## JEE Main 2020 Maths Paper With Solutions Shift 1 September 5

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

a)

b)

c) n = 9

d) n = 7

(12+n^{2})-(6+n)+n(2n-4) = 158

3n^{2}-5n+6-158 = 0

3n^{2}-5n-152 = 0

3n^{2-}24n+19n-152 = 0

(3n+19)(n-8) = 0

⇒ n = 8

⇒

**Answer: b**

**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 = (x_{1}^{2}+ x_{2}^{2}+ x_{4}^{2}+ x_{5}^{2}+ x_{6}^{2}+ x_{7}^{2})/7 -64

80×7 = x_{1}^{2}+ x_{2}^{2}+ x_{3}^{2}+…+ x_{7}^{2}

Now, x_{6}^{2}+ x_{7}^{2} = 560-(x_{1}^{2}+…x_{5}^{2})

x_{6}^{2}+x_{7}^{2} = 560-(4+16+100+144+196)

x_{6}^{2}+x_{7}^{2} = 100 ..(i)

Now, (x_{1}+x_{2}+…+x_{7})/7 = 8

x_{6}+x_{7} = 14

from (i) and (ii)

(x_{6}+x_{7})^{2}-2x_{6}x_{7} = 100

2x_{6}x_{7} = 96

⇒ x_{6}x_{7} = 48 ..(iii)

Now, |x_{6}-x_{7}| = √((x_{6}+x_{7})^{2}-4x_{6}x_{7)}

= √(196-192)

= 2

**Answer: d**

**4. **If 2^{10}+2^{9}×3^{1}+2^{8}×3^{2}+…..+2×3^{9}+3^{10 }= S-2^{11}, then S is equal to:

a) 3^{11}

b) (3^{11}/2)+2^{10}

c) 2×3^{11}

d) 3^{11}-2^{12}

Let S’ = 2^{10}+2^{9}×3^{1}+ 2^{8}×3^{2}+—–+2×3^{9}+3^{10} ..(i)

3×S’/2 = 2^{9}×3^{1}+2^{8}×3^{2}+…+3^{10}+3^{11}/2 ..(ii)

Equation (i) – (ii)

-S’/2 = 2^{10}-3^{11}/2

S’ = 3^{11}-2^{11}

Now S’= S-2^{11}

S = 3^{11}

**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)

t^{2}-84t+243 = 0

t^{2}-81t-3t+243 = 0

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

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

t = 81, 3

9^{sin2α} = 9^{2} or 3

sin2α = 1/2, 2 (rejected)

First term, a = 3^{2sin2α-1}

a = 1

Second term = 14

Common difference, d = 13

T_{6 }= a+5d

T_{6} = 1+5×13 = 66

**Answer: d**

**6.** If the common tangent to the parabolas, y^{2} = 4x and x^{2} = 4y also touches the circle, x^{2}+y^{2} = c^{2}, then c is equal to:

a) 1/2

b) 1/4

c) 1/√2

d) 1/2√2

y = mx+1/m

x^{2} = 4(mx+1/m)

x^{2} -4mx-4/m = 0

D = 0

16m^{2}+16/m = 0

16(m^{3}+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) =

C_{1} → C_{1}-C_{2}

C_{3} → C_{3}+C_{2}

=

C_{2} → C_{2}-C_{3}

=

= 1(2cos^{2} θ-8)+(8+2cos^{2} θ) -4sin^{2}θ

f(θ) = 4cos 2θ

**Answer: b**

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

2x_{1}-4x_{2}+λx_{3 }= 1

x_{1}-6x_{2}+x_{3} = 2

λx_{1}-10x_{2}+4x_{3} = 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)

D_{1} = -2(λ-3)

D_{2} = -2(λ+1)( λ-3)

D_{3} = –2(λ-3)

When λ = 3, then

D = D_{1} = D_{2} = D_{3} = 0

⇒ Infinite many solution

When λ = –2/3 then D_{1}, D_{2}, D_{3} none of them is zero so equations are inconsistent

λ = –2/3

**Answer: b**

**9.** If the point P on the curve, 4x^{2}+5y^{2} = 20 is farthest from the point Q(0, -4), then PQ^{2} is equal to:

a) 48

b) 29

c) 21

d) 36

Given ellipse is (x^{2}/5)+(y^{2}/4) = 1

Let point P is (√5 cos θ, 2 sin θ)

(PQ)^{2} = 5cos^{2} θ+ 4(sin θ + 2)^{2}

(PQ)^{2} = cos^{2}θ+ 16 sin θ + 20

(PQ)^{2} = -sin^{2}θ+ 16 sin θ + 21

= 85 -(sin θ – 8)^{2}

Will be maximum when sin θ = 1

(PQ)^{2}max = 85-49 = 36

**Answer: d**

**10.** The product of the roots of the equation 9x^{2}-18|x|+5 = 0 is :

a) 25/81

b) 5/9

c) 5/27

d) 25/9

Let |x| = t

9t^{2}-18t +5 = 0

9t^{2} -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_{e}13) 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^{-1}r

T_{1 }= tan^{-1}2-tan^{-1}1

T_{2} = tan^{-1}3-tan^{-1}2

T_{3} = tan^{-1}4-tan^{-1}3

T_{10 }= tan^{-1}11-tan^{-1}10

S = (tan^{-1}2-tan^{-1}1) + (tan^{-1 }3-tan^{-1}2) + (tan^{-1}4-tan^{-1}3) + ……… + (tan^{-1 }11-tan^{-1}10)

S = tan-^{1}11-tan^{-1}1

= 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 (k_{1}, k_{2}) 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 = -k_{2}

⇒ k_{2} = 1

f'(π^{–}) = f’(π^{+})

⇒ 0 = 0

So differentiable at x = 0

f’’(π^{–}) = f’’(π^{+})

2k_{1} = k_{2}

k_{1} = 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

CA^{2} = AB^{2}+BC^{2}

2^{2}x^{2}+2^{2}y^{2} = 32

x^{2}+y^{2} = 8

√(x^{2}+y^{2} ) = 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) e^{2}

⇒ 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) = x^{2}-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, 9x^{2}+16y^{2 }= 144, then PA+PB is equal to :

a) 6

b) 16

c) 9

d) 8

(x^{2}/16)+(y^{2}/9) = 1

e = √(1-9/16)

= √7/4

F_{1}(√7, 0), F_{2}(-√7, 0)

PF_{1}+PF_{2} = 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/x^{2})^{22} is 1540 is..

T_{r+1} = ^{22}C_{r} (x^{m})^{22-r}(1/x^{2})^{r}

= ^{22}C_{r} (x)^{22m-mr-2r}

Given ^{22}C_{r} = 1540 = ^{22}C_{19}

⇒ 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) = ^{4}C_{2} (2/6)^{2}(4/6)^{2}+ ^{4}C_{3 }(2/6)^{3}(4/6)^{1}+ ^{4}C_{4}^{ }(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

L_{1 }: 2x-y+3 = 0

L_{2} : 4x-2y+α = 0

L_{3} : 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**

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