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JEE Main 2022 June 27 – Shift 1 Maths Question Paper with Solutions

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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 area of the polygon, whose vertices are the non-real roots of the equation

\(\begin{array}{l}\overline{z}=iz^2\end{array} \)
is :

\(\begin{array}{l}(A)\ \frac{3\sqrt{3}}{4}\end{array} \)

\(\begin{array}{l}(B)\ \frac{3\sqrt{3}}{2}\end{array} \)

\(\begin{array}{l}(C)\ \frac{3}{2}\end{array} \)

\(\begin{array}{l}(D)\ \frac{3}{4}\end{array} \)

Answer (A)

Sol.

\(\begin{array}{l}\overline{z}=iz^2\end{array} \)

Let z = x + iy

x – iy = i(x2 – y2 + 2xiy)

x – iy = i(x2 – y2) – 2xy

∴ x = –2yx or x2 – y2 = –y

x = 0 or

\(\begin{array}{l}y=-\frac{1}{2}\end{array} \)

Case-I

x = 0

–y2 = –y

y = 0, 1

Case-II

\(\begin{array}{l}y=-\frac{1}{2}\end{array} \)

β‡’

\(\begin{array}{l}x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm\frac{\sqrt{3}}{2}\end{array} \)

\(\begin{array}{l}z=\left\{0,i,\frac{\sqrt{3}}{2}-\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2} \right\}\end{array} \)

Area of polygon

\(\begin{array}{l}=\frac{1}{2}\begin{vmatrix}0 & 1 & 1 \\\frac{\sqrt{3}}{2} & \frac{-1}{2} & 1 \\\frac{-\sqrt{3}}{2} & \frac{-1}{2} & 1 \\\end{vmatrix}\end{array} \)

\(\begin{array}{l}=\frac{1}{2}\left|-\sqrt{3}-\frac{\sqrt{3}}{2} \right|=\frac{3\sqrt{3}}{4}\end{array} \)

2. Let the system of linear equations x + 2y + z = 2, Ξ±x + 3y – z = Ξ±, –αx + y + 2z = –α be inconsistent. Then Ξ± is equal to :

\(\begin{array}{l}(A)\ \frac{5}{2}\end{array} \)

\(\begin{array}{l}(B)\ -\frac{5}{2}\end{array} \)

\(\begin{array}{l}(C)\ \frac{7}{2}\end{array} \)

\(\begin{array}{l}(D)\ -\frac{7}{2}\end{array} \)

Answer (D)

Sol. x + 2y + z = 2

Ξ±x + 3y – z = Ξ±

–αx + y + 2z = –α

\(\begin{array}{l}\Delta=\begin{vmatrix} 1& 2 & 1 \\\alpha & 3 & -1 \\-\alpha & 1 & 2 \\\end{vmatrix}=1\left ( 6+1 \right )-2\left ( 2\alpha-\alpha \right )+1\left ( \alpha+3\alpha \right )\end{array} \)

= 7 + 2Ξ±

\(\begin{array}{l}\Delta=0\Rightarrow\alpha=-\frac{7}{2}\end{array} \)

\(\begin{array}{l}\Delta_1=\begin{vmatrix}2 & 2 & 1 \\\alpha & 3 & -1 \\-\alpha & 1 & 2 \\\end{vmatrix}=14+2\alpha\neq0~\textup{for}~\alpha=-\frac{7}{2}\end{array} \)

∴ For no solution

\(\begin{array}{l}\alpha=-\frac{7}{2}\end{array} \)

3. If

\(\begin{array}{l}x=\displaystyle\sum\limits_{n=0}^\infty a^n,y=\displaystyle\sum\limits_{n=0}^\infty b^n, z=\displaystyle\sum\limits_{n=0}^\infty c^n,\end{array} \)
where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc≠ 0,

then :

(A) x, y, zare in A.P.

(B) x, y, zare in G.P.

(C)

\(\begin{array}{l}\frac{1}{x},\frac{1}{y},\frac{1}{z}\end{array} \)
are in A.P.

(D)

\(\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1-\left ( a+b+c \right )\end{array} \)

Answer (C)

Sol.

\(\begin{array}{l}x=\displaystyle\sum\limits_{n=0}^\infty a^n=\frac{1}{1-a};~y=\displaystyle\sum\limits_{n=0}^\infty b^n=\frac{1}{1-b};~z=\displaystyle\sum\limits_{n=0}^\infty c^n=\frac{1}{1-c}\end{array} \)

Now,

a, b, c→ AP

1 – a, 1 – b, 1 – cβ†’ AP

\(\begin{array}{l}\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\rightarrow\textup{HP}\end{array} \)

x, y, z→ HP

∴ [tex]\frac{1}{x},\frac{1}{y},\frac{1}{z}\rightarrow\textup{AP}[/tex]

4. Let

\(\begin{array}{l}\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}\end{array} \)
where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is

(A) 10

(B) 8

(C) 7

(D) 5

Answer (B)

Sol.

\(\begin{array}{l}\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}\end{array} \)

= bxdy + cy dy + a dy = ax dx – by dx + a dx

= cy dy + a dy – ax dx – a dx + b(x dy + y dx) = 0

\(\begin{array}{l}=c\int y~dy+a\int x~dx-a\int dx+b\int d\left ( xy \right )=0 \end{array} \)

\(\begin{array}{l}=\frac{cy^2}{2}+ay-\frac{ax^2}{2}-ax+bxy=k \end{array} \)

= ax2 – cy2 + 2ax – 2ay – 2bxy = k

Above equation is circle

β‡’ a = –c and b = 0

ax2 + ay2 + 2ax – 2ay = k

β‡’ x2 + y2 + 2x – 2y = Ξ»

\(\begin{array}{l}\left [ \lambda=\frac{k}{a} \right ]\end{array} \)

Passes through (2, 5)

4 + 25 + 4 – 10 = Ξ»β‡’Ξ» = 23

Circle ≑x2 + y2 + 2x – 2y – 23 = 0

Centre (–1, 1)

\(\begin{array}{l}r=\sqrt{\left ( -1 \right )^2+1^2+23=5}\end{array} \)

Shortest distance of (11, 6)

\(\begin{array}{l}=\sqrt{12^2+5^2}-5\end{array} \)

= 13 – 5

= 8

5. Let a be an integer such that

\(\begin{array}{l}\lim\limits_{x\rightarrow7}\frac{18-\left [ 1-x \right ]}{\left [ x-3a \right ]}\end{array} \)
exists, where [t] is greatest integer ≀ t. Then a is equal

to :

(A) –6

(B) –2

(C) 2

(D) 6

Answer (A)

Sol.

\(\begin{array}{l}\lim\limits_{x\rightarrow7}\frac{18-\left [ 1-x \right ]}{\left [ x-3a \right ]}\end{array} \)
exist &a∈I.

\(\begin{array}{l}=\lim\limits_{x\rightarrow7}\frac{17-\left [ -x \right ]}{\left [ x \right ]-3a}\end{array} \)
exist

\(\begin{array}{l}RHL =\lim\limits_{x\rightarrow7^+}\frac{17-\left [ -x \right ]}{\left [ x \right ]-3a}=\frac{25}{7-3a}~~~~\left [ a\neq\frac{7}{3} \right ]\end{array} \)

\(\begin{array}{l} LHL =\lim\limits_{x\rightarrow7^-}\frac{17-\left [ -x \right ]}{\left [ x \right ]-3a}=\frac{24}{6-3a}~~~~\left [ a\neq 2 \right ]\end{array} \)

For limit to exist

LHL = RHL

\(\begin{array}{l}\frac{25}{7-3a}=\frac{24}{6-3a}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}\end{array} \)

∴ a = -6

6. The number of distinct real roots of x4 – 4x + 1 = 0 is :

(A) 4

(B) 2

(C) 1

(D) 0

Answer (B)

Sol.

Ζ’(x) = x4 – 4x + 1 = 0

Ζ’β€²(x) = 4x3 – 4

= 4(x – 1) (x2 + 1 + x)

JEE Main 2022 June 27 Shift 1 Maths A6

β‡’ Two solution

7. The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to :

(A) 5

(B) 8

(C) 10

(D) 12

Answer (C)

Sol.

JEE Main 2022 June 27 Shift 1 Maths A7

\(\begin{array}{l}CD=\sqrt{\left ( 10+x^2 \right )^2-\left ( 10-x^2 \right )^2}=2\sqrt{10}\left|x \right|\end{array} \)

Area

\(\begin{array}{l}=\frac{1}{2}\times CD\times AB=\frac{1}{2}\times 2\sqrt{10}\left|x \right|\left ( 20-2x^2 \right )\end{array} \)

\(\begin{array}{l}A=\sqrt{10}\left|x \right|\left ( 10-x^2 \right )\end{array} \)

\(\begin{array}{l}\frac{dA}{dx}=\sqrt{10}\frac{\left|x \right|}{x}\left ( 10-x^2 \right )+\sqrt{10}\left|x \right|\left ( -2x \right )=0\end{array} \)

β‡’ 10 – x2 = 2x2

3x2 = 10

x = k

3k2 = 10

8. If

\(\begin{array}{l}\cos^{-1}\left ( \frac{y}{2} \right )=\textup{log}_e\left ( \frac{x}{5} \right )^5,\left|y \right|<2\end{array} \)
then :

(A) x2yβ€²β€² + xyβ€² – 25y = 0

(B) x2yβ€²β€² – xyβ€² – 25y = 0

(C) x2yβ€²β€² – xyβ€²+ 25y = 0

(D) x2yβ€²β€² + xyβ€²+ 25y = 0

Answer (D)

Sol.

\(\begin{array}{l}\cos^{-1}\left ( \frac{y}{2} \right )=\textup{log}_e\left ( \frac{x}{5} \right )^5,\left|y \right|<2\end{array} \)

Differentatingon both side

\(\begin{array}{l}-\frac{1}{\sqrt{1-\left (\frac{y}{2} \right )^2}}\times\frac{y^{‘}}{2}=\frac{5}{\frac{x}{5}}\times\frac{1}{5}\end{array} \)

\(\begin{array}{l}\frac{-xy^{‘}}{2}=5\sqrt{1-\left ( \frac{y}{2} \right )^2}\end{array} \)

Square on both side

\(\begin{array}{l}\frac{x^2y^{‘2}}{4}=25\left ( \frac{4-y^2}{4} \right )\end{array} \)

Diff on both side

\(\begin{array}{l}2xy^{‘2}+2y^{‘}y^{”}x^2=-25\times2yy^{‘}\end{array} \)

xyβ€² + yβ€²β€²x2 + 25y = 0

9. If

\(\begin{array}{l}\int\frac{\left ( x^2+1 \right )e^x}{\left ( x+1 \right )^2}dx=f\left ( x \right )e^x+C\end{array} \)
where C is a constant, then
\(\begin{array}{l}\frac{d^3f}{dx^3}\end{array} \)
at x = 1 is equal to :

\(\begin{array}{l}(A)\ -\frac{3}{4} \end{array} \)

\(\begin{array}{l}(B)\ \frac{3}{4} \end{array} \)

\(\begin{array}{l}(C)\ -\frac{3}{2} \end{array} \)

\(\begin{array}{l}(D)\ \frac{3}{2} \end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}I=\int\frac{e^x\left ( x^2+1 \right )}{\left ( x+1 \right )^2}dx=f\left ( x \right )e^x+c\end{array} \)

\(\begin{array}{l}I=\int\frac{e^x\left ( x^2-1+1+1 \right )}{\left ( x+1 \right)^2}dx\end{array} \)

\(\begin{array}{l}=\int e^x\left [ \frac{x-1}{x+1}+\frac{2}{\left ( x+1 \right )^2} \right ]dx\end{array} \)

\(\begin{array}{l}=e^x\left ( \frac{x-1}{x+1} \right )+c\end{array} \)

\(\begin{array}{l}\therefore f\left ( x \right ) =\frac{x-1}{x+1}\end{array} \)

\(\begin{array}{l}f\left ( x \right ) =1-\frac{2}{x+1}\end{array} \)

\(\begin{array}{l}f’\left ( x \right ) =2\left ( \frac{1}{x+1} \right )^2\end{array} \)

\(\begin{array}{l}f”\left ( x \right ) =-4\left ( \frac{1}{x+1} \right )^3\end{array} \)

\(\begin{array}{l}f”’\left ( x \right ) =\frac{12}{\left ( x+1 \right )^4}\end{array} \)

for x = 1

\(\begin{array}{l}f”’\left ( 1 \right ) =\frac{12}{2^4}=\frac{12}{16}=\frac{3}{4}\end{array} \)

10. The value of the integral

\(\begin{array}{l}\displaystyle\int\limits_{-2}^2\frac{\left|x^3+x \right|}{\left (e^{x\left|x\right|}+1 \right ) }dx\end{array} \)
is equal to:

(A) 5e2

(B) 3e–2

(C) 4

(D) 6

Answer (D)

Sol.

\(\begin{array}{l}I=\displaystyle\int\limits_{-2}^2\frac{\left|x^3+x \right|}{e^{x\left|x\right|}+1 }dx….(i)\end{array} \)

\(\begin{array}{l}I=\displaystyle\int\limits_{-2}^2\frac{\left|x^3+x \right|}{e^{-x\left|x\right|}+1 }dx….(ii)\end{array} \)

\(\begin{array}{l}2I=\displaystyle\int\limits_{-2}^2\left|x^3+x \right|dx\end{array} \)

\(\begin{array}{l}2I=2\displaystyle\int\limits_0^2\left ( x^3+x \right )dx\end{array} \)

\(\begin{array}{l}I=\displaystyle\int\limits_0^2\left ( x^3+x \right )dx\end{array} \)

\(\begin{array}{l}=\left.\begin{matrix}\frac{x^4}{4}+\frac{x^2}{2}\end{matrix}\right\]_0^2\end{array} \)

\(\begin{array}{l}=\left ( \frac{16}{4}+\frac{4}{2} \right )-0\end{array} \)

= 4 + 2 = 6

11. If

\(\begin{array}{l}\frac{dy}{dx}+\frac{2^{x-y}\left ( 2^y-1 \right )}{2^x-1}=0,x,y>0,y\left ( 1 \right ) =1\end{array} \)
, then y(2) is equal to :

(A) 2 + log2 3

(B) 2 + log3 2

(C) 2 – log3 2

(D) 2 – log2 3

Answer (D)

Sol.

\(\begin{array}{l}\frac{dy}{dx}+\frac{2^{x-y}\left ( 2^y-1 \right )}{2^x-1}=0\end{array} \)
, x, y> 0, y(1) = 1

\(\begin{array}{l}\frac{dy}{dx}=-\frac{2^{x}\left ( 2^y-1 \right )}{2^y\left ( 2^x-1 \right )}\end{array} \)

\(\begin{array}{l}\int\frac{2^y}{2^y-1}dy=-\int\frac{2^x}{2^x-1}dx\end{array} \)

\(\begin{array}{l}=\frac{\textup{log}_e\left ( 2^y-1 \right )}{\textup{log}_e2}=-\frac{\textup{log}_e\left (2^x-1 \right )}{\textup{log}_e2}+\frac{\textup{log}_ec}{\textup{log}_e2}\end{array} \)

= |(2y – 1)(2x – 1)| = c

∡ y(1) = 1

∴ c = 1

= |(2y – 1)(2x – 1)| = 1

For x = 2

|(2y – 1)3| = 1

\(\begin{array}{l}2^y-1=\frac{1}{3}\Rightarrow 2y=\frac{4}{3} \end{array} \)

Taking log to base 2.

∴ y = 2 – log2 3

12. In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If (Ξ±, Ξ²) is the centroid of Ξ”ABC, then 15(Ξ± + Ξ²) is equal to :

(A) 39

(B) 41

(C) 51

(D) 63

Answer (C)

Sol.

JEE Main 2022 June 27 Shift 1 Maths A12

\(\begin{array}{l}\left.\begin{matrix}2x+y=4 \\2x+6y=14\end{matrix}\right\}y=2, x=3\end{array} \)

B(1, 2)

Let C(k, 4 – 2k)

Now AB2 = AC2

52 + (–1)2 = (6 – k)2 + (–3 + 2k)2

β‡’ 5k2 – 24k + 19 = 0

(5k – 19)(k – 1) = 0

\(\begin{array}{l}\Rightarrow k=\frac{19}{5}\end{array} \)

\(\begin{array}{l}C\left ( \frac{19}{5},-\frac{18}{5} \right )\end{array} \)

Centroid (Ξ±, Ξ²)

\(\begin{array}{l}\alpha=\frac{6+1+\frac{19}{5}}{3}=\frac{18}{5}\end{array} \)

\(\begin{array}{l}\beta=\frac{1+2-\frac{18}{5}}{3}=-\frac{1}{5}\end{array} \)

Now 15(Ξ± + Ξ²)

\(\begin{array}{l}15\left ( \frac{17}{5} \right )=51\end{array} \)

13. Let the eccentricity of an ellipse

\(\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b, \end{array} \)
be
\(\begin{array}{l}\frac{1}{4} \end{array} \)
. If this ellipse passes through the point
\(\begin{array}{l}\left ( -4\sqrt{\frac{2}{5}},3 \right )\end{array} \)
, then a2 + b2 is equal to :

(A) 29

(B) 31

(C) 32

(D) 34

Answer (B)

Sol.

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

\(\begin{array}{l}\Rightarrow \frac{\left ( -4\sqrt{\frac{2}{5}} \right )^2}{a^2}+\frac{32}{b^2}=1 \end{array} \)

β‡’

\(\begin{array}{l}\frac{32}{5a^2}+\frac{9}{b^2}=1 \end{array} \)
…(i)

a2(1 – e2) = b2

\(\begin{array}{l}a^2\left ( 1-\frac{1}{16} \right )=b^2\end{array} \)

15a2 = 16b2 β‡’

\(\begin{array}{l}a^2=\frac{16b^2}{15} \end{array} \)

From (i)

\(\begin{array}{l}\frac{6}{b^2}+\frac{9}{b^2}=1 \Rightarrow b^2=15\end{array} \)
&a2 = 16

a2 + b2 = 15 + 16 = 31

14. If two straight lines whose direction cosines are given by the relations l + m – n = 0, 3l2 + m2 + cnl = 0 are parallel, then the positive value of cis :

(A) 6

(B) 4

(C) 3

(D) 2

Answer (A)

Sol. l + m – n = 0 β‡’n = l + m

3l2 + m2 + cnl = 0

3l2 + m2 + cl(l + m) = 0

= (3 + c)l2 + clm + m2 = 0

\(\begin{array}{l}=\left ( 3+c \right ) \left ( \frac{I}{m} \right )^2+c\left ( \frac{I}{m} \right )+1=0\end{array} \)

∡ Lines are parallel

D = 0

c2 – 4(3 + c) = 0

c2 – 4c – 12 = 0

(c – 4)(c + 3) = 0

c = 4 (as c> 0)

15. Let

\(\begin{array}{l}\vec{a}=\hat{i}+\hat{j}-\hat{k} \end{array} \)
and
\(\begin{array}{l}\vec{c}=2\hat{i}-3\hat{j}+2\hat{k} \end{array} \)
. Then the number of vectors
\(\begin{array}{l}\vec{b}\end{array} \)
such that
\(\begin{array}{l}\vec{b}\times\vec{c}=\vec{a}\end{array} \)
and
\(\begin{array}{l}\left|\vec{b} \right|\in\left\{1,2,\dots,10 \right\}\end{array} \)
is :

(A) 0

(B) 1

(C) 2

(D) 3

Answer (A)

Sol.

\(\begin{array}{l}\vec{a}=\hat{i}+\hat{j}-\hat{k}\end{array} \)

\(\begin{array}{l}\vec{c}=2\hat{i}-3\hat{j}+2\hat{k}\end{array} \)

Now,

\(\begin{array}{l}\vec{b}\times\vec{c}=\vec{a}\end{array} \)

\(\begin{array}{l}\vec{c}\cdot\left ( \vec{b}\times\vec{c} \right )=\vec{c}\cdot\vec{a}\end{array} \)

\(\begin{array}{l}\vec{c}\cdot\vec{a}=0\end{array} \)

\(\begin{array}{l}\Rightarrow \left ( \hat{i}+\hat{j}-\hat{k} \right )\left ( 2\hat{i}-3\hat{j}+2\hat{k} \right )=0\end{array} \)

= 2 – 3 – 2 = 0

β‡’ –3 = 0 (Not possible)

β‡’ No possible value of

\(\begin{array}{l}\vec{b} \end{array} \)
is possible.

16. Five numbers, x1, x2, x3, x4, x5 are randomly selected from the numbers 1, 2, 3,….., 18 and are arranged in the increasing order (x1<x2<x3<x4<x5). The probability that x2 = 7 and x4 = 11 is:

(A)

\(\begin{array}{l}\frac{1}{136} \end{array} \)
(B)
\(\begin{array}{l}\frac{1}{72} \end{array} \)

(C)

\(\begin{array}{l}\frac{1}{68} \end{array} \)
(D)
\(\begin{array}{l}\frac{1}{34} \end{array} \)

Answer (C)

Sol. Total cases=18C5

Favourable cases

\(\begin{array}{l}\begin{matrix}^6C_1 & ^3C_1 &^7C_1\\\left ( \textup{Select}~x_1 \right ) & \left ( \textup{Select}~x_3 \right )&\left ( \textup{Select}~x_5 \right )\\\end{matrix}\end{array} \)

\(\begin{array}{l}P=\frac{6\cdot3\cdot7}{^{18}C_5}=\frac{1}{68}\end{array} \)

17. Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:

(A)

\(\begin{array}{l}\frac{105}{16}\end{array} \)

(B)

\(\begin{array}{l}\frac{7}{16}\end{array} \)

(C)

\(\begin{array}{l}\frac{77}{36}\end{array} \)

(D)

\(\begin{array}{l}\frac{49}{16}\end{array} \)

Answer (C)

Sol. Given P(X = 3) = 5P(X = 4) and n = 7

β‡’

\(\begin{array}{l}^7C_3p^3q^4=5\cdot^7C_4p^4q^3\end{array} \)

β‡’ q = 5p and also p + q = 1

β‡’

\(\begin{array}{l}p=\frac{1}{6}~\textup{and}~q=\frac{5}{6}\end{array} \)

Mean

\(\begin{array}{l}=\frac{7}{6} \end{array} \)
and variance
\(\begin{array}{l}=\frac{35}{36} \end{array} \)

Mean + Variance

\(\begin{array}{l}=\frac{7}{6}+\frac{35}{36}=\frac{77}{36} \end{array} \)

18. The value of

\(\begin{array}{l}\cos\left ( \frac{2\pi}{7} \right )+\cos\left ( \frac{4\pi}{7} \right )+\cos\left ( \frac{6\pi}{7} \right )\end{array} \)
is equal to:

(A) –1

(B)

\(\begin{array}{l}-\frac{1}{2} \end{array} \)

(C)

\(\begin{array}{l}-\frac{1}{3} \end{array} \)

(D)

\(\begin{array}{l}-\frac{1}{4} \end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{\sin3\left ( \frac{\pi}{7} \right )}{\sin\frac{\pi}{7}}\cos\frac{\left ( \frac{2\pi}{7}+\frac{6\pi}{7} \right )}{2}\end{array} \)

\(\begin{array}{l}=\frac{\sin\left ( \frac{3\pi}{7}\right)\cdot\cos\left ( \frac{4\pi}{7} \right )}{\sin\left ( \frac{\pi}{7} \right )}\end{array} \)

\(\begin{array}{l}=\frac{2\sin\frac{4\pi}{7}\cos\frac{4\pi}{7}}{2\sin\frac{\pi}{7}} \end{array} \)

\(\begin{array}{l}=\frac{\sin\left ( \frac{8\pi}{7} \right )}{2\sin\frac{\pi}{7}}=\frac{-\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{-1}{2}\end{array} \)

19.

\(\begin{array}{l}\sin^{-1}\left ( \sin\frac{2\pi}{3} \right )+\cos^{-1}\left ( \cos\frac{7\pi}{6} \right )+\tan^{-1}\left ( \tan\frac{3\pi}{4} \right )\end{array} \)
is equal to:

(A)

\(\begin{array}{l}\frac{11\pi}{12}\end{array} \)

(B)

\(\begin{array}{l}\frac{17\pi}{12}\end{array} \)

(C)

\(\begin{array}{l}\frac{31\pi}{12}\end{array} \)

(D)

\(\begin{array}{l}-\frac{3\pi}{4}\end{array} \)

Answer (A)

Sol.

\(\begin{array}{l}\sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )+\cos^{-1}\left ( \frac{-\sqrt{3}}{2} \right )+\tan^{-1}\left ( -1 \right )\end{array} \)

\(\begin{array}{l}=\frac{\pi}{3}+\frac{5\pi}{6}-\frac{\pi}{4}\end{array} \)

\(\begin{array}{l}=\frac{4\pi+10\pi-3\pi}{12}=\frac{11\pi}{12}\end{array} \)

20. The boolean expression (~(p ∧q)) ∨q is equivalent to:

(A) qβ†’ (p ∧q)

(B) p→q

(C) p→ (p→q)

(D) pβ†’ (p∨q)

Answer (D)

Sol. Making truth table

p

q

p ∧ q

~p ∧ q

(~(p ∧ q)) ∨ q

p ∨ q

p β†’ q

p β†’ (p ∨ q)

T

T

T

F

T

T

T

T

T

F

F

T

T

T

F

T

F

T

F

T

T

T

T

T

F

F

F

T

T

F

T

T

Tautology

Tautology

\(\begin{array}{l}\therefore \left ( \sim \left ( p\wedge q \right ) \right )\vee q\equiv p\rightarrow\left ( p\vee q \right )\end{array} \)

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 ƒ :R→R be a function defined by

\(\begin{array}{l}f\left ( x \right )=\frac{2e^{2x}}{e^{2x}+e^x} \end{array} \)
Then
\(\begin{array}{l}f\left ( \frac{1}{100} \right )+f\left ( \frac{2}{100} \right )+f\left ( \frac{3}{100} \right )+\dots+f\left ( \frac{99}{100} \right )\end{array} \)
is equal to ________.

Answer (99)

Sol.

\(\begin{array}{l}f\left ( x \right )=\frac{2e^{2x}}{e^{2x}+e^x}~\textup{and}~f\left ( 1-x \right )=\frac{2e^{2-2x}}{e^{2-2x}+e^{1-x}}\end{array} \)

∴

\(\begin{array}{l}\frac{f\left ( x \right )+f\left ( 1-x \right )}{2}=1\end{array} \)

i.e. f(x) + f(1 – x) = 2

∴

\(\begin{array}{l}f\left ( \frac{1}{100} \right ) +f\left ( \frac{2}{100} \right )+\dots+f\left ( \frac{99}{100} \right )\end{array} \)

\(\begin{array}{l}=\displaystyle\sum\limits_{x=1}^{49}f\left ( \frac{x}{100} \right )+f\left ( 1-\frac{x}{100} \right )+f\left ( \frac{1}{2} \right )\end{array} \)

= 49 Γ— 2 + 1 = 99

2. If the sum of all the roots of the equation

\(\begin{array}{l}e^{2x} – 11e^x – 45e^{–x} +\frac{81}{2}=0\end{array} \)
is logep, then p is equal to _____.

Answer (45)

Sol. Let ex = t then equation reduces to

\(\begin{array}{l}t^2-11t-\frac{45}{t}+\frac{81}{2}=0 \end{array} \)

β‡’ 2t3 – 22t2 + 81t – 45 = 0 …(i)

if roots of

\(\begin{array}{l}e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0 \end{array} \)
are Ξ±, Ξ², Ξ³ then roots of (i) will be
\(\begin{array}{l}e^{\alpha_1}e^{\alpha_2}e^{\alpha_3}\end{array} \)
using product of roots

\(\begin{array}{l}e^{\alpha_1+\alpha_2+\alpha_3}=45\end{array} \)

β‡’ Ξ±1 + Ξ±2 + Ξ±3 = ln 45 β‡’ p = 45

3. The positive value of the determinant of the matrix A, whose

\(\begin{array}{l}Adj\left ( Adj\left ( A \right ) \right )=\begin{pmatrix}14 & 28 & -14 \\-14 & 14 & 28 \\28 & -14 & 14 \\\end{pmatrix}\end{array} \)
, is __________.

Answer (14)

Sol.

\(\begin{array}{l}\left|adj\left ( adj\left ( A \right ) \right ) \right|=\left|A \right|^{2^2}=\left|A \right|^4\end{array} \)

∴

\(\begin{array}{l}\left|A \right|^4=\begin{vmatrix}14 & 28 & -14 \\-14 & 14 & 28 \\28 & -14 & 14 \\\end{vmatrix}\end{array} \)

\(\begin{array}{l}=\left ( 14 \right )^3\begin{vmatrix}1 & 2 & -1 \\-1 & 1 & 2 \\2 & -1 & 1 \\\end{vmatrix}\end{array} \)

\(\begin{array}{l}=\left ( 14 \right )^3\left ( 3-2\left ( -5 \right )-1\left ( -1 \right ) \right )\end{array} \)

\(\begin{array}{l}\left|A \right|^4=\left ( 14 \right )^4\Rightarrow \left| A\right|=14\end{array} \)

4. The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _________.

Answer (56)

Sol. First we arrange 5 red cubes in a row and assume x1, x2, x3, x4, x5 and x6 number of blue cubes between them

Here, x1 + x2 + x3 + x4 + x5 + x6 = 11

and x2, x3, x4, x5β‰₯ 2

So x1 + x2 + x3 + x4 + x5 + x6 = 3

No. of solutions = 8C5 = 56

5. If the coefficient of x10 in the binomial expansion of

\(\begin{array}{l}\left ( \frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}} \right )^{60}\end{array} \)
is 5k.l, where l, k∈N and l is co-prime to 5, then k is equal to ___________.

Answer (5)

Sol.

\(\begin{array}{l}T_{r+1}=60_{C_r}\left ( x^{\frac{1}{2}} \right )^{60-r}\left ( x^{-\frac{1}{3}} \right )^r\left ( 5^{\frac{-1}{4}} \right )^{60-r}\left ( 5^\frac{1}{2} \right )^r\end{array} \)

for

\(\begin{array}{l}x^{10}\frac{60-r}{2}-\frac{r}{3}=10 \end{array} \)

β‡’ 180 – 3r – 2r = 60

β‡’ r = 24

∴ Coeff. of

\(\begin{array}{l}x^{10}=\frac{^{60}C_{24}}{5^9} 5^{12}=5^k~~I\end{array} \)

as l and 5 are coprime

k = 3 + exponent of 5 in 60C24

\(\begin{array}{l}=3+\left ( \left [\frac{60}{5} \right ]+\left [ \frac{60}{5^2} \right ]-\left [ \frac{24}{5} \right ]-\left [ \frac{24}{5^2} \right ]-\left [ \frac{36}{5} \right ]-\left [ \frac{36}{5^2} \right ] \right. \end{array} \)

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

6. Let

\(\begin{array}{l}A_1=\left\{\left ( x,y \right ):\left|x \right|\leq y^2,\left|x \right|+2y\leq8 \right\} \end{array} \)
and

\(\begin{array}{l}A_2=\left\{\left ( x,y \right ):\left|x \right|+ \left|y \right|\leq k \right\} \end{array} \)
. If 27(Area A1) = 5(Area A2), then k is equal to :

Answer (6*)

Sol.

JEE Main 2022 June 27 Shift 1 Maths NQA 6(i)

Required area (above x-axis)

\(\begin{array}{l}A_1=2\displaystyle\int\limits_0^4\left ( \frac{8-x}{2}-\sqrt{x} \right )dx \end{array} \)

\(\begin{array}{l}=2\left ( 16-\frac{16}{4}-\frac{8}{3/2} \right ) =\frac{40}{3}\end{array} \)

and

\(\begin{array}{l}A_2=4\left ( \frac{1}{2}\cdot k^2 \right )=2k^2\end{array} \)

JEE Main 2022 June 27 Shift 1 Maths NQA 6(ii)

∴

\(\begin{array}{l}27\cdot\frac{40}{3}=5\cdot\left ( 2k^2 \right ) \end{array} \)

β‡’ k = 6

* A1 is

JEE Main 2022 June 27 Shift 1 Maths NQA 6(iii)

Which tends to infinity if not mentioned above x-axis

7. If the sum of the first ten terms of the series

\(\begin{array}{l}\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\dots \end{array} \)
is
\(\begin{array}{l}\frac{m}{n}\end{array} \)
, where m and n are co-prime numbers, then m + n is equal to __________.

Answer (276)

Sol.

\(\begin{array}{l}T_r=\frac{r}{\left ( 2r^2 \right )^2+1} \end{array} \)

\(\begin{array}{l}=\frac{r}{\left ( 2r^2+1 \right )^2-\left ( 2r \right )^2} \end{array} \)

\(\begin{array}{l}=\frac{1}{4}\frac{4r}{\left ( 2r^2+2r+1\right )\left ( 2r^2-2r+1 \right )}\end{array} \)

\(\begin{array}{l}S_{10} = \frac{1}{4}\displaystyle\sum\limits_{r=1}^{10}\left ( \frac{1}{\left ( 2r^2-2r+1 \right )}-\frac{1}{\left ( 2r^2+2r+1 \right )} \right )\end{array} \)

\(\begin{array}{l}=\frac{1}{4}\left [1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\dots\dots+\frac{1}{181}-\frac{1}{221} \right ] \end{array} \)

β‡’

\(\begin{array}{l}S_{10}=\frac{1}{4}\cdot\frac{220}{221}=\frac{55}{221}=\frac{m}{n}\end{array} \)

∴ m + n = 276

8. A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is

2x – y + 4 = 0, then the area of R is ________.

Answer (16)

Sol.

JEE Main 2022 June 27 Shift 1 Maths NQA 8

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

∴

\(\begin{array}{l}a=\sqrt{\left ( 3-1 \right )^2+\left ( 6-2 \right )^2}=\sqrt{20} \end{array} \)

and

\(\begin{array}{l}\frac{b}{2}=\frac{4}{\sqrt{5}}\Rightarrow b=\frac{8}{\sqrt{5}} \end{array} \)

Area

\(\begin{array}{l}ab=2\sqrt{5}\cdot\frac{8}{\sqrt{5}}=16 \end{array} \)
.

9. A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola

\(\begin{array}{l}y=\left ( x-\frac{1}{4} \right )^2+\alpha \end{array} \)
where Ξ±> 0. Then (4Ξ± – 8)2 is equal to ___________.

Answer (63)

Sol.

JEE Main 2022 June 27 Shift 1 Maths NQA 9

Let the equation of circle be

\(\begin{array}{l}x\left ( x-\frac{1}{2} \right )+y^2+\lambda y=0\end{array} \)

β‡’

\(\begin{array}{l}x^2+y^2-\frac{1}{2}x+\lambda y=0 \end{array} \)

Radius

\(\begin{array}{l}=\sqrt{\frac{1}{16}+\frac{\lambda^2}{4}}=2\end{array} \)

β‡’

\(\begin{array}{l}\lambda^2=\frac{63}{4}\end{array} \)

β‡’

\(\begin{array}{l}\left ( x-\frac{1}{4} \right )^2+\left ( y+\frac{\lambda}{2} \right )^2=4 \end{array} \)

∡ This circle and parabola

\(\begin{array}{l}y-\alpha =\left ( x-\frac{1}{4} \right )^2\end{array} \)
touch each other, so

\(\begin{array}{l}\alpha=-\frac{\lambda}{2}+2\end{array} \)

β‡’

\(\begin{array}{l}\alpha-2=-\frac{\lambda}{2}\end{array} \)

β‡’

\(\begin{array}{l}\left ( \alpha-2 \right )^2=\frac{\lambda^2}{4}=\frac{63}{16}\end{array} \)

β‡’

\(\begin{array}{l}\left ( 4\alpha-8 \right )^2=63\end{array} \)

10. Let the mirror image of the point (a, b, c) with respect to the plane 3x – 4y + 12z + 19 = 0 be

(a – 6, Ξ², Ξ³). If a + b + c = 5, then 7Ξ² – 9Ξ³ is equal to __________.

Answer (137)

Sol.

\(\begin{array}{l}\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left ( 3a-4b+12c+19 \right )}{3^2+\left ( -4 \right )^2+12^2} \end{array} \)

\(\begin{array}{l}\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169} \end{array} \)

\(\begin{array}{l}\left ( x,y,z \right )\equiv\left ( a-6, \beta,\gamma \right )\end{array} \)

\(\begin{array}{l}\frac{\left ( a-b \right )-a}{3}=\frac{\beta-b}{-4}=\frac{\gamma-c}{12}=\frac{-6a+8b-24c-38}{169} \end{array} \)

\(\begin{array}{l}\frac{\beta-b}{-4}=-2\end{array} \)

β‡’ Ξ² = 8 + b

\(\begin{array}{l}\frac{\gamma-c}{12}=-2 \end{array} \)

β‡’ Ξ³ = -24 + c

\(\begin{array}{l}\frac{-6a+8b-24c-38}{169}=-2\end{array} \)

β‡’ 3a – 4b + 12c = 150….(1)

a + b + c = 5

β‡’ 3a + 3b + 3c = 15….(2)

Applying (1) – (2), we get;

-7b + 9c = 135

7b – 9c = -135

7Ξ² – 9Ξ³ = 7(8 + b) – 9 (–24 + c)

= 56 + 216 + 7b – 9c.

= 56 + 216 – 135 = 137.

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JEE Main 2022 June 27th Shift 1 Question Paper & Solutions

JEE Main 2022 June 27 Shift 1 Question Paper – Physics Solutions

JEE Main 2022 June 27 Shift 1 Question Paper – Chemistry Solutions

JEE Main 2022 June 27 Shift 1 Question Paper – Maths Solutions

Frequently Asked Questions – FAQs

Q1

How was the JEE Main 2022 June 27 Shift 1 Maths question paper?

The JEE Main 2022 June 27 Shift 1 Maths question paper was easy to moderate level difficulty. There were 8 easy questions, 11 medium questions and 2 difficult questions in the JEE Main 2022 June 27 Shift 1 Maths question paper.

Q2

Is there a negative marking in section B of the JEE Main 2022 June 27 Maths Shift 1 question paper?

No. There is no negative marking in section B of the JEE Main 2022 June 27 Maths Shift 1 question paper.

Q3

How many questions were asked from Class 11 in the JEE Main 2022 Maths June 27 Shift 1 question paper?

There were 10 questions from Class 11 in the JEE Main 2022 Maths June 27 Shift 1 question paper as per the memory-based question paper.

Q4

What is the overall difficulty level of the JEE Main 2022 Maths June 27 Shift 1 question paper?

The overall difficulty level of JEE Main 2022 June 27 Shift 1 Maths question paper is 1.71 out of 3

Q5

Were there any questions from Quadratic equations in the JEE Main 2022 June 27 Maths Shift 1 question paper?

Yes. Questions were there from Quadratic equations in the JEE Main 2022 June 27 Maths Shift 1 question paper, and those were easy.

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