JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1

SECTION A

Question 1: The coefficients a,b and c of the quadratic equation, ax² + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :

a. 1/54
b. 1/72
c. 1/36
d. 5/216

Answer: (d)

ax² + bx + c = 0

a,b,c ∈ {1,2,3,4,5,6}

n(s) = 6 × 6 × 6 = 216

D=0 ⇒ b² = 4ac

ac = b²/4, If b = 2, ac = 1 ⇒ a = 1,c = 1

If b = 4, ac = 4 ⇒ a = 1, c = 4

a = 4, c = 1

a = 2, c = 2

If b = 6, ac = 9 ⇒ a = 3, c = 3

Therefore, probability = 5/216

Question 2: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l² + m² – n² = 0. Then the value of sin⁴ α + cos⁴ α is :

a. 3/4
b. 1/2
c. 5/8
d. 3/8

Answer: (c)

Given that l+m=n ….(1)

l² + m² – n² = 0 ….(2)

Squaring equation (1)

l² + m² + 2lm = n² ….(3)

From equations (2) and (3)

lm=0 ⇒ l = 0 or m = 0

Case (1) : l = 0

⇒ m=n

⇒ l² + m² + n² = 0

⇒ m=n = ±1/(√2 )

∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2)

Case (2): m = 0

⇒ l = n

⇒ l² + m² + n² = 0

Question 3: The value of the integral

(where c is a constant of integration)

a. 1/18[9 – 2 sin⁶ θ – 3 sin⁴ θ – 6 sin² θ ]^(3/2) + c
b. 1/18[11 – 18 sin² θ + 9 sin⁴ θ – 2 sin⁶ θ ]^(3/2) + c
c. 1/18[11 – 18 cos² θ + 9 cos⁴ θ – 2 cos⁶ θ ]^(3/2) + c
d. 1/18[9 – 2 cos⁶ θ – 3 cos⁴ θ – 6 cos² θ ]^(3/2) + c

Answer: (c)

Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin²A

Question 4: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

a. 10(√3-1)
b. 10√3
c. 10
d. 10(√3+1)

Answer: (d)

x + y = √3h ….…….(1)

Also,

h/y = tan 45^o

h = y …….(2)

put in (1)

x + y = √3y

x = (√3 – 1)y

x/20=v (speed)

Therefore, time taken to reach

Foot from B

= y/V

= x/(√3-1)x . 20

= 10 (√3 + 1)

Question 5:

a. xyz = 4
b. xy – z = (x + y)z
c. xy + yz + zx = z
d. xy + z = (x+y)z

Answer: (d)

Question 6: The equation of the line through the point (0,1,2) and perpendicular to the line

a.

Question 7: The statement A→ (B → A) is equivalent to:

a. A → (A ᐱ B)
b. A → (A ᐯ B)
c. A → (A → B) 
d. A → (A ↔ B)

Answer: (b)

 A → (B → A) 

⇒ A → (∼ B ᐯ A)

⇒ ∼ A ᐯ (∼ B ᐯ A)

⇒ ∼ B ᐯ (∼ A ᐯ A)

⇒ ∼ B ᐯ t

= t (tautology)

From options:

(b) A → (A ᐯ B)

⇒ ∼ A ᐯ (A ᐯ B)

⇒ (∼ A ᐯ A) ᐯ B

⇒ t ᐯ B

⇒ t

Question 8: The integer k, for which the inequality x² – 2(3k – 1)x + 8k² – 7 >0 is valid for every x in R is :

a. 3
b. 2
c. 4
d. 0

Answer: (a)

D < 0

(2(3k-1))² – 4(8k² – 7) < 0

4(9k² – 6k + 1) – 4(8k² – 7) < 0

k² – 6 k + 8 < 0

(k–4)(k–2)<0

2< k < 4

then k = 3

Question 9: A tangent is drawn to the parabola y² = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?

a. (0,3)
b. (-6,0)
c. (4,5)
d. (5,4)

Answer: (d)

Equation of tangent : y = mx + 3/(2m)

m_T = (1/2) (Because perpendicular to line 2x + y = 1)

Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0

Substitute given options in the above equation and check.

We can find that (5, 4) does not lie on the line.

Hence option (d) is the answer.

Question 10: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?

a. f is one-one
b. If fog is one-one, then g is one-one
c. If g is onto, then fog is one-one
d. If f is onto, then f(n) = n for all n ∈ N

Answer: (c)

f(n + 1) = f(n) +f(1)

⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1)

General term = T_n = f(1) + (n-1)f(1) = nf(1)

⇒f(n) = nf(1)

Clearly f(n) is one-one.

For fog to be one-one, g must be one-one.

For f to be onto, f(n) should take all the values of natural numbers.

As f(x) is increasing, f(1)=1

⇒f(n)=n

If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.

Question 11: Let the lines (2 – i)z = (2 + i)

a. 3/√2
b. 3√2
c. 3/(2√2)
d. 1/(2√2)

Answer: (c)

(2–i)z = (2+i)

⇒(2–i)(x+ iy)=(2+i)(x–iy)

⇒2x–ix + 2iy + y=2x+ ix–2iy+y

⇒2ix –4iy=0

L₁ ∶ x–2y=0

⇒(2+i)z+(i –2)

⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0.

⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0

⇒2ix + 4iy – 4i =0

L₂ ∶ x+2y–2 = 0

Solve L₁ and L₂

4y=2 or y=1/2

∴ x = 1

Centre(1, 1/2)

L₃∶ iz +

⇒i(x+iy)+x–iy+1+i=0

⇒ix–y+x–iy+1+i=0

⇒(x–y+1)+i(x–y+1)=0

Radius = distance from (1, 1/2) to x – y + 1 = 0

r = (1-1/2+1)/ √2

r = 3/2√2

Question 12: All possible values of θ ∈ [0, 2π] for which sin2θ + tan 2θ > 0 lie in

a. 

Question 13: The image of the point (3,5) in the line x – y + 1 = 0, lies on :

a. (x – 2)² + (y – 4) ² =4
b. (x – 4) ² + (y + 2)² =16
c. (x – 4)² + (y – 4)² = 8
d. (x – 2)² + (y – 2)² =12

Answer: (a)

Image of P(3, 5) on the line x – y + 1 = 0 is

x = 4, y = 4

Image is (4, 4)

Which lies on (x – 2)² + (y – 4)² = 4

Question 14: If Rolle’s theorem holds for the function f(x) = x³ – ax² + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :

a. (–5, 8)
b. (5, 8)
c. (5, –8)
d. (–5, –8)

Answer: (b)

f(1) = f(2)

⇒ 1 – a + b –4 = 8 – 4a + 2b –4

3a – b = 7 …(1)

f’(x) = 3x² – 2ax + b

⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0

⇒–8a + 3b = –16 …(2)

a = 5, b = 8

Question 15: If the curves,

a. a + b = c + d
b. a- b = c – d
c. ab = (c+d)/(a+b)
d. a-c = b+d

Answer: (b)

diff :

Diff :

m₁m₂ = –1 ⇒

⇒ bdx² = – acy²…….(5)

(1)–(3) ⇒ (1/a – 1/c)x² + (1/b – 1/d) y² = 0

Solve above equation using 5

⇒ (c – a) – (d – b) = 0

⇒ c – a = d – b

⇒ c – d = a – b

Question 16:

a. 1/2
b. 1/e
c. 1
d. 0

Answer: (c)

Question 17: The total number of positive integral solutions (x, y, z) such that xyz = 24 is

a. 36
b. 45
c. 24
d. 30

Answer: (d)

x.y.z = 24

x.y.z=2³.3¹

x=2^{(a_1 )}⋅3^{(b_1 )}

y = 2^{(a_2 )}⋅3^{(b_2 )}

z = 2^{(a_3 )}⋅3^{(b_3 )}

a₁, a₂, a₃ ∈ {0,1,2,3}

b₁, b₂, b₃ ∈ {0,1}

Case 1: a₁ + a₂ + a₃ =3

Non negative solution = ^(3+3-1) C_(3-1) = ⁵C₂ = 10

Case 2: b₁ + b₂ + b₃ = 1

Non negative solution = ^(1+3-1) C_(3-1) = ³C₂ = 3

∴ Total solutions =10×3=30

Question 18: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is

a. (4, 5)
b. (5, 4)
c. (4, 4)
d. (5, 5)

Answer: (d)

Let x – 2 = t ⇒ dx = dt

and y + 4 = u ⇒dy = du

dy/dx = du/dt

du/dt = t + u/t ⇒ du/dt – u/t = t

Here, IF = 1/t

u. (1/t) = ∫ t.(1\t) dt

⇒ u/t = t + c

⇒ (y+4)/(y-2) = (x – 2) + c

Passing through (0, 0)

c = 0

⇒ (y + 4) = (x – 2)²

Question 19: The value of

a. (e+1)/3
b. (e-1)/3e
c. (e+1)/3e
d. 1/3e

Answer: (c)

Question 20: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

a. 1/8
b. 1/27
c. 3/4
d. 3/8

Answer:(a)

Probability of not getting intercepted = 2/3

Probability of missile hitting target = 3/4

Probability that all 3 hit the target = (2/3 x3/4)³ = 1/8

Section B

Question 1: Let A₁, A₂, A₃,…. be squares such that for each n ≥ 1, the length of the side of A_n equals the length of diagonal of A_n+1. If the length of A₁ is 12 cm, then the smallest value of n for which area of A_n is less than one is ____________.

Answer: (9)

Side lengths are in GP.

T_n = 12/(√2)^n-1

Area = 144/2^n-1 <1

=> 2^n-1 > 144

Smallest n = 9

Question 2: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A⁴ is equal to ___________

Answer: (64)

Question 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.

Answer: (2)

(√3)kx + ky – 4√3 = 0 …(1)

√3kx -ky= 4√3 k² …(2)

Adding equation (1) & (2)

2√3 kx = 4√3(k^2 + 1)

x = 2 (k + 1/k) ….(3)

Subtracting equation (1) & (2)

y = 2√3(1/k – k) ………(4)

x²/4 – y²/12 = 4

x²/16 – y²/48 = 1 – Hyperbola

e² = 1 + 48/16

or e = 2

Question 4:

Answer: 13

|I₂ + A|=|I₂ – A|

From equation (1)

∴a² + b² = 1

⇒13(a² + b²) = 13

Question 5: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = –1 and x = 1. If

Answer: (144)

f(x) = x⁶ + ax⁵ + bx⁴ + x³

f’(x) = 6x⁵ + 5ax⁴ + 4bx³ + 3x²

Roots 1 & – 1

Therefore, 6 + 5a + 4b + 3 = 0 & – 6 + 5a – 4b + 3 = 0 solving

a = -3/5 and b = -3/2

f(x) = x⁶ – (3/5)x⁵ – (3/2)x⁴ + x³

5.f(2) = 5 [64 – 96/5 – 24 + 8] = 144

Question 6: The number of points, at which the function f(x) = |2x + 1| – 3|x+2|+|x² + x–2|, x ∈ R is not differentiable, is ________.

Answer: 2

Check at 1, –2 and -1/2

Non-Differentiable at x = 1 and -1/2

Question 7: If the system of equations

kx + y + 2z = 1

3x – y – 2z = 2

–2x – 2y – 4z = 3

has infinitely many solutions, then k is equal to __________.

Answer: 21

D = D₁ = D₂ = D₃ = 0

Choose D₂ = 0

k(–8+6)–1(–12–4)+2(9+4)=0

–2k + 16 + 26 = 0

2k = 42

k = 21

Question 8: Let

Answer: 12

Question 9: Let A =

Answer: 7

Question 10: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.

Answer: 32