Students can instantly download the solved JEE Advanced 2021 Maths question paper for Paper 2 in PDF format for offline practice. We have also provided analysis videos of both Paper 1 and Paper 2 for the exam that was held on October 3rd. Students will further get helpful insights about the paper pattern, question types, weightage of marks and more. More significantly, engaging in this study exercise will keep every JEE aspirant in good stead as the exam approaches nearer and nearer.

**Question 1: Let; **

**S _{1 }= {(i, j, k) : i, j, k ∈ {1,2,…,10}} **

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

**S _{3} = {(i, j, k, l) : 1 ≤ i < j < k < l, i, j, k, l ∈ {1,2,…,10}}**

**S _{4} = {(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. n_{1} = 1000

b. n_{2} = 44

c. n_{3} = 220

d. n_{4}/12 = 420

Answer: (a, b, d)

Number of elements in S_{1} = 10 × 10 × 10 = 1000

Number of elements in S_{2} = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 44

Number of elements in S_{3} = ^{10}C_{4} = 210

Number of elements in S_{4} = ^{10}P_{4} = 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 – p^{2}/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 = (q^{2}+r^{2} – p^{2})/2qr

And (q^{2}+r^{2})/2 ≥ √(q^{2}. r^{2}) (AM ≥ GM)

⇒ (q^{2}+r^{2}) ≥ 2qr

So cos P ≥ (2qr – p^{2})/2qr

cos P ≥ 1 – p^{2}/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.

d.

Answer: (a, b, c)

f(0) = 1, ∫_{0}^{π/3} f(t) dt = 0

(a) Consider a function g(x) = ∫_{0}^{x }f(t)dt – sin 3x. g(x) is continuous and differentiable function

And g(0) = 0

g(π/3) = 0

By Rolle’s theorem g’(x) = 0 has at least one solution in (0, π/3)

f(x) – 3 cos 3x = 0 for some x ∈ (0, π/3)

(b) Consider a function

h(x) = ∫_{0}^{x} f(t)dt + cos 3x + 6x/π

h(x) is continuous and differentiable function and h(0) = 1

h(π/3) = 1

By Rolle’s theorem h’(x) = 0 for at least one x ∈ (0, π/3)

f(x) – 3 sin 3x + 6/π = 0 for some x ∈ (0, π/3)

(c)

(0/0 form)

By L’Hospital rule

=

=

= (0+2f(0))/(0-2)

= -1

(d)

=

=

= (1+0+1-0)/2

= 1

**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) = (x^{2}/2)e^{-x} + (e – ½)e^{-x}

b. f(x) = (-x^{2}/2)e^{-x }+ (e + ½)e^{-x}

c. f(x) = (e^{x}/2)(x-½) + (e – e^{2}/4)e^{-x}

d. f(x) = (e^{x}/2)(½ -x) + (e + e^{2}/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 – e^{2}/4

So, y = (e^{x}/2)(x-½) + (e – e^{2}/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 }= x^{2}/2 + C

y = e^{-x}x^{2}/2 + Ce^{-x}

y(1) = 1

1 = 1/2e + C/e

⇒ C = e – ½

y = e^{-x}x^{2}/2 + (e – ½)e^{-x}

**Question 5: Let O be the origin and **

**\(\begin{array}{l}\overrightarrow{OA}=2\hat{i}+2\hat{j}+\hat{k}\end{array} \)**

**,\(\begin{array}{l}\overrightarrow{OB}=\hat{i}-2\hat{j}+2\hat{k}\end{array} \) and \(\begin{array}{l}\overrightarrow{OC}=\frac{1}{2}(\overrightarrow{OB}-\lambda \overrightarrow{OA})\end{array} \) for some λ> 0. If \(\begin{array}{l}\left | \overrightarrow{OB}\times \overrightarrow{OC}\right |=\frac{9}{2}\end{array} \), then which of the following statements is (are) TRUE ?**

a. Projection of

b. Area of the triangle OAB is 9/2

c. Area of the triangle ABC is 9/2

d. The acute angle between the diagonals of the parallelogram with adjacent sides

Answer: (a, b, c)

=

Now,

So,

So, λ = 1 (since λ >0)

(a) Projection of vector OC on vector OA =

= ½ (-2-8+1)/3

= -3/2

(b) Area of triangle OAB =

(c) Area of the triangle ABC is =

=

= 9/2

(d) Acute angle between the diagonals of the parallelogram with adjacent sides

cos θ =

= 18/3√2 ×√90

θ ≠ π/3

**Question 6: Let E denote the parabola y ^{2} = 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 : y^{2} = 8x

P : (-2, 4)

Point P (-2, 4) lies on directrix (x = -2) of parabola y^{2} = 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/(t_{1}+t_{2}) = 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 y ^{2} ≤ 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 y^{2} = 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} + y^{2} = r^{2}

Solving it with parabola

y^{2} = 4 – x we get

(x – a)^{2} + 4 – x = r^{2}

x^{2} – x(2a + 1) + (a^{2} + 4 – r^{2}) = 0 …(1)

D = 0

⇒ 4r^{2} + 4a – 15 = 0

Clearly a ≥ r

So 4r^{2} + 4r – 15 ≤ 0

⇒ r_{max} = 3/2 = a

Radius of circle C is 3/2

From (1) x^{2 }– 4x + 4 = 0

⇒ x = 2 = α

**Question Stem for Question Nos. 9 and 10**

**Let f _{1} : (0, ∞) → R and f_{2 }: (0, ∞) → R be defined by f_{1}(x) = **

**\(\begin{array}{l}\int_{0}^{x}\prod_{j=1}^{21}(t-j)^{j}dt\end{array} \)**

**, x>0 and f\(\begin{array}{l}\prod_{i=1}^{n}a_{i}\end{array} \) denotes the product of a**

_{2}(x) = 98(x – 1)^{50}– 600(x – 1)^{49}+ 2450, x > 0, where, for any positive integer n and real number a_{1}, a_{2},…a_{n}._{1}, a_{2},..a_{n}. Let m_{i}and n_{i}, respectively, denote the number of points of local minima and the number of points of local maxima of function f_{i}, i = 1, 2, in the interval (0, ∞).

**Question 9: The value of 2m _{1} + 3n_{1} + m_{1}n_{1} is**

Answer: (57.00)

**Question 10: The value of 6m _{2} + 4n_{2} + 8m_{2}n_{2} is**

Answer: (06.00)

**Solution for Question 9 and 10**

f_{1}’(x) = (x – 1)(x – 2)^{2} (x – 3)^{3} ,…, (x – 20)^{20} (x – 21)^{21}

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

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

⇒ m_{1} = 6, n_{1} = 5

f_{2}(x) = 98(x – 1)^{50} – 600(x – 1)^{49} + 2450

f_{2}’(x) = 98 × 50(x – 1)^{49}– 600 × 49 × (x – 1)^{48}

= 98 × 50 × (x – 1)^{48} (x – 7)

f_{2}(x) has local minimum at x = 7 and no local maximum.

⇒ m_{2} = 1, n_{2} = 0

2m_{1} + 3n_{1} + m_{1}n_{1}

= 2 × 6 + 3 × 5 + 6 × 5

= 57

6m_{2} + 4n_{2} + 8m_{2}n_{2}

= 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 g_{1}(x) = 1, g_{2}(x) = |4x – π| and f(x) = sin^{2}x, for all x∈[π/8, 3π/8]. Define **

**\(\begin{array}{l}S_{i}=\int_{\frac{\pi }{8}}^{\frac{3\pi }{8}}f(x).g_{i}(x)dx\end{array} \)**

**, i = 1,2.**

**Question 11: The value of 16S _{1}/π is**

Answer: (2.00)

S_{1} = ∫_{π/8}^{3π/8} sin^{2}x . 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

=> 16S_{1}/π = 2

**Question 12:****The value of 48S _{2}/π^{2} is**

Answer: (1.50)

S_{2}= ∫_{π/8}^{3π/8} sin^{2}x |4x – π| dx

= ∫_{π/8}^{3π/8} 4 sin^{2}x |x – π/4| dx

Let x – π/4 = t

=> dx = dt

S_{2} = ∫_{-π/8}^{π/8} 4 sin^{2} (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/8}t dt + 0

S_{2} = [2t^{2}]_{0}^{π/8}

= π^{2}/32

48S_{2}/π^{2} = 3/2

**Question Paragraph: Let M = {(x, y) ∈ R × R : x ^{2} + y^{2}≤ r^{2}}, where r > 0. Consider the geometric progression a_{n} = 1/2^{n-1}, n = 1,2, 3… Let S_{0} = 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 – 1/2^{n})

For circles C_{n} to be inside M.

S_{n-1} + a_{n} < 1025/513

⇒ S_{n} < 1025/513

⇒ 1 – 1/2^{n} < 1025/1026

⇒ 1 – 1/1026

⇒ 2^{n} < 1026

⇒ n ≤ 10

∴ Number of circles inside be 10 = K

Clearly, alternate circles do not intersect each other i.e., C_{1}, C_{3}, C_{5}, C_{7}, C_{9} do not intersect each other as well as C_{2}, C_{4}, C_{6}, C_{8} and C_{10} 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 = (2 ^{199}-1)√2/2^{198}. 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 = (2^{199}-1)√2/2^{198}

Now, √2S_{n-1} + a_{n} < (2^{199}-1)√2/2^{198}

2√2(1 – 1/2^{n-1}) + 1/2^{n-1} < (2^{199}-1)/2^{198}

∴ 2√2 – √2/2^{n-2} + 1/2^{n-1} < 2√2 – √2/2^{198}

(1/2^{n-2})(½ – √2) < -√2/2^{198}

(2√2-1)/2. 2^{n-2} > √2/2^{198}

2^{n-2} < (2 – 1/√2) 2^{197}

n ≤ 199

∴ Number of circles = 199

Option (B) is correct.

**Question Paragraph: Let ψ _{1} = [0, ∞) → R, ψ_{2} = [0, ∞) → R, f:[0, ∞) → R and g:[0, ∞) → R be functions such that f(0) = g(0) = 0, **

**ψ _{1}(x) = e^{-x} + x, x≥0, **

**ψ _{2}(x) = x^{2} – 2x – 2e^{-x} + 2, x≥0**

**\(\begin{array}{l}f(x)=\int_{-x}^{x}(\left | t \right |-t^{2})e^{-t^{2}}dt, x> 0\end{array} \)**

**And **

**\(\begin{array}{l}g(x)=\int_{0}^{x^{2}}\sqrt{t}e^{-t}dt,x> 0\end{array} \)**

**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 ψ_{1}(x) = 1 + α x

c. For every x > 0, there exists a β ∈ (0, x) such that ψ_{2}(x) = 2x (ψ_{1}(β) -1)

d. f is an increasing function on the interval [0, 3/2]

Answer: (c)

Since,

Let t = u^{2}

⇒ dt = 2u du

So g(x) =

=

And

Therefore

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

Let t^{2} = 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

ψ_{1}(x) = e^{-x} + x

⇒ ψ_{1}’(x) = 1- e^{-x} <1 for x>1

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

∴ Option (b) is incorrect

Now ψ_{2}(x) = x^{2} – 2x – 2e^{-x} + 2

ψ_{2}’(x) = 2x – 2 + 2e^{-x}

∴ ψ_{2}’(x) = 2ψ_{1}(x)-2

From LMVT

[ψ_{2}(x) – ψ

_{2}(0)]/(x-0) = ψ

_{2}’(β) for β ∈(∞, x)

=> ψ_{2}(x) = 2x(ψ_{1}(β) – 1)

Option (c) is correct.

**Question 16: Which of the following statements is TRUE?**

a. ψ_{1}(x) ≤ 1, for all x>0

b. ψ_{2}(x) ≤ 0, for all x>0

c.

d. g(x)≤ (2/3) x

^{3}– (⅖)x

^{5}+ (1/7)x

^{7}, for all x∈(0, ½)

Answer: (d)

Since ψ_{1}(x) = e^{-x}+x

And for all x>0, ψ_{1}(x) > 1

∴ (a) is not correct

ψ_{2}(x) = x^{2} + 2 – 2(e^{-x} + x) >0 for x > 0

∴ (b) is not correct.

Now, √te^{-t} = √t(1 – t/1! + t^{2}/2! – t^{3}/3! + ….∞)

And √te^{-t} ≤ t^{1/2} – t^{3/2} + ½ t^{5/2}

∴

= (⅔) x^{3} – (⅔) x^{5} + (1/7) + (1/7) x^{7}

∴ Option (d) is correct

And

=

=

=

Therefore

=

∴ Option (c) is incorrect.

**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 x ^{2}/16 + y^{2}/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 **

**\(\begin{array}{l}l=\int_{0}^{10}\left [ \sqrt{\frac{10x}{x+1}} \right ]dx\end{array} \)**

**, then the value of 9I is;**

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

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