The JEE Main 2022 July 28 – Shift 1 Maths Question Paper with Solutions is given on this page. JEE Main 2022 answer keys are prepared by JEE specialists at BYJU’S. Students can solve these JEE Main 2022 question papers and find out the topics which they need to learn more. It will also help them to boost their confidence. JEE Main 2022 question paper with solutions is a perfect tool to crack the upcoming entrance exams. Go through the JEE Main 2022 July 28 – Shift 1 Maths Question Paper with Solutions given below.

JEE Main 2022 28th July Shift 1 Mathematics Question Paper and Solutions

SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. Let the solution curve of the differential equation

Answer (B)

Sol.

at x = 1, y = 0

⇒ C = 0

At x = 2,

2. Considering only the principal values of the inverse trigonometric functions, the domain of the function

Answer (B)

Sol.

⇒ 2x² – 4x + 5 ≥ 0 & -4x ≤ 1

x ∈ R & x ≥ -1/4 ]

3. Let the vectors

(A) A non-empty finite set

(B) Equal to N

Answer (C)

Sol.

4. Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation

Answer (A)

Sol.

5. Let the operations *, ◉ ∈ {∧, ∨}. If (p * q) ◉ (p ◉ ~q) is a tautology, then the ordered pair (*, ◉) is

(A) (∨, ∧)

(B) (∨, ∨)

(C) (∧, ∧)

(D) (∧, ∨)

Answer (B)

Sol. *, ◉ ∈ {∧, ∨}.

Now for (q * q)◉(p◉~q) is tautology

(A) (∨, ∧) : (p ∨ q) ∧ (p ∧ ~q) not a tautology

(B) (∨, ∨) : (p ∨ q) ∨ (p ∨ ~q)

= P ∨ T is tautology

(C) (∧ , ∧) : (p ∧ q) ∧ (p ∧ ~q)

= (p ∧ p) ∧ (q ∧ ~q) = p ∧ F not a tautology (Fallasy)

(D) (∧, ∨) : (p ∧ q) ∨ (p ∨ ~q) not a tautology

6. Let a vector

Answer (B)

Sol.

As given equation is identity

Coefficient of x² = coefficient of y² = coefficient of
xy = 0

7. For t ∈ (0, 2π), if ABC is an equilateral triangle with vertices A(sint, – cost), B(cost, sint) and C(a, b) such that its orthocentre lies on a circle with centre (1, 1/3), then (a² – b²) is equal to

Answer (B)

Sol. Let P(h, k) be the orthocentre of ΔABC.

Then

(orthocentre coincide with centroid)

∴ a = 3, b = 1

∴ a² – b² = 8

8. For α ∈ N, consider a relation R on N given by R = {(x, y) : 3x + αy is a multiple of 7}. The relation R is an equivalence relation if and only if

(A) α = 14

(B) α is a multiple of 4

(C) 4 is the remainder when α is divided by 10

(D) 4 is the remainder when α is divided by 7

Answer (D)

Sol.

Adding 3a + 7kb + (7k₂ – 3) c = 7 (k₁ + k₃)

3a + (7k₂ – 3) c = 7 m

∴ R is transitive

9. Out of 60% female and 40% male candidates appearing in an exam, 60% of candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability that the chosen candidate is a female, is

Answer (*) None of the given option is correct

Sol.

10. If y = y(x), x ∈ (0, π/2) be the solution curve of the differential equation

Answer (A)

Sol.

Solution of differential equation

11. If the tangents drawn at the points P and Q on the parabola y² = 2x – 3 intersect at the point R(0, 1), then the orthocentre of the triangle PQR is :

(A) (0, 1)

(B) (2, –1)

(C) (6, 3)

(D) (2, 1)

Answer (B)

Sol.

Equation of chord PQ

⇒ y × 1 = x – 3

⇒ x – y = 3

For point P & Q

Intersection of PQ with parabola P : (6, 3) Q : (2, –1)

Slope of RQ = –1 & Slope of PQ = 1

12. Let C be the centre of the circle

Answer (B)

Sol.

QR = 2r = 4[

13. The remainder 7²⁰²² + 3²⁰²² is divided by 5 is:

(A) 0

(B) 2

(C) 3

(D) 4

Answer (C)

Sol. Let E = 7²⁰²² + 3²⁰²²

= (15 – 1)¹⁰¹¹ + (10 – 1)¹⁰¹¹

= –1 + (multiple of 15) –1 + multiple of 10

= –2 + (multiple of 5)

Hence remainder on dividing E by 5 is 3.

14. Let the matrix

(A) 3²⁸

(B) 3³⁰

(C) 3³²

(D) 3³⁶

Answer (C)

Sol.

15. Let

Answer (C)

Sol.

So, Z₂ lies on imaginary axis or on real axis within [–1, 1]

16. The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

Answer (B)

Sol. Any point on x² + y² = 1, z = 0 is p(cosθ, sinθ, 0)

If foot of perpendicular of p on the plane 2x + 3y + z = 6 is (h, k, l) then

h = 2r + cosθ, k = 3r + sinθ, l = r

Hence, h – 2l = cosθ and k – 3l = sinθ

Hence (h – 2l)² + (k – 3l)² = 1

When l = 6 – 2h – 3k

Hence required locus is

17. If the minimum value of

(A) 32

(B) 64

(C) 128

(D) 256

Answer (C)

Sol.

18. Let α, β and γ be three positive real numbers.

(A) 0

(B) 3

(C) 9

(D) 27

Answer (A)

Sol.

∴ First and last term, second and second last and so on are equal in magnitude but opposite in sign.

19. Consider the sequence a₁, a₂, a₃, … such that a₁ = 1, a₂ = 2 and

(A) –30

(B) –31

(C) –60

(D) –61

Answer (C)

Sol.

For n = 1 a₃a₂ – a₁a₂ = 2

n = 2 a₄a₃ – a₃a₂ = 2

n = 3 a₅a₄ – a₄a₃ = 2

_:

n = na_{n + 2}a_{n + 1} – a_na_{n + 1} = 2

a_{n + 2} a_{n + 1} = 2n + a₁a₂

Now,

20. The minimum value of the twice differentiable function

Answer (A)

Sol.

Differentiate on both sides

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is α × 5⁶, then α is equal to _______.

Answer (7073)

Sol. If password is 6 character long, tehn

Total number of ways having atleast one number = 10⁶ – 5⁶

Similarly, if 7 character long = 10⁷ – 5⁷

and if 8-character long = 10⁸ – 5⁸

2. Let P(–2, –1, 1) and Q(56/17, 43/17, 111/17) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α, –1, β, where both α and β are integers of minimum absolute values, then α² + β² is equal to ___________.

Answer (450)

Sol.

d.r’s of RS = < α, –1, β >

as PQ and RS are diagonals of rhombus

for minimum integral value α = – 15 and β = 15

3. Let f : [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line
y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1) then the least number of points x ∈ (0, 1) at which f”(x) = 0, is ___________.

Answer (2)

Sol.

If a graph cuts y = 2x + 5 in (0, 1) twice then its concavity changes twice

∴ f”(x) = 0 at atleast two points.

4. If

Answer (10)

Sol. Put

Now put 1 + secθ = t²

5. Let

Answer (2)

Sol.

By (1) we get

Similarly If A² + AB + BA = 0 then

6. For p, q, ∈ R, consider the real valued function f(x) = (x – p)² – q, x ∈ R and q > 0, Let a₁, a₂, a₃ and a₄ be in an arithmetic progression with mean p and positive common difference. If |f(a_i)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is

Answer (50)

Sol. ∵ a₁, a₂, a₃, a₄ are in A.P and its mean is p.

∴ a₁ = p – 3d, a₂ = p –d, a₃ = p + d and a₄ = p + 3d

Where d > 0

either 9d² – q = d² – q

7. For the hyperbola H: x² – y² = 1 and the ellipse

(1) eccentricity of E be reciprocal of the eccentricity of H, and

(2) the line y = √(5/2) x + K be a common tangent of E and H.

Then 4(a² + b²) is equal to _______.

Answer (03.00)

Sol. The equation of tangent to hyperbola x² – y² = 1 within slope m is equal to

And for same slope m, equation of tangent to ellipse

∵ Equation (i) and (ii) are identical

∴ a²m² + b² = m² – 1

But equation of common tangent is

∴ 5a² + 2b² = 3 …(i)

⇒a² = 2b² …(ii)

From equation (i) and (ii):

∴ 4(a² + b²) = 3

8. Let x₁, x₂, x₃, …, x₂₀ be in geometric progression with x₁ = 3 and the common ratio 1/2. A new data is constructed replacing each x_i by (x_i – i)². If  x̄ is the mean of new data, then the greatest integer less than or equal to x̄ is _________.

Answer (142)

Sol. x₁, x₂, x₃, …, x₂₀ are in G.P.

9.

Answer (01)

Sol. Let x + 2cosx = a

x + 2 = b

as x → 0, a → 2 and b → 2

= e⁰

= 1

10. The sum of all real value of x for which

Answer (06)

Sol.

Either x² + x + 1 = 0 or 5x² – 7x + 19 = 3x² + 5x + 12 ⇒ 2x² – 12x + 7 = 0 ⇒ sum of roots = 6

For x² + x + 1 = 0, no real roots.

Download PDF of JEE Main 2022 July 28 Shift 1 Maths Paper & Solutions

JEE Main 2022 July 28th Shift 1 Paper Analysis