JEE Main 2021 question paper for Maths (March 16 – Shift 1) is provided here along with solutions that have been exclusively created by our subject experts. Candidates will find the Maths solved questions and these will help them to check the correct answers as well as analyze their performance. Students can download the JEE Main 2021 March 16th Shift 1 Maths Question Paper in a PDF format for studying and practicing offline to improve their speed and accuracy.

**Section A**

**Question 1: Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, b+2, c+2 is d, then which of the following is true?**

- a. b
^{2}= a^{2}+ c^{2}+ 3d^{2} - b. b
^{2}= 3 (a^{2}+ c^{2}) - 9d^{2} - c. b
^{2}= 3 (a^{2}+ c^{2}) + 9d^{2} - d. b
^{2}= 3 (a^{2}+ c^{2}+ d^{2})

Solution:

Answer: (b)

For a, b, c mean =

\(\bar{x}\)= [a + b + c] / 3\(\bar{x}\)= 2b / 3S.D. of a, b, c = d

d

^{2}= [a^{2}+ b^{2}+ c^{2}] / 3 - (4b^{2}/ 9)b

^{2}= 3a^{2}+ 3c^{2}– 9d^{2}

**Question 2: Let a vector **

\(\sqrt{3} \hat{i}+\hat{j}\)

by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices (ɑ, β), (0, β) and (0, 0) is equal to:- a. 1
- b. 1 / 2
- c. 1 / √2
- d. 2 √2

Solution:

Answer: (b)

(ɑ, β) = (2 cos 75

^{o}, 2 sin 75^{o})Area = (1 / 2) (2 cos 75°)(2 sin 75°)

= sin (150°)

= 1 / 2 square unit

**Question 3: If for a>0, the feet of perpendiculars from the points A (a, –2a, 3) and B (0, 4, 5) on the plane lx + my + nz = 0 are points C (0, –a, –1) and D respectively, then the length of line segment CD is equal to :**

- a. √41
- b. √55
- c. √31
- d. √66

Solution:

Answer: (d)

_{}Direction cosines of plane = λ (direction cosines of line AC)

Direction cosines of plane = λa, - λa, 4λ

Hence equation plane is: ax - ay + 4z = 0

The point C lies on the plane

a (0) − a (− a)+ 4 (−1) = 0 ⇒ a = 2 (∵ a > 0)

So plane is 2x − 2y + 4z = 0, C ≡ (0,−2,−1)

So for coordinates of D,

[x - 0] / 2 = [y - 4] / - 2 = [z - 5] / 4 = - {[2 (0) - 2 (4) + 4 (5)] / [2

^{2}+ 2^{2}+ 4^{2}]}D ≡ (- 1, 5, 3)

CD = √66 units

**Question 4: The range of a ∈ R for which the function f (x) = (4a – 3) (x + log _{e }5) + 2 (a – 7) cot (x / 2) sin^{2} (x / 2), x ≠ 2nπ, n ∈ N has critical points, is**

- a. [- 4 / 3, 2]
- b. [1, ∞)
- c. (- ∞, - 1]
- d. (– 3, 1)

Solution:

Answer: (a)

f (x) = (4a – 3) (x + ln 5) + 2(a – 7) [(cos (x / 2) / sin (x / 2)) * sin

^{2}(x / 2)]f (x) = (4a – 3) (x + log

_{e}5) + (a – 7) sin xf’ (x) = (4a – 3) + (a – 7) cos x = 0

cos x = - (4a - 3) / (a - 7)

- 1 ≤ - (4a - 3) / (a - 7) ≤ 1 (because - 1 ≤ cos x ≤ 1)

- 1 < [4a - 3] / [a - 7] ≤ 1

{[4a - 3] / [a - 7]} - 1 ≤ 0 and {[4a - 3] / [a - 7]} + 1 > 0

a.∈ [4 / 3, 7) and a ∈ (- ∞, 2) ⋃ (7, ∞)(- 4 / 3) ≤ a < 2

**Question 5: Let the functions f: R→R and g: R →R be defined as: **

\(g(x)=\left\{\begin{matrix} x^3, &x<1 \\ 3x-2& x\geq 1 \end{matrix}\right.\)

Then, the number of points in R where (fog)(x) is NOT differentiable is equal to: - a. 1
- b. 2
- c. 3
- d. 0

Solution:

Answer: (a)

\(fog(x))=\left\{\begin{matrix} x^3+2, &x<0 & \\ x^6,&0\leq x< 1 & \\ (3x-2)^2,& x\geq 1 & \end{matrix}\right.\)Clearly fog(x) is discontinuous at x = 0 then non-differentiable at x = 0

Now,

at x = 1

RHD = lim

_{h→0+}[f (1 + h) - f (1) ] / [h] = lim_{h→0+}{[3 (1 + h) - 2]^{2}- 1} / h = 6LHD = lim

_{h→0-}[f (1 - h) - f (1) ] / [- h] = lim_{h→0-}[(1 - h)^{6}- 1] / - h = 6Number of points of non-differentiability = 1

**Question 6:****Let a complex number z, |z| ≠ 1, satisfy log _{1/√2} [(|z| + 11) / (|z| - 1)^{2}] ≤ 2. Then, the largest value of |z| is equal to ____**

- a. 5
- b. 8
- c. 6
- d. 7

Solution:

Answer: (d)

[(|z| + 11) / (|z| - 1)

^{2}] ≥ 1 / 22|z| + 22 ≥ (|z| – 1)

^{2}2|z| + 22 ≥ |z|

^{2}– 2|z| + 1|z|

^{2}– 4|z| – 21 ≤ 0(|z| – 7) (|z| + 3) ≤ 0

|z| ≤ 7

|z|

_{max}= 7

**Question 7: A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade is:**

- a. 3 / 4
- b. 52 / 867
- c. 39 / 50
- d. 22 / 425

Solution:

Answer: (c)

\(P(\bar{S}_{missing}|both \ found \ spade)=\frac{P(\bar{S_m}\cap BFS)}{P(BFS)}\)= [1 - (13 / 52) * (13 / 51) (12 / 50)] / [(1 - (13 / 52) * {(13 / 51) (12 / 50) + (13 / 52) (12 / 51) * (11 / 50)}]

= 39 / 50

**Question 8: If n is the number of irrational terms in the expansion of [3 ^{1/4} + 5^{1/8}]^{60}, then (n - 1) is divisible by**

- a. 8
- b. 26
- c. 7
- d. 30

Solution:

Answer: (b)

T

_{r+1}=^{60}C_{r}(3^{1/4})^{60-r}(5^{1/8})^{r}It is rational if [60 - r] / 4, (r / 8), both are whole numbers, r ∈ {0,1,2,....60}

[60 - r] / 4 ∈ W ⇒ r ∈ {0, 4, 8, ……..60} and (r / 8) ∈ W ⇒ r ∈ {0,8,16,…56}

Common terms r ∈ {0,8,16,....56}

So 8 terms are rational

Then irrational terms = 61 – 8 = 53 = n

∴n – 1 = 52 = 13 × 2

^{2}The factors are 1,2,4,13,26,52.

**Question 9: Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, –1, 2) and (–2,1,–2) respectively. Let lines PR and QS intersect at T. If the vector TA is perpendicular to both PR and QS and the length of vector TA is √5 units, then the modulus of a position vector of A is:**

- a. √5
- b. √171
- c. √227
- d. √482

Solution:

Answer: (b)

P = 3i - j + 2k and Q = i + 2j - 4k

v

_{PR }= (4, - 1, 2) and v_{QS}= (- 2 , 1, - 2)L

_{PR}= (3i - j + 2k) + λ (4i - 1j + 2k)L

_{QS}: r = (i + 2j - 4k) + 𝝁 (- 2i + 1j - 2k)Now T on PR = (3 + 4λ, - 1 - λ, 2 + 2λ)

Similarly T on QS = (1 − 2𝝁, 2 + 𝝁,−4 − 2𝝁)

For λ and 𝝁:

3 + 4λ = 1 - 2𝝁 ⇒ 𝝁 + 2λ = - 1 ⇒ λ = 2

- 1 - λ = 2 + 𝝁 ⇒ 𝝁 + λ = - 3 ⇒ 𝝁 = - 5

And 2 + 2λ = - 4𝝁

T : (11, - 3, 6)

D.R. of TA = v

_{QS }x v_{PR}\(\begin{vmatrix} i &j &k \\ -2 &1 &-2 \\ 4 &-1 &2 \end{vmatrix}\)= 0i - 4j - 2kL

_{TA}: r = (11i - 3j + 6k) + λ (- 4j - 2k)Now A = (11,− 3 − 4λ, 6 − 2λ)

TA = √5

(4λ)

^{2}+ (2λ)^{2}= 516λ

^{2}+ 4λ^{2}= 5λ = ± 1 / 2

A: (11, –5, 5) or A: (11, –1, 7)

|A| = √(121 + 25 + 25) or |A| = √(121 + 1 + 49)

= √171 or √171

**Question 10: If the three normals drawn to the parabola, y ^{2 }= 2x pass through the point (a, 0) a ≠ 0, then ‘a’ must be greater than**

- a. 1
- b.1 / 2
- c. - 1 / 2
- d. –1

Solution:

Answer: (1)

Let the equation of the normal be y = mx – 2am – am

^{3}.Here 4a = 2

a = 1 / 2

y = mx – m – (1 / 2) m

^{3}It passes through A (a, 0) then

0 = am – m – (1 / 2) m

^{3}m = 0, m

^{2 }- 2 (a - 1) = 0For real values of m

2 (a – 1) > 0

a > 1

**Question 11: Let S _{k} = ∑_{r=1}^{k} tan^{-1}[(6^{r}) / (2^{r+1} + 3^{2r+1})]. Then lim_{k→∞} S_{k} =**

- a. tan
^{-1}(3 / 2) - b. cot
^{-1}(3 / 2) - c. π / 2
- d. tan
^{–1}(3)

Solution:

Answer: (2)

**Question 12: The number of roots of the equation, (81) ^{sin2x} + (81)^{cos2x} = 30 in the interval [0, π] is equal to : **

- a. 3
- b. 2
- c. 4
- d. 8

Solution:

Answer: (3)

(81)

^{sin2x}+ (81)^{cos2x}= 30(81)

^{sin2x}+ [81 / (81)^{sin2x}] = 30Let (81)

^{sin2x}= tt + (81 / t) = 30 ⇒ t

^{2}+ 81 = 30tt

^{2}– 30t + 81 = 0t

^{2}– 27t – 3t + 81 = 0(t – 3) (t – 27) = 0

t = 3, 27

(81)

^{sin2x}= 3, 3^{3}3

^{4sin2x}= 3^{1}, 3^{3}4 sin

^{2}x = 1, 3sin

^{2}x = (1 / 4), (3 / 4) in [0, π] sin x ≥ 0sin x = 1 / 2, √3 / 2

x = π / 6, 5π / 6, π / 3, 2π / 3

Number of solutions = 4

**Question 13: If y = y(x) is the solution of the differential equation, dy / dx + 2y tanx = sinx, y (π / 3) = 0, then the maximum value of the function y(x) over R is equal to :**

- a. 8
- b. 1 / 2
- c. - 15 / 4
- d. 1 / 8

Solution:

Answer: 4

dy / dx + 2y tanx = sinx

IF = e

^{ln (sec2x)}= sec^{2}xy sec

^{2}x = ∫tanx secx dx = secx + cNow, x = π / 3, y = 0, c = - 2

y = cosx - 2cos

^{2}xy = - 2 [cos

^{2}x - (1 / 2) cosx] = - 2 [(cosx - (1 / 4))^{2}- (1 / 16)]y = (1 / 8) - 2 [cosx - (1 / 4)]

^{2}y

_{max}= 1 / 8

**Question 14: Which of the following Boolean expression is a tautology?**

- a. (p ∧ q) ∧ (p → q)
- b. (p ∧ q) ∨ (p ∨ q)
- c. (p ∧ q) ∨ (p → q)
- d. (p ∧ q) → (p → q)

Solution:

Answer: (4)

**p****q****p ∧ q****p ∨ q****p → q****(p ∧ q) → (p → q)**T

T

T

T

T

T

F

T

F

T

T

T

T

F

F

T

F

T

F

F

F

F

T

T

**Question 15: Let **

\(A^8\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 8\\ 64 \end{bmatrix}\)

has :- a. No solution
- b. Exactly two solutions
- c. A unique solution
- d. Infinitely many solutions

Solution:

Answer: (1)

**Question 16: If for x ∈ (0, π / 2), log _{10}sin x + log_{10}cos x = –1 and log_{10}(sin x+cos x) = 1 / 2 (log_{10} n – 1), n>0, then the value of n is equal to:**

- a. 16
- b. 20
- c. 12
- d. 9

Solution:

Answer: (3)

log

_{10}sin x + log_{10}cos x = –1sinx * cosx = 1 / 10 --- (1)

log

_{10}(sinx + cosx) = (1 / 2) (log_{10}n - 1)(sinx + cosx) = (n / 10)

^{½}sin

^{2}x + cos^{2}x + 2 sinx cosx = n / 10 (squaring)1 + 2 (1 / 10) = n / 10 [using equation (1)]

n / 10 = 12 / 10

n = 12

**Question 17: The locus of the midpoints of the chord of the circle, x ^{2 }+ y^{2 }= 25 which is tangent to the hyperbola, x^{2} / 9 - y^{2} / 16 = 1 is :**

- a. (x
^{2}+ y^{2})^{2}- 16x^{2}+ 9y^{2}= 0 - b. (x
^{2}+ y^{2})^{2}- 9x^{2}+ 144y^{2}= 0 - c. (x
^{2}+ y^{2})^{2}- 9x^{2}- 16y^{2}= 0 - d. (x
^{2}+ y^{2})^{2}- 9x^{2}+ 16y^{2}= 0

Solution:

Answer: (4)

The tangent of hyperbola y = mx ± √(9m

^{2}- 16) …(i) which is a chord of circle with mid-point (h, k).So, the equation of chord T = S

_{1}hx + ky = h

^{2}+ k^{2}y = - (hx / k) + (h

^{2}+ k^{2}) / k ---- (ii)By (i) and (ii)

m = - h / k and √9m

^{2}- 16 = [h^{2}+ k^{2}] / k9 (h

^{2}/ k^{2}) - 16 = (h^{2}+ k^{2})^{2}/ k^{2}The locus 9x

^{2}– 16y^{2}= (x^{2}+ y^{2})^{2}.

**Question 18: Let [x] denote the greatest integer less than or equal to x. If for n ∈ N, (1 - x + x ^{3})^{n} = ∑_{j=0}^{3n} a_{j}x^{j}, then ∑_{j=0}^{3n/2} a_{2j} + 4 ∑_{j=0}^{(3n-1)/2} a_{2j+1} =**

- a. 1
- b. n
- c. 2
^{n–1} - d. 2

Solution:

Answer: (1)

(1 - x + x

^{3})^{n}= ∑_{j=0}^{3n}a_{j}x^{j}(1–x + x

^{3})^{n}= a_{0}+ a_{1}x + a_{2}x^{2}+ ….. + a_{3n}x^{3n}Put x = 1

1 = a

_{0}+ a_{1}+ a_{2}+ a_{3}+ a_{4}+ …… +a_{3n}…(1)Put x = –1

1 = a

_{0}– a_{1}+ a_{2}– a_{3}+ a_{4}…… (–1)^{3n}a_{3n}…(2)Add (1) + (2)

a

_{0}+ a_{2}+ a_{4}+ a_{6}+ …….. = 1Sub (1) – (2)

⇒ a

_{1}+ a_{3}+ a_{5}+ a_{7}+ …….. = 0Now ∑

_{j=0}^{3n/2}a_{2j}+ 4 ∑_{j=0}^{(3n-1)/2}a_{2j+1}== (a

_{0}+ a_{2}+ a_{4}+ …..) + 4(a_{1}+ a_{3}+ ….)= 1 + 4 × 0

= 1

**Question 19: Let P be a plane lx + my + nz = 0 containing the line, [1 - x] / 1 = [y + 4] / 2 = [z + 2] / 3. If plane P divides the line segment AB joining points A (–3, –6, 1) and B (2, 4, –3) in ratio k : 1 then the value of k is equal to : **

- a. 1.5
- b. 2
- c. 4
- d. 3

Solution:

Answer: (2)

Line lies on plane

– l + 2m + 3n = 0 ...(1)

Point on line (1,– 4, – 2) lies on plane

l – 4m – 2n = 0 …(2)

From (1) & (2)

–2m + n = 0

2m = n

l = 3n + 2m

l = 4n

l : m : n :: 4n : (n / 2) : n

l : m : n :: 8n : (n) : 2n

l : m : n :: 8 : 1 : 2

Now equation of plane is 8x + y + 2z = 0

R divide AB is ratio k : 1

R : [(- 3 + 2k) / (k + 1), (- 6 + 4k) / (k + 1), (1 - 3k) / (k + 1)] lies on the plane

8 [(- 3 + 2k) / (k + 1) + (- 6 + 4k) / (k + 1) + 2 (1 - 3k) / (k + 1)] = 0

–24 + 16 k – 6+ 4k + 2 – 6k = 0

–28 + 14k = 0

k = 2

**Question 20: The number of elements in the set x ∈ R : (|x| - 3) |x + 4| = 6) =**

- a. 2
- b. 1
- c. 3
- d. 4

Solution:

Answer: (1)

Case - 1: x ≤ – 4

(–x – 3)(–x – 4) = 6

(x + 3) (x + 4) = 6

x

^{2}+ 7x + 6 = 0x = –1 or – 6

But x ≤ – 4

x = – 6

Case - 2: x ∈ (– 4, 0)

(–x – 3)(x + 4) = 6

–x

^{2}– 7x – 12 – 6 = 0x

^{2}+ 7x + 18 = 0D < 0 No solution

Case - 3: x ≥ 0

(x – 3)(x + 4) = 6

x

^{2}+ x – 12 – 6 = 0x

^{2}+ x – 18 = 0x = (- 1 ± √73) / 2

Only x = (√73 - 1) / 2

Hence, 2 elements only

**Section - B**

**Question 21: Let f: (0, 2) → R be defined as f (x) = log _{2} [1 + tan (πx / 4)]. Then, lim_{n→∞} (2 / n) [f (1 / n) + f (2 / n) ……. f (1)] is equal to ______**

Solution:

Answer: (1)

**Question 22: The total number of 3 × 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AA ^{T} is 9, is equal to ____**

Solution:

Answer: (766)

\(AA^T=\begin{bmatrix} x &y &z \\ a&b &c \\ d &e &f \end{bmatrix}\begin{bmatrix} x&a &d \\ y&b &e \\ z&c &f \end{bmatrix}\\ =\begin{bmatrix} x^2+y^2+z^2 &ax+by+cz &dx+ey+fz \\ ax+by+cz& a^2+b^2+c^2 &ad+be+cf \\ dx+ey+fz &ad+be+cf & d^2+e^2+f^2 \end{bmatrix}\)Tr (AA

^{T}) = x^{2}+ y^{2}+ z^{2}+ a^{2}+ b^{2}+ c^{2}+ d^{2}+ e^{2}+ f^{2}= 9all→1 =1

one 3, rest = 0 (9! / 8!) = 9

two 2 , one 1 & rest 0 (9!) / (2! 6!) = 63 × 4 = 252

one 2 , five 1, rest 0 (9!) / (5! 3!) = 63 × 8 = 504

Total = 766

**Question 23: Let f:R→R be a continuous function such that f (x) + f (x + 1) = 2, for all x ∈ R. If I _{1} = ∫_{0}^{8} f (x) dx and I_{2} = ∫_{-1}^{3} f (x) dx, then the value of I_{1} + 2I_{2} is equal to ____**

Solution:

Answer: (16)

f (x) + f (x + 1) = 2 …..(i)

x→(x + 1)

f(x + 1) + f(x + 2) = 2 …..(ii)

By (i) & (ii)

f(x) – f(x + 2) = 0

f(x + 2) = f(x)

f(x) is periodic with T = 2

I

_{1 }= ∫_{0}^{2*4}f (x) dx = 4 ∫_{0}^{2}f (x) dxI

_{2}= ∫_{-1}^{3}f (x) dx = ∫_{0}^{4}f (x + 1) dx = ∫_{0}^{4}[2 - f (x)] dxI

_{2}= 8 - 2 ∫_{0}^{2}f (x) dxI

_{1 }+ 2I_{2}= 16

**Question 24: Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four-digit numbers, then the number of common terms in these two series is equal to ____**

Solution:

Answer: (3)

By observation

A.P : 11, 16, 21, 26 …….

G.P : 4, 8, 16, 32 …….

So common terms are 16, 256, 4096.

**Question 25: If the normal to the curve y (x) = ∫ _{0}^{x} (2t^{2} - 15t + 10) dt at a point (a, b) is parallel to the line x + 3y = – 5, a>1, then the value of |a + 6b| is equal to ____**

Solution:

Answer: (406)

y' (x) = (2x

^{2}– 15x + 10)At point (a, b) normal is

3 = (2a

^{2}– 15a + 10)2a

^{2}– 15a + 7 = 02a

^{2}– 14a – a + 7 = 02a (a – 7) – 1 (a – 7) = 0

a = 1 / 2 or 7,

Given a > 1, a = 7

Also P lies on curve,

b = ∫

_{0}^{a}(2t^{2}- 15t + 10) dtb = ∫

_{0}^{7}(2t^{2}- 15t + 10) dt6b = – 413

|a + 6b| = 406

**Question 26: If lim _{x→0} [ae^{x} - b cosx + ce^{-x}] / [x sinx] = 2, then a + b + c is equal to ______**

Solution:

Answer: (4)

a – b + c = 0 & a – c = 0 & (a / 2) + (b / 2) + (c / 2) = 2

a + b + c = 4

**Question 27: Let ABCD be a square of the side of unit length. Let a circle C _{1} centred at A with a unit radius is drawn. Another circle C_{2} which touches C_{1} and the lines AD and AB is tangent to it, is also drawn. Let a tangent line from point C to the circle C_{2} meet the side AB at E. If the length of EB is ɑ + √3β, where ɑ, β are integers, then ɑ + β is equal to ___**

Solution:

Answer: (1)

(i) √2r + r = 1

r = 1 / (√2 + 1)

r = √2 - 1

(ii) CC

_{2}= 2√2 - 2 = 2 (√2 - 1)From △CC

_{2}N = sin ɸ = (√2 - 1) / [2 (√2 - 1)]ɸ = 30

^{o}(iii) In △ACE are sine law

AE / sin ɸ = AC / sin 105

^{o}AE = (1 / 2) * (√2 / (√3 + 1)) * 2√2

AE = 2 / (√3 + 1) = √3 - 1

EB = 1 - (√3 - 1)

ɑ = 2, β = - 1

ɑ + β = 1

**Question 28: Let z and w be two complex numbers such that **

Solution:

Answer: (4)

Let z = x + iy

|z + i| = |z–3i|

y = 1

Now w = x

^{2}+ y^{2}– 2x – 2iy + 2w = x

^{2 }+ 1 – 2x –2i + 2Re(w) = x

^{2}– 2x + 3Re(w) = (x–1)

^{2}+ 2Re(w)

_{min}at x = 1z = 1 + i

Now w = 1 + 1 –2 – 2i + 2

w = 2(1–i) = 2√2 e

^{i(-π/4)}w

^{n}= 2√2 e^{i(-π/4)}If w

^{n}is real, n = 4

**Question 29: Let **

\(A=\begin{bmatrix} 2 &7 &\omega ^2 \\ -1&-\omega &1 \\ 0&-\omega &-\omega+1 \end{bmatrix}\)

where 𝛚 = (- 1 + i √3) / 2, and ISolution:

Answer: (36)

|P

^{-1}AP - I|^{2}= |(P

^{-1}AP - I) (P^{-1}AP - I)|= |P

^{-1}APP^{-1}AP - 2P^{-1}AP + I|= |P

^{-1}A^{2}P - 2P^{-1}AP + P^{-1}IP|= |P

^{-1}(A^{2}- 2A + I) P|= |P

^{-1}(A - I)^{2}P|= |P

^{-1}| |A - I|^{2}|P|= |A - I|

^{2}=

\(=\begin{bmatrix} 1 &7 &\omega ^2 \\ -1&-\omega-1 &1 \\ 0&-\omega &-\omega \end{bmatrix}^2\)= (1 (𝛚 (𝛚 + 1) + 𝛚) - 7𝛚 + 𝛚

^{2}. 𝛚)^{2}= (𝛚

^{2}+ 2𝛚 - 7𝛚 + 1)^{2}= (𝛚

^{2}- 5𝛚 + 1)^{2}= (- 6𝛚)

^{2}= 36𝛚

^{2}⇒ ɑ = 36

**Question 30: Let the curve y=y(x) be the solution of the differential equation, dy / dx = 2 (x + 1). If the numerical value of area bounded by the curve y = y(x) and the x-axis is 4√8 / 3, then the value of y(1) is equal to ___**

Solution:

Answer: (2)

y = x

^{2}+ 2x + cArea of rectangle ABCD = |(c - 1)√(1 - c)|

Area of parabola and x-axis = 2 (2 / 3 [(1 - c)

^{3/2}]) = 4√8 / 31 – c = 2

c = –1

Equation of f (x) = x

^{2}+ 2x – 1f (1) = 1 + 2 – 1 = 2

JEE Main 2021 Maths Paper March 16 Shift 1