JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 2

SECTION A

Question 1: Let a, b∈R. If the mirror image of the point P(a, 6, 9) with respect to the line (x – 3)/7 = (y – 2)/5 = (z – 1)/(-9) is (20, b, -a, -9), then |a + b| is equal to:

a. 86
b. 88
c. 84
d. 90

Answer: (b)

P(a, 6, 9), Q (20, b, -a-9)

Mid point of PQ=((a+20)/2, (b+6)/2, -a/2) lie on the line.

=> (a + 20 – 6)/14 = (b + 6 – 4)/10 = (-a – 2)/(-18)

=> (a + 14)/14 = (a + 2)/18

=> 18a + 252 = 14a + 28

=> 4a = -224

a = -56

(b + 2)/10 = (a + 2)/18

=> (b + 2)/10 = (-54)/18

=>(b + 2)/10 = -3

=> b = -32

|a + b| = |-56 – 32|

= 88

Question 2: Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x∈R. If |f(x) f'(x) f'(x) f”(x)| = 0, for all x∈R, then the value of f(1) lies in the interval:

a. (9, 12)
b. (6, 9)
c. (3, 6)
d. (0, 3)

Answer: (b)

Given f(x) f”(x) – f'(x)² =0

Let h(x) = f(x)/f'(x)

Then h'(x) = 0

⇒ h(x) = k

⇒ f(x)/f'(x) = k

⇒ f(x) = k f’(x)

⇒ f(0) = k f'(0)

⇒ k = 1/2

Now, f(x) = ½ f'(x)

⇒∫ 2dx = ∫ f'(x)/f(x) dx

⇒ 2x = ln |f(x)| + C

As f(0) = 1 ⇒ C = 0

⇒ 2x = ln |f(x)|

⇒ f(x) = ±e^2x

As f(0) = 1 ⇒ f(x) = e^2x

∴ f(1) = e² ≈ 7.38

Question 3: A possible value of tan (¼ sin^-1 √63/8) is:

a. 1/(2√2)
b. 1/√7
c. √7 – 1
d. 2√2 – 1

Answer: (b)

tan (¼ sin^-1 √63/8)

Let sin^-1(√63/8) = θ

sin θ = √63/8

cos θ = 1/8

2 cos²(θ/2) – 1 = 1/8

⇒ cos² θ/2 = 9/16

cos θ/2 = 3/4

⇒(1- tan² θ/4 )/(1 + tan² θ/4) = 3/4

tan θ/4 = 1/√7

Question 4: The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is:

a. 65/2⁷
b. 135/2⁹
c. 65/2⁸
d. 35/2⁷

Answer: (b)

Let A and B be two subsets.

For each x∈{1,2,3,4,5} , there are four possibilities:

x ∈ A∩B, x∈ A’∩B, x ∈ A∩B’, x ∈ A’∩B’

So, the number of elements in sample space =4⁵

Required probability

= (⁵C₂ × 3³)/4⁵

= (10×27)/2¹⁰

= 135/2⁹

Question 5: The vector equation of the plane passing through the intersection of the planes

a. 

Question 6: If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x − 1, then the co-ordinates of P are :

a. (–2, 8)
b. (1, 5)
c. (3, 13)
d. (2, 8)

Answer: (d)

Tangent at P is parallel to the given line.

dy/dx|_P = 4

⇒2x₁ = 4

⇒x₁ = 2

Required point is (2, 8)

Question 7: Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax² + bx + 1 = 0, then the value of α² + β² – αβ is:

a. 71/256
b. -69/256
c. 69/256
d. -71/256

Answer: (d)

2b = a + c

(2a + 2)/3 = 10/3 and (2b + c)/3 = 7/3

⇒ a = 4

2b + c = 7, 2b – c = 4 },solving

b = 11/4 and c = 3/2

∴ Quadratic equation is 4x²+ (11/4)x + 1 = 0

∴ The value of (α + β)² – 3αβ = (121/256) – (¾)

= -71/256

Question 8: The value of the integral,

a. –4
b. –5
c. -√2 – √3 – 1
d. -√2 – √3 + 1

Answer: (c)

Put x – 1 = t ; dx = dt

I = -6 + (√2 – 1) + 2√3 – 2√2 + 6 – 3√3

I = -1 – √2 – √3

Question 9: Let f: R → R be defined as

Let A = {x ∈ R ∶ f is increasing}. Then A is equal to :

a. (-5, -4) ∪ (4,∞)
b. (-5, ∞)
c. (-∞, -5) ∪ (4, ∞)
d. (-∞, -5) ∪ (-4, ∞)

Answer: (a)

f’(x) = {-55 ; x< -5 6(x² – x – 20) ; -5 < x < 4 6(x²– x – 6) ; x > 4

⇒ f’(x) = {-55 ; x< -5 6(x – 5)(x + 4) ; –5 < x < 4 6(x – 3)(x + 2) ; x > 4

Hence, f(x) is monotonically increasing in (-5, -4) ∪ (4, ∞)

Question 10: If the curve y = ax² + bx + c,x ∈ R passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:

a. a = 1, b = 1, c = 0
b. a = -1, b = 1, c = 1
c. a = 1, b = 0, c = 1
d. a = 1/2, b = 1/2 ,c = 1

Answer: (a)

2 = a + b + c

dy/dx = 2ax + b, (dy/dx)_(0,0) = 1

⇒ b = 1 and a + c = 1

Since (0, 0) lies on curve,

∴ c = 0, a = 1

TRICK: (0, 0) lies on the curve. Only option (1) has c = 0

Question 11: The negation of the statement ~p∧(p∨q) is∶

a. ~p∧q
b. p∧~q
c. ~p∨q
d. p∨∼q

Answer: (d)

Negation of ~p∧(p∨q) is

∼[~p∧(p∨q)]

≡ p∨ ∼(p∨q)

≡ p∨(∼p∧∼q)

≡ (p∨∼p)∧(p∨∼q)

≡ T∧(p∨∼q), where T is tautology.

≡ p∨∼q

Question 12: For the system of linear equations: x − 2y = 1, x − y + kz = −2, ky + 4z = 6,k ∈ R

Consider the following statements:

(A) The system has unique solution if k ≠ 2,k ≠ −2.

(B) The system has unique solution if k = -2.

(C) The system has unique solution if k = 2.

(D) The system has no-solution if k = 2.

(E) The system has infinite number of solutions if k ≠ -2.

Which of the following statements are correct?

a. (B) and (E) only
b. (C) and (D) only
c. (A) and (D) only
d. (A) and (E) only

Answer: (c)

x -2y + 0.z = 1

x – y + kz = -2

0.x + ky + 4z = 6

∆ =

= 4 – k²

For unique solution, 4 – k² ≠ 0

k ≠ ±2

For k = 2,

x – 2y + 0.z =1

x – y + 2z = -2

0.x + 2y + 4z = 6

Δ_x =

= -8 + 2 (-20)

⇒Δ_x= -48 ≠ 0

For k = 2,Δ_x ≠ 0

So, for k = 2, the system has no solution.

Question 13: For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2)?

a. x² + 9y² = 9
b. 2x² – 18y² = 9
c. y² = x/(6√3)
d. x² + y² = 7

Answer: (a)

Tangent to x² + 9y² = 9 at point (3√3/2, 1/2) is x(3√3)/2 + 9y(1/2) = 9

⇒3√3 x + 9y = 18

⇒x +√3 y = 2√3

⇒ Option (a) is true.

Question 14: The angle of elevation of a jet plane from a point A on the ground is 60⁰. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30⁰. If the jet plane is flying at a constant height, then its height is:

a. 1200√3 m
b. 1800√3 m
c. 3600√3 m
d. 2400√3 m

Answer: (a)

v = 432 × 1000/(60×60) m/sec

= 120 m/sec

Distance PQ = v × 20 = 2400 m

In ΔPAC

tan 60⁰ = h/AC

⇒ AC = h/(√3)

In ΔAQD

tan 30⁰ = h/AD

⇒AD = √3h

AD = AC + CD

⇒√3 h = h/√3 + 2400

⇒2h/√3 = 2400

⇒ h = 1200√3 m

Question 15: For the statements p and q, consider the following compound statements:

(a) (~q∧(p→q)) → ~p

(b) ((p∨q))∧~p) → q

Then which of the following statements is correct?

a. (a) is a tautology but not (b)
b. (a) and (b) both are not tautologies
c. (a) and (b) both are tautologies
d. (b) is a tautology but not (a)

Answer: (c)

(a) is tautology.

(b) is tautology.

∴ (a) and (b) both are tautologies.

Question 16: Let A and B be 3×3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A²B² − B²A²)X = O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has:

a. a unique solution
b. exactly two solutions
c. infinitely many solutions
d. no solution

Answer: (c)

A^T = A, B^T = –B

Let A²B² − B²A²= P

P^T = (A²B² – B²A²)^T

= (A²B²)^T – (B²A²)^T

= (B²)^T (A²)^T– (A²)^T (B²)^T

= B²A² – A²B

⇒ P is a skew-symmetric matrix.

∴ ay + bz = 0 …(1)

–ax + cz = 0 …(2)

–bx – cy =0 …(3)

From equation (1), (2), (3)

Δ = 0 and Δ₁ = Δ₂= Δ₃ = 0

∴ System of equations has infinite number of solutions.

Question 17: If n ≥ 2 is a positive integer, then the sum of the series ⁿ⁺¹C₂ + 2(²C₂ + ³C₂ + ⁴C₂ + …. + ⁿC₂) is

a. n(n + 1)² (n + 2)/12
b. n(n – 1)(2n + 1)/6
c. n(n + 1)(2n + 1)/6
d. n(2n + 1)(3n + 1)/6

Answer: (c)

²C₂ = ³C₃

Let S = ³C₃ + ³C₂ + ……. + ⁿC₂ = ⁿ⁺¹C₃ (∵ ⁿC_r + ⁿC_r–1 = ⁿ⁺¹C_r)

∴ ⁿ⁺¹C₂ + ⁿ⁺¹C₃ + ⁿ⁺¹C₃

= ⁿ⁺²C₃ + ⁿ⁺¹C₃

=(n + 2)!/3!(n – 1)! + (n + 1)!/3!(n – 2)!

=(n + 2)(n + 1)n/6 + (n + 1)(n)(n – 1)/6 = (n(n + 1)(2 + 1))/6

TRICK : Put n = 2 and verify the options.

Question 18: If a curve y = f(x) passes through the point (1, 2) and satisfies xdy/dx + y = bx⁴, then for what value of b,

a. 5
b. 62/5
c. 31/5
d. 10

Answer: (d)

dy/dx + y/x = bx³

I.F. =

= x

∴ yx = ∫bx⁴ dx = bx⁵/5 + c

The above curve passes through (1, 2).

2 = b/5 + c

Also,

⇒ (b/25) ×32 + c ln 2 – b/25 = 62/5

⇒ c = 0 and b = 10

Question 19: The area of the region: R{(x, y): 5x² ≤ y ≤ 2x² + 9} is:

a. 9√3 square units
b. 12√3 square units
c. 11√3 square units/
d. 6√3 square units

Answer: (b)

Required area

= 

=

= 

= 12√3

Question 20: Let f(x) be a differentiable function defined on [0,2] such that f’(x) =f ‘(2 – x) for all x∈(0, 2), f(0) = 1 and f(2) = e². Then the value of

a. 1 + e²
b. 1 – e²
c. 2(1 – e²)
d. 2(1+e²)

Answer: (a)

f'(x) = f'(2 – x)

On integrating both sides, we get

f(x) = -f(2 – x)+c

Put x = 0

f(0) + f(2) = c

⇒ c = 1+ e²

⇒ f(x) + f(2 – x) = 1 + e²

I =

= 1+e²

Section B

Question 1: The number of the real roots of the equation (x + 1)² + |x – 5| = 27/4 is ___.

Answer: 2

For x ≥ 5,

(x + 1)²+ (x – 5) = 27/4

⇒ x² + 3x – 4 = 27/4

⇒ x² + 3x – 43/4 = 0

⇒ 4x² + 12x – 43 = 0

x = (-12 ± √(144 + 688))/8

x = (-12 ± √832)/8

= (-12 ± 28.8)/8

= (-3 ± 7.2)/2

= (-3 + 7.2)/2, (-3 – 7.2)/2 (therefore, no solution)

For x < 5,

(x + 1)²– (x – 5) = 27/4

⇒ x² + x + 6 – 27/4=0

⇒ 4x² + 4x – 3=0

x = (-4 ± √(16 + 48))/8

x = (-4 ± 8)/8

⇒ x = -12/8, 4/8

∴ 2 real roots.

Question 2: The students S₁, S₂,… S₁₀ are to be divided into 3 groups A, B, and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___.

Answer: 31650

Number of ways

= ¹⁰C₁ [2⁹ – 2] + ¹⁰C₂ [2⁸– 2] + ¹⁰C₃ [2⁷ – 2]

= 2⁷ [¹⁰C₁×4 + ¹⁰C₂ ×2 + ¹⁰C₃] – 20 – 90 – 240

= 128 [40 + 90 + 120 ] – 350

= (128 × 250) – 350

= 10 [3165]

= 31650

Question 3: If a + α = 1, b + β = 2 and af(x) + αf(1/x) = bx + β/2, x ≠ 0 then the value of the expression [f(x) + f(1/x) ]/(x + 1/x)

Answer: 2

af(x) + αf(1/x) = bx + β/x …(i)

Replace x by 1/x

af(1/x) + αf(x) = b/x + βx …(ii)

(i) + (ii)

(a + α)[f(x) + f(1/x) ]

= (x + 1/x)(b + β)

⇒ ( f(x) + f(1/x))/(x + 1/x)

= 2/1

= 2

Question 4: If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ________.

Answer: 11

⇒

⇒ 10(9 + k²) – (81 + k² + 18k) < 1000

⇒ 90 + 10k² – k² – 18k – 81 < 1000

⇒ 9k²– 18k + 9 < 1000

⇒ (k – 1)² < 1000/9

⇒ k – 1< (10√10)/3

⇒ k < (10√10)/3 + 1

Maximum possible integral value of k is 11.

Question 5: Let λ be an integer. If the shortest distance between the lines x – λ = 2y – 1 = -2z and x = y + 2λ = z – λ is √7/2√2, then the value of |λ| is __

Answer: 1

(x – λ)/1 = (y – 1/2)/(1/2) = z/(-1/2)

(x – λ)/2 = (y-1/2)/1 = z/(-1) …(1) Point on line = (λ, 1/2, 0)

x/1 = (y + 2λ)/1 = (z – λ)/1 …(2) Point on line = (0, -2λ, λ)

= |-5λ – 3/2|/√14

= √7/(2√2) (Given)

⇒ |10λ + 3| = 7

⇒ λ = -1 as λ is an integer.

⇒ |λ| = 1

Question 6: If i = √-1. If [(-1 + i√3)²¹/(1 – i)²⁴ + (1 + i√3)²¹/(1 + i)²⁴ ] = k, and n = [|k| ] be the greatest integral part of |k|. Then 

Answer: 310

= 2⁹ e^i(20π) +2^{9 eiπ}

= 2⁹ + 2⁹ (-1)

= 0 = k

∴ n = 0

∑_(j=0)⁵ (j + 5)²– ∑_(j=0)⁵ (j + 5)

= [5² + 6² + 7² + 8² + 9² + 10² ] – [5 + 6 + 7 + 8 + 9 + 10]

= [(1²+ 2² +… + 10²) – (1² + 2²+ 3²+ 4²) ] – [(1 + 2 + 3 + … + 10) – (1 + 2 + 3 + 4) ]

= (385 – 30) -[55 – 10]

= 355 – 45

= 310

Question 7: Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (-5, 0). If the locus of the point P is a circle of radius r, then 4r² is equal to ___.

Answer: 56.25

Let P be (h, k), A(5, 0) and B(-5, 0)

Given PA = 3PB

⇒ PA² = 9PB²

⇒ (h-5)² + k² = 9[(h + 5)² + k²]

⇒ 8h² + 8k² + 100h + 200 = 0

∴ Locus of P is x² + y² + (25/2)x + 25 = 0

Centre = (-25/4, 0)

∴ r² = (-25/4)² – 25

= 625/16 – 25

= 225/16

∴4r² = 4×225/16

= 225/4

= 56.25

Question 8: For integers n and r, let

The maximum value of k for which the sum

exists, is equal to ____.

Answer: Bonus

(1+x)¹⁰ = ¹⁰C₀ + ¹⁰C₁x + ¹⁰C₂x² + …… + ¹⁰C₁₀x¹⁰

(1+x)¹⁵= ¹⁵C₀ + ¹⁵C₁x +¹⁵C₂x² + …… + ¹⁵C_k-1x^k-1 + ¹⁵C_k+1x^k+1 + …¹⁵C₁₅x¹⁵

Coefficient of x^k+1 in (1+x)²⁵

= ²⁵C_k+1

²⁵C_k + ²⁵C_k+1 = ²⁶C_k+1

For maximum value

as per given, k can be as large as possible.

Question 9: The sum of first four terms of a geometric progression (G.P.) is 65/12 and the sum of their respective reciprocals is 65/18. If the product of first three terms of the G.P. is 1, and the third term is α then 2α is____

Answer: 3

Let terms of GP are a, ar, ar², ar³

a + ar + ar² + ar³ = 65/12 …….(1)

1/a + 1/ar + 1/(ar²) + 1/ar³) = 65/18

⇒1/a (r³ + r² + r + 1)/r³) = 65/18 …….(2)

(1)/(2), we get

a²r³ = 18/12

= 3/2

Also, a³ r³ = 1

⇒ a(3/2) = 1

⇒a = 2/3

(4/9) r³ = 3/2

⇒ r³ = 3³/2³

⇒ r = 3/2

α = ar²

= 2/3.(3/2)² = 3/2

∴2α = 3

Question 10: If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x – 2)²+ (y – 3)² = 25 at the point (5, 7) is A, then 24A is equal to ___.

Answer: Question is wrong

Equation of normal at P is

(y – 7) = ((7 – 3)/(5 – 2))(x – 5)

⇒ 3y – 21 = 4x – 20

⇒ 4x – 3y + 1 = 0

⇒ M is (-1/4, 0)

Equation of tangent at P is

(y – 7) = (-¾) (x – 5)

⇒ 4y – 28 = -3x + 15

⇒ 3x + 4y = 43

⇒ N is (43/3, 0)

The question is wrong. The normal cuts at a point on the negative axis.