JEE Main 2021 February 25 – Shift 2 Maths Question Paper with Solutions

Question 5: If for the matrix,

a. 1
b. 3
c. 2
d. 4

Answer: (a)

Question 6: Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A × B. Then:

a. y = 273x
b. 2y = 91x
c. y = 91x
d. 2y = 273x

Answer: (b)

Number of elements in A = 3

Number of elements in B = 5

Number of elements in A × B = 15

Number of one-one function is x = 5 × 4 × 3

⇒x = 60

Number of one-one function is y = 15 × 14 × 13

⇒y = 15 × 4 × 14/4 × 13

⇒y = 60 × 7/2 × 13

⇒2y = (13)(7x)

⇒2y = 91x

Question 7: If the curve x² + 2y² = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is:

a. ??/2 + tan^–1(1/4)
b. ??/2 – ??????^–1(1/4)
c. ??/2 + ??????^–1(1/3)
d. ??/2 – ??????^–1(1/3)

Answer: (a)

Using homogenization

x² + 2y² = 2(1) ²

⇒x² + 2y² = 2(x + y) ²

⇒x² + 2y² = 2x² + 2y² + 4xy

⇒x² + 4xy = 0

for ax² + 2hxy + by² = 0, obtuse angle between lines θ is

tan θ = ±(2√(h²–ab))/(a+b)

⇒tan θ = ±4

⇒tan θ = –4

cot θ = -1/4

θ = cot^–1(–1/4)

θ = π – cot^–1 (1/4)

θ = π – (π/2– tan^–1(1/4) )

θ = π/2 + tan^-1(1/4)

Question 8:

a. log_e|x² + 5x – 7| + c
b. (1/4) log_e |x² + 5x – 7| + c
c. 4 log_e|x² + 5x – 7| + c
d. log_e √(x² + 5x – 7) + c

Answer: (c)

Question 9: A hyperbola passes through the foci of the ellipse x²/25 + y²/16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

a. x²/9 – y²/4 = 1
b. x²/9 – y²/16 = 1
c. x² – y² = 9
d. x²/9 – y²/25 = 1

Answer: (b)

For ellipse, e₁ = √(1–16/25)=3/5

Foci = (±3,0)

Let equation of hyperbola be x²/A² – y²/B² = 1, passes through (±3, 0)

A² =9, A=3, e₂=5/3

e₂² = 1 + B²/A²

25/9 = 1 + B²/9 ⇒B² = 16

x²/9 – y²/16 = 1

Question 10:

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

Answer: (c)

Question 11: In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from chest disorder. The probability that the selected person is a smoker and non-vegetarian is:

a. 7/45
b. 8/45
c. 14/45
d. 28/45

Answer: (d)

Based on Bayes’ theorem

Probability = ((160/400×35/100))/((160/400×35/100)+(100/400×20/100)+(140/400×10/100) )

= 5600/9000

= 28/45

Question 12: The following system of linear equations

2x + 3y + 2z = 9

3x + 2y + 2z = 9

x – y + 4z = 8

a. does not have any solution
b. has a unique solution
c. has a solution (α, β, γ) satisfying α + β² + γ³ = 12
d. has infinitely many solutions

Answer: (b)

Determinant of given system = -20 ≠ 0

It has unique solution.

Question 13: The minimum value of

a. a + 1/a
b. a + 1
c. 2a
d. 2√a

Answer: (d)

Using AM ≥ GM inequality, we get

Question 14: A function f(x) is given by f(x) = 5^x/(5^x+5), then the sum of the series f(1/20) + f(2/20) + f(3/20) + …… + f(39/20) is equal to:

a. 19/2
b. 49/2
c. 39/2
d. 29/2

Answer: (c)

f(x) = 5^x/(5 ^x + 5)….(i)

⇒ f(2–x) = 5^2–x/(5^2–x +5)

⇒f(2–x) = 5/(5^x+5)……(ii)

Adding equation(i)and(ii)

f(x)+f(2–x)=1

f(1/20)+f(39/20)=1

f(2/20)+f(38/20)=1

……

…..

f(19/20) + f(21/20)=1

and f(20/20) = f(1) = 1/2

Therefore, f(1/20) + f(2/20) + f(3/20) + …… + f(39/20) = 19 + 1/2 = 39/2

Question 15: Let α and β be the roots of x² – 6x – 2 = 0. If a_n = αⁿ – βⁿ for n ≥ 1, then the value of (a₁₀–2a₈)/(3a₉ ) is:

a. 4
b. 1
c. 2
d. 3

Answer: (c)

α and β be the roots of x²–6x–2=0.

So,

α²–6α–2=0⇒α²–2=6α

β²–6β–2=0⇒β²–2=6β

Now,

Question 16: Let A be a 3 × 3 matrix with det(A) = 4. Let R_i denote the i^th row of A. If a matrix B is obtained by performing the operation R₂→ 2R₂ + 5R₃ on 2A, then det(B) is equal to:

a. 64
b. 16
c. 80
d. 128

Answer: (a)

Question 17: The shortest distance between the line x – y = 1 and the curve x² = 2y is:

a. 1/2
b. 0
c. 1/(2√2)
d. 1/√2

Answer: (c)

Shortest distance must be along common normal

m₁ (slope of line x–y = 1) = 1 ⇒slope of perpendicular line =–1

m₂ = 2x/2 = x ⇒ m₂ = h ⇒slope of normal = –1/h

–1/h=–1 ⇒h=1

So, the point is (1,1/2)

Therefore, Shortest distance (D) =|(1–1/2–1)/√(1+1)|=1/(2√2)

Question 18: Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is:

a. 1/5
b. 2/9
c. 97/297
d. 122/297

Answer: (c)

Total cases

= ⁴C₁ × 9 × 9 × 9 – ³C₁ × 9 × 9

(as 4 digit number having 0 at thousands place have to be excluded)

For a number to have a remainder 2 when divided by 5 it’s unit digit should be 2 or 7

Case 1: when unit digit is 2

Number of four-digit number = ³C₁×9×9 – ²C₁×9

Case 2: when unit digit is 7

Number of four digits number = 8 × 9 × 9

So total number favorable cases = 3×9² – 2×9 + 8×9²

Required Probability = ((3×9×9)–(2×9)+(8×9×9))/((4×9³) – (3×9²))

= 97/297

Question 19: cosec[2 cot^–1( 5) + cos^–1(4/5) ] is equal to:

a. 75/56
b. 65/56
c. 56/33
d. 65/33

Answer: (b)

cosec[2 cot^–1( 5) + cos^–1(4/5) ]

Question 20: If 0 < x, y < π and cos x + cos y – cos (x + y) = 3/2, then sin x + cos y is equal to:

a. (1+√3)/2
b. (1–√3)/2
c. √3/2
d. 1/2

Answer: (a)

Section B

Question 1: If

Answer: 5

Applying L’Hospital Rule

Applying L’Hospital Rule

(-16)/(4a+4a)=(–16)/32=–1/2=b

a-2b = 4–2((–1)/2) = 4+1 = 5

Question 2: A line is a common tangent to the circle (x – 3)² + y² = 9 and the parabola y² = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a+c) is equal to ______.

Answer: 9

Sol. Circle: (x – 3)²+ y²= 9

Parabola: y² = 4x

Let common tangent equation be y = mx + a/m

⇒y = mx + 1/m

⇒m²x – my + 1 = 0

The above line is also tangent to circle

(x – 3)² + y²= 9

Therefore, the perpendicular from (3, 0) to line = 3

⇒|(3m² – 0 + 1)/√(m²+m⁴ )| = 3

⇒(3m² + 1)² = 9(m² + m⁴)

⇒m = ±1/√3

Tangent is

y = 1/√3x + √3 (it will be used)

=> m = 1/√3

or y = –1/√3x – √3 (rejected)

For Parabola, point of contact is (a/m² , 2a/m)= (3, 2√3) = (c, d)

Solving Circle (x – 3)² + y² = 9 & line equation y = (1/√3) x + √3

(x – 3)² + ((1/√3) x + √3) ² = 9

⇒x² + 9 – 6x + (1/3)x² + 3 + 2x = 9

⇒(4/3)x² – 4x + 3 = 0

⇒x = 3/2=a

∴ 2(a + c) = 2(3/2+3)= 9

Question 3: The value of

Answer: 19

x²-x-2=(x–2)(x+1)

Question 4: If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)²⁰²² is divided by 8 is ______.

Answer: 1

Let x = 4k + 3

(2020 + x)²⁰²²

= (2020 + 4k + 3)²⁰²²

= (2024 + 4k – 1)²⁰²²

= (4A – 1)²⁰²²

=²⁰²²C₀(4A)²⁰²²(–1)⁰ + ²⁰²²C₁(4A)²⁰²¹(–1)¹ + ……+²⁰²²C₂₀₂₁(4A)¹(–1)²⁰²¹+ ²⁰²²C₂₀₂₂(4A)⁰(–1)²⁰²²

Which will be of the form 8λ+1

So, Remainder is 1.

Question 5: A line L passing through origin is perpendicular to the lines

If the co-ordinates of the point in the first octant on L₂ at the distance of √17 from the point of intersection of L and L₁ are (a, b, c), then 18(a+b+c) is equal to ______.

Answer: 44

D.R. of L is parallel to (L₁ × L₂) ⇒ (-2, 3, -2)

Equation of l : x/2 = y/(–3) = z/2

Solving L & L₁

(2λ, –3λ, 2λ) = (µ + 3, 2µ – 1, 2µ + 4)

µ = – 1 , λ = 1

So, intersection point P(2, –3, 2)

Let, Q(2ν + 3, 2ν + 3, ν + 2) be required point on L₂

Question 6: A function f is defined on [–3,3] as

where [x] denotes the greatest integer ≤ x. The number of points, where f is not differentiable in (–3,3) is ______.

Answer: (5)

Points of non-differentiability in (–3, 3) are at x = –2, –1, 0, 1, 2.

i.e. 5 points.

Question 7: If the curves x = y⁴ and xy = k cut at right angles, then (4k)⁶ is equal to ______.

Answer: 4

Question 8: The total number of two digit numbers ‘n’, such that 3ⁿ+7ⁿ is a multiple of 10, is ______.

Answer: 45

Let n = 2t; t ∈ N

3ⁿ = 3^2t = (10 – 1)^t

=10p + (–1)^t

= 10p ± 1

If n = even then 7ⁿ + 3ⁿ will not be multiple of 10

So, if n is odd then only 7ⁿ + 3ⁿ will be a multiple of 10

Therefore, n = 11,13,15,………..,99

Therefore, the answer is 45

Question 9: Let

Answer: 2

Area of parallelogram =

=

(64)(3) = 16α² + 64 + 16α²

⇒ α² = 4

Now,

= 6 – α²

= 6 – 4

= 2

Question 10: If the curve y = y(x) represented by the solution of the differential equation (2xy² – y)dx + xdy = 0, passes through the intersection of the lines, 2x – 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to ______.

Answer:1

Given,

(2xy² – y)dx + xdy = 0

⇒ dy/dx + 2y² – y/x = 0

⇒ (–1/y²) . dy/dx + (1/y) (1/x) = 2

Let, 1/y = z

(–1/y²) dy/dx = dz/dx

⇒dz/dx + z(1/x) = 2

As it passes through P(2, 1)

Therefore, 2/1 = 4 + c

⇒c = –2

⇒x/y = x² – 2

Put x = 1

1/y = 1 – 2 = –1

⇒y(1) = –1

⇒|y(1)| = 1

Video Solutions- February 2021 Question Papers