JEE Main 2021 February 26 – Shift 1 Maths Question Paper with Solutions

SECTION A

Question 1: The number of seven-digit integers with the sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is:

a. 77
b. 42
c. 35
d. 82

Answer: (a)

Case – I: 1, 1 , 1, 1, 1, 2, 3

Number of ways = 7! / 5! = 42

Case – II: 1, 1, 1, 1, 2, 2, 2

Number of ways = 7! / [4! * 3!] = 35

Total number of ways = 42 + 35 = 77

Question 2: The maximum value of the term independent of ‘t’ in the expansion of [(tx^1/5 + {(1 – x)^1/10 / t}]¹⁰ where x ∈ (0, 1) is :

a. 10! / [√3 (5!)²]
b. [2 . 10!] / [3 (5!)²]
c. [10!] / [3 (5!)²]
d. [2 . 10!] / [3√3 (5!)²]

Answer: (d)

T_r+1 = ¹⁰C_r (tx^1/5)^10-r {[(1 – x)^1/10] / t}^r

= ¹⁰C_r t ^10-2r x^[10-r]/5 (1 – x)^r/10

10 – 2r = 0

r = 5

T₆ = ¹⁰C₅ x √1 – x

dT₆ / dx = ¹⁰C₅ [√(1 – x) – (x/2√(1 – x)] = 0

1 – x = x / 2

x = 2 / 3

Max (T₆) = {[10!] / [5! 5!] * [2] / [3√3]}

Question 3: The value of ∑_n=1¹⁰⁰ ∫_n-1ⁿ e^x-[x] dx, where [x] is the greatest integer ≤ x, is :

a. 100 (e – 1)
b. 100e
c. 100 (1 – e)
d. 100 (1 + e)

Answer: (a)

∑_n=1¹⁰⁰ ∫_n-1ⁿ e^x-[x] dx

= ∫₀¹ e^{x} dx + ∫₁² e^{x} dx + ∫₂³ e^{x} dx + …….. ∫₉₉¹⁰⁰ e^{x} dx {because {x} = x – [x]}

= e^x |₀¹ + e^(x-1)|₁² + e^(x-2)|₂³ + …… + e^(x-99)|₉₉¹⁰⁰

= (e – 1) + (e – 1) + (e – 1) + …… + (e – 1)

= 100 (e – 1)

Question 4: The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria has increased by 20% in 2 hours. If the population of bacteria is 2000 after k / log_e (6 / 5) hours, then (k / log_e 2)² is equal to :

a. 4
b. 2
c. 16
d. 8

Answer: (a)

Let x be the number of bacteria at time t.

dx / dt ∝ x

dx / dt = λx

∫₁₀₀₀^x dx / x = ∫₀^t λ dt

(ln x) – (ln 1000) = λt

ln (x / 1000) = λt

Put t = 2, x = 1200

(ln 12 / 10) = 2λ

λ = (1 / 2) ln (6 / 5)
Now, ln (x / 1000) = (t / 2) ln (6 / 5)

x = 1000e^{t/2 ln (6 / 5)}

Given, x = 2000 at t = k / log_e (6 / 5)

2000 = 1000 e^{[k/2 ln (6 / 5)] * [ln (6 / 5)]}

2 = e^k/2 ⇒ ln 2 = k / 2

k / ln 2 = 2

Question 5: If

a. (1 / 2) |

Question 6: In an increasing geometric series, the sum of the second and the sixth term is 25 / 2 and the product of the third and fifth term is 25. Then, the sum of 4^th, 6^th and 8^th terms is equal to :

a. 35
b. 30
c. 26
d. 32

Answer: (a)

Let a be the first term and r be the common ratio.

ar + ar⁵ = 25 / 2 and ar² x ar⁴ = 25

a²r⁶ = 25

ar³ = 5

a = 5 / r³ …(1)

(5r / r³) + (5r⁵ / r³) = 25 / 2

(1 / r²) + r² = 5 / 2

Put r² = t

2t² – 5t + 2 = 0

2t² – 4t – t + 2 = 0

(2t – 1) (t – 2) = 0

t = 1 / 2, 2

r² = 1 / 2, 2

r = √2 as the G.P. is increasing.

ar³ + ar⁵ + ar⁷

= ar³( 1 + r² + r⁴)

= 5 [1 + 2 + 4]

= 35

Question 7: Consider the three planes

P₁ : 3x + 15y + 21z = 9,

P₂ : x – 3y – z = 5, and

P₃ : 2x + 10 y + 14z = 5

Then, which one of the following is true ?

a. P₁ and P₃ are parallel.
b. P₂ and P₃ are parallel.
c. P₁ and P₂ are parallel.
d. P₁, P₂ and P₃ all are parallel.

Answer: (a)

P₁ = x + 5y + 7z = 3

P₂ = x – 3y – z = 5

P₃ = x + 5y + 7z = 5/2

P₁||P₃

Question 8: The sum of the infinite series 1 + (2 / 3) + (7 / 3²) + (12 / 3³) + (17 / 3⁴) + (22 / 3⁵) + …… is equal to :

a. 9 / 4
b. 15 / 4
c. 13 / 4
d. 11 / 4

Answer: (c)

s = 1 + (2 / 3) + (7 / 3²) + (12 / 3³) + (17 / 3⁴) + (22 / 3⁵) + ……

(s / 3) = (1 / 3) + (2 / 3²) + (7 / 3³) + ……. infinity

_________________________________

(2s / 3) = 1 + (1 / 3) + (5 / 3²) + (5 / 3³) + ….. infinity

2s / 3 = (4 / 3) + (5 / 3) [(1 / 3) / (1 – (1 / 3)] = (4 / 3) + (5 /6) = 13 / 6

s = 13 / 4

Question 9: The value of 

a. –2
b. (a + 1) (a + 2) (a + 3)
c. 0
d. (a + 2) (a + 3) (a + 4)

Answer: (a)

Question 10: If (sin^-1 x) / a = (cos^-1 x) / b) = (tan^-1 y) / c; 0 < x < 1, then the value of cos (πc / [a + b]) is:

a. [1 – y²] / 2y
b. [1 – y²] / [1 + y²]
c. 1 – y²
d. [1 – y²] / y √y

Answer: (b)

(sin^-1 x) / a = (cos^-1 x) / b) = (tan^-1 y) / c

(sin^-1 x) / a = (cos^-1 x) / b) = [sin^-1 x + cos^-1 x] / (a + b) = π / [2 (a + b)]

Now, [tan^-1 y / c] = π / [2 (a + b)]

2 tan^-1 y = (πc / (a + b))

cos (πc / [a + b]) = cos (2 tan^-1 y) = [1 – y²] / [1 + y²]

Question 11: Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A² is 1, then the possible number of such matrices is :

a.6
b. 1
c. 4
d. 12

Answer: (c)

Let A =

a² + 2b² + c² = 1

a = 1, b = 0, c = 0

a = 0, b = 0, c = 1

a = –1, b = 0, c = 0

c = –1, b = 0, a = 0

Question 12: The intersection of three lines x – y = 0 and x + 2y = 3 and 2x + y = 6 is a:

a. Equilateral triangle
b. Right angled triangle
c. Isosceles triangle
d. None of the above

Answer: (c)

AB = AC = √5, BC = √2

Hence, the triangle is isosceles.

Question 13: The maximum slope of the curve y = (1 / 2) x⁴ – 5x³ + 18x² – 19x occurs at the point:

a. (2, 9)
b. (2, 2)
c. (3, 21 / 2)
d. (0, 0)

Answer: (b)

dy / dx = 2x³ – 15x² + 36x – 19

Let f (x) = 2x³ – 15x² + 36x – 19

f’ (x) = 6x² – 30x + 36 = 0

x² – 5x + 6 = 0

x = 2, 3

f’’ (x) = 12x – 30

f’’ (x) < 0 for x = 2

So, at x = 2, slope is maximum.

y = 8 – 40 + 72 – 38

= 72 – 70

= 2

Maximum slope occurs at (2, 2).

Question 14: Let f be any function defined on R and let it satisfy the condition:

|f (x) – f (y)|≤ |(x – y)²|, ∀ x, y ∈ R. If f (0) = 1, then :

a. f (x) < 0, ∀ x ∈ R
b. f (x) can take any value in R
c. f (x) = 0, ∀ x ∈ R
d. f (x) > 0, ∀ x ∈ R

Answer: (d)

|f (x) – f (y)| ≤ |(x – y)²|, ∀ x, y ∈ R

|[f (x) – f (y)] / [(x – y)]| ≤ |x – y|

lim_x→y |[f (x) – f (y)] / [(x – y)]| ≤ 0

|f’ (y)| ≤ 0

f’ (y) = 0

f (y) = c

Since f (0) = 1, f (y) = 1, ∀ y ∈ R

Question 15: The value of ∫_-π/2^π/2 cos² x / [1 + 3^x] dx is:

a. 2π
b. 4π
c. π / 2
d. π / 4

Answer: (d)

Let I = ∫_-π/2^π/2 cos² x / [1 + 3^x] dx

I = ∫_-π/2^π/2 [cos² x] / [1 + 3^-x] dx

= ∫_-π/2^π/2 [3^x cos² x] / [1 + 3^x] dx

2I = ∫_-π/2^π/2 cos² x dx

I = ∫₀^π/2 cos² x dx

= π / 4

Question 16: The value of is :

a. 3 / 4
b. 2 / √3
c. 4 / 3
d. 2 / 3

Answer: (c)

Question 17: A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to the probability of getting 9 heads, then the probability of getting 2 heads is:

a. 15 / 2¹²
b. 15 / 2¹³
c. 15 / 2¹⁴
d. 15 / 2⁸

Answer: (b)

P (x = 9) = P (x = 7)

⇒ ⁿC₉ (1 / 2)^n-9 (1 / 2)⁹ = ⁿC₇ (1 / 2)^n-7 (1 / 2)⁷

⇒ ⁿC₉ (1 / 2)ⁿ = (1 / 2)ⁿ x ⁿC₇

⇒ n = 9 + 7 = 16

P (x = 2) = ¹⁶C₂ (1 / 2)¹⁴ (1 / 2)²

= ¹⁶C₂ (1 / 2)¹⁶

= 15 / 2¹³

Question 18: If (1, 5, 35), (7, 5, 5), (1, λ, 7) and (2 λ, 1, 2) are coplanar, then the sum of all possible values of λ is:

a. − 44 / 5
a. 39 / 5
a. − 39 / 5
a. 44 / 5

Answer: (d)

Let P (1, 5, 35), Q (7, 5, 5), R (1, λ, 7) and S (2 λ, 1, 2)

⇒ {−33λ + 165 − 112} + 5(λ − 5)(2λ − 1) = 0

⇒ 53 − 33λ + 5 {2λ² − 11λ + 5} = 0

⇒ 10λ² − 88λ + 78 = 0

5λ² − 44λ + 39 = 0

∴ λ₁ + λ₂ = 44 / 5

Question 19: Let R = {P,Q)|P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1,–1) is the set :

a. S = {(x, y)|x² + y² = 1}
b. S = {(x, y)|x² + y² = 4}
c. S = {(x, y)|x² + y² = √2}
d. S = {(x, y)|x² + y² = 2}

Answer: (d)

P(a, b) , Q(c, d), OP = OQ

a² + b² = c² + d²

R(x ,y), S = (1, –1) OR = OS ( equivalence class)

x² + y² = 2

Question 20: In the circle given below, let OA = 1 unit, OB = 13 unit and PQ perpendicular to OB. Then, the area of the triangle PQB (in square units) is :

a. 26 √3
b. 24 √2
c. 24 √3
d. 26 √2

Answer: (c)

OC = 13 / 2 = 6.5

AC = CO – AO

= 6.5 – 1

= 5.5

In △PAC

PA = √(6.5² – 5.5²)

PA = √12

PQ = 2PA = 2 √12

Now, area of △PQB = (1 / 2) * PQ * AB

= (1 / 2) * 2 √12 * 12

= 12 √12

= 24 √3

Section-B

Question 1: The area bounded by the lines y = |x – 1| – 2 is ______.

Answer: 4

NTA Ans. (8)

Required area area of △PQR

Area = (1 / 2) * 4 * 2 = 4

This is a bonus question as the second curve is also not given.

If the second curve is x – axis, then answer will be 4

Question 2: The number of integral values of ‘k’ for which the equation 3 sin x + 4 cos x = k + 1 has a solution, k ∈ R is ______.

Answer: 11

3 sin x + 4 cos x = k + 1

⇒ −5 ≤ k + 1 ≤ 5

⇒ −6 ≤ k ≤ 4

−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4 → 11 integral values

Question 3: Let m, n ∈ N and gcd (2, n) = 1. If 30 ³⁰C₀ + 29 ³⁰C₁ + ….. + 2 ³⁰C₂₈ + 1 ³⁰C₂₉ = n . 2^m, then n + m =

Answer: 45

Let S = ∑_r=0³⁰ (30 – r) ³⁰C_r

= 30 ∑_r=0³⁰³⁰C_r – ∑_r=0³⁰ r ³⁰C_r

= 30 * 2³⁰ – ∑_r=1³⁰ r (30 / r) ²⁹C_r-1

= 30 * 2³⁰ – 30 * 2²⁹

= 30 * 2²⁹ (2 – 1)

= 15 * 2³⁰

n = 15 and m = 30

n + m = 45

Question 4: If y = y (x) is the solution of the equation e^siny cosy (dy / dx) + e^siny cosx = cos x, y (0) = 0; then 1 + y (π / 6) + (√3 / 2) y (π / 3) + (1 / √2) y (π / 4) is equal to ______.

Answer: 1

e^siny cosy (dy / dx) + e^siny cosx = cos x

Put e^siny = t

e^siny * cos y (dy / dx) = dt / dx

Then, dt / dx + t cos x = cos x

IF = e^{∫cosx dx} = e^sinx

Solution of differential equation

t . e^sinx = ∫e^sinx cosx dx

e^siny . e^sinx = e^sinx + c

At x = 0, y = 0

1 = 1 + c ⇒ c = 0

sin y + sin x = sin x

y = 0

y (π / 6) = 0, y (π / 3) = 0, y (π / 4) = 0

The required answer is 1 + 0 + 0 + 0 = 1.

Question 5: The number of solutions of the equation log₄ (x – 1) = log₂ (x – 3) is ______.

Answer: 1

(1 / 2) log₂ (x − 1) = log₂ (x − 3)

⇒ x − 1 = (x − 3)²

⇒ x² − 6𝑥 + 9 = x − 1

⇒ x² − 7𝑥 + 10 = 0

⇒ x = 2, 5

x = 2 Not possible as log₂ (x − 3) is not defined.

∴ Number of solution = 1

Question 6: If √3 (cos² x) = (√3 – 1) cos x + 1 the number of solutions of the given equation when x ∈ [0, π / 2] is _______.

Answer: 1

√3 t² – (√3 – 1) t – 1 = 0, where t = cos x

Now, t = {(√3 – 1) ± [√4 + 2 √3]} / 2 √3

t = cos x = 1 or – 1 / √3 (rejected as x ∈ [0, π / 2])

cos x = 1

Number of solution = 1

Question 7: Let (λ, 2, 1) be a point on the plane which passes through the point (4, – 2, 2). If the plane is perpendicular to the line joining the points (– 2, – 21, 29) and (– 1, – 16, 23), then (λ / 11)² – (4λ / 11) – 4 is equal to ______.

Answer: 8

AB is perpendicular to PQ.

4 – λ – 20 – 6 = 0

λ = – 22

Now, (λ / 11) = – 2

(λ / 11)² – (4λ / 11) – 4 = 4 + 8 – 4 = 8

Question 8: The difference between degree and order of a differential equation that represents the family of curves given by y² = a (x + (√a / 2)), a > 0 is ______.

Answer: 2

y² = a (x + (√a / 2))

Differentiating w.r.t.

2yy’ = a

y² = 2yy’ [x + (√2yy’ / 2)]

y = 2y’ [(x + √yy’ / 2)]

y – 2xy’ = √2y’ √yy’

Degree = 3 and Order = 1

Degree – Order = 3 – 1 = 2

Question 9: The sum of 162^th power of the roots of the equation x³ – 2x² + 2x – 1 = 0 is _________.

Answer: 3

Let roots of x³ – 2x² + 2x – 1 = 0 be ɑ, β, γ.

(x³ – 1) – (2x² – 2x) = 0

(x – 1) (x² – x + 1) = 0

x = 1, – 𝞈, – 𝞈²

Now, ɑ¹⁶² + β¹⁶² + γ¹⁶²

= 1 + (𝞈)¹⁶² + (𝞈²)¹⁶²

= 1 + (𝞈³)⁵⁴ + (𝞈³)¹⁰⁸ = 3

Question 10: The value of the integral ∫₀^π |sin 2x| dx is _______.

Answer: 2

I = ∫₀^π |sin 2x| dx

= 2 ∫₀^π/2 |sin 2x| dx = 2 ∫₀^π/2 sin 2x dx

= 2 [- cos (2x) / 2]₀^π/2

= 2