## JEE Main 2020 Maths Paper With Solutions Shift 2 September 4

**1. **Suppose the vectors x_{1}, x_{2} and x_{3} are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b_{1}, b_{2} and b_{3 }respectively. If x_{1} =

_{2}=

_{3}=

_{1}=

_{2}=

_{3}=

a) 2

b) 1/2

c) 3/2

d) 4

Using AX = B

A =

a_{1}+a_{2}+a_{3} = 1

a_{4}+a_{5}+a_{6} = 0

a_{7}+a_{8}+a_{9} = 0

2a_{2}+a_{3} = 0

2a_{5}+a_{6} = 2

2a_{8}+a_{9} = 0

a_{3} = 0, a_{6} = 0, a_{9} = 2

a_{8} = -1

a_{5 }= 1

a_{2} = 0

⇒ a_{1} = 1

a_{4} = -1

a_{7} = -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})

= √(4λ^{2}+9λ^{2}+36λ^{2})

= 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} (2tf^{2}(x)-2x^{2}f(t).f’(t)) = 0

⇒ 2xf^{2}(x)-2x^{2}f(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 } ≥ 2^{1}×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)-(k^{2}-1)/2 = -4

(k^{2}-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 = x^{2}-1 below the x-axis, is:

a) 2/3√3

b) 4/3

c) 1/3√3

d) 4/3√3

Area = 2a(a^{2}-1)

A = 2a^{3}-2a

dA/da = 6a^{2}-2 = 0

d^{2}A/da^{2} = 12a

at a = -1/√3

d^{2}A/da^{2 }= -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+5}C_{r-1}: ^{n+5}C_{r}:^{ n+5}C_{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} = ^{11}C_{6} = 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_{e}x)^{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

t^{2}/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, x^{2}-x+2λ = 0 and α and γ are the roots of the equation, 3x^{2}-10x+27λ = 0, then βγ/λ is equal to:

a) 27

b) 9

c) 18

d) 36

x^{2}-x+2λ = 0 (α, β) ..(i)

3x^{2}-10x+27λ = 0 (α, γ) ..(ii)

Multiply (i) by 3

3x^{2}-3x+6λ = 0 ..(iii)

Equation (ii)-(iii)

-7x+21λ = 0

α = 3λ

Put in equation (i)

9λ^{2}-3λ+2λ = 0

9λ^{2}-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 30^{0}. If the angle of depression of the image of C in the lake from the point P is 60^{0}, then PC (in m) is equal to :

a) 200√3

b) 400√3

c) 400

d) 100

(h-200)/x = tan 30^{0}

(h+200)/x = tan 60^{0}

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

h+200 = 3h-600

2h = 800

h = 400

(h-200)/PC = sin 30^{0}

PC = 400 m

**Answer: c **

**17.** Let

_{i}contains 10 elements and each Y

_{i}contains 5 elements. If each element of the set T is an element of exactly 20 of sets X

_{i}’s and exactly 6 of sets Y

_{i}’s, then n is equal to :

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

e^{2} = 1-b^{2}/a^{2}

⇒ 1/4 = 1-(b^{2}/4)

(b^{2}/4) = 3/4

⇒ b^{2} = 3

Elipse (x^{2}/4)+(y^{2}/3) =1

P(1,β)

x = 1

(1/4)+(β^{2}/3) = 1

β^{2}/3 = 3/4

⇒ β = 3/2

⇒ P(1, 3/2)

Equation of normal (a^{2}x/x_{1})-(b^{2}y/y_{1}) = a^{2}-b^{2}

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

4x-2y = 1

**Answer: b**

**19. **Let a_{1}, a_{2}, …, a_{n} be a given A.P. whose common difference is an integer and** **Sn = a_{1}+a_{2}+ …. +a_{n}. If a_{1} = 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)

a_{1} = 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 S_{20} = (20/2){2+19×13}

= 10×249

= 2490

a_{20} = 1+19×13

= 248

(S_{20}, a_{20}) = (2490, 248)

**Answer: d**

**20. **The circle passing through the intersection of the circles, x^{2}+y^{2}-6x = 0 and x^{2}+y^{2}-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)

S_{1}+λ(S_{1}-S_{2}) = 0

x^{2} + y^{2}-6x+λ(4y-6x) = 0

x^{2}+y^{2}-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 x^{2}+y^{2}+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

^{2}-n), (nN, n>1) are three consecutive terms of a G.P., then n is equal to

=

= n/2

⇒ 1+2+…n-1 = n(n-1)/2

⇒ n/2, n(n-1)/2, 10(n^{2}-n) → G.P

⇒ n^{2}(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

^{6}C_{4}×1×3^{2} = 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 x^{2}+y^{2 }= 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 :

Class |
10-20 |
20-30 |
30-40 |

Frequency |
2 |
x |
2 |

is 50, then x is equal to

400/(4+x) = 50

x = 4

**Answer: 4**

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