JEE Main 2020 Maths Paper With Solutions Shift 1 Sept 5

1. If the volume of a parallelopiped, whose coterminuous edges are given by the vectors

a)

2. A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be:

a) 63
b) 54
c) 38
d) 36

C → Coffee and T → Tea

n(C) = 73, n(T) = 65

n(coffee) = 73/100

n(tea) = 65/100

n(T∩C) = x/100

n(C∪T) = n(C)+n(T)-x

= 73+65-x

⇒ x

⇒ 73-x

⇒ x

⇒ 65-x

⇒ x

38

x = 36

x

Answer: d

3. The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2,4,10,12,14, then the absolute difference of the remaining two observations is:

a) 1
b) 4
c) 3
d) 2

Var(x) =

16 = (x₁²+ x₂²+ x₄²+ x₅²+ x₆²+ x₇²)/7 -64

80×7 = x₁²+ x₂²+ x₃²+…+ x₇²

Now, x₆²+ x₇² = 560-(x₁²+…x₅²)

x₆²+x₇² = 560-(4+16+100+144+196)

x₆²+x₇² = 100 ..(i)

Now, (x₁+x₂+…+x₇)/7 = 8

x₆+x₇ = 14

from (i) and (ii)

(x₆+x₇)²-2x₆x₇ = 100

2x₆x₇ = 96

⇒ x₆x₇ = 48 ..(iii)

Now, |x₆-x₇| = √((x₆+x₇)²-4x₆x₇₎

= √(196-192)

= 2

Answer: d

4. If 2¹⁰+2⁹×3¹+2⁸×3²+…..+2×3⁹+3¹⁰ = S-2¹¹, then S is equal to:

a) 3¹¹
b) (3¹¹/2)+2¹⁰
c) 2×3¹¹
d) 3¹¹-2¹²

Let S’ = 2¹⁰+2⁹×3¹+ 2⁸×3²+—–+2×3⁹+3¹⁰ ..(i)

3×S’/2 = 2⁹×3¹+2⁸×3²+…+3¹⁰+3¹¹/2 ..(ii)

Equation (i) – (ii)

-S’/2 = 2¹⁰-3¹¹/2

S’ = 3¹¹-2¹¹

Now S’= S-2¹¹

S = 3¹¹

Answer: a

5. If 3^{2 sin2α-1},14 and 3^{4-2 sin2α} are the first three terms of an A.P. for some α , then the sixth term of this A.P. is:

a) 65
b) 81
c) 78
d) 66

28 = 3^2sin2α-1+3^{4-2 sin2α}

28 = 9^sin2α/3 + 81/9^sin2α

Let 9^sin2α = t

28 = (t/3)+(81/t)

t²-84t+243 = 0

t²-81t-3t+243 = 0

t(t -81)-3(t-81) = 0

(t -81)(t-3) = 0

t = 81, 3

9^sin2α = 9² or 3

sin2α = 1/2, 2 (rejected)

First term, a = 3^2sin2α-1

a = 1

Second term = 14

Common difference, d = 13

T₆ = a+5d

T₆ = 1+5×13 = 66

Answer: d

6. If the common tangent to the parabolas, y² = 4x and x² = 4y also touches the circle, x²+y² = c², then c is equal to:

a) 1/2
b) 1/4
c) 1/√2
d) 1/2√2

y = mx+1/m

x² = 4(mx+1/m)

x² -4mx-4/m = 0

D = 0

16m²+16/m = 0

16(m³+1)/m = 0

m = -1

⇒ y+x = -1

Now, |-1/√2| = |c|

c =

Answer: c

7. If the minimum and the maximum values of the function f: [π/4, π/2] → R defined by f(θ) =

a) (0,4)
b) (-4,0)
c) (-4,4)
d) (0,2√2)

f(θ) =

C₁ → C₁-C₂

C₃ → C₃+C₂

=

C₂ → C₂-C₃

=

= 1(2cos² θ-8)+(8+2cos² θ) -4sin²θ

f(θ) = 4cos 2θ

Answer: b

8. Let λ∈R. the system of linear equations

2x₁-4x₂+λx₃ = 1

x₁-6x₂+x₃ = 2

λx₁-10x₂+4x₃ = 3

is inconsistent for:

a) exactly two values of λ
b) exactly one negative value of λ.
c) every value of λ.
d) exactly one positive value of λ.

D =

= 2(3λ+ 2)( λ-3)

D₁ = -2(λ-3)

D₂ = -2(λ+1)( λ-3)

D₃ = –2(λ-3)

When λ = 3, then

D = D₁ = D₂ = D₃ = 0

⇒ Infinite many solution

When λ = –2/3 then D₁, D₂, D₃ none of them is zero so equations are inconsistent

λ = –2/3

Answer: b

9. If the point P on the curve, 4x²+5y² = 20 is farthest from the point Q(0, -4), then PQ² is equal to:

a) 48
b) 29
c) 21
d) 36

Given ellipse is (x²/5)+(y²/4) = 1

Let point P is (√5 cos θ, 2 sin θ)

(PQ)² = 5cos² θ+ 4(sin θ + 2)²

(PQ)² = cos²θ+ 16 sin θ + 20

(PQ)² = -sin²θ+ 16 sin θ + 21

= 85 -(sin θ – 8)²

Will be maximum when sin θ = 1

(PQ)²max = 85-49 = 36

Answer: d

10. The product of the roots of the equation 9x²-18|x|+5 = 0 is :

a) 25/81
b) 5/9
c) 5/27
d) 25/9

Let |x| = t

9t²-18t +5 = 0

9t² -15t -3t+5 = 0

(3t -5)(3t-1) = 0

|x| = 5/3, 1/3

⇒ x = 5/3, -5/3, 1/3, -1/3

⇒ Product = 25/81

Answer: a

11. If y = y(x) is the solution of the differential equation ((5+e^x)/(2+y))(dy/dx)+e^x = 0 satisfying y(0) = 1, then a value of y(log_e13) is:

a) 1
b) 0
c) 2
d) -1

((5+e^x)/(2+y))(dy/dx) = -e^x

∫dy/(2+y) = ∫-e^-x/(e^x+5) dx

ln (y+2) = -ln(e^x+5) + C

(y+2) (e^x+5) = C

Since y(0) = 1

⇒ C = 18

y+2 = 18/e^x+5

at x = ln 13

y+2 = 18/(13+5) = 1

y = -1

Answer: d

12. If S is the sum of the first 10 terms of the series tan^-1(1/3) + tan^-1(1/7) + tan^-1(1/13) + tan^-1(1/21) +…, then tan(S) is equal to:

a) 5/11
b) 5/6
c) -6/5
d) 10/11

S =

T_r =

T_r = tan^-1(r+1)-tan^-1r

T₁ = tan^-12-tan^-11

T₂ = tan^-13-tan^-12

T₃ = tan^-14-tan^-13

T₁₀ = tan^-111-tan^-110

S = (tan^-12-tan^-11) + (tan^-1 3-tan^-12) + (tan^-14-tan^-13) + ……… + (tan^-1 11-tan^-110)

S = tan-¹11-tan^-11

= tan^-1 [(11-1)/(1+11)]

⇒ tan S = 10/12

= 5/6

Answer: b

13. The value of

a) π/2
b) π/4
c) π
d) 3π/2

I =

=

⇒ 2I = π

I = π/2

Answer: a

14. If (a, b, c) is the image of the point (1,2,-3) in the line, (x+1)/2 = (y-3)/-2 = z/-1 then a+b+c is:

a) 2
b) 3
c) -1
d) 1

Vector PM and (

⇒ (2λ-2)2+(1-2λ)(-2)+(3-λ) (-1) = 0

⇒ 4λ-4+4 λ-2+λ-3 = 0

⇒ 9λ = 9

⇒ λ = 1

⇒ M(1,1,-1)

Now, P’ = 2M-P

= 2(1,1,-1)-(1, 2, -3)

= (1, 0, 1)

a+b+c = 2

Answer: a

15. If the function

is twice differentiable, then the ordered pair (k₁, k₂) is equal to:

a) (1,1)
b) (1,0)
c) (1/2, -1)
d) (1/2, 1)

f(x) is continuous and differentiable

f(π^–) = f(π) = f(π⁺)

-1 = -k₂

⇒ k₂ = 1

f'(π^–) = f’(π⁺)

⇒ 0 = 0

So differentiable at x = 0

f’’(π^–) = f’’(π⁺)

2k₁ = k₂

k₁ = 1/2

Answer: d

16. If the four complex numbers z,

a) 2
b) 4
c) 4√2
d) 2√2

Let z = x+iy

CA² = AB²+BC²

2²x²+2²y² = 32

x²+y² = 8

√(x²+y² ) = 2√2

Answer: d

17.If where c is a constant of integration, then g(0) is equal to :

a) 2
b) e
c) 1
d) e²

⇒ g(0) = 2

Answer: a

18. The negation of the Boolean expression x ↔ y is equivalent to:

a) (x˄y) ˄(∼x˅∼y)
b) (x˄y) ˅ (∼x˄∼y)
c) (x˄∼y) ˅ (∼x˄∼y)
d) (∼x˄y) ˅ (∼x˄∼y)

As we know

p ↔ q = (p→ q) ˄ (q → p)

∼(p ↔ q) = (p˄∼q) ˅ (∼p˄q)

⇒ so, ∼(x ↔ ∼y) = (x˄y) ˅ (∼x˄∼y)

Answer: b

19. If α is positive root of the equation, p(x) = x²-x-2 = 0, then

a) 1/2
b) 3/√2
c) 3/2
d) 1/√2

Answer: b

20. If the co-ordinates of two points A and B are (√7, 0) and (-√7, 0) respectively and P is any point on the conic, 9x²+16y² = 144, then PA+PB is equal to :

a) 6
b) 16
c) 9
d) 8

(x²/16)+(y²/9) = 1

e = √(1-9/16)

= √7/4

F₁(√7, 0), F₂(-√7, 0)

PF₁+PF₂ = 2a

PA+PB = 2×4 = 8

Answer: d

21. The natural number m, for which the coefficient of x in the binomial expansion of (x^m+1/x²)²² is 1540 is..

T_r+1 = ²²C_r (x^m)^22-r(1/x²)^r

= ²²C_r (x)^22m-mr-2r

Given ²²C_r = 1540 = ²²C₁₉

⇒ r = 19

Since 22m-rm-2r = 1

⇒ m = (2r+1)/(22-r)

m = 13 (At r = 19)

Answer: 13

22. Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

P(at 2, 3 or 4) = ⁴C₂ (2/6)²(4/6)²+ ⁴C₃ (2/6)³(4/6)¹+ ⁴C₄ (2/6)⁴

= 6×(1/9)×(4/9)+4×(1/27)×(2/3)+(1/81)

= 33/81

= 11/27

⇒ nP ⇒11

Answer: 11

23. Let f(x) = x[x/2], for -10<x<10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to

f(x) = x[x/2]

x∈ (-10, 10)

x/2 ∈(-5,5) → 9 integers

Check continuity at x = 0

f(0) = 0

f(0⁺) = 0

f(0^–) = 0

continuous at x = 0.

Function will be discontinuous when

x/2 =

-4, -3, -2, -1, 0, 1, 2, 3, 4

8 points of discontinuity

Answer: 8

24. The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ’SYLLABUS’ such that two letters are distinct and two letters are alike, is

SS, Y, LL, A, B, U

ABCC type words

Answer: 240

25. If the line, 2x-y+3 = 0 is at a distance 1/√5 and 2/√5 from the lines 4x-2y+α = 0 and 6x-3y+β = 0, respectively, then the sum of all possible values of α and β is

L₁ : 2x-y+3 = 0

L₂ : 4x-2y+α = 0

L₃ : 6x-3y+β = 0

|(α/2)-3|/√5 = 1/√5

⇒ (α/2)-3 = 1, -1

⇒ α = 8, 4

|(β/3)-3|/√5 = 2/√5

⇒ (β/3)-3 = 2, -2

⇒ β = 15, 3

sum of all possible values of α and β = 8+4+15+3 = 30

Answer: 30