JEE Main 2020 Maths Paper With Solutions Shift 2 Sept 4

1. Suppose the vectors x₁, x₂ and x₃ are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. If x₁ =

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

Using AX = B

A =

a₁+a₂+a₃ = 1

a₄+a₅+a₆ = 0

a₇+a₈+a₉ = 0

2a₂+a₃ = 0

2a₅+a₆ = 2

2a₈+a₉ = 0

a₃ = 0, a₆ = 0, a₉ = 2

a₈ = -1

a₅ = 1

a₂ = 0

⇒ a₁ = 1

a₄ = -1

a₇ = -1

A =

A = 2×1 = 2

Answer: a

2. If a and b are real numbers such that (2+α)⁴ = a+bα, where α = (-1+i√3)/2 then a+b is equal to:

a) 33
b) 57
c) 9
d) 24

(2+α)⁴ = a+bα

(2+(-1+i√3)/2)⁴ = a+bα

b√3/2 = 9√3/2

⇒ b = 9

a = 0

a+b = 9

Answer: c

3. The distance of the point (1, -2, 3) from the plane x-y+z = 5 measured parallel to the line (x/2) = (y/3) = (z/-6) is:

a) 1/7
b) 7
c) 7/5
d) 1

Equation of line through (1,-2,3) whose d.r.s. are (2,3,-6)

(x-1)/2 = (y+2)/3 = (z-3)/-6 = λ

Any point on the line (2λ+1, 3λ-2, -6λ+3)

Put in (x-y+z = 5)

2λ+1- 3λ+2-6λ+3 = 5

-7λ = -1

λ = 1/7

Distance = √((2λ)²+(3λ)²+(6λ)²)

= √(4λ²+9λ²+36λ²)

= 7λ

= 1 unit

Answer: d

4. Let f: (0, ∞)→ (0, ∞) be a differentiable function such that f(1) = e and

a) e
b) 2e
c) 1/e
d) 1/2e

f(1) = e ..(i)

L’ Hospital rule

⇒ lim_t→x (2tf²(x)-2x²f(t).f’(t)) = 0

⇒ 2xf²(x)-2x²f(x).f’(x) = 0

⇒ 2xf(x){f(x)-xf’(x)} = 0

⇒ f’(x)/f(x) = 1/x

ln f(x) = ln x+ln c

⇒ f(x) = cx ..(ii)

If x = 1

f(1) = c(1)

f(1) = c

From equation (i) and (ii)

c = e ..(iii)

From (iii)

f(x) = ex

⇒ y = ex or y = cx

if f(x) = 1

⇒ x = 1/e

Answer: c

5. Contrapositive of the statement :

‘If a function f is differentiable at a, then it is also continuous at a’, is:

a) If a function f is not continuous at a, then it is not differentiable at a.
b) If a function f is continuous at a, then it is differentiable at a.
c) If a function f is continuous at a, then it is not differentiable at a.
d) If a function f is not continuous at a, then it is differentiable at a.

Contrapositive of p → q = ∼q → ∼p

Answer: a

6. The minimum value of 2^sinx+2^cosx is:

a) 2^1-√2
b) 2^1-1/√2
c) 2^-1+√2
d) 2^-1+1/√2

Using A.M. ≥ G.M.

y = 2^sinx+2^cosx

(2^sinx+2^cosx)/2 ≥ √(2^sinx+cosx)

2^sinx+cosx ≥ 2¹×2^{(sinx+cosx)/2}

2^sinx+cosx ≥ 2^{(2+sinx+cosx)/2}

⇒ (2^{sin x}+2^{cos x})_minimum = 2^(2-√2)/2

= 2^1-1/√2

Answer: b

7. If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to -4, then a value of k is:

a) -2
b) √15
c) √14
d) -4

m_PQ = (4-3)/(1-k)

⇒

Midpoint of PQ = ((k+1)/2, 7/2)

equation of perpendicular bisector

y-7/2 = (k-1)(x-(k+1)/2)

for y intercept put x = 0

y = (7/2)-(k²-1)/2 = -4

(k²-1)/2 = 15/2

⇒ k = 4 and k = -4

Answer: d

8. The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x²-1 below the x-axis, is:

a) 2/3√3
b) 4/3
c) 1/3√3
d) 4/3√3

Area = 2a(a²-1)

A = 2a³-2a

dA/da = 6a²-2 = 0

d²A/da² = 12a

at a = -1/√3

d²A/da² = -4√3 <0

So area is maximum at a = -1/√3

A_max = (-2/3√3)+(2/√3)

= (-2+6)/3√3

= 4/3√3 sq. units

Answer: d

9. The integral

a) 9/2
b) -1/18
c) -1/9
d) 7/18

= (1/2)[9×0-(1/3)×(1/3)×1]

= -1/18

Answer: b

10. If the system of equations

x+y+z = 2

2x+4y–z = 6

3x+2y+ λz = μ

has infinitely many solutions, then

a) λ-2μ = -5
b) 2λ+μ = 14
c) λ+2μ = 14
d) 2λ-μ = 5

D = 0

⇒

(4λ+2)-1(2λ+3)+1(4-12) = 0

4λ+2-2λ-3-8 = 0

2λ = 9

⇒ λ = 9/2

D_x =

⇒ μ = 5

Now check option

2λ+μ = 14

Answer: b

11. In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six The game stops as soon as either of the players wins. The probability of A winning the game is :

a) 5/31
b) 31/61
c) 30/61
d) 5/6

sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

P(sum 7) = 6/36

sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}

P(sum 6) = 5/36

P(A_win) =

= (5/36)+(31/36)×(30/36)×(5/36)+…

= (5/36)÷(1-(31×30)/(36×36)

= (5×36)/(36×36-31×30)

= 5×36/366

= 30/61

Answer: c

12. If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1+x)ⁿ⁺⁵ are in the ratio 5:10:14, then the largest coefficient in this expansion is :

a) 792
b) 252
c) 462
d) 330

T_r:T_r+1:T_r+2

ⁿ⁺⁵C_r-1: ⁿ⁺⁵C_r: ⁿ⁺⁵C_r+1 = 5:10:14

(n+5)!/(r-1)!(n+6-r)! : (n+5)!/r!(n+5-r)! = 5/10

r/(n+6-r) = 1/2

(r+1)!(n+4-r)!/r!(n+5-r)! = 5/7

2r = n+6-r

3r = n+6 ..(i)

(r+1)/(n+5-r) = 5/7

7r+7 = 5n + 25-5r

12r = 5n+18 ..(ii)

From eq.(i) and (ii)

4(n+6) = 5n+18

n = 6, r = 4

Then we can find (1+x)¹¹

Largest coefficient in the expansion (1+x)¹¹ = ¹¹C₆ = 462

Answer: c

13. The function

a) both continuous and differentiable on R-{-1}
b) continuous on R-{-1} and differentiable on R-{-1,1}
c) continuous on R-{1} and differentiable on R-{-1,1}
d) both continuous and differentiable on R-{1}

At x = 1, f(1) = π/2

f(1⁺) = 0

discontinuous ⇒ non diff.

At x = -1

f(-1) = 0

f(-1^–) = (1/2){+1-1} = 0

cont. at x = -1

At x = -1, f’(-1) = -1/2

And f’(-1⁺) = 1/2

So, at x = -1 it is non differentiable.

Answer: c

14. The solution of the differential equation (dy/dx)-((y+3x)/log_e(y+3x)) + 3 = 0 is: (where c is a constant of integration)

a) x-log_e(y+3x) = C
b) x-(1/2)(log_e(y+3x))² = C
c) x-2log_e(y+3x) = C
d) y+3x-(1/2)(log_ex)² = C

(dy/dx)-((y+3x)/ln(y+3x))+3 = 0

⇒ (dy/dx)+3 = (y+3x)/ln(y+3x)

Let ln(y + 3x) = t

(1/y+3x)×((dy/dx)+3) = dt/dx

(y+3x)dt/dx = (y+3x)/t

⇒ tdt = dx

t²/2 = x+C

(1/2)(ln(y+3x))² = x+C

Answer: b

15. Let λ ≠ 0 be in R. If α and β are the roots of the equation, x²-x+2λ = 0 and α and γ are the roots of the equation, 3x²-10x+27λ = 0, then βγ/λ is equal to:

a) 27
b) 9
c) 18
d) 36

x²-x+2λ = 0 (α, β) ..(i)

3x²-10x+27λ = 0 (α, γ) ..(ii)

Multiply (i) by 3

3x²-3x+6λ = 0 ..(iii)

Equation (ii)-(iii)

-7x+21λ = 0

α = 3λ

Put in equation (i)

9λ²-3λ+2λ = 0

9λ²-λ = 0

⇒ λ = 1/9

⇒ α = 1/3

αβ = 2/9

⇒ β = 2/3

αγ = 1

⇒ γ = 3

βγ/λ ⇒ (2/3)×3÷1/9

= 18

Answer: c

16. The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30⁰. If the angle of depression of the image of C in the lake from the point P is 60⁰, then PC (in m) is equal to :

a) 200√3
b) 400√3
c) 400
d) 100

(h-200)/x = tan 30⁰

(h+200)/x = tan 60⁰

(h+200)/(h-200) = 3

h+200 = 3h-600

2h = 800

h = 400

(h-200)/PC = sin 30⁰

PC = 400 m

Answer: c

17. Let

a) 15
b) 30
c) 50
d) 45

Number of elements of set T =

⇒ 50×10/20 = n×5/6

⇒ (50/2)×(6/5) = n

⇒ n = 30

Answer: b

18. Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. If P(1, β ), β> 0 is a point on this ellipse, then the equation of the normal to it at P is :

a) 8x-2y = 5
b) 4x-2y = 1
c) 7x-4y = 1
d) 4x-3y = 2

e = 1/2

x = a/e = 4

⇒ a = 2

e² = 1-b²/a²

⇒ 1/4 = 1-(b²/4)

(b²/4) = 3/4

⇒ b² = 3

Elipse (x²/4)+(y²/3) =1

P(1,β)

x = 1

(1/4)+(β²/3) = 1

β²/3 = 3/4

⇒ β = 3/2

⇒ P(1, 3/2)

Equation of normal (a²x/x₁)-(b²y/y₁) = a²-b²

(4x/1)-(3y/3/2) = 4-3

4x-2y = 1

Answer: b

19. Let a₁, a₂, …, a_n be a given A.P. whose common difference is an integer and Sn = a₁+a₂+ …. +a_n. If a₁ = 1, a_n = 300 and 15 ≤ n ≤ 50, then the ordered pair (S_n-4, a_n-4) is equal to:

a) (2480,248)
b) (2480,249)
c) (2490,249)
d) (2490,248)

a₁ = 1, a_n = 300, 15≤ n ≤50

300 = 1+(n-1)d

(n-1) = 299/d

n-1 = integer, d has to be a factor of 299.

So, d = 23 or 13

if d = 23

then n-1 = 13

n = 14 (reject)

or d = 13

n-1 = 23

n = 24 is possible (15 ≤ n <50)

Or S₂₀ = (20/2){2+19×13}

= 10×249

= 2490

a₂₀ = 1+19×13

= 248

(S₂₀, a₂₀) = (2490, 248)

Answer: d

20. The circle passing through the intersection of the circles, x²+y²-6x = 0 and x²+y²-4y = 0, having its centre on the line, 2x-3y+12 = 0, also passes through the point:

a) (-1,3)
b) (1,-3)
c) (-3,6)
d) (-3,1)

S₁+λ(S₁-S₂) = 0

x² + y²-6x+λ(4y-6x) = 0

x²+y²-6x(1+ λ)+4λy = 0

Centre (3(1+λ ), -2λ) put in 2x-3y+12 = 0

6+6λ+6λ+12 = 0

12λ = -18

λ = -3/2

Circle is x²+y²+3x-6y = 0

Check options

(-3,6) is the point.

Answer: c

21. Let {x} and [x] denote the fractional part of x and the greatest integer ≤ x respectively of a real number x. If

=

= n/2

⇒ 1+2+…n-1 = n(n-1)/2

⇒ n/2, n(n-1)/2, 10(n²-n) → G.P

⇒ n²(n-1)²/4 = (n/2)×10×n×(n-1)

⇒ n-1 = 20

⇒ n = 21

Answer: 21

22. A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

⁶C₄×1×3² = 15×9 = 135

Answer: 135

23. If

=

= 1+4

= 5

Similarly

= 8

= 5

⇒ 5+8+5 = 18

Answer: 18

24. Let PQ be a diameter of the circle x²+y² = 9. If α and β are the lengths of the perpendiculars from P and Q on the straight line, x+y = 2 respectively, then the maximum value of αβ is:

α =

β =

αβ =

⇒ αβ =

⇒ αβ =

αβ_max = (9+5)/2

= 7

Answer: 7

25. If the variance of the following frequency distribution :

is 50, then x is equal to

400/(4+x) = 50

x = 4

Answer: 4