JEE aspirants will find solutions for all the questions asked in JEE Main 2021 Mathematics paper for February 26th Shift 2 exam on this page. The set of solutions have been prepared by BYJU’s expert teachers. The solutions have been crafted in a detailed manner. Practising these questions will help every JEE aspirant to understand the overall exam pattern and prepare accordingly.

**SECTION A**

**Question 1: Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P (ɑ, β, γ) is the foot of the perpendicular from (3, 2, 1) on L, then the value of 21 (ɑ + β + γ) equals:**

- a. 142
- b. 68
- c. 136
- d. 102

Solution:

Answer: (d)

Equation of the line is x + 2y + z - 6 = 0 = y + 2z = 4

[x + 2] / 3 = [y - 4] / - 2 = z / 1 = λ

Dr’s of PQ: 3λ - 5, - 2λ - 2, λ - 1

Dr’s of y lines are (3, - 2, 1)

Since PQ is perpendicular to the line.

3 (3λ - 5) - 2 (- 2λ + 2) + 1 (λ - 1) = 0

λ = 10 / 7

P (16 / 7, 8 / 7, 10 / 7)

21 (ɑ + β + γ) = 21 (34 / 7) = 102

**Question 2: The sum of the series ∑ _{n=1}^{∞}∞[n^{2 }+ 6n + 10] / (2n + 1)! is equal to:**

- a. (41/ 8) e + (19 / 8) e
^{-1}- 10 - b. (- 41/ 8) e + (19 / 8) e
^{-1}- 10 - c. (41/ 8) e - (19 / 8) e
^{-1}- 10 - d. (41/ 8) e + (19 / 8) e
^{-1}+ 10

Solution:

Answer: (c)

∑

_{n=1}^{∞}∞[n^{2 }+ 6n + 10] / (2n + 1)!Put 2n + 1 = r, where r = 3, 5, 7…..

n = [r - 1] / 2

**Question 3: Let f(x) be a differentiable function at x = a with f’ (a) = 2 and f (a) = 4. Then lim _{x→a} [x f (a) - a f (x)] / [x - a] equals:**

- a. 2a + 4
- b. 2a - 4
- c. 4 - 2a
- d. a + 4

Solution:

Answer: (c)

By L-H rule

L = lim

_{x→a}[f (a) - a f’ (x)] / [1]L = 4 - 2a

**Question 4: Let A (1, 4) and B (1, - 5) be two points. Let P be a point on the circle (x - 1) ^{2} + (y - 1)^{2} = 1 such that (PA)^{2} + (PB)^{2} have maximum value, then the points P, A and B lie on:**

- a. a parabola
- b. a straight line
- c. a hyperbola
- d. an ellipse

Solution:

Answer: (b)

PA

^{2}= cos^{2}θ + (sin θ - 3)^{2}= 10 - 6 sin θPB

^{2}= cos^{2}θ + (sin θ + 6)^{2}= 37 + 12 sin θPA

^{2}+ PB^{2}|_{max}= 47 + 6 sin θ|_{max}θ = π / 2

The points P, A and B lie on the line x = 1.

**Question 5: If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x ^{2 }+ y^{2 }= 1 is a circle of the radius r, then r is equal to :**

- a. 14
- b. 12
- c. 1
- d. 13

Solution:

Answer: (b)

P ≡ (2h - 3), (2k - 2) → on circle

[h - (3 / 2)]

^{2}+ (k - 1)^{2}= 1 / 4Radius = 1 / 2

**Question 6: Let the slope of the tangent line to a curve at any point P (x, y) be given by [xy ^{2} + y] / x. If the curve intersects the line x + 2y = 4 at x = - 2, then the value of y, for which the point (3, y) lies on the curve, is:**

- a. - 18 / 11
- b. - 18 / 19
- c. - 4 / 3
- d. 18 / 35

Solution:

Answer: (b)

dy / dx = [xy

^{2}+ y] / x[x dy - y dx] / y

^{2}= x dx- d (x / y) = d (x

^{2}/ 2)(- x / y) = (x

^{2}/ 2) + cThe curve intersects the line x + 2y = 4 at x = - 2.

So, - 2 + 2y = 4

y = 3

The curve passes through (-2, 3).

(2 / 3) = 2 + c

c = - 4 / 3

The curve is (- x / y) = (x

^{2}/ 2) - (4 / 3).It also passes through (3, y).

(- 3 / y) = (9 / 2) - (4 / 3)

y = - 18 / 19

**Question 7: Let A _{1} be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A_{2} be the area of the region bounded by the curves y = sinx, y = cosx, the x-axis and x = π / 2 in the first quadrant. Then,**

- a. A
_{1}= A_{2}and A_{1}+ A_{2}= √2 - b. A
_{1}: A_{2}= 1 : 2 and A_{1}+ A_{2}= 1 - c. 2A
_{1}= A_{2}and A_{1}+ A_{2}= 1 + √2 - d. A
_{1}: A_{2}= 1 : √2 and A_{1}+ A_{2}= 1

Solution:

Answer: (d)

A

_{1}+ A_{2}= ∫_{0}^{π/2}cosx dx = sin x|_{0}^{π/2}= 1A

_{1 }= ∫_{0}^{π/4}(cosx - sinx) dx = (sinx + cosx)_{0}^{π/4}= √2 - 1A

_{2}= 1 - (√2 - 1) = 2 - √2A

_{1}/ A_{2}= (√2 - 1) / (√2 (√2 - 1)) = 1 / √2

**Question 8: If 0 < a, b < 1, and tan ^{-1} a + tan^{-1} b = π / 4, then the value of (a + b) - [a^{2} + b^{2}] / 2 + [a^{3} + b^{3} ] / 3 - [a^{4} + b^{4}] / 4 + …. is**

- a. log
_{e}2 - b. log
_{e}(e / 2) - c. e
- d. e
^{2}- 1

Solution:

Answer: (a)

tan

^{-1}{[a + b] / [1 - ab]} = (π / 4)a + b = 1 - ab

(1 + a) (1 + b) = 2

(a + b) - [a

^{2}+ b^{2}] / 2 + [a^{3}+ b^{3}] / 3 +…infinity= [a - (a

^{2}/ 2) + (a^{3}/ 3) …..] + [b - (b^{2}/ 2) + (b^{3}/ 3) …..]= log

_{e}(1 + a) + log_{e}(1 + b)= log

_{e}(1 + a) (1 + b) = log_{e}2

**Question 9: Let F _{1} (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F_{2} (A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then:**

- a. F
_{1}is not a tautology but F_{2}is a tautology - b. F
_{1}is a tautology but F_{2}is not a tautology - c. F
_{1}and F_{2}both area tautologies - d. Both F
_{1}and F_{2}are not tautologies

Solution:

Answer: (a)

F

_{1 }(A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~AUsing the set theory

(A ∩ B′) ∪ (C′ ∩ (A ∪ B)) ∪ A′ = (A − A ∩ B) ∪ (S − A) ∪ [(S − C) ∩ (A ∪ B)]

= (S − A ∩ B) ∪ [A ∪ B − C ∩ (A ∪ B)]

= S − A ∩ B ∩ C.

Hence not a tautology.

Now F

_{2 }(A, B) = (A ∨ B) ∨ (B → ~A) = (A ∨ B) ∨ (~B ∨ A)Using set theory (A ∪ B) ∪ (B′ ∪ A) = (A ∪ B) ∪ (S − A ∪ B) = S

Hence it is a tautology.

**Question 10: Consider the following system of equations:**

**x + 2y - 3z = a**

**2x + 6y - 11 z = b**

**x - 2y + 7z = c,**

**Where a, b and c are real constants. Then the system of equations:**

- a. has a unique solution when 5a = 2b + c
- b. has an infinite number of solutions when 5a = 2b + c
- c. has no solution for all a, b and c
- d. has a unique solution for all a, b and c

Solution:

Answer: (b)

= 20a - 2(7b + 11c) - 3(- 2b - 6c)

= 20a - 14b - 22c + 6b + 18c

= 20a - 8b - 4c

= 4(5a - 2b - c)

\(\Delta _{2} = \begin{vmatrix} 1 & a & -3 \\ 2& b & -11\\ 1& c & 7 \end{vmatrix}\)= 7b+ 11c - a(25) - 3(2c - b)

= 7b+ 11c - 25a - 6c + 3b

= - 25a + 10b+ 5c

= - 5(5a - 2b- c)

\(\Delta _{3} = \begin{vmatrix} 1 & 2 & a\\ 2& 6 & b\\ 1& -2 & c \end{vmatrix}\)= 6c + 2b- 2(2c - b) - 10a

= - 10a + 4b+ 2c

= - 2(5a - 2b- c)

For infinite solution,

Δ = Δ

_{1}= Δ_{2}= Δ_{3}= 0⇒ 5a = 2b + c

**Question 11: A seven-digit number is formed using the digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is:**

- a. 6 / 7
- b. 4 / 7
- c. 3 / 7
- d. 1 / 7

Solution:

Answer: (3)

n (S) = 7! / [2! 3! 2!]

n (E) = 6! / [2! 2! 2!]

P (E) = n (E) / n (S) = [6! / 7!] * [[2! 3! 2!] / [2! 2! 2!]] = 3 / 7

**Question 12: **If vectors a_{1} = xi - j + k and a_{2} = i + yj + zk are collinear, then a possible unit vector parallel to the vector xi + yj + zk is :

- a. (1 / √2) (- j + k)
- b. (1 / √2) (i - j)
- c. (1 / √3) (i - j + k)
- d. (1 / √3) (i + j - k)

Solution:

Answer: (3)

(x / 1) = - 1 / y = 1 / z = λ (let)

Unit vector parallel to xi + yj + zk = ± [λi - (1 / λ) j + (1 / λ) k] / √λ

^{2}+ (2 / λ^{2})For λ = 1, it is ± [i - j + k] / √3

**Question 13: For x > 0, if f (x) = ∫ _{1}^{x} log_{e} t / [1 + t] dt, then f (e) + f (1 / e) is equal to:**

- a. 1 / 2
- b. -1
- c. 1
- d. 0

Solution:

Answer: (a)

**Question 14: Let f : R → R be defined as**

**If f (x) is continuous on R, then a + b equals:**

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

Solution:

Answer: (b)

If f is continuous at x = - 1, then f (- 1

^{-}) = f (- 1)2 = |a - 1 + b|

|a + b - 1| = 2 --- (i)

Similarly f (1

^{-}) = f (1)|a + b + 1| = 0

a + b = - 1

**Question 15: Let A = {1, 2, 3 ……. , 10} and f : A → A be defined as **

- a. 10
^{5} - b.
^{10}C_{5} - c. 5
^{5} - d. 5!

Solution:

Answer: (a)

g (f (x)) = f (x)

g (x) = x, when x is even.

5 elements in A can be mapped to any 10

So, 10

^{5}× 1 = 10^{5}

**Question 16: A natural number has prime factorization given by n = 2 ^{x}3^{y}5^{z}, where y and z are such that y + z = 5 and y^{-1} + z^{-1} = 5 / 6, y > z. Then the number of odd divisors of n, including 1, is: **

- a. 11
- b. 6x
- c. 12
- d. 6

Solution:

Answer: (c)

y + z = 5 . . . (1)

(1 / y) + (1 / z) = 5 / 6

yz = 6

Also, (y - z)

^{2}= (y + z)^{2}- 4yzy - z = ± 1 --- (2)

From (1) and (2), y = 3 or 2 and z = 2 or 3

For calculating the odd divisor of p = 2

^{x}3^{y}5^{z}, x must be 0.P = 2

^{0}3^{3}5^{2}Total odd divisors must be (3 + 1) (2 + 1) = 12.

**Question 17: Let f (x) = sin ^{-1} x and g (x) = [x^{2} - x - 2] / [2x^{2} - x - 6]. If g (2) = lim_{x→2 }g (x), then then the domain of the function fog is:**

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

Solution:

Answer: (a)

g (2) = lim

_{x→2}[(x - 2) (x + 1)] / [(2x + 3) (x - 2)] = 3 / 7For domain of f (g (x))

|(x

^{2}- x - 2) / (2x^{2}- x - 6)| ≤ 1 [because of f (x) is [- 1, 1])(x + 1)

^{2}≤ (2x +3)^{2}3x

^{2}+ 10x + 8 ≥ 0(3x + 4) (x + 2) ≥ 0

x belongs to (- ∞, - 2] ⋃ [- 4 / 3, ∞]

**Question 18: If the mirror image of the point (1, 3, 5) with respect to the plane 4x - 5y + 2z = 8 is (ɑ, β, γ), then 5 (ɑ + β + γ) equals:**

- a. 47
- b. 39
- c. 43
- d. 41

Solution:

Answer: (a)

Image of (1, 3, 5) in the plane 4x - 5y + 2z = 8 is (ɑ, β, γ).

[ɑ - 1] / 4 = [β - 3] / - 5 = [γ - 5] / 2 = - 2 [4 (1) - 5 (3) + 2 (5) - 8] / [4

^{2}+ 5^{2}+ 2^{2}] = 2 / 5ɑ = 1 + 4 (2 / 5) = 13 / 5

β = 3 - 5 (2 / 5) = 1 = 5 / 5

γ = 5 + 2 (2 / 5) = 29 / 5

5 (ɑ + β + γ) = 5 [(13 / 5) + (5 / 5) + (29 / 5)] = 47

**Question 19: Let f (x) = ∫ _{0}^{x} e^{t} f (t) dt + e^{x} be a differentiable function for all x ∈ R. Then f (x) equals**

- a. 2e
^{ex-1}- 1 - b. e
^{(e^x-1)} - c. 2e
^{e^x}- 1 - d. e
^{e^x}- 1

Solution:

Answer: (a)

Given, f (x) = ∫

_{0}^{x}e^{t}f (t) dt + e^{x}---- (1)f’ (x) = e

^{x}. f (x) + e^{x}[using Newton Leibnitz Theorem][f’ (x)] / [f (x) + 1] = e

^{x}ln (f (x) + 1) = e

^{x}+ cPut x = 0, ln 2 = 1 + c

ln [f (x) + 1] = e

^{x}+ ln 2 - 1f (x) + 1 = 2 e

^{ex-1}f (x) = 2 e

^{ex-1}- 1

**Question 20: The triangle of the maximum area that can be inscribed in a given circle of radius ‘r’ is:**

- a. A right-angle triangle having two of its sides of length 2r and r.
- b. An equilateral triangle of height 2r / 3.
- c. isosceles triangle with base equal to 2r.
- d. An equilateral triangle having each of its side of length √3r.

Solution:

Answer: (d)

The triangle of the maximum area that can be inscribed in a circle is an equilateral triangle.

Let △ABC be inscribed in the circle,

Now, in △OBD

OD = r cos 60

^{o}= r / 2Height = AD = 3r / 2

Again in △ABD

Now sin 60

^{o}= (3r / 2) / ABAB = √3r

**Section - B**

**Question 1: The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _______**

Solution:

Answer: 1000

Since the required number has G.C.D with 18 as 3. It must be an odd multiple of ‘3’ but not a multiple of ‘9’.

(i) Now, a 4-digit number which is an odd multiple of ‘3’ are,

1005, 1011, 1017, ⋯ , 9999 → 1499

(ii) The 4-digit number which is an odd multiple of 9 are,

1017, 1035, ⋯ , 9999 → 499

Required numbers = 1499 − 499 = 1000

**Question 2: Let ɑ and β be two real numbers such that ɑ + β = 1 and ɑ β = - 1. Let P _{n} = (ɑ)^{n} + (β)^{n}, P_{n-1} = 11 and P_{n+1} = 29 for some integer n≥ 1. Then, the value of P_{n}^{2} is______________.**

Solution:

Answer: 324

Given, ɑ + β = 1 and ɑ β = - 1

A quadratic equation with roots ɑ, β is x

^{2}- x - 1 = 0.ɑ

^{2}= ɑ + 1Multiplying both sides by ɑ

^{n-1}ɑ

^{n+1}= ɑ^{n}+ ɑ^{n-1}--- (1)Similarly,

β

^{n+1}= β^{n}+ β^{n-1}--- (2)Adding (1) and (2)

ɑ

^{n+1}+ β^{n+1}= (ɑ)^{n}+ (β)^{n}+ [ɑ^{n-1}+ β^{n-1}]P

_{n+1}= P_{n}+ P_{n-1}29 = P

_{n}+ 11P

_{n}= 18P

_{n}^{2}= 18^{2}= 324

**Question 3: Let X _{1}, X_{2}, …. X_{18} be eighteen observation such that ∑_{i=1}^{18} (X_{i} - ɑ) = 36 and ∑_{i=1}^{18} (X_{i} - β)^{2} = 90, where ɑ and β are distinct real numbers. If the standard deviation of these observations is 1, then the value of |ɑ - β| is ___________.**

Solution:

Answer: 4

Given, ∑

_{i=1}^{18}(X_{i}- ɑ) = 36∑x

_{i}- 18ɑ = 36∑x

_{i}= 18 (ɑ + 2) ---- (1)∑

_{i=1}^{18}(X_{i}- β)^{2}= 90∑x

_{i}^{2}+ 18β^{2}- 2β ∑x_{i}= 90∑x

_{i}^{2}+ 18β^{2}- 2β * 18 (ɑ + 2) = 90 (using equation (1))∑x

_{i}^{2}= 90 - 18β^{2}+ 36β (ɑ + 2)Now, 𝜎

^{2}= 1 ⇒ (1 / 18) ∑x_{i}^{2}- (∑x_{i}/ 18)^{2}= 1(1 / 18) [90 - 18β

^{2}+ 36ɑβ + 72β] - [18 (ɑ + 2) / 18]^{2}= 15 − β

^{2}+ 2ɑβ + 4β − (ɑ + 2)^{2}= 1⇒ 5 − β

^{2}+ 2ɑβ + 4β − ɑ^{2}− 4 − 4ɑ = 1⇒ ɑ

^{2}− β^{2}+ 2ɑβ + 4β − 4ɑ = 0⇒ (ɑ − β)(ɑ − β + 4) = 0

⇒ ɑ − β = −4

Hence |ɑ − β| = 4 (ɑ ≠ β).

**Question 4: In I _{m,n} = ∫_{0}^{1} x^{m-1} (1 - x)^{n-1} dx, for m, n ≥ 1 and ∫_{0}^{1} [x^{m-1} + x^{n-1}] / [(1 + x)^{m+n}] dx = ɑI_{m,n}, ɑ belongs to R, then ɑ =**

Solution:

Answer: 1

**Question 5: Let L be a common tangent line to the curves 4x ^{2} + 9y^{2} = 36 and (2x)^{2} + (2y)^{2} = 31. Then the square of the slope of the line L is ______________.**

Solution:

Answer: 3

E : (x

^{2}/ 9) + (y^{2}/ 4) = 1C : x

^{2}+ y^{2}= 31 / 4The equation of the tangent to the ellipse is y = mx ± √(9m

^{2}+ 4) --- (1)The equation of the tangent to the circle is y = mx ± √[(31 / 4) m

^{2}+ (31 / 4)] --- (2)Comparing equation (1) & (2), 9m

^{2}+ 4 = (31 / 4) m^{2}+ (31 / 4)36m

^{2}+ 16 = 31m^{2}+ 315m

^{2}= 15m

^{2}= 3

**Question 6: If the matrix **

\(\begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 1 \end{bmatrix}\)

for some real numbers ɑ and β, then β -ɑ is equal to_________.Solution:

Answer: 4

**Question 7: If the arithmetic mean and the geometric mean of the p ^{th} and q^{th} terms of the sequence - 16, 8, - 4, 2, … satisfy the equation 4x^{2} - 9x + 5 = 0, then p + q is equal to ______________.**

Solution:

Answer: 10

Given, 4x

^{2}- 9x + 5 = 0(x - 1) (4x - 5) = 0

AM = 5 / 4, GM = 1 (AM ≥ GM)

Again for the series, - 16, 8, - 4, 2, …

p

^{th }term t_{p}= - 16 (- 1 / 2)^{p-1}q

^{th }term t_{q}= - 16 (- 1 / 2)^{q-1}AM = [t

_{p}+ t_{q}] / 2 = 5 / 4GM = √(t

_{p}t_{q}) = 116

^{2}(- 1 / 2)^{p+q-2}= 1(- 2)

^{8}= (- 2)^{p+q-2}p + q = 10

**Question 8: Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, - 3) and (4, - 2 √2), and given that a - 2√2b = 3, then (a ^{2} + b^{2} + ab) is equal to______________.**

Solution:

Answer: 9

Let the equation of normal is Y - y = (- 1 / m) (X - x), where m = dy / dx

As it passes through (a, b)

b - y = (- 1 / m) (a - x) = - dx / dy [a - x]

(b - y) dy = (x - a) dx

by - (y

^{2}/ 2) = (x^{2}/ 2) - ax + c --- (i)It passes through (3, - 3) & (4, - 2 √2)

-3b - (9 / 2) = (9 / 2) - 3a + c

3a - 3b - c = 9 --- (ii)

Also - 2√2b - 4 = 8 - 4a + c

4a - 2 √2b - c = 12 --- (iii)

Also a - 2 √2b = 3 --- (iv) (given)

From (ii) - (iii), - a + (2√2 - 3) b = - 3 ---- (v)

From (iv) + (v), b = 0, a = 3

a

^{2}+ b^{2}+ ab = 9

**Question 9: Let z be those complex number which satisfies |z + 5| ≤ 4 and z (1 + i) + **

Solution:

Answer: 48

Given, |z + 5| ≤ 4

(x + 5)

^{2}+ y^{2}≤ 16Also, z (1 + i) +

\(\bar{z}\)(1 - i) ≥ - 10x - y ≥ - 5 ---- (2)

From (1) and (2), the locus of z is the shaded region in the diagram.

|z + 1| represents the distance of ‘z’ from Q (- 1, 0).

Clearly, p is the required position of z when |z + 1| is maximum.

P ≡ (- 5 - 2 √2, - 2 √2)

PQ

^{2}= 32 + 16 √2ɑ = 32

β = 16

ɑ + β = 48

**Question 10: Let a be an integer such that all the real roots of the polynomial 2x ^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to _______.**

Solution:

Answer: 2

Let f (x) = 2x

^{5}+ 5x^{4}+ 10x^{3}+ 10x^{2}+ 10x + 10f’ (x) = 10 (x

^{4}+ 2x^{3}+ 3x^{2}+ 2x + 1)= 10x

^{2}[x^{2}+ (1 / x^{2}) + 2 [x + (1 / x)] + 3]= 10x

^{2}[(x + (1/x))^{2}+ 2 (x + (1 / x)) + 1]= 10x

^{2}[[x + (1 / x)] + 1]^{2}> 0, for all x belongs to Rf (x) is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

By observation, f (- 1) = 3 > 0

f (- 2) = - 64 + 80 - 80 + 40 - 20 + 10

= - 34 < 0

f (x) has at least one root in (- 2, - 1) = (a, a + 1)

|a| = 2

JEE Main 2021 Maths Paper February 26 Shift 2