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:
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∞∞[n2 + 6n + 10] / (2n + 1)! is equal to:
Solution:
Answer: (c)
∑n=1∞∞[n2 + 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 limx→a [x f (a) - a f (x)] / [x - a] equals:
Solution:
Answer: (c)
By L-H rule
L = limx→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:
Solution:
Answer: (b)
PA2 = cos2 θ + (sin θ - 3)2 = 10 - 6 sin θ
PB2 = cos2 θ + (sin θ + 6)2 = 37 + 12 sin θ
PA2 + PB2|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, x2 + y2 = 1 is a circle of the radius r, then r is equal to :
Solution:
Answer: (b)
P ≡ (2h - 3), (2k - 2) → on circle
[h - (3 / 2)]2 + (k - 1)2 = 1 / 4
Radius = 1 / 2
Question 6: Let the slope of the tangent line to a curve at any point P (x, y) be given by [xy2 + 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:
Solution:
Answer: (b)
dy / dx = [xy2 + y] / x
[x dy - y dx] / y2 = x dx
- d (x / y) = d (x2 / 2)
(- x / y) = (x2 / 2) + c
The 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) = (x2 / 2) - (4 / 3).
It also passes through (3, y).
(- 3 / y) = (9 / 2) - (4 / 3)
y = - 18 / 19
Question 7: Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 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,
Solution:
Answer: (d)
A1 + A2 = ∫0π/2 cosx dx = sin x|0π/2 = 1
A1 = ∫0π/4 (cosx - sinx) dx = (sinx + cosx)0π/4 = √2 - 1
A2 = 1 - (√2 - 1) = 2 - √2
A1 / A2 = (√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) - [a2 + b2] / 2 + [a3 + b3 ] / 3 - [a4 + b4] / 4 + …. is
Solution:
Answer: (a)
tan-1 {[a + b] / [1 - ab]} = (π / 4)
a + b = 1 - ab
(1 + a) (1 + b) = 2
(a + b) - [a2 + b2] / 2 + [a3 + b3] / 3 +…infinity
= [a - (a2 / 2) + (a3 / 3) …..] + [b - (b2 / 2) + (b3 / 3) …..]
= loge (1 + a) + loge (1 + b)
= loge (1 + a) (1 + b) = loge 2
Question 9: Let F1 (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F2 (A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then:
Solution:
Answer: (a)
F1 (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A
Using 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 F2 (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:
Solution:
Answer: (b)
= 20a - 2(7b + 11c) - 3(- 2b - 6c)
= 20a - 14b - 22c + 6b + 18c
= 20a - 8b - 4c
= 4(5a - 2b - c)
= 7b+ 11c - a(25) - 3(2c - b)
= 7b+ 11c - 25a - 6c + 3b
= - 25a + 10b+ 5c
= - 5(5a - 2b- c)
= 6c + 2b- 2(2c - b) - 10a
= - 10a + 4b+ 2c
= - 2(5a - 2b- c)
For infinite solution,
Δ = Δ1 = Δ2 = Δ3 = 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:
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 a1 = xi - j + k and a2 = i + yj + zk are collinear, then a possible unit vector parallel to the vector xi + yj + zk is :
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] / √λ2 + (2 / λ2)
For λ = 1, it is ± [i - j + k] / √3
Question 13: For x > 0, if f (x) = ∫1x loge t / [1 + t] dt, then f (e) + f (1 / e) is equal to:
Solution:
Answer: (a)
Question 14: Let f : R → R be defined as
If f (x) is continuous on R, then a + b equals:
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
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, 105 × 1 = 105
Question 16: A natural number has prime factorization given by n = 2x3y5z, 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:
Solution:
Answer: (c)
y + z = 5 . . . (1)
(1 / y) + (1 / z) = 5 / 6
yz = 6
Also, (y - z)2 = (y + z)2 - 4yz
y - z = ± 1 --- (2)
From (1) and (2), y = 3 or 2 and z = 2 or 3
For calculating the odd divisor of p = 2x3y5z, x must be 0.
P = 203352
Total odd divisors must be (3 + 1) (2 + 1) = 12.
Question 17: Let f (x) = sin-1 x and g (x) = [x2 - x - 2] / [2x2 - x - 6]. If g (2) = limx→2 g (x), then then the domain of the function fog is:
Solution:
Answer: (a)
g (2) = limx→2 [(x - 2) (x + 1)] / [(2x + 3) (x - 2)] = 3 / 7
For domain of f (g (x))
|(x2 - x - 2) / (2x2 - x - 6)| ≤ 1 [because of f (x) is [- 1, 1])
(x + 1)2 ≤ (2x +3)2
3x2 + 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:
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] / [42 + 52 + 22] = 2 / 5
ɑ = 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) = ∫0x et f (t) dt + ex be a differentiable function for all x ∈ R. Then f (x) equals
Solution:
Answer: (a)
Given, f (x) = ∫0x et f (t) dt + ex ---- (1)
f’ (x) = ex . f (x) + ex [using Newton Leibnitz Theorem]
[f’ (x)] / [f (x) + 1] = ex
ln (f (x) + 1) = ex + c
Put x = 0, ln 2 = 1 + c
ln [f (x) + 1] = ex + ln 2 - 1
f (x) + 1 = 2 eex-1
f (x) = 2 eex-1 - 1
Question 20: The triangle of the maximum area that can be inscribed in a given circle of radius ‘r’ is:
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 60o = r / 2
Height = AD = 3r / 2
Again in △ABD
Now sin 60o = (3r / 2) / AB
AB = √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 Pn = (ɑ)n + (β)n, Pn-1 = 11 and Pn+1 = 29 for some integer n≥ 1. Then, the value of Pn2 is______________.
Solution:
Answer: 324
Given, ɑ + β = 1 and ɑ β = - 1
A quadratic equation with roots ɑ, β is x2 - x - 1 = 0.
ɑ2 = ɑ + 1
Multiplying both sides by ɑn-1
ɑn+1 = ɑn + ɑn-1 --- (1)
Similarly,
βn+1 = βn + βn-1 --- (2)
Adding (1) and (2)
ɑn+1 + βn+1 = (ɑ)n + (β)n + [ɑn-1 + βn-1]
Pn+1 = Pn + Pn-1
29 = Pn + 11
Pn = 18
Pn2 = 182 = 324
Question 3: Let X1, X2, …. X18 be eighteen observation such that ∑i=118 (Xi - ɑ) = 36 and ∑i=118 (Xi - 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=118 (Xi - ɑ) = 36
∑xi - 18ɑ = 36
∑xi = 18 (ɑ + 2) ---- (1)
∑i=118 (Xi - β)2 = 90
∑xi2 + 18β2 - 2β ∑xi = 90
∑xi2 + 18β2 - 2β * 18 (ɑ + 2) = 90 (using equation (1))
∑xi2 = 90 - 18β2 + 36β (ɑ + 2)
Now, 𝜎2 = 1 ⇒ (1 / 18) ∑xi2 - (∑xi / 18)2 = 1
(1 / 18) [90 - 18β2 + 36ɑβ + 72β] - [18 (ɑ + 2) / 18]2 = 1
5 − β2 + 2ɑβ + 4β − (ɑ + 2)2 = 1
⇒ 5 − β2 + 2ɑβ + 4β − ɑ2 − 4 − 4ɑ = 1
⇒ ɑ2 − β2 + 2ɑβ + 4β − 4ɑ = 0
⇒ (ɑ − β)(ɑ − β + 4) = 0
⇒ ɑ − β = −4
Hence |ɑ − β| = 4 (ɑ ≠ β).
Question 4: In Im,n = ∫01 xm-1 (1 - x)n-1 dx, for m, n ≥ 1 and ∫01 [xm-1 + xn-1] / [(1 + x)m+n] dx = ɑIm,n, ɑ belongs to R, then ɑ =
Solution:
Answer: 1
Question 5: Let L be a common tangent line to the curves 4x2 + 9y2 = 36 and (2x)2 + (2y)2 = 31. Then the square of the slope of the line L is ______________.
Solution:
Answer: 3
E : (x2 / 9) + (y2 / 4) = 1
C : x2 + y2 = 31 / 4
The equation of the tangent to the ellipse is y = mx ± √(9m2 + 4) --- (1)
The equation of the tangent to the circle is y = mx ± √[(31 / 4) m2 + (31 / 4)] --- (2)
Comparing equation (1) & (2), 9m2 + 4 = (31 / 4) m2+ (31 / 4)
36m2 + 16 = 31m2 + 31
5m2 = 15
m2 = 3
Question 6: If the matrix
Solution:
Answer: 4
Question 7: If the arithmetic mean and the geometric mean of the pth and qth terms of the sequence - 16, 8, - 4, 2, … satisfy the equation 4x2 - 9x + 5 = 0, then p + q is equal to ______________.
Solution:
Answer: 10
Given, 4x2 - 9x + 5 = 0
(x - 1) (4x - 5) = 0
AM = 5 / 4, GM = 1 (AM ≥ GM)
Again for the series, - 16, 8, - 4, 2, …
pth term tp = - 16 (- 1 / 2)p-1
qth term tq = - 16 (- 1 / 2)q-1
AM = [tp + tq] / 2 = 5 / 4
GM = √(tptq) = 1
162 (- 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 (a2 + b2 + 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 - (y2 / 2) = (x2 / 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
a2 + b2 + 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 + y2 ≤ 16
Also, z (1 + i) +
x - 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)
PQ2 = 32 + 16 √2
ɑ = 32
β = 16
ɑ + β = 48
Question 10: Let a be an integer such that all the real roots of the polynomial 2x5 + 5x4 + 10x3 + 10x2 + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to _______.
Solution:
Answer: 2
Let f (x) = 2x5 + 5x4 + 10x3 + 10x2 + 10x + 10
f’ (x) = 10 (x4 + 2x3 + 3x2 + 2x + 1)
= 10x2 [x2 + (1 / x2) + 2 [x + (1 / x)] + 3]
= 10x2 [(x + (1/x))2 + 2 (x + (1 / x)) + 1]
= 10x2 [[x + (1 / x)] + 1]2 > 0, for all x belongs to R
f (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