and C such that the centroid of the triangle is (1,2,4), then the equation of the plane is
A) x + 2y + 4z = 12
B) 4x + 2y + z = 12
C) x + 2y + 4z = 3
D) 4x + 2y + z = 3
Answer: B) 4x + 2y + z = 12
Explanation: Let the equation of the plane is x / α + y / β + z / γ = 1
Then, A(α, 0, 0), B(0, β, 0) and C(0, 0, γ) are the points on the coordinate axes. Since, the centroid of the triangle is (1,2,4).
∴ α / 3 = 1 ⇒ α = 3 β / 3 = 2 ⇒ β = 6 and γ / 3 = 4 ⇒ γ = 12
∴ The equation of the plane is x / 3 + y / 6 + z / 12 = 1 ⇒ 4x + 2y + z = 12
106. The volume of the tetrahedron included between the plane 3x + 4y - 5z - 60 = 0 and the coordinate planes is
A) 60
B) 600
C) 720
D) 400
Answer: B) 600
Explanation:
107. ∫2Ï€ 0 (sin x + |sin x|)dx is equal to
A) 0
B) 4
C) 8
D) 1
Answer: B) 4
Explanation: ∫2Ï€ 0 (sin x + |sin x|)dx = ∫2Ï€ 0 (sin x + sin x)dx + ∫2Ï€ Ï€ (sin x - sin x)dx = 2 ∫ Ï€ 0 sin x dx + 0 = -2 ∫Ï€ 0 cos x dx = -2(cos Ï€ - cos 0) = -2(-1-1) = 4
108. The value of ∫ √2 0 [x²] dx, where [⋅] is the greatest integer function, is
A) 2 - √2
B) 2 + √2
C) √2 - 1
D) √2 - 2
Answer: C) √2 - 1
Explanation: ∫ √2 0 [x²]dx = ∫¹ 0 [x²]dx + ∫√2 ₁ [x²]dx = ∫¹ 0 0 dx + ∫√2 ₁ 1dx = [x]₁ √2 = √2 - 1
109. If
A) 2n / m + 1 - n / m + 1 ⋅ l(m + 1 , n - 1)
B) n / m + 1 ⋅ l(m + 1 , n - 1)
C) 2n / m + 1 + n / m + 1 l ⋅ (m + 1 , n - 1)
D) m / n + 1 ⋅ l(m + 1 , n - 1)
Answer: A) 2n / m + 1 - n / m + 1 ⋅ l(m + 1 , n - 1)
Explanation: We have, l(m,n) = l = ∫¹ 0 tm(1 + t)n dt ⇒ l(m,n) = [(1 + t)n ⋅ tm+1 / m + 1]¹ 0 - n / m + 1 ∫¹ 0 (1+t)n-1 ⋅ tm+1 ⋅ dt = 2n / m + 1 - n / m + 1 ⋅ l(m + 1 , n - 1)
110. The area in the first quadrant between x² + y² = Ï€² and y = sin x is
A) Ï€³ - 8 / 4
B) Ï€³ / 4
C) Ï€³ - 16 / 4
D) Ï€³ - 8 / 2
Answer: A) Ï€³ - 8 / 4
Explanation:
111. The area bounded by y = xe|x| and lines |x| = 1 , y = 0 is
A) 4 sq units
B) 6 sq units
C) 1 sq unit
D) 2 sq units
Answer: D) 2 sq units
Explanation:
112. The solution of dy / dx = x² + y² + 1 / 2xy , satisfying y(1) = 0 is given by
A) hyperbola
B) circle
C) ellipse
D) parabola
Answer: A) hyperbola
Explanation: Given differential equation is dy / dx = x² + y² + 1 / 2xy ⇒ 2xydy = (x² + 1)dx + y²dx ⇒ xd(y²) - y²dx / x² = (x² + 1 / x²) ⋅ dx ⇒ ∫ d (y² / x) = ∫ ( 1 + 1 / x²)dx ⇒ y² / x = x - 1 / x + C ⇒ y² = (x² - 1 + Cx)
When x = 1, y = 0 Then, 0 = 1 - 1 + C⇒ C = 0
∴ The solution is x² - y² = 1 i.e., hyperbola.
113. If
A) k ⋅ ex²/2
B) k ⋅ ey²/2
C) k ⋅ ex²
D) k ⋅ exy/2
Answer: A) k ⋅ ex²/2
Explanation: Given, x ⋅ dy / dx + y = x ⋅ Æ’(xy) / Æ’'(xy) i.e., d / dx(xy) = x Æ’(x,y) / Æ’'(x,y)
⇒ Æ’'(xy) / Æ’(xy) d(xy) = x dx ⇒ ∫ Æ’'(xy) / Æ’(xy) d(xy) = ∫ x dx
⇒ log[Æ’(xy)] = x² / 2 + C
⇒ Æ’(xy) = e(x² / 2 + C) = e x² / 2 ⋅ eC = k ⋅ ex² / 2
114. The differential equation of the rectangular hyperbola, where axes are the asymptotes of the hyperbola, is
A) y(du/dx) = x
B) x(dy/dx) = -y
C) x(dy/dx) = y
D) xdy + ydx = c
Answer: B) x(dy/dx) = -y
Explanation: The differential equation of the rectangular hyperbola xy = c² is y + x dy / dx = 0 ⇒ x dy / dx = -y
115. The length of longer diagonal of the parallelogram constructed on 5a + 2b and a - 3b, if it is given that |a| = 2√2 , |b| = 3 and the angle between a and b is Ï€ / 4 , is
A) 15
B) √113
C) √593
D) √369
Answer: C) √593
Explanation: Given that, |a| = 2√2 , |b| = 3
The longer vector is 5a + 2b + a - 3b = 6a - b Length of one diagnol = |6a -b| = √36a² + b² - 2 * 6|a|⋅ |b| ⋅ cos 45° = √ 36 * 8 + 9 - 12 * 2√2 * 3 * 1 / √2 = √ 288 + 9 - 12 * 6 = √225 = 15
Other diagnol is 4a + 5b.
Its length is = √(4a)² + 5b)² + 2 * |4a| |5b| cos 45° = √ 16 * 8 + 25 * 9 + 40 * 6 = √593
116. If r = αb * c + βc * a + γa * b and [a b c] = 2, then α + β + γ is equal to
A) r ⋅ [b * c + c * a + a * b]
B) 1/2 r ⋅ (a + b + c)
C) 2r ⋅ (a + b + c)
D) 4
Answer: B) 1/2 r ⋅ (a + b + c)
Explanation: Now, r ⋅ a = α(a ⋅ b * c) + β(a ⋅ c * a) + γ(a ⋅ a * b) = α [abc] + 0 + 0 Similarly, r ⋅ b = β[abc] and r ⋅ c = γ[abc] ∴ 1/2 r ⋅ (a + b + c) = 1/2 (r ⋅ a + r ⋅ b + r ⋅ c) = 1/2 (α + β + γ) [a b c] = 1/2(α + β + γ) * 2 = α + β + γ
117. If a, b, c are three non-coplanar vectors and p, q, r are reciprocal vectors, then (la + mb + nc) ⋅ (lp + mq + nr) is equal to
A) l + m + n
B) l³ + m³ + n³
C) l² + m² + n²
D) None of these
Answer: C) l² + m² + n²
Explanation: ∴ p,q and r are reciprocal vectors of a,b and c respectively. ∴ p ⋅ r = 1, p ⋅ b = 0 = p ⋅ c q.a = 0, q.b = 1, q.c = 0 r.a = 0, r.b = 0, r.c = 1 ∴ (la + mb + nc) ⋅ (lp + mq + nr) = l² + m² + n²
118. If the integers m and n are chosen at random from 1 to 100, then the probability that a number of the form 7n + 7m is divisible by 5, equals to
A) 1/4
B) 1/2
C) 1/8
D) 1/3
Answer: A) 1/4
Explanation: Let l = 7n + 7m, then we observe that 7¹, 7², 7³ and 74 ends in 7, 9, 3 and 1, respectively.
Thus, 7i ends in 7,9,3 or 1 according to as i is of the form 4K + 1, 4K + 2, 4K - 1 or 4K respectively. If S is the sample space, then n(S) = (100)². 7m + 7n is divisible by 5,
if (i) m is of the form 4K + 1 and n is of the form 4K - 1 or
(ii) m is of the form 4K + 2 and n is of the form 4K or
(iii) m is of the form 4K - 1 and n is of the form 4K + 1 or
(iv) m is of the form 4K and n is of the form 4K + 1.
Thus , number of favourable ordered pairs (m,n) = 4 * 25 * 25
Hence, required probability = 4 * 25 *25 / (100)² = 1/4
119. Let X denote the sum of the numbers obtained when two fair dice are rolled. The variance and standard deviation of X are
A) 31 / 6 and √31 / 6
B) 35 / 6 and √35 / 6
C) 17 / 6 and √17 / 6
D) 31 / 6 and √35 / 6
Answer: B) 35 / 6 and √35 / 6
120. A four-digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is
A) zero
B) 1/3
C) 1/4
D) None of these
Answer: D) None of these
Explanation: Given numbers are 1,2,3,4.
Possibilities for ten's digit = 3
Possibilities for hundred's place digit = 2
Possibilities for thousand' place's digit = 1
∴ Number of favourable outcomes = 2 * 3 * 2 * 1 = 12
Number of numbers formed by 1,2,3,4 (without repetitions) = 4!
∴ Required probability = 12 / 4 * 3 * 2 = 1/2
ALL THE BEST!
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