## JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1

**Question 1.**Let y = y(x) be the solution of the differential equation, xy’-y = x

^{2}(xcosx+sinx),x > 0. If y(π) = π, then y’’(π/2)+y(π/2) is equal to:

a) 2+(π/2)+(π

^{2}/4)

b) 2+(π/2)

c) 1+(π/2)

d) 1+(π/2)+ (π

^{2}/4)

xy’-y = x^{2}(x cosx + sinx) x > 0, y(π) =π

y’-(1/x)y = x(x cosx+sin x)

I.F =

= e^{– ln x}

= 1/x

y(1/x) = ∫(1/x)x(x cos x+sin x) dx

(y/x) = ∫(x cos x+ sin x) dx

(y/x) = ∫ (d/dx)(x sin x) dx

(y/x) = x sin x+C

y = x^{2} sin x+Cx

x = π, y = π

π = πC

C = 1

y = x^{2} sin x+x

y(π/2) = (π^{2}/4)+(π/2)

y’ = 2x sinx+x^{2}cosx+1

y’’ = 2sinx+2xcosx+2x cosx-x^{2}sinx

y’’(π/2) = 2-(π^{2}/4)

y(π/2)+ y’’(π/2) = 2+π/2

**Answer:**(b)

**Question 2.**The value of

a)

^{51}C

_{7}–

^{30}C

_{7}

b)

^{51}C

_{7}+

^{30}C

_{7}

c)

^{50}C

_{7}–

^{30}C

_{7}

d)

^{50}C

_{6}–

^{30}C

_{6}

= ^{50}C_{6}+^{49}C_{6}+^{48}C_{6}+…+^{31}C_{6}+^{30}C_{6}

Add and subtract ^{30}C_{7}

Using ^{n}C_{r}+^{ n}C_{r-1} = ^{n+1}C_{r}

^{30}C_{6}+^{30}C_{7 }= ^{31}C_{7}

^{31}C_{6}+^{31}C_{7 }= ^{32}C_{7}

Similarly solving

^{51}C_{7}–^{30}C_{7}

**Answer:** (a)

**Question 3.**Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]

^{2}+2[x+2]-7 = 0 has:

a) exactly four integral solutions.

b) infinitely many solutions.

c) no integral solution.

d) exactly two solutions.

^{2}+2[x+2]-7 = 0 [x]

^{2 }+2[x]-3 = 0

let [x] = y

y^{2}+3y-y -3 = 0

(y -1)(y+3) = 0

[x] = 1 or [x] = -3x ∈ [1,2) or x ∈ [-3,-2)

**Answer:** (b)

**Question 4.**Let P(3, 3) be a point on the hyperbola, (x

^{2}/a

^{2})-(y

^{2}/b

^{2}) = 1. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a

^{2}, e

^{2}) is equal to:

a) (9, 3)

b) (9/2, 2)

c) (9/2, 3)

d) (3/2, 2)

(x^{2}/a^{2})-(y^{2}/b^{2}) = 1

Point P(3, 3) on hyperbola.

(9/a^{2})-(9/b^{2}) = 1 ..(i)

Equation of normal (a^{2}x/3)+(b^{2}y)/3 = a^{2}e^{2}

At x axis y = 0

a^{2}x/3 = a^{2}e^{2}

x = 3e^{2} = 9

3e^{2} = 9

e^{2} = 3

e = √3

e^{2} = 1+b^{2}/a^{2} = 3

b^{2} = 2a^{2} ..(ii)

Put in equation 1

(9/a^{2})-(9/2a^{2}) = 1

(9/2a^{2}) = 1

a^{2} = 9/2

(a^{2}, e^{2}) = (9/2,3)

**Answer:** (c)

**Question 5.**Let (x

^{2}/a

^{2})+(y

^{2}/b

^{2}) = 1 (a>b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, Φ(t) = (5/12)+t-t

^{2}, then a

^{2}+b

^{2}is equal to:

a) 135

b) 116

c) 126

d) 145

L.R = 2b^{2}/a = 10 ..(i)

Φ(t) = (5/12)-(t-1/2)^{2}+(1/4)

= (8/12)-(t-1/2)^{2}

Φ(t)_{max} = 2/3 = e

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

= 4/9

b^{2}/a^{2} = 5/9

From (i)

b^{2}/a.a = 5/9

5/a = 5/9

a = 9

b^{2} = 45

a^{2}+b^{2} = 81+45 = 126

**Answer:** (c)

**Question 6.**Let

a) (-π/6)+(1/2)+(√3/4)

b) (π/6)+(1/2)-(√3/4)

c) (-π/12)+(1/2)+(√3/4)

d) (π/12)+(1/2)-(√3/4)

Substituting x = tan^{2}t

dx = 2tan t sec^{2}t dt

f(x) =

f(x) = 2∫sin^{2}t dt

x = 3 t = π/3

x = 1 t = π/4

**Answer:** (d)

**Question 7.**If 1+(1-2

^{2}×1)+(1-4

^{2}×3)+(1-6

^{2}×5)+……+(1-20

^{2}×19) = α – 220β, then an ordered pair (α, β) is equal to:

a) (10,97)

b) (11,103)

c) (11,97)

d) (10,103)

T_{n} = 1-(2n)^{2}(2n-1)

= 1-4n^{2}(2n-1)

= 1-8n^{3}+4n^{2}

S_{n} =

= n-∑8n^{3}+∑4n^{2}

= n-8×(n^{2}(n+1)^{2}/4)+4n(n+1)(2n+1)/6

S_{10} = 10-2×100×121 +(2/3)×10×11×21

= 10-24200+1540

= 10-22660

Sum of series = 11-220×103

= α – 220β

α = 11

β = 103

**Answer: **(b)

**Question 8.**The integral

a) tan x-(x sec x/(x sin x+cos x))+C

b) sec x-(x tan x/(x sin x+cos x))+C

c) sec x+(x tan x/(x sin x+cos x))+C

d) tan x+(x sec x/(x sin x+cos x))+C

**= **tan x-(x sec x/(x sin x+cos x))+C

**Answer: **(a)

**Question 9.**Let f(x)= |x-2| and g(x) = f(f(x)) ,x∈0,4]. Then

a) 1/2

b) 0

c) 1

d) 3/2

f(x) = |x-2| =

g(x) = ||x-2|-2| =

=

=

= [(1/2)×2×2+1+(1/2)×1×1]-[(1/2)×2×2+(1/2)×1×1)]

= (7/2)-(5/2)

= 1

**Answer: **(c)

**Question 10.**Let x

_{0}be the point of local maxima of f(x) =

_{0}is:

a) -22

b) -4

c) -30

d) 14

=

= x(x^{2}-2)+2(-2x+7)+ 3(4-7x)

= x^{3}-2x- 4x+14+12-21x

f(x) = x^{3}– 27x+26

f’(x) = 3x^{2}-27 = 0

x = 3

f’’(x) = 6x

at x = 3, f” (3) = 6×3= 18 > 0

at x = -3, f” (-3) = 6×3 = -18 < 0

Max at x_{0} = -3

So

= 6+6-3-14+6+3-21+4-9

= 25-47

= -22

**Answer:** (a)

**Question 11.**A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If BAC = 90

^{0}and ar(ΔABC) = 5√5 s units, then the abscissa of the vertex C is:

a) 1+√5

b) 1+2√ 5

c) 2√5-1

d) 2+√5

AB = √(4+1) = √5

(1/2)×√5×x = 5√5

x = 10

m_{AB } = -1/2

m_{AC} = 2 = tanθ

Since sinθ = 2/√5

cosθ = 1/√5

by parametric co-ordinates

a = 1+ xcosθ = 1+10×1/√5

= 1+2√5

**Answer:** (b)

**Question 12.**Let f be a twice differentiable function on (1,6). If f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4, for all x ∈ (1,6), then:

a) f(5)+f’(5) ≥28

b) f’(5)+f’’(5) ≤20

c) f(5) ≤10

d) f(5)+f’(5) ≤26

f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4,

x∈(1,6)

adding (1) and (2) we get,

f(5)+f’(5) ≥28

**Answer: **(a)

**Question 13.**Let α and β be the roots of x

^{2}-3x+p = 0 and 1/γ and 1/δ be the roots of x

^{2}-6x+q = 0. If α, β, γ, δ form a geometric progression. Then ratio (2q+p): (2q-p) is:

a) 33 :31

b) 9 : 7

c) 3 : 1

d) 5 : 3

Roots of x^{2}-3x+p = 0 are α and β.

The roots of x^{2}-6x+q = 0 are γ and δ.

α + β = 3

γ + δ = 6

α = a , β = ar, γ = ar^{2}, δ = ar^{3}

a(1+r) = 3 …(i)

ar^{2}(1+r) = 6 …(ii)

Divide (ii) by (i)

r^{2 }= 2

α.β = p = a^{2}r

γ.δ = q = a^{2}r^{5}

(2q+p)/(2q-p) = (2r^{4}+1)/( 2r^{4}-1)

= (2×2^{2}+1)/( 2×2^{2}-1) = 9/7

**Answer:** (b)

**Question 14.**Let u = (2z+i)/(z-ki), z = x+iy and k>0. If the curve represented by Re(u) +Im(u) =1 intersects the y-axis at the points P and Q where PQ =5, then the value of k is:

a) 4

b) 1/2

c) 2

d) 3/2

u = (2z+i)/(z-ki)

z = x+iy

Re(u)+Img(u) = 1

2x^{2}+(2y+1)(y -k )+x+2xk = x^{2} + (y-k)^{2}

at y – axis, x = 0

(2y +1)(y- k) = (y- k )^{2}

2y^{2 }+y- 2yk- k = y^{2}+k^{2}-2yk

Roots of y^{2}+y-(k +k^{2}) = 0 are y_{1} and y_{2}

Diff. of roots = 5

√(1+4k+4k^{2}) = 5

4k^{2}+4k = 24

k^{2}+k-6 = 0

(k +3)(k-2) = 0

k = 2

**Answer: **(c)

**Question 15.**If A =

^{5}=

a) a

^{2}-d

^{2}= 0

b) a

^{2}-c

^{2}= 1

c) 0 ≤a

^{2}+b

^{2}≤1

d) a

^{2}-b

^{2}= 1/2

A =

A^{2} =

a = d = cos (5θ)

b = c = i sin (5θ)

a^{2}-b^{2} = cos^{2}5θ + sin^{2}5θ

= 1

**Answer: **(d)

**Question 16.**The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is:

a) 3

b) 9

c) 7

d) 5

(5+7+10+12+14+15+x+y)/8 = 10

x+y = 17 ..(i)

Variance = (5^{2}+7^{2}+10^{2}+12^{2}+14^{2}+15^{2}+x^{2}+y^{2})/8 -100 = 13.5

(739+x^{2}+y^{2})/8 – 100 = 13.5

x^{2}+y^{2} = 169 ..(ii)

∴ x = 12, y = 5

|x-y| = 7

**Answer:** (c)

**Question 17.**A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:

a) 37

b) 29

c) 65

d) 55

n(B) ≤n(A∪B) ≤n(U)

76 ≤76+63-x ≤100

-63 ≤-x ≤-39

63 ≥x ≥ 39

**Answer** (d)

**Question 18.**Given the following two statements

(S_{1}): (q˅p)→(p ↔ ~q) is a tautology.

(S_{2}): ~q ˄ (~p ↔ q) is a fallacy. Then:

a) only (S_{1}) is correct.

b) both (S_{1}) and (S_{2}) are correct.

c) only (S_{2}) is correct

d) both (S_{1}) and (S_{2}) are not correct.

**Answer** (d)

**Question 19.**Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is:

a) 5

b) 20/3

c) 10/3

d) 6

Using similar triangle concept

tan θ_{1} = 10/x = H/x_{1}

x_{1} = Hx/10

tan θ_{2} = 15/x = H/x_{2}

x_{2} = Hx/15

Since x_{1}+x_{2} = x

(Hx/10)+( Hx/15) = x

15H+10H = 150

H = 150/25 = 6 m

**Answer:** (d)

**Question 20.**If (a+√2b cos x)( a-√2b cos y) = a

^{2}-b

^{2}, where a>b>0, then dy/dx at (π/4, π/4) is:

a) (a+b)/(a-b)

b) (a-2b)/(a+2b)

c) (a-b)/(a+b)

d) (2a+b)/(2a-b)

(a+√2b cos x)( a-√2b cos y) = a^{2}-b^{2}

Differentiating both sides w.r.t.y

-√2b sin x (dx/dy)(a-√2b cos y)+ (a+√2b cos x)(√2b sin y) = 0

x = y = π/4

-b(dx/dy) (a-b)+(a+b)b = 0

dx/dy = (a+b)/(a-b)

**Answer:** (a)

**Question 21.**Suppose a differentiable function f(x) satisfies the identity f(x+y) = f(x)+f(y)+xy

^{2}+x

^{2}y,for all real x and y. If

f(x+y) = f(x)+f(y)+xy^{2}+x^{2}y

x = y = 0

f(0) = 2f(0)

f(0) = 0

Now,

f’(x) = 1+0+x^{2}

f’(x) = 1+x^{2}

f’(3) = 10

**Answer:** (10)

**Question 22.**If the equation of a plane P, passing through the intersection of the planes, x+4y-z+7 = 0 and 3x+y+5z = 8 is ax+by+6z = 15 for some a, b∈R, then the distance of

**t**he point (3,2,-1) from the plane P is ….. units.

p_{1}+λp_{2} = 0

(x +4y-z+7)+λ(3x+y+5z-8) = ax+by+6z-15

(1+3λ)/a = (4+λ)/b = (-1+5λ)/6 = (7-8λ)/-15

∴ 15-75λ = 42-48λ

-27 = 27λ

= -1

∴ plane is -2x+3y-6z+15 = 0

d = |-6+6+6+15)/√(4+9+36)|

= 3 units

**Answer: **(3)

**Question 23.**If the system of equations

x-2y+3z = 9

2x+y+z = b

x-7y+az = 24, has infinitely many solutions, then a-b is equal to:

D = 0

1(a +7)+2(2a -1)+3(-14-1) = 0

a+7+4a-2-45 = 0

5a = 40

a = 8

D_{1} =

9(8+7)+2(8b-24)+3(-7b-24) = 0

135+16b-48-21b-72 = 0

15 = 5b

b = 3

Hence a – b = 8-3 = 5

**Answer:** (5)

**Question 24.**Let (2x

^{2}+3x+4)

^{10}=

_{7}/a

_{13}is equal to:

Given (2x^{2}+3x+4)^{10} =

Replace x by 2/x in above identity:

Now, comparing coefficient of x^{7} from both sides

(take r = 7 in L.H.S. and r = 13 in R.H.S)

2^{10}a_{7} = a_{13}2^{13 }

a_{7}/a_{13} = 2^{3} = 8

**Answer:** (8)

**Question 25.**The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4 is:

Probability of hitting, P(H) = 1/10

probability of missing, P(M) = 9/10

We have, 1-(probability of all shots result in failure ) >1/4

= 1-P(M)^{n} > 1/4

= 1-(9/10)^{n} > 1/4

(9/10)^{n }<3/4

n≥3

**Answer **(3)

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