JEE Advanced Question Paper 2021 Maths Paper 1

Question 1: Consider a triangle Δ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is;

a. x² + y² – 3x + y = 0
b. x² + y² + x + 3y = 0
c. x² + y² + 2y – 1 = 0
d. x² + y² + x + y = 0

Answer: b

As we know mirror image of the orthocenter lies on the circumcircle.

Image of (1, 1) in x-axis is (1, -1)

Image of (1, 1) in x + y + 1 = 0 is (-2, -2).

∴ The required circle will be passing through both (1, -1) and (-2, -2).

∴ Only x² + y² + x + 3y = 0 satisfy both.

Question 2: The area of the region {(x, y): 0 ≤ x ≤ 9/4, 0 ≤ y ≤ 1, x ≥ 3y, x + y ≥ 2}

a. 11/32
b. 35/96
c. 37/96
d. 13/32

Answer: a

A rough sketch of the required region is;

∴ Required area is area of ΔACD + Area of ΔABC

i.e (¼) + (3/32) = 11/32 sq.units.

Question 3: Consider three sets E₁ = {1, 2, 3}, F₁ = {1, 3, 4} and G₁ = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E₁, and let S₁ denote the set of these chosen elements. Let E₂ = E₁ – S₁ and F₂ = F₁ ⋃ S₁. Now two elements are chosen at random, without replacement, from the set F₂ and let S₂ dentoe the set of these chosen elements.

Let G₂ = G₁ ⋃ S₂. Finally, two elements are chosen at random, without replacement from the set G₂ and let S₃ denote the set of these chosen elements. Let E₃ = E₂ ⋃ S₃. Given that E₁ = E₃, let p be the conditional probability of the event S₁ = {1, 2}. Then the value of p is;

a. 1/5
b. 3/5
c. 1/2
d. 2/5

Answer: a

We will follow the tree diagram,

P(E₁ = E₃) = ⅓ [(½ × 1/10) + (½ × 0) + (½ × 1/10) + (½ × ⅙) + (⅔ × 1/10) + (⅓ × 0)]

= ⅓ (¼)

Required probability = ⅓ (½ × 1/10)/(⅓ × ¼ )

= 1/5

Question 4: Let θ₁, θ₂, …., θ₁₀ be positive valued angles (in radian) such that θ₁+ θ₂+ ….+ θ₁₀ = 2π. Define the complex numbers z₁ =

P: |z₂ – z₁| + |z₃ – z₂| + …. +|z₁₀ – z₉| + |z₁ – z₁₀| ≤ 2π

Q: |z₂² – z₁²| + |z₃² – z₂²| + …. +|z₁₀² – z₉²| + |z₁² – z₁₀²| ≤ 4π

Then,

a. P is TRUE and Q is FALSE
b. Q is TRUE and P is FALSE
c. Both P and Q are TRUE
d. Both P and Q are FALSE

Answer: c

|z₂ – z₁| = length of line AB ≤ length of arc AB

|z₃ – z₂| = length of line BC ≤ length of arc BC

∴ Sum of length of these 10 lines ≤ Sum of length of arcs (i.e. 2π)

(As θ₁+ θ₂+ ….+ θ₁₀ = 2π)

∴ |z₂ – z₁| + |z₃ – z₂| + …. + |z₁ – z₁₀| ≤ 2π

And |z_k² – z²_k-1| = |z_k – z_k-1| |z_k + z_k-1|

As we know |z_k + z_k-1| ≤ |z_k| + |z_k-1| ≤ 2

|z₂² – z₁²| + |z₃² – z₂²| + …. + |z₁² – z₁₀²| ≤ 2(|z₂ – z₁| + |z₃ – z₂| + …. + |z₁ – z₁₀|) ≤ 2(2π)

∴ Both P and Q are true.

Question Stem for Question Nos. 5 and 6

Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, …, 100}. Let p₁ be the probability that the maximum of chosen numbers is at least 81 and p₂ be the probability that the minimum of chosen numbers is at most 40.

Question 5: The value of (625/4)p₁ is

Answer: 76.25

For p₁, we need to remove the cases when all three numbers are less than or equal to 80.

So p₁ = 1 – (80/100)³

= 61/125

So (625/4)p₁ = (625/4)× 61/125

= 305/4

= 76.25

Question 6: The value of (125/4)p₂ is

Answer: 24.50

For p₂, we need to remove the cases when all three numbers are greater than 40.

So p₂ = 1 – (60/100)³

= 98/125

So (125/4)p₂ = (125/4)×98/125

= 24.50

Question Stem for Question Nos. 7 and 8

Let α, β and γ be real numbers such that the system of linear equations

x + 2y + 3z = α

4x + 5y + 6z = β

7x + 8y + 9z = γ – 1 is consistent.

Let |M| represent the determinant of the matrix.

M =

Let P be the plane containing all those (α, β, γ) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.

Question 7: The value of |M| is

Answer: 1

Question 8: The value of D is

Answer: 1.50

Solution for Q.7 and Q.8

∆ =

Given system of equation will be consistent even if α = β = γ-1 = 0, i.e. equations will form homogeneous system.

So α = 0, β = 0, γ = 1

M =

= -1×(-1)

= 1

As given equations are consistent

x + 2y + 3z – α = 0 …P₁

4x + 5y + 6z – β = 0 …P₂

7x + 8y + 9z – (γ – 1) = 0 …P₃

For some scalar λ and μ

μP₁ + λP₂ = P₃

μ(x + 2y + 3z – α) + λ(4x + 5y + 6z – β) = 7x + 8y + 9z – (γ – 1)

comparing coefficients

μ+ 4λ = 7, 2μ + 5λ = 8, 3μ + 6λ = 9

λ = 2 and μ = -1 satisfy all these conditions

comparing constant terms,

αμ – βλ = – (γ – 1)

α – 2β + γ = 1

So equation of plane is

x – 2y + z = 1

distance from (0, 1, 0) = |(-2-1)/√6| = 3/√6

D = (3/√6)² = 9/6 = 3/2 = 1.50

Question Stem for Question Nos. 9 and 10

Consider the lines L₁ and L₂ defined by

L₁ : x√2 + y – 1 = 0 and L₂ : x√2 – y + 1 = 0

For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L₁ and the distance of P from L₂ is λ². The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is √270.

Let the perpendicular bisector of RS meet C at two distinct points R’ and S’. Let D be the square of the distance between R’ and S’.

Question 9: The value of λ² is

Answer: 9

Question 10: The value of D is

Answer: 77.14

Solution for Q.9 and Q.10

C: |(x√2 + y – 1)/√3| |(x√2 – y + 1)/√3| = λ²

⇒ C: |2x² – (y-1)²| = 3λ²

C cuts y-1 = 2x at R(x₁, y₁) and S(x₂, y₂)

So |2x² – 4x²| = 3λ²

⇒ x = ±√(3/2) |λ|

So |x₁ – x₂| = √6|λ| and |y₁ – y₂| = 2| x₁ – x₂| = 2√6|λ|

Since RS² = (x₁ – x₂)² + (y₁-y₂)²

⇒ 270 = 30λ²

⇒ λ² = 270/30 = 9

Since slope of RS = 2 and midpoint of RS is ((x₁+x₂)/2, (y₁+y₂)/2) ≡ (0, 1)

So R’S’ ≡ y-1 = -½ x

Solving y-1 = -½ x with ‘C’ we get x² = (12/7) λ²

⇒ |x₁-x₂| = 2√(12/7) |λ| and |y₁-y₂| = ½ |x₁-x₂| = √(12/7) |λ|

Hence D = (R’S’)² = (x₁-x₂)² + (y₁-y₂)² = (12/7)9×5 ≈ 77.14

Question 11: For any 3 × 3 matrix M, let |M| denote the determinant of M. Let E =

P =

And F =

If Q is a nonsingular matrix of order 3 × 3, then which of the following statements is(are) TRUE?

a. F = PEP and P² =

Question 12: Let F: R→ R be defined by f(x) = (x²-3x-6)/(x²+2x+4). Then which of the following statements is (are) TRUE?

a. f is decreasing in the interval (-2, -1)
b. f is increasing in the interval (1, 2)
c. f is onto
d. Range of f is [-3/2, 2]

Answer: (a, b)

f(x) = (x²-3x-6)/(x²+2x+4)

⇒ f’(x) = 5x(x+4)/(x²+2x+4)²

⇒ f(x) has local maxima at x = -4 and minima at x = 0

Range of f(x) is [-3/2, 11/6]

Question 13: Let E, F and G be three events having probabilities P(E) = 1/8, P(F) = ⅙ and P(G) = ¼, and P(E⋂F⋂G) = 1/10. For any event H, if H^c denotes its complement, then which of the following statements is(are) TRUE?

a. P(E⋂F⋂G^c) ≤ 1/40
b. P(E^c⋂F⋂G) ≤ 1/15
c. P(E⋃F⋃G) ≤ 13/24
d. P(E^c⋂F^c⋂G^c) ≤ 5/12

Answer: (a, b, c)

Let P(E⋂F) = x, P(F⋂G) = y and P(E⋂G) = z

Clearly x, y, z ≥ 1/10

Since x + z ≤ 27/120

⇒ x, z ≤ 15/120

x+y ≤ 32/120

⇒ x, y ≤ 20/120

And y + z ≤ 42/120 ⇒ y, z ≤ 30/120

Now P(E⋂F⋂G^c) = x – 12/120 ≤ 3/120 = 1/40

P(E^c⋂F⋂G) = y – 12/120 ≤ 80/120 = 1/15

P(E⋃F⋃G) ≤ P(E) + P(F) + P(G) = 13/24

And P(E^c⋂F^c⋂G^c) = 1 – P(E⋃F⋃G) ≥ 11/24 ≥ 5/12

Question 14: For any 3 × 3 matrix M, let |M| denote the determinant of M. Let I be the 3 × 3 identify matrix. Let E and F be two 3 × 3 matrices such that (I – EF) is invertible. If G = (I – EF)^–1, then which of the following statements is(are) TRUE?

a. |FE| = |I – FE| |FGE|
b. (I – FE) (I + FGE) = I
c. EFG = GEF
d. (I – FE) (I – FGE) = I

Answer: (a, b, c)

I – EF = G^–1

G – GEF = I …(1)

And G – EFG = I …(2)

Clearly GEF = EFG (option C is correct)

Also (I – FE)(I + FGE) = I – FE + FGE – FE + FGE

= I – FE + FGE – F(G – I)E

= I – FE + FGE – FGE + FE

= I (option B is correct and D is incorrect)

Now, (I – FE)(I – FGE) = I – FE – FGE + F(G – I)E

= I – 2FE

(I – FE)(- FGE) = – FE

|I – FE||FGE| = |FE|

Question 15: For any positive integer n, let S_n: (0, ∞) → R be defined by S_n(x) = ∑_k=1ⁿ cot^-1((1 + k(k+1)x²)/x) where for any x ∈R , cot^–1(x) ∈ (0, π) and tan^–1( x)∈ (-π/2, π/2). Then which of the following statements is (are) TRUE?

a. S₁₀(x) = π/2 – tan^-1(1 + 11x²)/10x, for all x >0
b. lim _n→∞ cot(S_n(x)) = x, for all x >0
c. The equation S₃(x) = π/4 has a root in (0, ∞)
d. tan(S_n(x)) ≤1/2, for all n≥1 and x > 0

Answer: (a, b)

S_n(x) =

=

= tan^-1(n+1)x – tan^-1x

= tan^-1(nx/(1+(n+1)x²))

Now,

(a) S₁₀(x) = tan^-1(10x/(1+11x²)) = π/2 – tan^-1((1+11x²)/10x)

(b) lim_{n→ ∞}cot (S_n(x)) = cot (tan^-1x/x²) = x

(c) S₃(x) = π/4 => 3x/(1+4x²) = 1

⇒ 4x²-3x+1=0 has no real root.

(d) for x = 1, tan(S_n(x)) = n/(n+2) which is greater than 1/2 for n≥3 so this option is incorrect.

Question 16: For any complex number w = c + id, let arg(w) ∈ (-π, π], where i = √-1 . Let α and β be real numbers such that for all complex numbers z = x + iy satisfying arg(z+α)/(z+β) = π/4, the ordered pair (x, y) lies on the circle x² + y² + 5x – 3y + 4 = 0. Then which of the following statements is (are) TRUE?

a. α = -1
b. αβ = 4
c. αβ = -4
d. β = 4

Answer:(b, d)

Circle x² + y² + 5x – 3y + 4 = 0 cuts the real axis (x-axis) at (-4, 0), (-1, 0)

Clearly α = 1 and β = 4

Question 17: For x ∈ R, the number of real roots of the equation 3x² – 4|x² – 1| + x – 1 = 0 is

Answer: 4

3x² – 4|x² – 1| + x – 1 = 0

Let x ∈ [-1, 1]

⇒ 3x² – 4(-x²+1) + x – 1 = 0

⇒ 3x² + 4x² – 4 + x – 1 = 0

⇒ 7x² + x – 5 = 0

⇒ x = -1±√(1+140)/2

Both values are acceptable.

Let x ∈ (-∞, -1) ⋃ (1, ∞)

x²-4(x²-1)+x-1 = 0

⇒ x²-x-3 = 0

⇒ x = 1±√(1+12)/2

Again both are acceptable.

Hence, total number of solution = 4.

Question 18: In a triangle ABC, let AB = √23, and BC = 3 and CA = 4. Then the value of (cot A + cot C)/cot B is;

Answer: 2

With standard notations

Given c = √23, a = 3 and b = 4

Now (cot A + cot C)/cot B = (cos A/sin A + cos C/sin C)/(cos B/sin B)

= ((b²+c²-a²)/2bc sin A + (a²+b²-c²)/2ab sin C)/((c²+a²-b²)/2ac sin B)

= ((b²+c²-a²)/4∆ + (a²+b²-c²)/4∆)/((c²+a²-b²)/4∆)

= 2b²/(a²+c²-b²)

= 2

Question 19: Let

Answer: 7

Given

a.so,

Let

4K² – 2K = 0

⇒

Or

Therefore

= 49

⇒