JEE Main 2022 July 25 – Shift 1 Maths Question Paper with Solutions

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JEE Main 2022 July 25th 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. The total number of functions, f : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6} such that f(1) + f(2) = f(3), is equal to

(A) 60

(B) 90

(C) 108

(D) 126

Answer (B)

Sol. Case 1:If f(3) = 3 then f(1) and f(2) take 1 OR 2

No. of ways = 2⋅6 = 12

Case 2: If f(3) = 5 then f(1) and f(2) take 2 OR 3

OR 1 and 4

No. of ways = 2⋅6⋅2 = 24

Case 3: If f(3) = 2 then f(1) = f(2) = 1

No. of ways = 6

Case 4: If f(3) = 4 then f(1) = f(2) = 2

No. of ways = 6

OR f(1) and f(2) take 1 and 3

No. of ways = 12

Case 5: If f(3) = 6 then f(1) = f(2) = 3 ⇒ 6 ways

OR f(1) and f(2) take 1 and 5 ⇒ 12 ways

OR f(2) and f(1) take 2 and 4 ⇒ 12 ways

2. If α, β, γ, δ are the roots of the equation x⁴ + x³ + x² + x + 1 = 0, then α²⁰²¹ + β²⁰²¹ + γ²⁰²¹ + δ²⁰²¹ is equal to

(A) –4

(B) –1

(C) 1

(D) 4

Answer (B)

Sol.

From properties of n^th root of unity

3. For n ∈ N let

(A) 0

(B) 2

(C) 3

(D) 4

Answer (*)

Sol.

Similarly T_n represents circle with centre C₂(2, –3) and radius

As S_n ∩ T_n = φ

C₁C₂>r₁ + r₂ OR C₁C₂< |r₁ – r₂|

OR

n = 1, 2, 3, 4 n may take infinite values.

4. The number of q∈ (0, 4π) for which the system of linear equations

3(sin 3θ) x – y + z = 2

3(cos 2θ) x + 4y + 3z = 3

6x + 7y + 7z = 9

has no solution, is

(A) 6

(B) 7

(C) 8

(D) 9

Answer (B)

Sol.

= 3sin3θ(7) + 1(21cos2θ – 18) + 1(21cos2θ – 24)

Δ = 21sin3θ + 42cos2θ – 42

For no solution

sin3θ + 2cos2θ = 2

⇒ sin3θ = 2⋅2sin²θ

⇒ 3sinθ – 4sin³θ = 4sin²θ

⇒ sinθ(3 – 4sinθ – 4sin²θ) = 0

5. If

(A) 4

(B) –8

(C) –4

(D) 8

Answer (C)

Sol.

∴ α = –1

Now,

Now,

6. If the absolute maximum value of the function

(A) α = 0

(B) α = –3

(C) α∈ (–1, 0)

(D) α∈ (–3, –1]

Answer (B)

Sol. Given,

f₁(x) = x² – 2x + 7

and positive also

So, f₂(x) is also decreasing and positive in {–3, 0}

∴ absolute maximum value of f(x) occurs at x = –3

∴ α = -3

7. The curve y(x) = ax³ + bx² + cx + 5 touches the x-axis at the point P(–2, 0) and cuts the y-axis at the point Q, where y′ is equal to 3. Then the local maximum value of y(x) is

Answer (A)

Sol. f(x) = y = ax³ + bx² + cx + 5 …(i)

Touches x-axis at P(–2, 0)

Touches x-axis at P(–2, 0) also implies

y = f(x) cuts y-axis at (0, 5)

Given,

From (iii), (iv) and (v)

f′(x) = 0 at x = –2 and x = 1

By first derivative test x = 1 in point of local maximum

Hence local maximum value of f(x) is f(1)

8. The area of the region given by

Answer (B)

Sol. A = {(x, y) : x² ≤ y ≤  min {x + 2, 4 – 3x}

So, area of the required region

9. For any real number x, let [x] denote the largest integer less than equal to x. Let f be a real valued function defined on the interval [–10, 10] by

Then the value of

(A) 4

(B) 2

(C) 1

(D) 0

Answer (A)

Sol.

Graph of f(x)

So,

10. The slope of the tangent to a curve C : y = y(x) at any point (x, y) on it is

Answer (B)

Sol.

It is given that the curve passes through

Now if

11. The general solution of the differential equation (x – y²)dx + y(5x + y²)dy = 0 is :

Answer (A)

Sol.

Let y² = t

Now substitute, t = vx

12. A line, with the slope greater than one, passes through the point A(4, 3) and intersects the line
x – y – 2 = 0 at the point B. If the length of the line segment AB is

(A) 2x + y = 9

(B) 3x – 2y = 7

(C) x + 2y = 6

(D) 2x – 3y = 3

Answer (C)

Sol.

Let inclination of required line is θ,

So the coordinates of point B can be assumed as

Which satisfices x – y – 2 = 0

By squaring

Which also satisfies x + 2y = 6

13. Let the locus of the centre (α, β), β> 0, of the circle which touches the circle x² +(y – 1)² = 1 externally and also touches the x-axis be L. Then the area bounded by L and the line y = 4 is :

Answer (C)

Sol.

Radius of circle S touching x-axis and centre (α, β) is |β|. According to given conditions

α² + (β – 1)² = (|β| + 1)²

α² + β² – 2β + 1 = β² + 1 + 2|β|

α² = 4β as β> 0

∴ Required louse is L : x² = 4y

14. Let P be the plane containing the straight line

(B) 96

(D) 54

Answer (C*)

Sol. Let <a, b, c> be direction ratios of plane containing lines

∴ 2a + 3b + 5c = 0 …(i)

and 3a + 7b + 8c = 0 …(ii)

from eq. (i) and (ii)

∴ D.R^S. of plane are < 11, 1, –5>

Let D.R^S of plane P be <a₁, b₁, c₁> then.

11a₁ + b₁ – 5c₁ = 0 …(iii)

and 9a₁ – b₁ – 5c₁ = 0 …(iv)

From eq. (iii) and (iv) :

∴ D.A⁵. of plane P are < 1, –1, 2>

Equation plane P is : 1(x – 3) –1(y + 4) +2(z –7) = 0

⇒ x – y + 2z – 21 = 0

Distance from point (2, –5, 11) is

15. Let ABC be a triangle such that

Then

(A) Both (S1) and (S2) are true

(B) Only (S1) is true

(C) Only (S2) is true

(D) Both (S1) and (S2) are false

Answer (C*)

Sol.

then

then

For

Hence (S1) is not correct

For (S2) : from (i)

∴ S(2) is correct.

16. If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Answer (C)

Sol. If n is number of trails, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24 …(i)

np.npq = 128 …(ii)

and q = 1 – p …(iii)

from eq. (i), (ii) and (iii) :

∴ Required probability

17. If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x² + αx + β> 0, for all x ∈ R, is :

Answer (A)

Sol. For x² + αx + β> 0 ∀ x∈R to hold, we should have α² – 4β<0

If α = 1, β can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If α = 2, β can be 2, 3, 4, 5, 6 i.e., 5 choices

If α = 3, β can be 3, 4, 5, 6 i.e., 4 choices

If α = 4, β can be 5 or 6 i.e., 2 choices

If α = 6, No possible value for βi.e., 0 choices

Hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0

= 17

Total possible choices for α and β = 6 × 6 = 36

Required probability

18. The number of solutions of |cos x| = sinx, such that –4π ≤ x ≤ 4π is :

(A) 4

(B) 6

(C) 8

(D) 12

Answer (C)

Sol. Number of solutions of the equation |cosx| = sinx for x ∈ [-4π, 4π] will be equal to 4 times the number of solutions of the same equation for x∈[0, 2π].

Graphs of y = |cosx| and y = sinx are as shown below.

Hence, two solutions of given equation in [0, 2π]

⇒ Total of 8 solutions in [–4π, 4π].

19. A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :

Answer (A)

Sol.

From (i) and (ii)

On simplification,

Hence height of the tower

20. Which of the following statements is a tautology?

Answer (D)

Sol. Truth Table

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

Answer (17)

Sol. Here

We get A² = A and similarly, for

We get B² = – B⇒B³ = B

∴ Aⁿ + (ωB)ⁿ = A + (ωB)ⁿ for n∈ N

For ωⁿ to be unity n shall be multiple of 3 and for Bⁿ to be B.n shell be 3, 5, 7, … 99

∴ n = {3, 9, 15,….. 99}

Number of elements = 17.

2. The letters of the work ‘MANKIND’ are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word ‘MANKIND’ is ______.

Answer (1492)

Sol. Arranging letter in alphabetical order A D I K M N N for finding rank of MANKIND making arrangements of dictionary we get

A ………….→

D ………….→ 360

I ………….→ 360

K ………….→ 360

M A D ………….→

M A I ………….→ 12

M A K ………….→ 12

M A N D ………….→ 3! = 6

M A N I ………….→ 6

M A N K D ………….→ 2

M A N K I D ………….→ 1

M A N K I N D ………….→ 1

∴ Rank of MANKIND = 1440 + 36 + 12 + 2 + 2

= 1492

3. If the maximum value of the term independent of t in the expansion of

Answer (6006)

Sol. General Term

for term independent on t

2(15 – r) – r = 0

⇒ r = 10

∴ T₁₁ = ¹⁵C₁₀x(1 – x)

Maximum value of x (1 – x) occur at

4. Let a, b be two non-zero real numbers. If p and r are the roots of the equation x² – 8ax + 2a = 0 and q and s are the roots of the equation x² + 12bx + 6b = 0, such that

Answer (38)

Sol. ∵ Roots of 2ax² – 8ax + 1 = 0 are

Let

So sum of roots 2α – 2β = 4 and 2α + 2β = – 2

Clearly

Now product of roots,

So,

5. Let a₁ = b₁ = 1, a_n = a_{n – 1} + 2 and b_n = a_a + b_{n – 1} for every natural number n ≥ 2. Then

Answer (27560)

Sol. a₁ = b₁ = 1

a_n = a_{n – 1} + 2 (for n≥ 2) b_n = a_n + b_{n – 1}

a₂ = a₁ + 2 = 1 + 2 = 3 b₂ = a₂ + b₁ = 3 + 1 = 4

a₃ = a₂ + 2 = 3 + 2 = 5 b₃ = a₃ + b₂ = 5 + 4 = 9 a₄ = a₃ + 2 = 5 + 2 = 7 b₄ = a₄ + b₃ = 7 + 9 = 16

a₁₅ = a₁₄ + 2 = 29 b₁₅ = 225

6. Let

Answer (3)

Sol.

∴ Non-diff at

7. If

Answer (5)

Sol.

8. Let the equation of two diameters of a circle x² + y² – 2x + 2fy + 1 = 0 be 2px – y = 1 and 2x + py = 4p. Then the slope m ∈ (0, ∞) of the tangent to the hyperbola 3x² – y² = 3 passing through the centre of the circle is equal to _______.

Answer (2)

Sol. x² + y² – 2x + 2fy + 1 = 0 [entre = (1, –f]

Diameter 2px – y = 1 …(i)

2x + py = 4p …(ii)

∴ Tangent should pass through (2, 3) for 3x² – y² = 3

Substitute (2, 3)

9. The sum of diameters of the circles that touch (i) the parabola 75x² = 64(5y – 3) at the point (8/5, 6/5) and (ii) the y-axis is equal to _______.

Answer (10)

Sol.

Equation of tangent to the parabola at

⇒ 120x = 160y

⇒ 3x = 4y

Equation of circle touching the given parabola at P can be taken as

If this circle touches y-axis then

⇒ D = 0

Radius = 1 or 4

Sum of diameter = 10

10. The line of shortest distance between the lines

Answer (*)

Sol.

Where,

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