Aspirants preparing for JEE Advanced exam can refer to previous year question papers with answers provided by BYJU’S. Referring to last year papers help aspirants for the kind of challenges that the actual exam might throw up and help them in evaluation of performance and preparation level. JEE Advanced Question Paper 2019 Maths Paper 1 solutions are given below. Students can download to practise offline.
Question 1: Let M =
Solution:
Let M =
M^{2} = αM + βI
a_{11} = sin θ – 1 – sin^{2} θ – cos^{2} θ – cos^{2} θ sin^{2} θ = β + α sin^{4} θ sin^{8} θ – 2 – cos^{2} θsin^{2} θ
= β + α sin^{4} θ
a_{21} = sin^{4} θ + cos^{2} θ sin^{4} θ + cos^{4} θ + cos^{6} θ = α (1 + cos^{2} θ)
(1 + cos^{2} θ) α = sin^{4} θ (1 + cos^{2} θ) + cos^{4} θ (1 + cos^{2} θ)
α = sin^{4} θ + cos^{4} θ = 1 – 2 sin^{2} θ cos^{2} θ = 1 – ((sin^{2} θ)/2)
α_{min} = ½
β = sin^{8} θ – 2 – cos^{2} θsin^{2} θ - sin^{8} θ – cos^{4} θsin^{4} θ
= -2 - (sin^{2} 2θ)/4 - (sin^{4} 2θ)/16
β_{min} = -2 – 1/16 [4t^{2} + t^{4}]
= -2 – 1/16 [t^{2} + 2]^{2} + ¼
= -7/4 – 1/16(9) = -37/16
Therefore, α + β = -37/16 + 1/2 = -29/16
Question 2: A line y = mx + 1 intersects the circle (x - 3)^{2} + (y + 2)^{2} = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate -3/5, then which one of the following options is correct?
Solution:
y = mx + 1
(x - 3)^{2} + ((mx +1) + 2)^{2} = 25
=> x^{2} (1+ m^{2} ) + 6(m -1) x - 7 = 0
1 + m^{2} = 5m – 5
Or m = 2, 3
Question 3: Let S be the set of all complex numbers z satisfying |z - 2 + i| ≥ √5. If the complex number z_{0} is such that 1/|z_{0} - 1| is the maximum of the set
Solution:
|z - 2 + i| ≥ √5
P is along
Question 4: The area of the region {(x, y) : xy ≤ 8, 1 ≤ y ≤ x^{2}} is
Solution:
{(x, y) : xy ≤ 8, 1 ≤ y ≤ x^{2}}
8/x = x^{2}
= 16 log_{e} 2 – 14/3
Question 5: There are three bags B_{1}, B_{2} and B_{3}. The bag B_{1} contains 5 red and 5 green balls, B_{2} contains 3 red and 5 green balls, and B_{3} contains 5 red and 3 green balls, Bags B_{1}, B_{2} and B_{3} have probabilities 3/10, 3/10 and 4/10 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
Solution:
Question 6: Define the collections E_{1}, E_{2}, E_{3} ,..... of ellipses and {R_{1}, R_{2}, R_{3}, …..} of rectangles as follows:
R_{1}: Rectangle of largest area, with sides parallel to the axes, inscribed in E_{1};
E_{n}: Ellipse
Then which of the following options is/are correct?
Solution:
A = 6 cosθ . 4 sin θ = 12 sin 2θ -> max
θ = π/4
E_{2} = a_{2} = 3 cos(π/4) = 3/√2
b_{2} = 2 sin θ = √2
So, e_{18} = 1 – b_{2}/a_{2} = √5/3
e of all ellipses -> same
Difference of f from conic in equation a_{q}e
Length of latus rectum = [2x4]/4(√2) = 1/6
Question 7: Let
Solution:
(a)
adj M =
b – 6 = -5
=>b = 1
and ab – 1 = 1
or a = 2
Now,
|M| = -2
And a + b = 2
(b) |adj M^{2}|= |M^{2}|^{2} = |M|^{4} = 16
(c) (adj M)^{-1} + adj M^{-1} = 2 (adj M)^{-1}
= 2 M^{-1} M
= 2 x (-1/2) x M
= -M
(d)
This implies,
β + 2γ = 1
α + 2β + 3γ = 2
3α + β + γ = 1
α - β + γ = 3
Solving all the equations, we have α = 1, β = -1 and γ = 1
Question 8:
Then which of the following options is/are correct?
Solution:
Range: -∞, +1
Therefore, Not monotonic.
f(x) = x^{2} – x + 1
f’(x) = 2x - 1
Max at x = 0 and 1
x ≥ 3
f(3) = 1 log 1 – 3 + 10/3 = 1/3
f(∞) -> ∞
Loc. Max at x = 1
Question 9: Let α and β be the roots of x^{2} – x – 1 = 0, with α > β. For all positive integers n, define
b_{1} = 1 and b_{n} = a_{n-1} + a_{n+1}, n ≥ 2.
Then which of the following options is/are correct?
Solution:
x^{2} – x – 1 = 0
α = [1+ √5]/2 and β = [1- √5]/2
As, b_{1} = 1 and b_{n} = a_{n-1} + a_{n+1}, n ≥ 2
a_{1} + a_{2} + …+ a_{n} =
=> a_{n} + a_{n+1} = a_{n+2}
=>
= a_{n+2} – [α^{2} - β^{2}]/[α -β]
= a_{n+2} – [α + β]
= a_{n+2} – 1
Question 10: Let denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to at a point P intersect the y-axis at Y_{P}. If PY_{P} has length 1 for each point P on , then which of the following is options is/are correct?
Solution:
y - y_{1} = m(x – x_{1})
y_{p} - y_{1} = -mx_{1}
=> 1 = x_{1}^{2} + m^{2} x_{1}^{2}
1 = x^{2}[1 + (dy/dx)^{2}]
Or (dy/dx)^{2} = 1/x^{2} – 1
dy/dx = ± √[1-x^{2}]/x
Here x = sinθ, dx = cos θ dθ
=>
And the correct differential equation will be xy' + √[1-x^{2}] = 0
Question 11 In a non-right-angle triangle ΔPQR, let p, q, r denote the lengths of the sides opposite to the angles at P, Q, R respectively. The median from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at 0. If p = √3, q = 1, and the radius of the circumcircle of the ΔPQR equals 1, then which of the following options is/are correct?
Solution:
(sin P)/√3 = (sin Q)/1 = 1/2R = ½ ; [Given R = 1]
∠P = π/3 or 2π/3
∠Q = π/6 or 5π/6
p>q=> ∠P = ∠Q
If ∠P = π/3 and ∠Q = π/6 => ∠R = π/2
Therefore, ∠P = 2π/3 and ∠Q = ∠R = π/6
Question 12: Let L_{1} and L_{2} denotes the lines
respectively. If L_{3} is a line which is perpendicular to both L_{1} and L_{2} and cuts both of them, then which of the following options describe(s) L_{3}?
Solution:
L_{1} : [x-1]/-1 = [y-0]/2 = [z-0]/2
L_{2} : x/2 = y/-1 = z/2
L_{3} : x/a = y/b = z/c
L_{3} = L_{1} x L_{2}
Similarly,
A on L_{1} : (-λ + 1, 2λ, 2λ)
B on L_{2} : (2μ, -μ, 2μ)
AB Δrs : (2μ, + λ - 1 -μ - 2λ, 2μ - 2λ) = (6, 6, -3) or (2, 2, -1)
=>
λ = [3k+1]/3 and μ = -4k – 2/3
Now, 2 μ - 2 λ + k = 0
Putting μ and λ values, we have
k = -2/9=> μ = 2/9 and λ = 1/9
So, A: (8/9, 2/9, 2/9) and B : (4/9, -2/9, 4/9)
Mid point is (2/3, 0, 1/3)
For equation of L_{3};
vector A is the position vector of any point on L_{3}.
Answer: a, b, d
Question 13: If
Solution:
27I^{2} = 4
Question 14: Let the point B be the reflection of the point A(2, 3) with respect to the line 8x – 6y – 23 = 0. Let _{A} and _{B} be circle of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circle _{A} and _{B} such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is ________.
Solution:
ΔAPC and ΔBQC are similar.
BC/AC = 1/2 => 2(AC - AB) = AC
AC = 2AB = 10
Question 15: Let AP (a; d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d > 0. If AP(1; 3) ∩ AP (2; 5) ∩ AP (3; 7) = AP (a; d) then a + d equals_________
Solution:
I: 1, 4, 7, 10, 13, 16 ….. 32
II: 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52
III: 3, 10, 17, 24, 31, 38, 45, 52
52 <-- a + d --> LCM of 3, 5, 7 = 105
So, a + d = 157
Question 16: Let S be the sample space of all 3 × 3 matrices with entries from the set {0, 1}. Let the events E_{1} and E_{2} be given by
E_{1} = {A ϵ S : det A = 0} and
E_{2} = {A ϵ S : sum of entries of A is 7}.
If a matrix is chosen at random from S, then the conditional probability P(E_{1}|E_{2}) equals …….
Solution:
S: 2^{9}
E_{2} : Sum of entries 7, we need 7 - 1s and 2 0’s
Total E_{2} = 9!/7!2! = 36
Per |A| to be zero, both zeroes should be in the same row/column.
Therefore, 3 x 3 x 2 = 18 cases
Question 17: Three lines are given by
Let the lines cut the plane x + y + z 1 at the points A, B and C respectively. If the area of the triangle ABC is Δ then the value of (6Δ)^{2} equals __________
Solution:
1^{st} line: x = λ, y = 0, z = 0
x + y + z = 1 => λ = 1
A(1, 0, 0)
2^{nd} line: x = μ, y = μ, z = 0
2μ = 1 or μ = ½
B(1/2, ½, 0)
Parallels, C(1/3, 1/3, 1/3)
Question 18: Let ω ≠ 1 be a cube root of unity. Then the minimum of the set {|a + bω + cω^{2}|^{2} ; Where, a, b, c distinct non-zero integers }equals ________
Solution:
| a + bω + cω^{2}|^{2}
= (a + bω + cω^{2}) (a + cω + bω^{2})
= (a^{2} + b^{2} + c^{2} - ab – bc – ca)
[because, conjugate of (a + bω + cω^{2}) = (a + cω + bω^{2})]
= > (1/2)[(a-b)^{2} + (b-c)^{2} + (c-a)^{2}] = ½(1+1+4) = 3
[For minimum value a = 1, b = 2 and c = 3]