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## JEE Main 2022 Maths Question Paper and Solutions – June 24th Shift 2

**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 x * y = x^{2} + y^{3} and (x * 1) * 1 = x * (1 * 1). Then a value of **

**\(\begin{array}{l}2\ sin^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)\end{array} \)**

**is**

**Answer (B)**

**Sol. **Given *x* * *y* = *x*^{2} + *y*^{3} and (*x* * 1) * 1 = *x* * (1 * 1)

So, (*x*^{2} + 1) * 1 = *x* * 2

β (*x*^{2} + 1)^{2} + 1 = *x*^{2} + 8

β *x*^{4} + 2*x*^{2} + 2 = *x*^{2} + 8

β (*x*^{2})^{2} + *x*^{2} β 6 = 0

β΄ (*x*^{2} + 3)(*x*^{2} β 2) = 0

β΄ *x*^{2} = 2

Now,

**2. The sum of all the real roots of the equation ( e^{2}^{x} β 4)(6e^{2}^{x }β 5e^{x} + 1) = 0 is**

(A) log* _{e}*3

(B) βlog* _{e}*3

(C) log* _{e}*6

(D) βlog* _{e}*6

**Answer (B)**

**Sol. **Given equation : (*e*^{2}* ^{x}* β 4)(6

*e*

^{2}

*β 5*

^{x }*e*+ 1) = 0

^{x}β *e*^{2}* ^{x}* β 4 = 0 or 6

*e*

^{2}

*β 5*

^{x}*e*+ 1 = 0

^{x}β e^{2x} = 4 or 6(*e ^{x}*)

^{2}β 3

*e*β 2

^{x}*e*+ 1 = 0

^{x}β 2*x* = ln4 or (3*e ^{x}* β 1)(2

*e*β 1) = 0

^{x}or

Sum of all real roots = ln2 β ln3 β ln2

= βln3

**3. Let the system of linear equations **

*x* + *y* + *az* = 2

**3 x + y + z = 4 **

*x* + 2*z* = 1

**have a unique solution ( x*, y*, z*). If (Ξ±, x*), (y*, Ξ±) and (x*, βy*) are collinear points, then the sum of absolute values of all possible values of Ξ± is **

(A) 4

(B) 3

(C) 2

(D) 1

**Answer (C)**

**Sol. **Given system of equations

*x* + *y* + *az* = 2 β¦(i)

3*x* + *y* + *z* = 4 β¦(ii)

*x* + 2*z* = 1 β¦(iii)

Solving (i), (ii) and (iii), we get

*x* = 1, *y* = 1, *z* = 0 (and for unique solution *a* β β3)

Now, (Ξ±, 1), (1, Ξ±) and (1, β1) are collinear

β Ξ±(Ξ± + 1) β 1(0) + 1(β1 β Ξ±) = 0

β Ξ±^{2} β 1 = 0

β΄ Ξ± = Β±1

β΄ Sum of absolute values of Ξ± = 1 + 1 = 2

**4. Let x, y > 0. If x^{3}y^{2} = 2^{15}, then the least value of 3x + 2y is **

(A) 30

(B) 32

(C) 36

(D) 40

**Answer (D)**

**Sol. ***x*, *y* > 0 and *x*^{3}*y*^{2} = 2^{15}

Now,

So, by A.M G.M inequality

β΄ Least value of 3*x* + 4*y* = 40

**5. Let **

**\(\begin{array}{l}f(x)\left\{\begin{matrix} \frac{\sin(x-[x])}{x-[x]},& x \in (-2,-1)\\ \max{\left\{2x,3[\left|x \right|]\right\}},&\left|x \right|<1 \\ 1& ,\text{otherwise}& \\\end{matrix}\right. \end{array} \)**

**Where [ t] denotes greatest integer t. If m is the number of points where f is not continuous and n is the number of points where f is not differentiable, then the ordered pair (m, n) is **

(A) (3, 3)

(B) (2, 4)

(C) (2, 3)

(D) (3, 4)

**Answer (C)**

**Sol. **

It clearly shows that *f*(*x*) is discontinuous

At *x* = β1, 1 also non differentiable and at *x* = 0,

L.H.D

R.H.D

β΄ *f*(*x*) is not differentiable at *x* = 0

β΄ *m* = 2, *n* = 3

** 6. The value of the integral**

**\(\begin{array}{l}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{(1+e^x)(sin^6x+cos^6x)}\end{array} \)**

**is equal to**

(A) 2Ο

(B) 0

(C) Ο

(D) Ο/2

**Answer (C)**

**Sol. **

From equation (i) & (ii)

Now,

When x = 0, t = 0

When x = Ο/4, t ββ

**7. **

**\(\begin{array}{l}\displaystyle \lim_{n\to \infty}\left(\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n^2+9)(n+3)}……+\frac{n^2}{(n^2+n^2)(n+n)}\right)\end{array} \)**

**is equal to **

**Answer (A)**

**Sol. **

**8. A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with **

(A) Length of latus rectum 3

(B) Length of latus rectum 6

**Answer (A)**

**Sol. **According to the question,

Let P(x,y)

β΄ c = 3

β΄ Required parabola *y*^{2} = 3*x* and L.R = 3

**9. Let the maximum area of the triangle that can be inscribed in the ellipse **

**\(\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{4}=1, a>2,\end{array} \)**

**having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the**

*y*-axis, be 6β3. Then the eccentricity of the ellipse is**Answer (A)**

**Sol. **

Given ellipse

β΄ Let *A*(ΞΈ) be the area of Ξ*ABB*β²

Then

For maxima *A*β²(ΞΈ) = 0

But for maximum area

β *a* = 4

**10. Let the area of the triangle with vertices A(1, Ξ±), B(Ξ±, 0) and C(0, Ξ±) be 4 sq. units. If the points (Ξ±, βΞ±), (βΞ±, Ξ±) and (Ξ±^{2}, Ξ²) are collinear, then Ξ² is equal to**

(A) 64

(B) β8

(C) β64

(D) 512

**Answer (C)**

**Sol. **β΅ *A*(1, Ξ±), *B*(Ξ±, 0) and *C*(0, Ξ±) are the vertices of Ξ*ABC* and area of Ξ*ABC* = 4

Now, (Ξ±, βΞ±), (βΞ±, Ξ±) and (Ξ±^{2}, Ξ²) are collinear

**11. The number of distinct real roots of the equation x^{7} β 7x β 2 = 0 is**

(A) 5

(B) 7

(C) 1

(D) 3

**Answer (D)**

**Sol. **Given equation *x*^{7} β 7*x* β 2 = 0

Let *f*(*x*) = *x*^{7} β 7*x* β 2

*f*β²(*x*) = 7*x*^{6} β 7 = 7(*x*^{6} β 1)

and *f*β²(*x*) = 0 β *x* = +1

and *f*(β1) = β1 + 7 β 2 = 5 > 0

*f*(1) = 1 β 7 β 2 = β8 < 0

So, roughly sketch of *f*(*x*) will be

So, number of real roots of *f*(*x*) = 0 and 3.

**12. A random variable X has the following probability distribution : **

X | 0 | 1 | 2 | 3 | 4 |

P(X) | k | 2k | 4k | 6k | 8k |

**The value of P(1 < X < 4 | x β€ 2) is equal to**

**Answer (A)**

**Sol. **

β΅ *x* is a random variable

β΄ *k* + 2*k* + 4*k* + 6*k* + 8*k* = 1

Now,

** 13. The number of solutions of the equation **

**\(\begin{array}{l}cos\left ( x+\frac{\pi}{3} \right)cos\left (\frac{\pi}{3}-x\right)=\frac{1}{4}cos^22x,x\in[-3\pi,3\pi]\end{array} \)**

**is:**

(A) 8

(B) 5

(C) 6

(D) 7

**Answer (D)**

**Sol. **

β΄ Number of solutions = 7

**14. If the shortest distance between the lines **

**\(\begin{array}{l}\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}~\text{and}~ \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\end{array} \)**

**is 1/β3, then the sum of all possible values of Ξ» is :**

(A) 16

(B) 6

(C) 12

(D) 15

**Answer (A)**

**Sol. **Let

β΄ Shortest distance

**15. Let the points on the plane P be equidistant from the points (β4, 2, 1) and (2, β2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is **

**Answer (C)**

**Sol. **

Let *P*(*x*, *y*, *z*) be any point on plane *P*_{1}

Then

And *P*_{2} : 2*x* + *y* + 3*z* = 1

β΄ angle between *P*_{1} and *P*_{2}

**16. **

**\(\begin{array}{l}\text{Let}\ \hat{a}\ \text{and}\ \hat{b}\ \text{be two unit vectors such that}\end{array} \)**

**Β\(\begin{array}{l}\left|(\hat{a}+\hat{b})+2(\hat{a}\times\hat{b}) \right|=2. \end{array} \) If ΞΈ β (0, Ο) is the angle between \(\begin{array}{l}\hat{a}\ \text{and}\ \hat{b},\ \text{then among the statements}:\end{array} \)**

**(S1): **

**(S2):** The projection of

(A) Only (S1) is true

(B) Only (S2) is true

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

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

**Answer (C)**

**Sol. **

(S1) is correct

(S2) is correct.

**17. If **

**\(\begin{array}{l}y=tan^{-1}(sec~ x^3-tan~x^3),\frac{\pi}{2}<x^3<\frac{3\pi}{2},\end{array} \)**

**then**

(A) *xy*β²β² + 2*y*β² = 0

(C) *x*^{2}*y*β³ β 6*y* + 3Ο = 0

(D) *xy*β³ β 4*y*β² = 0

**Answer (B)**

**Sol. **

β΄ *y* = tan^{β1} (secΞΈ β tanΞΈ)

*y*β³ = β 3*x*

** 18. Consider the following statements:**

* A *: Rishi is a judge.

* B *: Rishi is honest.

* C *: Rishi is not arrogant.

**The negation of the statement βif Rishi is a judge and he is not arrogant, then he is honestβ is**

(A) *B* β (*A* β¨ *C*)

(B) (~ *B*) β§ (*A* β§ *C*)

(C) *B* β ((~ *A*) β¨ (~ C))

(D) *B* β (*A* β§ *C*)

**Answer (B)**

**Sol. **β΅given statement is

(*A* β§ *C*) β *B*

Then its negation is

~ {(*A* β§ *C*) β *B*}

or ~ {~ (*A* β§ *C*) β¨ *B*}

β΄ (*A* β§ *C*) β§ (~ *B*)

or (~ *B*) β§ (*A* β§ *C*)

**19. The slope of normal at any point ( x, y), x > 0, y > 0 on the curve y = y(x) is given by **

**\(\begin{array}{l}\frac{x^2}{xy-x^2y^2-1}.\end{array} \)**

**If the curve passes through the point (1, 1), then**

*e*Β·*y*(*e*) is equal to(B) tan(1)

(C) 1

**Answer (D)**

**Sol. **

β΅ *y*(1) = 1 β tan^{β1} (*xy*) = ln*x* + tan^{β1}(1)

Put *x* = *e* and *y* = *y*(*e*) we get

tan^{β1} (*e* Β· *y*(*e*)) = 1 + tan^{β1} 1.

tan^{β1} (*e* Β· *y*(*e*)) β tan^{β1} 1 = 1

**20. Let Ξ»* be the largest value of Ξ» for which the function f_{Ξ»}(x) = 4Ξ»x^{3} β 36Ξ»x^{2} + 36x + 48 is increasing for all x β β. Then f_{Ξ»}* (1) + f_{Ξ»}* (β 1) is equal to :**

(A) 36

(B) 48

(C) 64

(D) 72

**Answer (D)**

**Sol. **β΅* f*_{Ξ»}(*x*) = 4Ξ»*x*^{3} β 36Ξ»*x*^{2} + 36*x* + 48

For *f*_{Ξ»}(*x*) increasing : (6Ξ»)^{2} β 12Ξ» β€ 0

Now,

= 72.

**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, β00.33, β00.30, 30.27, β27.30) using the mouse andw the on-screen virtual numeric keypad in the place designated to enter the answer.

**1. **

**\(\begin{array}{l}\text{Let}\ S={z\in\mathbb{C}:\left|z-3 \right|}\leq1\ \text{and}\ z(4~+~3{i})+{\overline{z}(4-3i)\leq 24}.\end{array} \)**

**If Ξ± +**

*i*Ξ² is the point in*S*which is closest to 4*i*, then 25(Ξ± + Ξ²) is equal to ______.**Answer (80)**

**Sol. **Here |*z* β 3| < 1

β (*x* β 3)^{2} + *y*^{2} < 1

And

β΄ Coordinate of *P* = (3 β cosΞΈ, sinΞΈ)

** 2. Let **

**\(\begin{array}{l}S = \begin{Bmatrix}\begin{pmatrix}-1 &a\\0 &b \\\end{pmatrix},a, b \in (1,2,3,….100)\\\end{Bmatrix}\end{array} \)**

**and let\(\begin{array}{l}\bigcap_{n=1}^{100}T_n\ \text{is}—–.\end{array} \)**

*T*_{n}= {*A*β*S*:*A*^{n}^{(}^{n}^{ + 1)}=*I*}. Then the number of elements in**Answer (100)**

**Sol. **

*b*= 1 and

*a*β {1,β¦.., 100}

Here, *n*(*n* + 1) is always even.

β΄ *T*_{1}, *T*_{2}, *T*_{3}, β¦, *T*_{n }are all *I* for *b* = 1 and each value of *a*.

**3. The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____. **

**Answer (576)**

**Sol. **Sum of all given numbers = 31

Difference between odd and even positions must be 0, 11 or 22, but 0 and 22 are not possible.

β΄ Only difference 11 is possible

This is possible only when either 1, 2, 3, 4 is filled in odd position in some order and remaining in other order. Similar arrangements of 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even positions.

β΄ Total possible arrangements = (4! Γ 3!) Γ 4

= 576

**4. The sum of all the elements of the set {Ξ± β {1, 2, β¦, 100} : HCF(Ξ±, 24) = 1} is**

**Answer (1633)**

**Sol.** The numbers upto 24 which gives g.c.d. with 24 equals to 1 are 1, 5, 7, 11, 13, 17, 19 and 23.

Sum of these numbers = 96

There are four such blocks and a number 97 is there upto 100.

β΄ Complete sum

= 96 + (24 Γ 8 + 96) + (48 Γ 8 + 96) + (72 Γ 8 + 96) + 97

= 1633

**5. The remainder on dividing 1 + 3 + 3 ^{2} + 3^{3} + β¦ + 3^{2021} by 50 ____ is **

**Answer (4)**

**Sol.**

= 50 *k*_{1} + 4

β΄ Remainder = 4

**6. The area (in sq. units) of the region enclosed between the parabola y^{2} = 2x and the line x + y = 4 is _______. **

**Answer (18)**

**Sol.**

The required area

= 18 square units

**7. Let a circle C : (x β h)^{2} + (y β k)^{2} = r^{2}, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ____. **

**Answer (7)**

**Sol.**

** **

Here, *OM*^{2} = *OP*^{2} β *PM*^{2}

β΄ *r*^{2} β 2*r* β 3 = 0

β΄ *r* = 3

β΄ Equation of circle is

β΄ *h* = 1, *k* = 3, *r* = 3

β΄ *h* + *k* + *r* = 7

**8. In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability 3/4and the remaining 6 questions correctly with probability ΒΌ. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is **

**\(\begin{array}{l}\frac{27k}{4^{10}},\end{array} \)**

**then**

*k*is equal to**Answer (479)**

**Sol. **Student guesses only two wrong. So there are three possibilities

(i) Student guesses both wrong from 1^{st} section

(ii) Student guesses both wrong from 2^{nd} section

(iii) Student guesses two wrong one from each section

Required probabilities

9. Let the hyperbola

**\(\begin{array}{l}H:\frac{x^2}{a^2}-y^2=1\end{array} \)**

**and the ellipse\(\begin{array}{l}E:3x^2+4y^2=12\end{array} \)be such that the length of latus rectum of \(\begin{array}{l}12\left(e_H^2+e_E^2\right)\ \text{is equal to}——–.\end{array} \)Β **

*H*is equal to the length of latus rectum of*E*. If*e*and_{H}*e*are the eccentricities of_{E}*H*and*E*respectively, then the value of**Answer (42)**

**Sol. **

β΄ Length of latus rectum

Length of latus rectum

**10. Let P_{1} be a parabola with vertex (3, 2) and focus (4, 4) and P_{2} be its mirror image with respect to the line x + 2y = 6. Then the directrix of P_{2} is x + 2y = _______.**

**Answer (10)**

**Sol. **Focus = (4, 4) and vertex = (3, 2)

β΄ Point of intersection of directrix with axis of parabola = *A* = (2, 0)

Image of *A*(2, 0) with respect to line

*x* + 2*y* = 6 is *B*(*x*_{2}, *y*_{2})

Point *B* is point of intersection of direction with axes of parabola *P*_{2}.

β΄ *x* + 2*y* = Ξ» must have point

β΄ *x* + 2*y* = 10

### Download PDF of JEE Main 2022 June 24 Shift 2 Maths Paper & Solutions

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## Frequently Asked Questions β FAQs

### How many questions are there in JEE Main 2022 June 24 Shift 2 Maths question paper?

There are 30 questions in JEE Main 2022 June 24 Shift 2 question paper.

### What is the difficulty level of JEE Main 2022 24th June Shift 2 Maths question paper?

The difficulty level of JEE Main 2022 June 24 Shift 2 Maths question paper is 1.90 out of 3. The questions were difficult comparing Physics and Chemistry papers.

### How many questions were asked from class 12 in JEE Main 2022 Shift 2 June 24 Maths question paper?

11 questions were asked from class 12 in Maths paper as per the memory-based question paper.

### What is the difficulty level of questions asked from the chapter Sequences and Series?

The questions from the chapter Sequences and Series were easy.

### How were the questions from Integrals in JEE Main Shift 2 June 24 Maths question paper 2022?

The questions from Integrals were quite difficult.

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