JEE Main 2021 March 16th Shift 2 Maths Question Paper is provided here along with solutions that have been prepared by our subject experts. Students can go through the solutions or download the solved paper in a PDF format for offline practice to improve their problem solving skills. While referring to these JEE Main 2021 Maths questions and solutions students will also be able to learn about the different types of problems asked in the entrance exam and at the same time develop better speed and acccuracy to solve them.
Section A
Question 1: The least value of z where z is a complex number which satisfies the inequality exp [{(z + 3) (z – 1)} / {z + 1}] log_{e} 2 ≥ log_{√2} 5 √7 + 9i, i = √1, is equal to:
a. 2
b. 3
c. 8
d. √5
Answer: (b)
2^{[{(z + 3) (z – 1)} / {z + 1}]} ≥ 2^{3} ⇒ [{(z + 3) (z – 1)} / {z + 1}] ≥ 3
z^{2} + 2 z – 3 ≥ 3 z + 3
z^{2} – z – 6 ≥ 0
(z – 3) (z + 2) ≥ 0
z_{min} = 3
Question 2: Let f : S → S where S = (0, ∞) be a twice differentiable function such that f (x + 1) = x f (x). If g : S → R be defined as g (x) = log_{e }f (x), then the value of g”(5) – g”(1) is equal to:
a. 197 / 144
b. 187 / 144
c. 205 / 144
d. 1
Answer: (c)
f(x+1) = xf(x)
g(x+1) = log_{e}(f(x+1))
g(x+1) = log_{e }x + log_{e }f(x)
g(x+1) – g(x) = log_{e }x
g’’(x+1) – g’’(x) = – 1 / x^{2}
g’’(2) – g’’(1) = –1
g’’(3) – g’’(2) = – 1 / 4
g’’(4) – g’’(3) = – 1 / 9
g’’(5) – g’’(4) = – 1 / 16
g’’(5) – g’’(1) = [1 + (1 / 4) + (1 / 9) + (1 / 16)]
g’’(5) – g’’(1) = [144 + 36 + 16 + 9] / [16 * 9] = 205 / 144
Question 3: If y = y (x) is the solution of the differential equation (dy / dx) + (tan x) y = sinx, 0 ≤ x ≤ π / 3, with y(0) = 0, then y (π / 4) equal to:
a. log_{e }2
b. (1 / 2) log_{e} 2
c. (1 / 2 √2) log_{e }2
d. (1 / 4) log_{e} 2
Answer: (c)
I.F. = e^{∫tanx dx}
= e^{ln secx}
= sec x
Solution of the equation:
y(sec x) = ∫sin x sec x dx
y / (cosx) = ln (secx) + c
Put x = 0, c = 0
∴ y = cos x ln(sec x)
Put x = π / 4
y = (1 / √2) ln √2 = (1 / 2√2) ln 2
y = (ln 2) / (2√2)
Question 4: If the foot of the perpendicular from point (4, 3, 8) on the line L_{1} : [x – a] / l = [y – 2] / 3 = [z – b] / 4, l ≠ 0 is (3, 5, 7), then the shortest distance between the line L_{1} and line L_{2} : [x – 2] / 3 = [y – 4] / 4 = [z – 5] / 5 is equal to:
a. √2 / 3
b. 1 / √3
c. 1 / 2
d. 1 / √6
Answer: (d)
(3, 5, 7) lies on given line L_{1}
[3 – a] / l = 3 / 3 = [7 – b] / 4 [7 – b] / 4 = 1 ⇒ b = 3 [3 – a] / l = 1 ⇒ 3 – a = lA (4, 3, 8)
B (3, 5, 7)
DR’s of AB = (1, –2, 1)
AB ⊥ line L_{1}
(1)(l) + (–2)(3) + 4(1) = 0
⇒ l = 2
a = 1
a = 1, b = 3, l = 2
[x – 1] / 2 = [y – 2] / 3 = [z – 3] / 4 [x – 2] / 3 = [y – 4] / 4 = [z – 5] / 5SD =
Question 5: If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression 3 + [x – 11] / [(y – 19)^{2} (z – 12)^{2}] + [y – 19] / [(x – 11)^{2} (z – 12)^{2}] + [z – 12] / [(z – 11)^{2} (y – 19)^{2}] – [x + y + z] / [14 (x – 11) (y – 19) (z – 12)] is equal to :
a. 3
b. 0
c. 39
d. – 45
Answer: (a)
Equation of plane is x + y + z = 42 or, (x − 11)+ (y − 19)+ (z −12) = 0.
Now, [x – 11] / [(y – 19)^{2} (z – 12)^{2}] + [y – 19] / [(x – 11)^{2} (z – 12)^{2}] + [z – 12] / [(z – 11)^{2} (y – 19)^{2}]
= [(x – 11)^{3} + (y – 19)^{3} + (z – 12)^{3}] / [(x – 11)^{2} (y – 19)^{2} (z – 12)^{2}]
= [3 (x – 11) (y – 19) (z – 12)] / [(x – 11)^{2} (y – 19)^{2} (z – 12)^{2}] [If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc]
= 3 / [(x – 11) (y – 19) (z – 12)]
The given expression is equal to 3 + 3 / [(x – 11) (y – 19) (z – 12)] – 3 / [(x – 11) (y – 19) (z – 12)] = 3
Question 6: Consider the integral I = ∫_{0}^{10} {[x] e^{[x]}} / [e^{x1}] dx, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
a. 45 (e – 1)
b. 45 (e + 1)
c. 9(e – 1)
d. 9(e + 1)
Answer: (a)
I = ∫_{0}^{10} {[x] e^{[x]}} / [e^{x1}] dx
= ∫_{0}^{10} [x] e^{[x]+1x}dx
= ∫_{1}^{2} e^{2x }dx + ∫_{2}^{3} 2 e^{3x} dx + ∫_{3}^{4} 3 e^{4x }dx + ……… ∫_{9}^{10} 9 e^{10x} dx
= – {(1 – e) + 2(1 – e) + 3(1 – e) +……+ 9(1 – e)}
= 45(e – 1)
Question 7: Let A (–1, 1), B (3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A_{1} and A_{2} be the areas of ΔABC and ΔPQC respectively, such that A_{1} = 3A_{2}, then the value of m is equal to:
a. 4 / 15
b. 1
c. 2
d. 3
Answer: (b)
A_{1} = Area of ΔABC =
A_{1} = 13 / 2
Equation of line AC is y – 1 = ( 1 / 3) (x + 1)
Solving it with line y = mx, we get P [2 / (3m + 1), 2m / (3m + 1)]
Equation of line BC is y – 0 = 4(x – 2)
Solving it with line y = mx, we get Q [ 8 / (m – 4), – 8m / (m – 4)]
A_{2} = Area of ΔPQC =
⇒ 12m^{2} = ± (3m^{2} – 11m – 4)
Taking +ve sign,
9m^{2} + 11m + 4 = 0 (Rejected ∵ m is imaginary)
Taking –ve sign,
15m^{2} – 11m – 4 = 0
m = 1, – 4 / 15
m = 1 as m > 0
Question 8: Let f be a realvalued function, defined on R – {–1, 1} and given by f(x) = 3 log_{e} (x – 1) / (x + 1) – 2 / (x – 1). Then in which of the following intervals, function f (x) is increasing?
a. ( ∞, – 1) ⋃ ([1 / 2, ∞) – {1})
b. ( 1, 1 / 2]
c. (–∞, ∞) – {–1, 1}
d. ( ∞, 1 / 2] – { 1}
Answer: (a)
f’ (x) = [(x + 1) / (x – 1) * (x + 1 – (x – 1)) / (x + 1)^{2}] 3 + 2 / (x – 1)^{2} = 6 / (x – 1) (x + 1) + 2 / (x – 1)^{2}
= [4(2x – 1)]/[(x + 1)(x – 1)^{2}]
x ∈ ( ∞, – 1) ⋃ ([1 / 2, ∞) – {1})
Question 9: Let the lengths of intercepts on xaxis and yaxis made by the circle x^{2} + y^{2} + ax + 2ay + c = 0, (a < 0) be 2 √2 and 2 √5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:
a. √10
b. √6
c. √11
d. √7
Answer: (b)
2 √(a^{2} / 4 – c) = 2√2
√(a^{2} – 4c) = 2√2
a^{2} – 4c = 8 — (1)
2√(a^{2} – c) = 2√5
a^{2} – c = 5 — (2)
(2) – (1), we get
3c = –3 ⇒ c = –1
a^{2} = 4 ⇒ a = –2
x^{2} + y^{2} – 2x – 4y – 1 = 0
Equation of tangent 2x – y + λ = 0
∴ p = r
(0 – 0 + λ) / √5 = √6
λ = ± √30
∴ Tangents are 2x – y ± √30 = 0
Distance from origin = √30 / √5 = √6
Question 10: Let A denote the event that a 6digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then the probability of event A is equal to:
a. 4 / 9
b. 9 / 56
c. 3 / 7
d. 11 / 27
Answer: (a)
Total case = 6∟6
Fav. case = (0, 1, 2, 3, 4, 5) + (0, 1, 2, 4, 5, 6) + (1, 2, 3, 4, 5, 6)
= 5 ∟5 + 5 ∟5 + ∟6
= 1920
Probability = 1920 / 6∟6 = 4 / 9
Question 11: Let α ∈ R be such that the function is continuous at x = 0, where {x} = x – [x], [x] is the greatest integer less than or equal to x. Then:
a. α = π / 4
b. No such α exists
c. α = 0
d. α = π / √2
Answer: (b)
Question 12: The maximum value of f (x) =
a. √7
b. √5
c. 5
d. 3 / 4
Answer: (b)
C_{1 }→ C_{1} + C_{2}
R_{1} → R_{1} – R_{2}
= (–1)[2sin2x – cos2x] = cos2x – 2sin2x
Maximum value = √5
Question 13: Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β – α) is equal to:
a. 1890
b. 795
c. 717
d. 1173
Answer: (c)
α = ^{6}C_{1 }^{7}C_{1 }^{9}C_{1} + ^{5}C_{1 }^{7}C_{1 }^{9}C_{1} + ^{5}C_{1 }^{6}C_{1 }^{9}C_{1} + ^{5}C_{1 }^{6}C_{1 }^{7}C_{1}
= 378 + 315 + 270 + 210
= 1173
β = ^{5}C_{1 }^{6}C_{1 }^{7}C_{1 }^{9}C_{1} = 1890
⇒ β – α = 1890 – 1173 = 717
Question 14: Let C be the locus of the mirror image of a point on the parabola y^{2} = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is:
a. 2x + y = 5
b. x + 2y = 4
c. x + 3y = 5
d. x – y = 1
Answer: (d)
Image of y^{2} = 4x w.r.t. y = x is x^{2} = 4y
Tangent from (2, 1)
xx_{1} = 2(y + y_{1})
2x = 2(y + 1)
x = y + 1
Question 15: Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin^{1} (3x / 5) + sin^{1} (4x / 5) = sin^{1} x =
a. 1
b. 2
c. 3
d. 0
Answer: (c)
Taking sin on both sides, (3x / 5) √[1 – (16x^{2} / 25)] + (4x / 5) √[1 – (9x^{2} / 25)] = x
3x √(25 – 16x^{2}) = 25x – 4x √(25 – 9x^{2})
x = 0 or 3√(25 – 16x^{2}) = 25 – 4√(25 – 9x^{2})
9 (25 – 16x^{2}) = 625 – 200 √(25 – 9x^{2}) + 16 (25 – 9x^{2})
200 √(25 – 9x^{2}) = 800
√(25 – 9x^{2}) = 4
x^{2} = 1
x = ± 1
Total number of solutions = 3
Question 16: Let C_{1} be the curve obtained by the solution of the differential equation 2xy (dy / dx) = y^{2} – x^{2}, x > 0. Let curve C_{2} be the solution of 2xy / (x^{2} – y^{2}) = dy / dx. If both the curves pass through (1, 1), then the area enclosed by the curves C_{1} and C_{2} is equal to :
a. (π / 2) – 1
b. (π / 4) + 1
c. π – 1
d. π + 1
Answer: (a)
dy / dx = [y^{2} – x^{2}] / 2xy
Put y = vx
v + x (dv / dx) = [v^{2}x^{2} – x^{2}] / 2vx^{2} = [v^{2} – 1] / 2v
x (dv / dx) = [v^{2} – 1 – 2v^{2}] / 2v = – (v^{2} + 1) / 2v
2v / (v^{2} + 1) dv = – dx / x
ln (v^{2} + 1) = – ln x + ln c ⇒ v^{2} + 1 = c / x
(y^{2} / x^{2}) + 1 = c / x ⇒ x^{2} + y^{2} = cx
It passes through (1, 1)
∴ x^{2 }+ y^{2 }– 2x=0
Similarly, for second differential equation dy / dx = [x^{2} – y^{2}] / 2xy
equation of curve is x^{2 }+ y^{2 } – 2y = 0
Now the required area is
= [(1 / 4) * π * 1^{2} – (1 / 2) (1) (1)] * 2
= [(π / 2) – 1] square units
Question 17: Let a = i + 2j – 3k and b = 2i – 3j + 5k. If r x a = b x r, r . (ɑi + 2j + k) = 3 and r . (2i + 5j – ɑk) = – 1, ɑ ∈ R, then the value of ɑ + r^{2} =
a. 11
b. 15
c. 9
d. 13
Answer: (b)
r x a = – r x b
r x (a + b) = 0 (a + b = 3i – j + 2k)
r  (a + b)
r = λ (a + b)
∵ r . (2i + 5j – ɑk) = – 1
λ [3i – j + 2k] . [2i + 5j – ɑk] = – 1
⇒ λ(6 – 5 – 2α) = –1
λ(1 – 2α) = –1 …(1)
r (ɑi + 2j + k) = 3
λ (3i – j + 2k) . (ɑi + 2j + k) = 3
⇒ λ[3α – 2 + 2] = 3 ⇒ λα = 1 …(2)
Solving (1) & (2)
λ [1 – (2 / λ)] = – 1
λ – 2 = –1 ⇒ λ = 1 and α = 1
r = 3i – j + 2k
α + r^{2} = 1 + 14 = 15
Question 18: Let P(x) = x^{2} + bx + c be a quadratic polynomial with real coefficients such that ∫_{0}^{1} P (x) dx and P(x) leaves remainder 5 when it is divided by (x – 2). Then the value of 9 (b+c) is equal to :
a. 7
b. 11
c. 15
d. 9
Answer: (a)
(x – 2) Q (x) + 5 = x^{2} + bx + c
Put x = 2
5 = 2b + c + 4 …(1)
∫_{0}^{1} (x^{2} + bx + c) dx = 1
(1 / 3) + (b / 2) + c = 1
(b / 2) + c = 2 / 3 …(2)
Solving (1) & (2)
b = 2 / 9
c = 5 / 9
9 (b + c) = 7
Question 19: If the points of intersections of the ellipse (x^{2} / 16) + (y^{2} / b^{2}) = 1 and the circle x^{2} + y^{2} = 4b, b > 4 lie on the curve y^{2} = 3x^{2}, then b is equal to :
a. 5
b. 6
c. 12
d. 10
Answer: (c)
(x^{2} / 16) + (y^{2} / b^{2}) = 1 …(1)
x^{2} + y^{2} = 4b …(2)
y^{2} = 3x^{2} …(3)
From eqns. (2) and (3), x^{2} = b and y^{2} = 3b
From eqn. (1), (b / 16) + (3b / b^{2} ) = 1
b^{2} + 48 = 16b
b = 12
Question 20: Let A = {2, 3, 4, 5, ……, 30} and ‘⩭’ be an equivalence relation on A×A, defined by (a, b) ⩭ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
a. 7
b. 5
c. 6
d. 8
Answer: (a)
a = bc
(a, b) R (4, 3) ⇒ 3a = 4b
a = (4 / 3) b
b must be multiple of 3
b can be 3, 6, 9, ……, 30
Also, a must be less than or equal to 30.
(a, b) = (4, 3), (8, 6), (12, 9) (16, 12), (20, 15), (24, 18), (28, 21)
⇒ 7 ordered pairs
Section B
Question 21: Let
Answer: (28)
Question 22: In ΔABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of ΔABC is 30 cm^{2} and R and r are respectively the radii of circumcircle and incircle of ΔABC, then the value of 2R + r (in cm) is equal to ______
Answer: (15)
Area = (1 / 2) * 5 * 12 * sin θ = 30
sin θ = 1 ⇒ θ = π / 2
ABC is rightangled at A.
r = (s – a) tan (A / 2)
r = (s – a)
2R + r = s ( a = 2R)
2R + r = 30 / 2 = 15
Question 23: Consider the statistics of two sets of observations as follows:
Size 
Mean 
Variance 

Observation I 
10 
2 
2 
Observation II 
n 
3 
1 
If the variance of the combined set of these two observations is 17 / 9, then the value of n is equal to ______
Answer: (5)
For group1: ∑x_{i} / 10 = 2 ⇒ ∑x_{i} = 20
∑x_{i}^{2} / 10 – (2^{2}) = 2
∑x_{i}^{2} = 60
For group2 : ∑y_{i} / n = 3
∑y_{i} = 3n
∑y_{i}^{2} / n – 3^{2} = 1
∑y_{i}^{2} = 10n
Now, combined variance
𝜎^{2} = [∑(x_{i}^{2} + y_{i}^{2}) / (10 + n)] – [∑ (x_{i} + y_{i}) / (10 + n)]^{2}
(17 / 9) = [60 + 10n] / [10 + n] – [20 + 3n]^{2} / [10 + n]^{2}
17 (n^{2} + 20n + 100) = 9 (n^{2} + 40n + 200)
8n^{2} – 20n – 100 = 0
2n^{2} – 5n – 25 = 0
n = 5
Question 24: Let S_{n}(x) = _{}up to nterms, where a >1. If S_{24}(x) = 1093 and S_{12}(2x) = 265, then value of a is equal to______
Answer: (16)
S_{n} (x) = 2 log_{a} x + 3 log_{a} x + 6 log_{a} x + 11 log_{a} x + ……..
S_{n} (x) = log_{a} x (2 + 3 + 6 + 11 + …… )
S_{r} = 2 + 3 + 6 + 11
General term, T_{r} = r^{2} – 2r + 3
S_{n} (x) = ∑_{r=1}^{n} x (r^{2} – 2r + 3)
S_{24} (x) = x ∑_{r=1}^{24} (r^{2} – 2r + 3)
1093 = 4372 log_{a }x
log_{a }x = 1 / 4
x = a^{1/4}
S_{12} (2x) = log_{a} (2x) ∑_{r=1}^{12} (r^{2} – 2r + 3)
265 = 530 log_{a}(2x)
log_{a}(2x) = 1 / 2
2x = a^{1/2}
After solving (i) and (ii), we get
a^{1/4} = 2
a = 16
Question 25: Let n be a positive integer. Let A = ∑_{k=0}^{n} (1)^{k}^{n}C_{k} [(1 / 2)^{k} + (3 / 4)^{k} + (7 / 8)^{k} + (15 / 16)^{k} + (31 / 32)^{k}]. If 63A = 1 – (1 / 2^{30}), then n is equal to ______
Answer: (6)
A = (1 / 2)^{n} + (1 / 4)^{n} + (1 / 8)^{n} + (1 / 16)^{n} + (1 / 32)^{n}
= (1 / 2^{n}) + (1 / 2^{2n}) + (1 / 2^{3n}) + (1 / 2^{4n}) + (1 / 2^{5n})
= (1 / 2^{n}) [1 – (1 / 2^{n})]^{5} / [1 – (1 / 2^{n})]
A = [2^{5n} – 1] / [2^{5n} (2^{n} – 1)]
63A = [63 (2^{5n} – 1)] / [2^{5n} (2^{n} – 1)]
= (63 / (2^{n} – 1)) [1 – (1 / 2^{5n})] = 1 – (1 / 2^{30})
n = 6
Question 26: Let f : R → R and g : R → R be defined as f (x) =
Answer: (1)
Given g(f(x)) is continuous.
So, at x = –a & at x = 0
1 = b + 1 & (a–1)^{2} + b = b
b = 0 & a = 1
⇒ a + b = 1
Question 27: If the distance of the point (1, – 2, 3) from the plane x + 2y – 3z + 10 = 0 measured parallel to the line, [x – 1] / 3 = [2 – y] / m = [z + 3] / 1 is √7 / 2, then the value of m is equal to_____
Answer: (2)
[x – 1] / 3 = [y + 2] / – m = [z – 3] / 1 = λPoint Q (3λ + 1, – mλ – 2, λ + 3) lies on the plane.
(3λ + 1) + 2 ( mλ – 2) – 3 (λ + 3) + 10 =0
3λ – 2mλ – 3λ + 1 – 4 – 9 + 10 = 0
– 2mλ = 2
mλ = –1 ⇒ λ = – 1 / m
Q [( 3 / m) + 1, – 1, ( 1 / m) + 3]
PQ = √(7/2)
√( 3 / m)^{2} + 1 + ( 1 / m)^{2} = √(7/2)
[10 + m^{2}] / m^{2} = 7 / 2⇒ 20 + 2m^{2} = 7m^{2}
m^{2} = 4 ⇒ m = 2
Question 28: Let 1 / 16, a and b be in G.P. and 1/a, 1/b, 6 be in A.P., where a, b, > 0. Then 72 (a + b) is equal to ______
Answer: (14)
a^{2} = b / 16 and 2 / b = (1 / a) + 6
Solving, we get a = 1 / 12 or a = – 1 / 4 [rejected]
If a = 1 / 12 ⇒ b = 1 / 9
72 (a + b) = 72 [(1 / 12) + (1 / 9)] = 14
Question 29: Let
Answer: (1)
XB = A
(1 / √3)
b_{1} – b_{2} = √3a_{1}
3a_{1}^{2} = b_{1}^{2} + b_{2}^{2} – 2b_{1}b_{2}
b_{1} + kb_{2} = √3a_{2}
3a_{2}^{2} = b_{1}^{2} + k^{2}b_{2}^{2} + 2kb_{1}b_{2}
Adding both equations, 3 (a_{1}^{2} + a^{2}) = 2b_{1}^{2} + (k^{2} + 1) b_{2}^{2} + 2b_{1}b_{2} (k – 1)
(k^{2} – 1) b_{2}^{2} + 2b_{1}b_{2} (k – 1) = 0
(k – 1) [(k + 1) b_{2}^{2} + 2b_{1}b_{2} = 0
k = 1
Question 30: For real numbers α, β, γ and δ, if where C is an arbitrary constant, then the value of 10 (α + βγ + δ) is equal to______.
Answer: (6)
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