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## JEE Main 2022 June 28th Maths Shift 2 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 R_{1} = {(a, b) β N Γ N : |a β b| β€ 13} and R_{2} = {(a, b) β N Γ N : |a β b| β 13}. Then on N:**

(A) Both *R*_{1} and *R*_{2} are equivalence relations

(B) Neither *R*_{1} nor *R*_{2} is an equivalence relation

(C) *R*_{1} is an equivalence relation but *R*_{2} is not

(D) *R*_{2} is an equivalence relation but *R*_{1} is not

**Answer (B)**

**Sol. ***R*_{1} = {(*a*, *b*) β *N* Γ *N* : |*a* β *b*| β€ 13} and

*R*_{2} = {(*a*, *b*) β *N* Γ *N* : |*a* β *b*| β 13}

In *R*_{1}: β΅ |2 β 11| = 9 β€ 13

β΄ (2, 11) β *R*_{1} and (11, 19) β *R*_{1} but

(2, 19) β *R*_{1}

β΄ *R*_{1} is not transitive

Hence *R*_{1} is not equivalence

In *R*_{2 }: (13, 3) β *R*_{2} and (3, 26) β *R*_{2} but

(13, 26) β *R*_{2} (β΅ |13 β 26| = 13)

β΄ *R*_{2} is not transitive

Hence *R*_{2} is not equivalence.

**2. Let f(x) be a quadratic polynomial such that f(β2) + f(3) = 0. If one of the roots of f(x) = 0 is β1, then the sum of the roots of f(x) = 0 is equal to:**

**Answer (A)**

**Sol. **β΅ *x* = β1 be the roots of *f*(*x*) = 0

β΄ let *f*(*x*) = *A*(*x* + 1)(*x* β *b*) β¦(i)

Now, *f*(β2) + *f*(3) = 0

β *A*[β1(β2 β *b*) + 4(3 β *b*)] = 0

β΄ Second root of *f*(*x*) = 0 will be 14/3.

β΄ Sum of roots

**3. The number of ways to distribute 30 identical candies among four children C_{1}, C_{2}, C_{3} and C_{4} so that C_{2} receives atleast 4 and atmost 7 candies, C_{3} receives atleast 2 and atmost 6 candies, is equal to:**

(A) 205

(B) 615

(C) 510

(D) 430

**Answer (D)**

**Sol. **By multinomial theorem, no. of ways to distribute 30 identical candies among four children *C*_{1}, *C*_{2} and *C*_{3}, *C*_{4}

= Coefficient of *x*^{30} in (*x*^{4} + *x*^{5} + β¦ + *x*^{7}) (*x*^{2} + *x*^{3} +β¦+ *x*^{6}) (1 + *x* + *x*^{2}β¦)^{2}

= Coefficient of *x*^{24} in (1 β *x*^{4} β *x*^{5} + *x*^{9}) (1 β *x*)^{β4}

= ^{27}*C*_{24} β ^{23}*C*_{20} β ^{22}*C*_{19} + ^{18}*C*_{15} = 430

**4. The term independent of x in the expansion of **

**\(\begin{array}{l}\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}, x \neq 0\end{array} \)**

**is:**

**Answer (B)**

**Sol. **

So, term independent from *x* in given expression

**5. If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is:**

(A) 21

(B) 22

(C) 23

(D) 24

**Answer (C)**

**Sol. **a, *A*_{1}, *A*_{2 }β¦β¦.. *A _{n}*, 100

Let *d* be the common difference of above A.P. then

β 7*a* + 8*d* = 100 β¦(i)

and *a* + *n* = 33 β¦(ii)

and 100 = *a* + (*n* + 1)*d*

β 800 = 8*a* + 7*a*^{2} β 338*a* + 3400

β 7*a*^{2} β 330*a* + 2600 = 0

β΄ *n* = 23

**6. Let f,g : R β R be functions defined by**

** **

**\(\begin{array}{l}f(x)=\left\{\begin{array}{ll}{[x],} & x<0 \\ |1-x|, & x \geq 0\end{array}\right.$ and $g(x)= \begin{cases}e^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geq 0\end{cases}\end{array} \)**

** Where [ x] denotes the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :**

(A) one point

(B) two points

(C) three points

(D) four points

**Answer (B)**

**Sol. **

Graph of y = g(x)

So, *x* = 0, 2 are the two points where *fog* is discontinuous.

**7. Let f : R β R be a differentiable function such that **

**\(\begin{array}{l}f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0 \textup{and} f^{\prime}\left(\frac{\pi}{2}\right)=1\end{array} \)**

**and let\(\begin{array}{l}g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \operatorname{sec~t} f(t)\right) d t\end{array} \) \(\begin{array}{l}\text{for}\ x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)\ \text{Then}\ \lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g(x)\ \text{is equal to}\end{array} \)Β **

(A) 2

(B) 3

(C) 4

(D) β3

**Answer (B)**

**Sol. **Given :

Now,

**8. Let f : R β R be a continuous function satisfying f(x) + f(x + k) = n, for all x β R where k > 0 and n is a positive integer. If **

**\(\begin{array}{l}l_{1}=\int_{0}^{4 n k} f(x) d x \quad and \quad I_{2}=\int_{-k}^{3 k} f(x) d x\end{array} \)**

**, then**

**Answer (C)**

**Sol. ***f* : *R* β *R* and f(x) + f(x + k) = n β x β R

*x* β *x* + *k*

*f(x + k) + f(x + 2k) = n*

β΄ *f(x + 2k) = f(x)*

So, period of *f*(*x*) is 2*k*

Now,

x = t + k β dx = dt (in second integral)

Now,

**9. The area of the bounded region enclosed by the curve **

**\(\begin{array}{l}y=3-\left|x-\frac{1}{2}\right|-|x+1|\end{array} \)**

**and the**

*x*-axis is**Answer (C)**

**Sol. **

_{}

Area of shaded region (required area)

**10. Let x = x(y) be the solution of the differential equation **

**\(\begin{array}{l}2 y e^{\frac{x}{y^{2}}} d x+\left(y^{2}-4 x e^{\frac{x}{y^{2}}}\right) d y=0\end{array} \)**

**such that**

*x*(1) = 0. Then,*x*(*e*) is equal to(A) e log_{e}(2)

(B) -e log_{e}(2)

(C) e^{2} log_{e}(2)

(D) -e^{2} log_{e}(2)

**Answer (D)**

**Sol. **Given differential equation

Now, using *x*(1) = 0, *c* = 2

So, for *x*(*e*), Put *y* = *e* in (i)

**11. Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tanx(cosx β y). If the curve passes through the point (Ο/4, 0) then the value of **

**\(\begin{array}{l}\int_{0}^{\pi / 2} y d x\end{array} \)**

**is equal to :**

**Answer (B)**

**Sol. **

β΄ Solution of D.E. will be

β΅ Curve passes through

**12. Let a triangle be bounded by the lines L _{1} : 2x + 5y = 10; L_{2} : β4x + 3y = 12 and the line L_{3}, which passes through the point P(2, 3), intersects L_{2} at A and L_{1} at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to**

**Answer (B)**

**Sol. ***L*_{1} : 2*x* + 5*y* = 10

*L*_{2} : β 4*x* + 3*y* = 12

Solving *L*_{1} and *L*_{2} we get

Now, Let

and

So, 3*x*_{1} + *x*_{2} = 8 and 10 *x*_{1} β *x*_{2} = β 5

So,

**13. Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola **

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

**Let\(\begin{array}{l}\mathrm{e}^{2}=\frac{11}{14} l\ \text{and}\ \left(\mathrm{e}^{\prime}\right)^{2}=\frac{11}{8} l^{\prime}\end{array} \) then the value of 77**

*e*β² and*l*β² respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If*a*+ 44*b*is equal to :(A) 100

(B) 110

(C) 120

(D) 130

**Answer (D)**

**Sol. **H :

*I*be the length of LR)

and

(*I*β² be the length of LR of conjugate hyperbola)

By (i) and (ii)

7a = 4b

then by (i)

β 44*b* = 65 and 77*a* = 65

β΄ 77*a* + 44*b* = 130

**14. Let, **

**\(\begin{array}{l}\vec{a}=\alpha \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}\end{array} \)**

**, where Ξ± β R. If the area of the parallelogram whose adjacent sides are represented by the vectors\(\begin{array}{l}\vec{a}\ \text{and}\ \vec{b}\ is\ \sqrt{15\left(\alpha^{2}+4\right)}\ \text{, then the value of}\ 2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}\end{array} \) is equal to :**

(A) 10

(B) 7

(C) 9

(D) 14

**Answer (D)**

**Sol. **

Now

β (2 + Ξ±)^{2} + (Ξ± β 2)^{2 }+ (Ξ±^{2} + 4)^{2 }= 15(Ξ±^{2} + 4)

β Ξ±^{4} β 5Ξ±^{2 }β 36 = 0

β΄ Ξ± = Β±3

Now,

**15. If vertex of a parabola is (2, β1) and the equation of its directrix is 4 x β 3y = 21, then the length of its latus rectum is :**

(A) 2

(B) 8

(C) 12

(D) 16

**Answer (B)**

**Sol. **Vertex of Parabola : (2, β1)

and directrix : 4*x* β 3*y* = 21

Distance of vertex from the directrix

β΄ length of latus rectum = 4*a* = 8

**16. Let the plane ax + by + cz = d pass through (2, 3, β5) and is perpendicular to the planes 2x + y β 5z = 10 and 3x + 5y β 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :**

(A) 18

(B) 20

(C) 24

(D) 22

**Answer (D)**

**Sol. **Equation of plane through point (2, 3, β5) and perpendicular to planes 2*x* + *y* β 5*z* = 10 and

3*x* + 5*y* β 7*z* = 12 is

β΄ Equation of plane is (*x* β 2) (β 7 + 25) β (*y* β 3) (β 14 + 15) + (*z* + 5) Β· 7 = 0

β΄ 18*x* β *y* + 7*z* + 2 = 0

β 18*x* β *y* + 7*z* = β 2

β΄ β 18*x* + *y* β 7*z* = 2

On comparing with *ax* + *by* + *cz* = *d* where *d* > 0 is

*a* = β 18, *b* = 1, *c* = β 7, *d* = 2

β΄ *a* + 7*b* + *c* + 20*d* = 22

**17. The probability that a randomly chosen one-one function from the set { a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) β f(c) = f(d) is :**

**Answer (D)**

**Sol. **Number of one-one function from {*a*, *b*, *c*, *d*} to set {1, 2, 3, 4, 5} is

The required possible set of value

(*f*(*a*), *f*(b), *f*(*c*), *f*(*d*)) such that *f*(*a*) + 2*f*(*b*) β *f*(*c*) = *f*(*d*) are (5, 3, 2, 1), (5, 1, 2, 3), (4, 1, 3, 5), (3, 1, 4, 5), (5, 4, 3, 2) and (3, 4, 5, 2)

β΄ *n*(*E*) = 6

β΄ Required probability

**18. The value of **

**\(\begin{array}{l}\lim _{n \rightarrow \infty} 6 \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}\end{array} \)**

**is equal to :**

(A) 1

(B) 2

(C) 3

(D) 6

**Answer (C)**

**Sol. **

**19. Let **

**\(\begin{array}{l}\vec{a}\ \text{be a vector which is perpendicular to the vector}\end{array} \)**

**Β\(\begin{array}{l}3 \hat{i}+\frac{1}{2} \hat{j}+2 \hat{k}.\ \text{If }\ \vec{a} \times(2 \hat{i}+\hat{k})=2 \hat{i}-13 \hat{j}-4 \hat{k}\end{array} \), then the projection of the vector on the vector \(\begin{array}{l}2 \hat{i}+2 \hat{j}+\hat{k}\ \text{is}:\end{array} \)Β **

(A) 1/3

(B) 1

(C) 5/3

(D) 7/3

**Answer (C)**

**Sol. **Let

and

and

β΄ *a*_{2} = 2 β¦(ii)

and *a*_{1} β 2*a*_{3} = 13 β¦(iii)

From eq. (i) and (iii) : *a*_{1} = 3 and *a*_{3} = β5

**20. If **

**\(\begin{array}{l}cot~\alpha = 1\ \text{and}\ sec ~\beta = -\frac{5}{3}\ \text{where}\ \pi<\alpha<\frac{3\pi}{2}\ \text{and}\ \frac{\pi}{2}<\beta<\pi,\end{array} \)**

**then the value of tan(Ξ± + Ξ²) and the quadrant in which Ξ± + Ξ² lies, respectively are :**

(A) -1/7 and IV^{th} quadrant

(B) 7 and I^{st} quadrant

(C) β 7 and IV^{th} quadrant

(D) 1/7 and I^{st} quadrant

**Answer (A)**

**Sol.** β΅ cot Ξ± = 1,

then tan Ξ± = 1

and

then

*i*.

*e*. fourth quadrant

**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 and the on-screen virtual numeric keypad in the place designated to enter the answer.

**1. Let the image of the point P(1, 2, 3) in the line **

**\(\begin{array}{l}L: \frac{x-6}{3}=\frac{y-1}{2}=\frac{z-2}{3}\end{array} \)**

**be**

*Q*. Let*R*(Ξ±, Ξ², Ξ³) be a point that divides internally the line segment*PQ*in the ratio 1 : 3. Then the value of 22(Ξ± + Ξ² + Ξ³) is equal to ________.**Answer (125)**

**Sol.** The point dividing *PQ* in the ratio 1 : 3 will be mid-point of *P* & foot of perpendicular from *P* on the line.

** β΄** Let a point on line be Ξ»

as *P*β² is foot of perpendicular

(3Ξ» + 5)3 + (2Ξ» β 1)2 + (3Ξ» β 1)3 = 0

β 22Ξ» + 15 β 2 β 3 = 0

Mid-point of PPβ

β 22(Ξ± + Ξ² + Ξ³) = 62 + 23 + 40 = 125

**2. Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is __________.**

**Answer (0)**

**Sol.** According to given data

So for any *x _{i}* , (

*x*β 62)

_{i}^{2}β€ 140

β *x _{i} > *50 β

*i*= 1, 2, 3, β¦7

So no student is going to score less than 50.

**3. If one of the diameters of the circle **

**\(\begin{array}{l}x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0\end{array} \)**

**is a chord of the circle\(\begin{array}{l}(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}\end{array} \), then the value of **

*r*^{2}is equal to _______.**Answer (10)**

**Sol. **For

Radius

β Diameter = 2β6

If this diameter is chord to

β *r*^{2} = 6 + 4 = 10

β *r*^{2} = 10

**4. If **

**\(\begin{array}{l}\lim _{x \rightarrow 1} \frac{\sin \left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2\end{array} \)**

**, then the value of (**

*a*β*b*) is equal to _______.**Answer (11)**

**Sol. **

So *f*(*x*) = 2*x*^{3} β 7*x*^{2} + *ax* +*b* = 0 has *x* = 1 as repeated root, therefore *f*(1) = 0 and *fΒ *β²(1) = 0 gives

*a* + *b* + 5 and *a* = 8

So, *a* β *b* = 11

**5. Let for n = 1, 2, β¦, 50, S_{n} be the sum of the infinite geometric progression whose first term is n^{2} and whose common ratio is **

**\(\begin{array}{l}\frac{1}{(n+1)^{2}}\ \text{. Then the value of}\end{array} \)**

**\(\begin{array}{l}\frac{1}{26}+\sum_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)\end{array} \) is equal to ________. **

**Answer (41651)**

**Sol. **

Now

= 1 + 25 Γ 17 (101 β 3)

= 41651

**6. If the system of linear equations **

** 2 x β 3y = Ξ³ + 5,**

** Ξ± x + 5y = Ξ² + 1, where Ξ±, Ξ², Ξ³ β R has infinitely many solutions, then the value of |9Ξ± + 3Ξ² + 5Ξ³| is equal to _______.**

**Answer (58)**

**Sol. **If 2*x* β 3*y* = Ξ³ + 5 and Ξ±*x* + 5*y* = Ξ² + 1 have infinitely many solutions then

So |9Ξ± + 3Ξ² + 5Ξ³| = |β30 β 28| = 58

**7. Let **

**\(\begin{array}{l}A=\left(\begin{array}{cc}1+i & 1 \\ -i & 0\end{array}\right)\ \text{where}\ i=\sqrt{-1}.\end{array} \)**

**Β Then, the number of elements in the set\(\begin{array}{l}\left\{n \in\{1,2, \ldots, 100\}: A^{n}=A\right\}\end{array} \) is ________.**

**Answer (25)**

**Sol. **

So *A*^{5} = *A*, *A*^{9} = *A* and so on.

Clearly *n* = 1, 5, 9, ….., 97

Number of values of *n* = 25

**8. Sum of squares of modulus of all the complex numbers z satisfying **

**\(\begin{array}{l}\bar{z}=iz^2 + z^2 – z\end{array} \)**

**is equal to ________.**

**Answer (2)**

**Sol. **Let *z* = *x* + *iy*

So 2*x* = (1 + *i*)(*x*^{2} β *y*^{2} + 2*xyi*)

β 2*x* = *x*^{2} β *y*^{2} β 2*xy* …(i) and

*x*^{2} β *y*^{2} + 2*xy* = 0 …(ii)

From (i) and (ii) we get

*x* = 0 or

When *x* = 0 we get *y* = 0

When

So there will be total 3 possible values of *z*, which are

Sum of squares of modulus

= 2

**9. Let S = {1, 2, 3, 4}. Then the number of elements in the set { f : S Γ S β S : f is onto and f(a,b)=f(b,a) **

*β₯*

*a β (a, b)βΒ S*is _____.**Γ S****Answer (37)**

**Sol. **There are 16 ordered pairs in *S* Γ *S*. We write all these ordered pairs in 4 sets as follows.

*A* = {(1, 1)}

*B* = {(1, 4), (2, 4), (3, 4) (4, 4), (4, 3), (4, 2), (4, 1)}

*C* = {(1, 3), (2, 3), (3, 3), (3, 2), (3, 1)}

*D* = {(1, 2), (2, 2), (2, 1)}

All elements of set *B* have image 4 and only element of *A* has image 1.

All elements of set *C* have image 3 or 4 and all elements of set *D* have image 2 or 3 or 4.

We will solve this question in two cases.

**Case I :** When no element of set *C* has image 3.

Number of onto functions = 2 (when elements of set *D* have images 2 or 3)

**Case II :** When atleast one element of set *C* has image 3.

Number of onto functions = (2^{3} β 1)(1 + 2 + 2)

= 35

Total number of functions = 37

**10. The maximum number of compound propositions, out of p β¨ r β¨ s, p β¨ r β¨ ~s, p β¨ ~q β¨ s, ~p β¨ ~r β¨ s, ~p β¨ ~r β¨ ~s, ~p β¨ q β¨ ~s, q β¨ r β¨ ~s, q β¨ ~r β¨ ~s, ~p β¨ ~q β¨ ~s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to ____________ .**

**Answer (9)**

**Sol. **There are total 9 compound propositions, out of which 6 contain ~*s*. So if we assign *s* as false, these 6 propositions will be true.

In remaining 3 compound propositions, two contain *p* and the third contains ~*r*. So if we assign *p* and *r* as true and false respectively, these 3 propositions will also be true.

Hence maximum number of propositions that can be true are 9.

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## Frequently Asked Questions β FAQs

### How was the JEE Main 2022 June 28 Shift 2 Maths question paper?

JEE Main 2022 June 28 Shift 2 Maths question paper was of moderate difficulty. It had 7 easy questions, 12 medium questions and 4 difficult questions as per the memory-based question paper.

### How many marks will be reduced for an incorrect answer in section A of JEE Main 2022 Shift 2 June 28 Maths question paper?

1 mark will be reduced for every incorrect answer in section A of JEE Main 2022 Shift 2 June 28 Maths question paper.

### What is the overall difficulty level of the JEE Main 2022 June 28 Maths Shift 2 question paper?

The overall difficulty level of the JEE Main 2022 June 28 Maths Shift 2 question paper is 1.87 out of 3.

### How was the class-wise distribution of questions in the JEE Main 2022 Shift 2 June 28 Maths question paper?

There were 13 questions from Class 11 and 10 questions from Class 12 in the JEE Main 2022 Shift 2 June 28 Maths question paper as per the memory-based question paper.

### What is the difficulty level of questions asked from Vector algebra in the JEE Main 2022 Maths Shift 2 June 28 question paper?

The questions asked from Vector algebra in the JEE Main 2022 Maths Shift 2 June 28 question paper were easy.

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