Here, candidates will find JEE Main 2021 Maths shift 2 question paper with solutions. Candidates who appeared in the JEE Main Maths shift 2 exam 2021 can also watch the below video to check the correct answers and analyze their performance. The candidates will also get detailed solutions along with answers in the JEE Main 2021 February 25 Shift 2 Maths question paper PDFs.

**SECTION A**

**Question 1: A plane passes through the points A(1,2,3), B(2,3,1) and C(2,4,2). If O is the origin and P is (2,–1,1), then the projection of vector(OP) on this plane is of length:**

- a. √(2/5)
- b. √(2/3)
- c. √(2/11)
- d. √(2/7)

Solution:

Answer: (c)

A(1,2,3),B(2,3,1),C(2,4,2),O(0,0,0)

Equation of plane passing through A, B, C will be

(x – 1)(–1 + 4) – (y – 2)(–1 + 2) + (z – 3)(2 – 1) = 0

(x – 1)(3) – (y – 2)(1) + (z – 3)(1) = 0

3x – 3 – y + 2 + z – 3 = 0

3x – y + z – 4 = 0, is the required plane equation

Now, given O(0,0,0) & P(2,–1,1)

Plane is 3x – y + z – 4 = 0

Let O’ & P’ be the foot of perpendiculars.

**Question 2: The contrapositive of the statement “If you will work, you will earn money” is:**

- a. If you will not earn money, you will not work
- b. You will earn money, if you will not work
- c. If you will earn money, you will work
- d. To earn money, you need to work

Solution:

Answer: (a)

Contrapositive of p → q is ~q →~p

p: you will work

q: you will earn money

~q: you will not earn money

~p: you will not work

~q →~p: if you will not earn money, you will not work

**Question 3: If α, β∈ R are such that 1 – 2i (here i ^{2} = –1) is a root of z^{2} + αz + β = 0, then (α – β) is equal to:**

- a. 7
- b. -3
- c. 3
- d. -7

Solution:

Answer: (d)

1-2i is a root of z

^{2}+ αz + β = 0.(1-2i)

^{ 2}+ α(1-2i)+β=0⇒1-4-4i+α-2iα+β=0

⇒(α+β-3)-i(4+2α)=0

⇒α+β-3=0 & 4+2α=0

So, α=-2, β=5

Therefore, α-β=-7

**Question 4: If **

- a. 1/(I
_{2}+ I_{4}) , 1/(I_{3}+ I_{5}), 1/(I_{4}+ I_{6}) are in G.P. - b. 1/(I
_{2}+ I_{4}) , 1/(I_{3}+ I_{5}), 1/(I_{4}+ I_{6}) are in A.P. - c. I
_{2}+ I_{4}, I_{3}+ I_{5}, I_{4}+ I_{6}) are in A.P. - d. I
_{2}+ I_{4}, I_{3}+ I_{5}, I_{4}+ I_{6}) are in G.P.

Solution:

Answer: (b)

\(I_{n+2}+I_n = \int_{\pi/4}^{\pi/2}cot^nx . cosec^2x dx=

\(\frac{-(cotx)^{n+1}}{n+1}]_{\pi/4}^{\pi/2}\)I

_{n+2}+ I_{n}= 1/(n+1)I

_{2}+ I_{4}=1/3, I_{3}+ I_{5}=1/4 and I_{4}+ I_{6}=1/5So, 1/(I

_{2}+ I_{4}) , 1/(I_{3}+ I_{5}) and 1/(I_{4}+ I_{6}) are in A.P.

**Question 5: If for the matrix, **

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

Solution:

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

Solution:

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 x ^{2} + 2y^{2} = 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. 𝜋/2 + tan
^{–1}(1/4) - b. 𝜋/2 – 𝑡𝑎𝑛
^{–1}(1/4) - c. 𝜋/2 + 𝑡𝑎𝑛
^{–1}(1/3) - d. 𝜋/2 – 𝑡𝑎𝑛
^{–1}(1/3)

Solution:

Answer: (a)

Using homogenization

x

^{2}+ 2y^{2}= 2(1)^{ 2}⇒x

^{2}+ 2y^{2}= 2(x + y)^{ 2}⇒x

^{2}+ 2y^{2}= 2x^{2}+ 2y^{2}+ 4xy⇒x

^{2}+ 4xy = 0for ax

^{2}+ 2hxy + by^{2}= 0, obtuse angle between lines θ istan θ = ±(2√(h

^{2}–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}|x^{2}+ 5x – 7| + c - b. (1/4) log
_{e }|x^{2}+ 5x – 7| + c - c. 4 log
_{e}|x^{2}+ 5x – 7| + c - d. log
_{e }√(x^{2 }+ 5x – 7) + c

Solution:

Answer: (c)

**Question 9: A hyperbola passes through the foci of the ellipse x ^{2}/25 + y^{2}/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. x
^{2}/9 – y^{2}/4 = 1 - b. x
^{2}/9 – y^{2}/16 = 1 - c. x
^{2}– y^{2}= 9 - d. x
^{2}/9 – y^{2}/25 = 1

Solution:

Answer: (b)

For ellipse, e

_{1}= √(1–16/25)=3/5Foci = (±3,0)

Let equation of hyperbola be x

^{2}/A^{2}– y^{2}/B^{2}= 1, passes through (±3, 0)A

^{2 }=9, A=3, e_{2}=5/3e

_{2}^{2}= 1 + B^{2}/A^{2}25/9 = 1 + B

^{2}/9 ⇒B^{2 }= 16x

^{2}/9 – y^{2}/16 = 1

**Question 10: **

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

Solution:

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

Solution:

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 + β
^{2}+ γ^{3}= 12 - d. has infinitely many solutions

Solution:

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

Solution:

Answer: (d)

Using AM ≥ GM inequality, we get

\(\frac{a^{ax}+\frac{a}{a^{ax}}}{2}\geq \left ( a^{ax}.\frac{a}{a^{ax}} \right )^{\frac{1}{2}} \Rightarrow a^{ax} + a^{1-ax}\geq 2\sqrt{a}\)

**Question 14: A function f(x) is given by f(x) = 5 ^{x}/(5^{x}+5), then the sum of the series **

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

Solution:

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 x ^{2} – 6x – 2 = 0. If a_{n} = α^{n} – β^{n} for n ≥ 1, then the value of (a_{10}–2a_{8})/(3a_{9} ) is:**

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

Solution:

Answer: (c)

α and β be the roots of x

^{2}–6x–2=0.So,

α

^{2}–6α–2=0⇒α^{2}–2=6αβ

^{2}–6β–2=0⇒β^{2}–2=6βNow,

**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 R_{2}→ 2R_{2} + 5R_{3} on 2A, then det(B) is equal to:**

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

Solution:

Answer: (a)

**Question 17: The shortest distance between the line x – y = 1 and the curve x ^{2} = 2y is:**

- a. 1/2
- b. 0
- c. 1/(2√2)
- d. 1/√2

Solution:

Answer: (c)

Shortest distance must be along common normal

m

_{1}(slope of line x–y = 1) = 1 ⇒slope of perpendicular line =–1m

_{2}= 2x/2 = x ⇒ m_{2}= 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

Solution:

Answer: (c)

Total cases

=

^{4}C_{1}× 9 × 9 × 9 –^{3}C_{1}× 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 =

^{3}C_{1}×9×9 -^{2}C_{1}×9Case 2: when unit digit is 7

Number of four digits number = 8 × 9 × 9

So total number favorable cases = 3×9

^{2}- 2×9 + 8×9^{2}Required Probability = ((3×9×9)–(2×9)+(8×9×9))/((4×9

^{3}) - (3×9^{2}))= 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

Solution:

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

Solution:

Answer: (a)

**Section B**

**Question 1: If **

Solution:

Answer: 5

\(\lim_{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}\)Applying L'Hospital Rule

\(\lim_{x \rightarrow 0} \frac{a-4e^{4x}}{a(e^{4x}-1)+ax(4e^{4x})}\)So for limit to exist,a=4Applying L'Hospital Rule

\(\lim_{x \rightarrow 0} \frac{-16e^{4x}}{a(4e^{4x}) +a(4e^{4x})+ax(16e^{4x})}\)(-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} + y^{2} = 9 and the parabola y^{2} = 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 ______.**

Solution:

Answer: 9

Sol. Circle: (x – 3)

^{2}+ y^{2}= 9Parabola: y

^{2}= 4xLet common tangent equation be y = mx + a/m

⇒y = mx + 1/m

⇒m

^{2}x – my + 1 = 0The above line is also tangent to circle

(x - 3)

^{2}+ y^{2}= 9Therefore, the perpendicular from (3, 0) to line = 3

⇒|(3m

^{2}– 0 + 1)/√(m^{2}+m^{4})| = 3⇒(3m

^{2}+ 1)^{2}= 9(m^{2}+ m^{4})⇒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/m

^{2}, 2a/m)= (3, 2√3) = (c, d)Solving Circle (x – 3)

^{2}+ y^{2}= 9 & line equation y = (1/√3) x + √3(x – 3)

^{2}+ ((1/√3) x + √3)^{ 2}= 9⇒x

^{2}+ 9 – 6x + (1/3)x^{2}+ 3 + 2x = 9⇒(4/3)x

^{2}– 4x + 3 = 0⇒x = 3/2=a

∴ 2(a + c) = 2(3/2+3)= 9

**Question 3: The value of **

Solution:

Answer: 19

x

^{2}-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 ______.**

Solution:

Answer: 1

Let x = 4k + 3

(2020 + x)

^{2022}= (2020 + 4k + 3)

^{2022}= (2024 + 4k - 1)

^{2022}= (4A – 1)

^{2022}=

^{2022}C_{0}(4A)^{2022}(–1)^{0}+^{2022}C_{1}(4A)^{2021}(–1)^{1}+ ……+^{2022}C_{2021}(4A)^{1}(–1)^{2021}+^{2022}C_{2022}(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 L_{2} at the distance of √17 from the point of intersection of L and L_{1} are (a, b, c), then 18(a+b+c) is equal to ______.

Solution:

Answer: 44

D.R. of L is parallel to (L

_{1}× L_{2}) ⇒ (-2, 3, -2)Equation of

*l*: x/2 = y/(–3) = z/2Solving L & L

_{1}(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 L

_{2}

**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 ______.**

Solution:

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 = y ^{4} and xy = k cut at right angles, then (4k)^{6} is equal to ______.**

Solution:

Answer: 4

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

Solution:

Answer: 45

Let n = 2t; t ∈ N

3

^{n }= 3^{2t }= (10 - 1)^{t}=10p + (–1)

^{t}= 10p ± 1

If n = even then 7

^{n}+ 3^{n}will not be multiple of 10So, if n is odd then only 7

^{n}+ 3^{n}will be a multiple of 10Therefore, n = 11,13,15,...........,99

Therefore, the answer is 45

**Question 9: Let **

\(\vec{a}\)

and \(\vec{a}\)

is 8√3 square units, then \(\vec{a} . \vec{b}\)

is equal to ______Solution:

Answer: 2

Area of parallelogram =

\(|\vec{a} \times \vec{b}|\)=

\(|(\hat{i} + \alpha \hat{j} + 3 \hat{k}) \times (3\hat{i} - \alpha \hat{j} + \hat{k})|\)(64)(3) = 16α

^{2}+ 64 + 16α^{2}⇒ α

^{2}= 4Now,

\(\vec{a}.\vec{b}\)= 3 - α^{2}+ 3= 6 – α

^{2}= 6 – 4

= 2

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

Solution:

Answer:1

Given,

(2xy

^{2}– y)dx + xdy = 0⇒ dy/dx + 2y

^{2}– y/x = 0⇒ (–1/y

^{2}) . dy/dx + (1/y) (1/x) = 2Let, 1/y = z

(–1/y

^{2}) dy/dx = dz/dx⇒dz/dx + z(1/x) = 2

As it passes through P(2, 1)

[Point of intersection of 2x - 3y = 1 and 3x + 2y = 8]

Therefore, 2/1 = 4 + c

⇒c = –2

⇒x/y = x

^{2}– 2Put x = 1

1/y = 1 – 2 = –1

⇒y(1) = –1

⇒|y(1)| = 1

JEE Main 2021 Maths Paper February 25 Shift 2