JEE Advanced Question Paper 2021 Maths Paper 2

Question 1: Let;

S₁ = {(i, j, k) : i, j, k ∈ {1,2,…,10}}

S₂ = {(i, j) : 1 ≤ i < j + 2 ≤ 10,i, j ∈ {1,2,…,10}}

S₃ = {(i, j, k, l) : 1 ≤ i < j < k < l, i, j, k, l ∈ {1,2,…,10}}

S₄ = {(i, j,k,l ) : i, j, k and l are distinct elements in {1,2,…,10}}.

If the total number of elements in the set S_r is n_r, r = 1, 2,3,4, then which of the following statements is (are) TRUE?

a. n₁ = 1000
b. n₂ = 44
c. n₃ = 220
d. n₄/12 = 420

Answer: (a, b, d)

Number of elements in S₁ = 10 × 10 × 10 = 1000

Number of elements in S₂ = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 44

Number of elements in S₃ = ¹⁰C₄ = 210

Number of elements in S₄ = ¹⁰P₄ = 210 × 4! = 5040

Question 2: Consider a triangle PQR having sides of lengths p, q, and r opposite to the angles P, Q, and R, respectively. Then which of the following statements is (are) TRUE?

a. cos P ≥ 1 – p²/2qr
b. cos R ≥ ((q-r)/(p+q))cos P + ((p-r)/(p+q))cos Q
c. (q+r)/p < 2√(sin Q sin R)/sin P
d. if p<q and p<r, then cos Q > p/r and cos R > p/q

Answer: (a, b)

(a) cos P = (q²+r² – p²)/2qr

And (q²+r²)/2 ≥ √(q². r²) (AM ≥ GM)

⇒ (q²+r²) ≥ 2qr

So cos P ≥ (2qr – p²)/2qr

cos P ≥ 1 – p²/2qr

(b) ((q-r) cos P + (p-r) cos Q)/(p+q) = ((q cos P + p cos Q) – r(cos P + cos Q))/(p+q)

= r(1- cos P – cos Q)/(p+q)

= (r(q-p cos R) – (p-q cos R))/(p+q)

= ((r-p-q)+(p+q)cos R)/(p+q)

= cos R + (r-q-p)/(p+q) ≤ cos R (since r < p+q)

(c) (q+r)/p = (sin Q + sin R)/sin P ≥ 2√(sin Q sin R)/sin P

(d) If p < q and q < r

So, p is the smallest side, therefore one of Q or R can be obtuse

So, one of cos Q or cos R can be negative

Therefore, cos Q > p/r and cos R > p/q cannot hold always.

Question 3: Let f: [-π/2, π/2] → R be a continuous function such that f(0) = 1 and ∫₀^π/3 f(t) dt = 0. Then which of the following statements is (are) TRUE?

a. The equation f(x) – 3 cos 3x = 0 has at least one solution in (0, π/3)
b. The equation f(x) – 3 sin 3x = -6/π has at least one solution in (0, π/3)
c.

Question 4: For any real numbers α and β, let y_α,β(x), x ∈ R, be the solution of the differential equation dy/dx + αy = xe^βx, y(1) = 1. Let S = {y_α,β(x), α, β ∈ R } . Then which of the following functions belong(s) to the set S?

a. f(x) = (x²/2)e^-x + (e – ½)e^-x
b. f(x) = (-x²/2)e^-x + (e + ½)e^-x
c. f(x) = (e^x/2)(x-½) + (e – e²/4)e^-x
d. f(x) = (e^x/2)(½ -x) + (e + e²/4)e^-x

Answer: (a, c)

dy/dx + αy = xe^βx

Integrating factor (I.F) = e^∫αdx = e^αx

So, the solution is y.e^αx = ∫xe^βx e^αx dx

y.e^αx = ∫xe^(β+α)x dx

If α + β ≠ 0

ye^αx = x e^(α+β)x/(α+β) – e^(α+β)x/(α+β)² + C

y = [e^βx/(α+β)][x – 1/(α+β)] + Ce^-αx …(i)

Put α = β = 1 in (i)

y = (e^x/2)(x – ½) + Ce^-x

y(1) = 1

1 = (e/2)(½) +C/e

⇒ C = e – e²/4

So, y = (e^x/2)(x-½) + (e – e²/4)e^-x

If α + β = 0 and α = 1

dy/dx + y = xe^-x

I.F = e^x

ye^x = ∫x dx

ye^x = x²/2 + C

y = e^-xx²/2 + Ce^-x

y(1) = 1

1 = 1/2e + C/e

⇒ C = e – ½

y = e^-xx²/2 + (e – ½)e^-x

Question 5: Let O be the origin and

a. Projection of

Question 6: Let E denote the parabola y² = 8x. Let P = (-2, 4), and let Q and Q’ be two distinct points on E such that the lines PQ and PQ’ are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE?

a. The triangle PFQ is a right-angled triangle
b. The triangle QPQ’ is a right-angled triangle
c. The distance between P and F is 5√2
d. F lies on the line joining Q and Q’

Answer: (a, b, d)

E : y² = 8x

P : (-2, 4)

Point P (-2, 4) lies on directrix (x = -2) of parabola y² = 8x

So, ∠QPQ’ = π/2 and chord QQ’ is a focal chord and segment PQ subtends a right angle at the focus.

Slope of QQ’ = 2/(t₁+t₂) = 1

Slope of PF = -1

PF = 4√2

Question Stem for Question Nos. 7 and 8

Consider the region R = {(x,y)∈ R×R : x ≥ 0 and y² ≤ 4 – x. Let F be the family of all circles that are contained in R and have centres on the x-axis. Let C be the circle that has the largest radius among the circles in F. Let (α, β) be a point where circle C meets the curve y² = 4 – x.

Question 7: The radius of the circle C is

Answer: (1.50)

Question 8: The value of α is

Answer: (2.00)

Sol: For comprehension Question 7 and Question 8

Let the circle be,

(x – a)² + y² = r²

Solving it with parabola

y² = 4 – x we get

(x – a)² + 4 – x = r²

x² – x(2a + 1) + (a² + 4 – r²) = 0 …(1)

D = 0

⇒ 4r² + 4a – 15 = 0

Clearly a ≥ r

So 4r² + 4r – 15 ≤ 0

⇒ r_max = 3/2 = a

Radius of circle C is 3/2

From (1) x² – 4x + 4 = 0

⇒ x = 2 = α

Question Stem for Question Nos. 9 and 10

Let f₁ : (0, ∞) → R and f₂ : (0, ∞) → R be defined by f₁(x) =

Question 9: The value of 2m₁ + 3n₁ + m₁n₁ is

Answer: (57.00)

Question 10: The value of 6m₂ + 4n₂ + 8m₂n₂ is

Answer: (06.00)

Solution for Question 9 and 10

f₁’(x) = (x – 1)(x – 2)² (x – 3)³ ,…, (x – 20)²⁰ (x – 21)²¹

Checking the sign scheme of f₁’(x) at x = 1, 2, 3, …, 21, we get

f₁(x) has local minima at x = 1, 5, 9, 13, 17, 21 and local maxima at x = 3, 7, 11, 15, 19

⇒ m₁ = 6, n₁ = 5

f₂(x) = 98(x – 1)⁵⁰ – 600(x – 1)⁴⁹ + 2450

f₂’(x) = 98 × 50(x – 1)⁴⁹– 600 × 49 × (x – 1)⁴⁸

= 98 × 50 × (x – 1)⁴⁸ (x – 7)

f₂(x) has local minimum at x = 7 and no local maximum.

⇒ m₂ = 1, n₂ = 0

2m₁ + 3n₁ + m₁n₁

= 2 × 6 + 3 × 5 + 6 × 5

= 57

6m₂ + 4n₂ + 8m₂n₂

= 6 × 1 + 4 × 0 + 8 × 1 × 0

= 6

Question Stem for Question Nos. 11 and 12

Let g_i = [π/8, 3π/8] → R, i = 1, 2 and f: [π/8, 3π/8] → R be the functions such that g₁(x) = 1, g₂(x) = |4x – π| and f(x) = sin²x, for all x∈[π/8, 3π/8]. Define

Question 11: The value of 16S₁/π is

Answer: (2.00)

S₁ = ∫_π/8^3π/8 sin²x . 1 dx

= ½ ∫_π/8^3π/8(1 – cos 2x)dx

= ½ (x – sin 2x/x)_π/8^3π/8

= ½ (π/4 – 0)

= π/8

=> 16S₁/π = 2

Question 12:The value of 48S₂/π² is

Answer: (1.50)

S₂= ∫_π/8^3π/8 sin²x |4x – π| dx

= ∫_π/8^3π/8 4 sin²x |x – π/4| dx

Let x – π/4 = t

=> dx = dt

S₂ = ∫_-π/8^π/8 4 sin² (π/4 + t)|t| dt

= ∫_-π/8^π/8 2(1 – cos 2(π/4 + t) |t| dt

= ∫_-π/8^π/8 (2 + 2 sin 2t) |t| dt

= 2∫_-π/8^π/8|t|dt + 2∫_-π/8^π/8|t| sin 2t dt

= 4∫₀^π/8t dt + 0

S₂ = [2t²]₀^π/8

= π²/32

48S₂/π² = 3/2

Question Paragraph: Let M = {(x, y) ∈ R × R : x² + y²≤ r²}, where r > 0. Consider the geometric progression a_n = 1/2^n-1, n = 1,2, 3… Let S₀ = 0 and, for n ≥ 1, let S_n denote the sum of the first n terms of this progression. For n ≥ 1 , let C_n denote the circle with center (S_n–1, 0) and radius a_n, and D_n denote the circle with center (S_n–1, S_n–1) and radius a_n.

Question 13: Consider M with r = 1025/513. Let k be the number of all those circles C_n that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then,

a. k + 2l = 22
b. 2k + l = 26
c. 2k + 3l = 34
d. 3k + 2l = 40

Answer: (d)

a_n = 1/2^n-1

And S_n = 2(1 – 1/2ⁿ)

For circles C_n to be inside M.

S_n-1 + a_n < 1025/513

⇒ S_n < 1025/513

⇒ 1 – 1/2ⁿ < 1025/1026

⇒ 1 – 1/1026

⇒ 2ⁿ < 1026

⇒ n ≤ 10

∴ Number of circles inside be 10 = K

Clearly, alternate circles do not intersect each other i.e., C₁, C₃, C₅, C₇, C₉ do not intersect each other as well as C₂, C₄, C₆, C₈ and C₁₀ do not intersect each other hence maximum of 5 set of circles do not intersect each other.

∴ l = 5

∴ 3K + 2l = 40

∴ Option (D) is correct

Question 14: Consider M with r = (2¹⁹⁹-1)√2/2¹⁹⁸. The number of all those circles D_n that are inside M is;

a. 198
b. 199
c. 200
d. 201

Answer: (B)

Since r = (2¹⁹⁹-1)√2/2¹⁹⁸

Now, √2S_n-1 + a_n < (2¹⁹⁹-1)√2/2¹⁹⁸

2√2(1 – 1/2^n-1) + 1/2^n-1 < (2¹⁹⁹-1)/2¹⁹⁸

∴ 2√2 – √2/2^n-2 + 1/2^n-1 < 2√2 – √2/2¹⁹⁸

(1/2^n-2)(½ – √2) < -√2/2¹⁹⁸

(2√2-1)/2. 2^n-2 > √2/2¹⁹⁸

2^n-2 < (2 – 1/√2) 2¹⁹⁷

n ≤ 199

∴ Number of circles = 199

Option (B) is correct.

Question Paragraph: Let ψ₁ = [0, ∞) → R, ψ₂ = [0, ∞) → R, f:[0, ∞) → R and g:[0, ∞) → R be functions such that f(0) = g(0) = 0,

ψ₁(x) = e^-x + x, x≥0,

ψ₂(x) = x² – 2x – 2e^-x + 2, x≥0

And

Question 15: Which of the following statements is TRUE?

a. f(√ln 3) + g(√ln 3) = ⅓
b. For every x > 1, there exists an α ∈ (1, x) such that ψ₁(x) = 1 + α x
c. For every x > 0, there exists a β ∈ (0, x) such that ψ₂(x) = 2x (ψ₁(β) -1)
d. f is an increasing function on the interval [0, 3/2]

Answer: (c)

Since,

Let t = u²

⇒ dt = 2u du

So g(x) =

=

And

Therefore

From equation (i) + (ii) : f(x) + g(x) =

Let t² = P

⇒ 2t dt = dP

f(x) + g(x) =

f(x) + g(x) =

∴ f(√ln 3) + g(√ln 3) = 1-e^{-ln 3}

= 1-⅓

= ⅔

∴ Option (a) is incorrect.

From equation (ii) : f’(x) =

Since f(x) is increasing in (0, 1)

∴ Option (d) is incorrect

ψ₁(x) = e^-x + x

⇒ ψ₁’(x) = 1- e^-x <1 for x>1

Then for α ∈ (1, x), ψ₁(x) = 1+ αx does not true for α > 1.

∴ Option (b) is incorrect

Now ψ₂(x) = x² – 2x – 2e^-x + 2

ψ₂’(x) = 2x – 2 + 2e^-x

∴ ψ₂’(x) = 2ψ₁(x)-2

From LMVT

=> ψ₂(x) = 2x(ψ₁(β) – 1)

Option (c) is correct.

Question 16: Which of the following statements is TRUE?

a. ψ₁(x) ≤ 1, for all x>0
b. ψ₂(x) ≤ 0, for all x>0
c.

Question 17: A number is chosen at random from the set {1, 2, 3……, 2000}. Let p be the probability that the number is a multiple of 3 or a multiple of 7. Then the value of 500p is;

Answer: (214)

E = a number which is multiple of 3 or multiple of 7

n(E) = (3, 6, 9, …….., 1998) + (7, 14, 21, ……….., 1995) – (21, 42, 63, ……… 1995)

n(E) = 666+ 285 – 95

n(E) = 856

n(E) = 2000

P(E) = 856/2000

P(E) × 500 = 856/4 = 214

Question 18: Let E be the ellipse x²/16 + y²/9 = 1. For any three distinct points P, Q and Q’ on E, let M (P, Q) be the mid-point of the line segment joining P and Q, and M(P, Q’) be the mid-point of the line segment joining P and Q’. Then the maximum possible value of the distance between M (P, Q) and M(P, Q’), as P, Q and Q’ vary on E, is

Answer: (4)

Let P(α), Q(θ), Q’(θ’)

M = ½ (4 cos α + 4 cos θ), ½ (3 sin α + 3 sin θ)

M’ = ½ (4 cos α + 4 cos θ’), ½ (3 sin α + 3 sin θ’)

MM’ = ½ √((4 cos θ – 4 cos θ’)² + (3 sin θ – 3 sin θ’)²)

MM’ = ½ distance between Q and Q’

MM’ is not depending on P

Maximum of QQ’ is possible when QQ’ = major axis

QQ’ = 2(4) = 8

MM’ = ½ (QQ’)

MM’ = 4

Question 19: For any real number x, let [x] denote the largest integer less than or equal to x. If

Answer: (182.00)

y = 10x/(x+1), 0 ≤ x≤10

xy+y = 10x

x = y/(10-y)

0≤ y/(10-y) ≤ 10

y/(10-y) ≥ 0 and (y/(10-y)) – 10 ≤ 0

y/(y-10) ≤ 0 and (11y-100)/(y-10) ≥ 0

y ∈ [0, 10) and y ∈ (-∞, 100/11] ⋃ (10, ∞)

y ∈ [0, 100/11]

√y ∈ [0, 10/√11]

⇒ [√y] = {0, 1, 2, 3}

Case 1:

0≤10x/(x+1) < 1

10x/(x+1) ≥ 0 and 10x/(x+1) – 1 < 0

x∈ (-∞, -1) ⋃ [0, ∞) and x∈ (-1, 1/9)

x∈ [0, 1/9) then [√(10x/(x+1))] = 0

Case 2:

1 ≤ 10x/(x+1) < 4

10x/(x+1) – 1 ≥ 0 and 10x/(x+1) – 4 < 0

(9x-1)/(x+1) ≥ 0 and (6x-4)/(x+1) < 0

x∈ (-∞, -1)⋃ [1/9, ∞) and x∈ (-1, ⅔)

x∈[1/9, ⅔), [√(10x/(x+1))] = 1

Case 3:

4 ≤ (10x/x+1) < 9

10x/(x+1) – 4 ≥ 0 and 10x/(x+1) < 9

(6x-4)/(x+1) ≥ 0 and (x-9)/(x+1) < 0

x∈ (-∞, -1)⋃ [2/3, ∞) and x∈ (-1, 9)

x∈ [⅔, 9); [√(10x/(x+1))] = 2

Case 4:

x∈ [9, 10]

⇒ [√(10x/(x+1))] = 3

l =

l = (⅔ – 1/9) +2(9-⅔) + 3(10-9)

l = 5/9 + 50/3 + 3

9l = 182