**Question 1.**A line parallel to the straight line 2x-y = 0 is tangent to the hyperbola (x

^{2}/4)-(y

^{2}/2) = 1 at the point (x

_{1}, y

_{1}). Then x

_{1}

^{2}+5y

_{1}

^{2}is equal to:

a) 6

b) 10

c) 8

d) 5

T: (xx_{1}/4)-(yy_{1}/2) = 1..(1)

t: 2x-y = 0 is parallel to T

T: 2x-y = λ ..(2)

Now compare (1) and (2)

x_{1} ÷(4/2) = y_{1} ÷(2/1) = 1/λ

x_{1} = 8/λ and y_{1} = 2/λ

(x_{1}, y_{1}) lies on hyperbola

(64/4λ^{2}) -(4/2λ^{2}) = 1

14 = λ^{2}

Now x_{1}^{2}+5y_{1}^{2} = (64/λ^{2}) +5(4/λ^{2})

= 84/14

= 6

**Answer:**(a)

**Question 2.**The domain of the function f(x) = sin

^{-1 }[(|x|+5)/(x

^{2}+1)] is (-∞, -a]U[a,∞). Then a is equal to:

a) (√17-1)/2

b) √17/2

c) (1+√17)/2

d) (√17/2)+1

-1≤ (|x|+5)/x^{2}+1 ≤1

-x^{2}-1≤|x|+5≤ x^{2}+1

Case 1

-x^{2}-1≤ |x|+5

This inequality is always right for all x∈R.

Case 2:

|x|+5 ≤ x^{2}+1

x^{2}-|x| ≥4

x^{2}-x-4 ≥0

(|x|-(1+√17)/2)(|x|-(1-√17)/2) ≥0

|x|≤(1-√17/2) (Not possible)

Or |x| ≥ (1+√17/2)

x∈(-∞, (-1-√17)/2] U[(1+√17)/2, )

a = (1+√17)/2

**Answer:** (c)

**Question 3.**If a function f(x) defined by

be continuous for some a, b, c ∈ R and f’(0)+f’(2) = e, then the value of a is:

a) 1/(e^{2}-3e+13)

b) e/(e^{2}-3e-13)

c) e/(e^{2}+3e+13)

d) e/(e^{2}-3e+13)

f(x) is continuous

at x =1 At x = 1, b = ce – ae^{2}

at x =3

9c = 9a+6c

c = 3a

Now f ’(0) +f ‘(2) = e

a-b+4c = e

a-e (3a-ae)+4×3a = e

a-3ae+ae^{2}+12a = e

13a-3ae+ae^{2 }= e

a = e/(13-3e+e^{2})

**Answer:** (d)

**Question 4.**The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in

a) (-∞, -9]∪[3, ∞)

b) [-3, ∞)

c) (-∞, -9]

d) (-∞, -3]∪[9, ∞)

(a/r).a.ar = 27

a = 3

(a/r)+a+ar = S

(1/r)+1+r = S/3

r+(1/r) = (S/3)-1

r+(1/r) ≥ 2 or r+(1/r) ≤ -2

(S/3) ≥ 3 or (S/3) ≤ -1

S≥ 9 or S ≤ -3

S∈(-∞, -3]∪[9,∞)

**Answer:** (d)

**Question 5.**If R = {(x,y):x,y ∈ Z, x

^{2}+3y

^{2}≤ 8} is a relation on the set of integers Z, then the domain of R

^{-1}is:

a) {-1, 0, 1}

b) {-2, -1, 1, 2}

c) {0,1}

d) {-2,-1,0,1,2}

3y^{2} ≤ 8-x^{2}

R:{(0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1), (2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}

R: {-2, -1, 0, 1, 2} ->{-1,0,-1}

Hence R^{-1}:{-1,0,1}-> {-2, -1, 0, 1, 2}

**Answer:** (a)

**Question 6.**The value of (1+sin (2π/9)+i cos(2π/9)/(1+sin (2π/9)-i cos(2π/9))

^{3}is:

a) (-1/2)(1-i√3)

b) (1/2)(1-i√3)

c) (-1/2)(√3-i)

d) (1/2)(√3-i)

= (e^{i(5π/36)+i(5π/36)})^{3}

= (e^{2i(5π/36)})^{3}

= e^{i(30π/36)}

= e^{i(5π/6)}

= cos (5π/6)+ i sin 5π/6

= (-√3/2)+(i/2)

**Answer:** (c)

**Question 7.**Let P(h,k) be a point on the curve y = x

^{2}+7x+2, nearest to the line, y = 3x-3. Then the equation of the normal to the curve at P is:

a) x+3y-62 = 0

b) x-3y-11 = 0

c) x-3y+22 = 0

d) x+3y+26 = 0

C: y = x^{2}+7x+2

Let P : (h, k) lies on

Curve k = h^{2}+7h+2 …(1)

Now for shortest distance

Slope of tangent line at point P = slope of line L

dy/dx at P(h,k) = m_{L}

(d/dx)(x^{2}+7x+2) _{at P(h,k) }= 3

(2x+7)_{at P(h,k)} = 3

2h+7 = 3

h = -2

from (1) k = -8

P: (-2, -8)

equation of normal to the curve is perpendicular to L: 3x-y = 3

N: x+3y = λ

Pass(-2,-8)

λ = -26

N: x+3y+26 = 0

**Answer: **(d)

**Question 8.**Let A be a 2×2 real matrix with enries from {0,1} and A ≠ 0. Consider the following two statements:

(P) If A ≠I_{2} , then A = -1

(Q) If A =1 , then tr(A) = 2,

where I_{2} denotes 2×2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:

a) Both (P) and (Q) are false

b) (P) is true and (Q) is false

c) Both (P) and (Q) are true

d) (P) is false and (Q) is true

P: A =

_{2}

And A ≠0 and A = 1 (False)

Q: A =

then Tr(A) = 2 (true)

**Answer: **(d)

**Question 9.**Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is:

a) 4/17

b) 8/17

c) 2/5

d) 2/3

Let P(I) represent probability of selecting box I and P(II) represent probability of selecting box II.

P(I) = 1/2

P(II) = 1/2

In Box I, prime numbers are {2,3,5,7,11,13,17,19,23,27}

In Box II, prime numbers are {31,37,41,43,47}

Let A be event that selected number on card is non prime.

P(A) = P(I).P(A/I)+P(II).P(A/II)

= (1/2)×(20/30)+(1/2)×(15/20)

= 17/24

Now, P(I/A) = P(II).P(A/I)÷P(A)

= (1/2)×(20/30)÷ (17/24)

= (1/3)×(24/17)

= 8/17

**Answer: **(b)

**Question 10.**If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to :

a) 12

b) -12

c) -24

d) 6

p’(1) = 0 and p’(2) = 0

p’(x) = a(x-1)(x-2)

p(x) = a[(x^{3}/3)-(3x^{2}/2)+2x]+b

p(1) = 8

a[(1/3)-(3/2)+2]+b = 8 ..(i)

p(2) = 4

a[(8/3)-(3×4/2)+2×2]+b = 4 ..(ii)

From equation (i) and (ii)

a = 24 and b = -12

p(0) = b = -12

**Answer:** (b)

**Question 11.**The contrapositive of the statement “If I reach the station in time, then I will catch the train” is:

a) If I will catch the train, then I reach the station in time.

b) If I do not reach the station in time, then I will catch the train.

c) If I do not reach the station in time, then I will not catch the train.

d) If I will not catch the train, then I do not reach the station in time.

Statement p and q are true

Statement, then the contra positive of the implication

p->q = (~q) ->(~p)

**Answer:** (d)

**Question 12.**Let α and β be the roots of the equation, 5x

^{2}+6x-2 = 0. If S

_{n}= α

^{n}+β

^{n}, n = 1,2,3,.. then:

a) 5S

_{6}+6S

_{5}+2S

_{4}= 0

b) 6S

_{6}+5S

_{5}= 2S

_{4}

c) 6S

_{6}+5S

_{5}+2S

_{4}= 0

d) 5S

_{6}+6S

_{5}= 2S

_{4}

5x^{2}+6x-2 = 0

Put x = α

5α^{2}+6α-2 = 0

6α-2 = -5α^{2} ..(i)

Similarly

6β-2 = -5β^{2} ..(ii)

S_{6} = α^{6}+β^{6}

S_{5} = α^{5}+β^{5}

S_{4} = α^{4}+β^{4}

Now 6S_{5}-2S_{4} = 6α^{5}+6β^{5}-2α^{4}-2β^{4}

= α^{4}(6α-2)+β^{4}(6β-2)

= α^{4}(-5α^{2})+β^{4}(-5β^{2})

= -5(α^{6}+β^{6})

= -5S_{6}

Hence 6S_{5}+5S_{6}= 2S_{4 }

**Answer: **(d)

**Question 13.**If the tangent to the curve y = x+sin y at a point (a,b) is parallel to the line joining (0, 3/2) and (1/2, 2), then:

a) b = (π/2)+a

b) |a+b| = 1

c) |b-a| = 1

d) b = a

(dy/dx)_{p(a,b)} = (2-3/2)/(1/2 -0)

1+cos b = 1. p: (a,b) lies on curve

cos b = 0

b = a+sin b

b = a

b-a =

|b-a| = 1

**Answer:** (c)

**Question 14.**Area (in sq. units) of the region outside (x/2)+ (y/3) = 1 and inside the ellipse (x

^{2}/4)+(y

^{2}/9) = 1 is:

a) 3(π-2)

b) 6(π-2)

c) 6(4-π)

d) 3(4-π)

Given (x/2)+ (y/3) = 1

A = 4((πab/4)-(1/2)×2×3)

= π×2×3-12

= 6(π-2)

**Answer: **(b)

**Question 15.**If |x| <1, |y| <1, and xy, then the sum to infinity of the following series – (x+y)+(x

^{2}+xy+y

^{2})+(x

^{3}+x

^{2}y+xy

^{2}+y

^{3})+… is:

a) (x+y+xy)/(1-x)(1-y)

b) (x+y-xy)/(1-x)(1-y)

c) (x+y+xy)/(1+x)(1+y)

d) (x+y-xy)/(1+x)(1+y)

(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+….. ∞

= (1/x-y){(x^{2}-y^{2})+(x^{3}-y^{3})+(x^{4}-y^{4})+…..∞

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

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

= ((x+y)-xy)(x-y)/(1-x)(1-y)(x-y)

= (x+y-xy)/(1-x)(1-y)

**Answer: **(b)

**Question 16.**Let α>0, β>0, be such that α

^{3}+β

^{2}= 4. If the maximum value of the term independent of x in the binomial expansion of (αx

^{1/9}+βx

^{-1/6})

_{10}is 10k, then k is equal to:

a) 176

b) 336

c) 352

d) 84

For term independent of x

T_{r+1} = ^{10}C_{r}(αx^{1/9 })^{10-r}(βx^{-1/6})^{r}

= ^{10}C_{r}α^{10-r}β^{r}.x^{(10-r)/9}.x^{-r/6}

Since (10-r)/9-(r/6) = 0

r = 4

T_{5} = ^{10}C_{r}α^{6.β4}

AM≥GM

Now

Take 4^{th} power

(4/4)^{4} ≥ (α^{6}β^{4}/2^{4})

α^{6}.β^{4} ≤ 2^{4}

^{10}C_{4}α^{6}.β^{4}≤ ^{10}C_{4}2^{4}

T_{5} ≤ ^{10}C_{4}2^{4}

T_{5} ≤ 10!2^{4}/(6!4!)

T_{5} ≤ (10×9×8×7×2^{4})/(4×3×2×1)

Maximum value of T_{5} = 10×3×7×16 = 10k

k = 3×7×16

k = 336

**Answer:** (b)

**Question 17.**Let S be the set of all λ ∈ R for which the system of linear equations

2x-y+2z = 2

x-2y+λz = -4

x+λy+z=4

has no solution. Then the set S

a) is an empty set.

b) is a singleton.

c) contains more than two elements.

d) contains exactly two elements.

For no solution

** Δ= **0

**Δ = **

2(-2-λ^{2})+1(1-λ)+2(λ+2) = 0

-4-2λ^{2}+1-λ+2λ+4 = 0

2λ^{2}-λ-1 = 0

λ= 1 or -1/2

For two values of λ, equations has no solution.

**Answer** (d)

**Question 18.**Let X = {x∈N: 1≤x≤17} and Y = {ax+b:x∈X and a,b∈R, a>0}. If mean and variance of elements of Y are 17 and 216 respectively then a+b is equal to:

a) -27

b) 7

c) -7

d) 9

X :{1,2,…17}

Y: {ax+b:x∈X and a,b∈R, a>0}

Given var(Y) = 216

(Ʃy_{1}^{2})/n – mean^{2} = 216

(Ʃy_{1}^{2})/17 – 289 = 216

(Ʃy_{1}^{2}) = 8585

(a+b)^{2} +(2a+b)^{2}+….+(17a+b)^{2} = 8585

105a^{2} +b^{2}+18ab=505 ..(1)

Now Ʃy_{1} = 17×17

a(17×9) + 17.b = 17×17

9a+b = 17 ….(2)

from equation (1) & (2)

a = 3 and b = -10

a+b = -7

**Answer** (c)

**Question 19.**Let y = y(x) be the solution of the differential equation, (2+sinx)/(y+1) (dy/dx) = -cos x, y>0, y(0) = 1. If y(π) = a, and (dy/dx) at x = π is b, then the ordered pair (a,b) is equal to:

a) (2,2/3)

b) (1,1)

c) (2,1)

d) (1,-1)

∫(dy/y+1) = ∫-cos x/(2+sin x)dx

ln |y+1| = -ln |2+sin| x +k

Put (0,1)

k = ln 4

Now C: (y+1)(2+sin x) = 4

y(π) = a (a+1)(2+0) = 4

⇒ a = 1

(dy/dx)_{x=π} = b

⇒ b = -(-1)(2+0)/(1+1)

= 1

(a,b) = (1,1)

**Answer:** (b)

**Question 20.**The plane passing through the points (1,2,1), (2,1,2) and parallel to the line, 2x = 3y, z = 1 also passes through the point:

a) (0,-6,2)

b) (0,6,-2)

c) (-2,0,1)

d) (2,0,-1)

Given lines are 2x = 3y, z = 1

Dr of line (3,2,0)

Plane : -2(x-1)+3(y-2)+5(z -1) = 0

Plane : -2x+3y+5z+2-6-5=0

Plane : 2x-3y-5z = -9

**Answer:** (c)

**Question 21.**The number of integral values of k for which the line, 3x+4y = k intersects the circle, x

^{2}+y

^{2}-2x-4y+4 = 0 at two distinct points is…

c: (1,2) and r = 1

cp < r

|(3×1+4×2-k)/5| < 1

|11-k| < 5

-5<k-11<5

6<k<16

k = 7,8,9,..,15

So total 9 value of k.

**Answer:** (9)

**Question 22.**Let

a^{2}+b^{2}-2a.b+a^{2}+c^{2}-2a.c = 8

2a^{2}+b^{2}+c^{2}-2a.b -2a.c=8

a.b+a.c = -2

Now

= 2a^{2}+4b^{2}+4c^{2}+4a.b+4a.c

= 2+4+4+4(-2)

= 2

**Answer: **(2)

**Question 23.**If the letters of the word ’MOTHER’ be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word ’MOTHER’ is….

**Answer:** (309)

**Question 24.**If

Given

1+2+3+…+n = 820

Ʃn = 820

n(n+1)/2 = 820

n = 40

**Answer:** (40)

**Question 25.**The integral

=

=

=

= [(1/2)-0-(1/4)-0]+(1-(1/4))-(1-(1/2))+1

= (1/2)-(1/4)+(3/4)-(1/2)+1

= 3/2

**Answer **(3/2)

## Video Lessons – Maths

## JEE Main 2020 Maths Paper With Solutions Sep 2 Shift 1

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