Students can understand the question pattern and the difficulty level of questions from each chapter. Practising these solutions will help them to self analyse their preparation. Students can easily download the solutions in PDF format for free.

**1.**If A = {x∈ R∶ |x| <2} and B = {x∈ R∶ |x-2| ≥3} then :

a. A – B = [-1,2]

b. B – A = R-(-2,5)

c. A⋃B = R-(2,5)

d. A∩B = (-2,-1)

A = {x: x ∈ (-2,2)}

B = {x: x ∈ (∞,-1] ⋃ [5, ∞)}

A∩B = {x: x ∈ (-2,-1] }

B – A = {x: x ∈ (∞,-2] ⋃ [5, ∞)}

A – B = {x: x ∈ (-1,2)}

A⋃B = {x: x ∈ (∞,2] ⋃ [5, ∞)}

**Answer: (b)**

**2.**If 10 different balls has to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

a. 965/2

^{10}

b. 945/2

^{10}

c. 945/2

^{11}

d. 965/2

^{11}

Total ways to distribute 10 balls in 4 boxes is = 4^{10}

Total ways of placing exactly 2 and 3 balls in any two of these boxes is = ^{4}C_{2}×^{10}C_{5}×(5!/(2!3!))×2×2^{5 }

P(E) = 945/2^{10}

**Answer: (b)**

**3.**If x = 2 sin θ – sin 2 θ and y = 2 cos θ – cos 2 θ, θ ∈ [0, 2π], then d

^{2}y/dx

^{2}at θ =π is:

a. -3/8

b. 3/4

c. 3/2

d. -3/4

dx/d θ = 2 cos θ -2 cos2 θ

dy/d θ = -2 sin θ + 2 sin2 θ

dy/dx = cot (3 θ/2)

d^{2}y/dx^{2} = -(3/2) cosec^{2}(3 θ/2)(d θ/dx)

(

**Answer : None of the above option satisfies the answer.**

**4.**Let f and g be differentiable functions on R, such that fog is the identity function. If for some a, b ∈ R, g’(a) = 5 and g(a) = b, then f'(b) is equal to :

a. 2/5

b. 5

c. 1

d. 1/5

f(g(x) = x

f’(g(x))g’(x) = 1

put x = a

f’(g(a))g’(a) = 1

f’(b) ×5 = 1

f’(b) = 1/5

**Answer: (d)**

**5**. In the expansion of

_{1}is the least value of the term independent of x when (π/8) ≤ θ ≤ (π/4) and l

_{2}is the least value of the term independent of x when (π/16) ≤ θ ≤ (π/8), then the ratio l

_{2}: l

_{1}is equal to :

a. 16:1

b. 8:1

c. 1:8

d. 1:16

For term independent of x, 16-2r = 0

⇒ r = 8

T_{9} =

l_{1 }= ^{16}C_{8} 2^{8} at θ = π/4

l_{2} =

l_{2}/l_{1} = 16:1

**Answer: (a)**

**6.**Let a,b ∈R, a ≠ 0, such that the equation, ax

^{2}-2bx+5 = 0 has a repeated root α, which is also a root of the equation x

^{2}-2bx-10 = 0. If β is the root of this equation, then α

^{2}+β

^{2}is equal to:

a. 24

b. 25

c. 26

d. 28

ax^{2}-2bx+5 = 0 has both roots as α

2α = 2b/a

α= b/a

And α^{2} = 5/a

b^{2 }= 5a (a≠0) ..(i)

α+β = 2b

And αβ = -10

α= b/a is also a root of x^{2} -2bx-10 = 0

b^{2} -2ab^{2}-10a^{2} = 0

b^{2} = 5a

5a-10a^{2} -10a^{2} = 0

a = 1/4

b^{2} = 5/4

α^{2} = 20

β^{2} = 5

α^{2}+β^{2} = 25

**Answer: **(b)

**7.**Let a function f:[0,5] R, be continuous, f(1) = 3 and F be defined as:

F(x) =

Then for the function F, the point x = 1 is

a. a point of inflection.

b. a point of local maxima

c. a point of local minima

d. not a critical point

F'(x) = x^{2}g(x)

Put x = 1

F'(1) = g(1) = 0 ..(i)

Now F’’(x) = 2xg(x) + g’(x)x^{2}

F’’(1) = 2g(1) + g’(1) {∵g’(x) = f(x)}

F’’(1) = f(1) = 3 ..(ii)

From (1) and (2), F(x) has local minimum at x = 1.

**Answer: (c) **

**8.**Let [t] denotes the greatest integer ≤ t and

^{2}] sin πx is discontinuous, when x is equal to

a. √(A+1)

b. √A

c. √(A+5)

d. √(A+21)

f(x)=[x^{2}]sin π x

It is continuous ∀x ∈ Z as sinπx 🡪 as 🡪 Z.

f(x) is discontinuous at points where [x^{2}] is discontinuous i.e. x^{2}∈Z with an exception that f(x) is continuous as x is an integer.

∴ Points of discontinuity for f(x) would be at

x = ±√2, ±√3, ±√5,……

Also, it is given that

A = 4

√(A+5) = 3

√(A+1) = √5

√(A+21) =5

√A = 2

Points of discontinuity for f(x) is x = √5

**Answer: (a)**

**9.**Let a-2b+c = 1. If f(x) =

a. f(-50) = 501

b. f(-50) = -1

c. f(50) = 1

d. f(-50) = -501

Given f(x) =

a -2b+c = 1

Applying R_{1} 🡪 R_{1 }-2R_{2}+ R_{3}

f(x) =

Using a-2b+c = 1

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

f(x) = 1

f(50) = 1

f(-50) = 1

**Answer: (c)**

**10.**Given:

and g(x)= (x− 1/2)^{2}, x ∈ R. Then the area (in sq. units) of the region bounded by the curves y = f(x) and y = g(x) between the lines 2x = 1 to 2x = √3 is :

a. (√3/4) – (1/3)

b. (1/3) + (√3/4)

c. (1/2) + (√3/4)

d. (1/2) – (√ 3/4)

The area between f(x) and g(x) from x = 1/2 to x = √3/2 :

Points of intersection of f(x) and g(x):

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

x = √3/2 ,-√3/2

Required area =

=

=

= (√3/4) – (1/3)

**Answer: **(a)

**11.**The following system of linear equations

7x+6y-2z = 0

3x+4y+2z = 0

x-2y-6z = 0, has

a. infinitely many solutions, (x,y,z) satisfying y = 2z

b. infinitely many solutions (x,y,z) satisfying x = 2z

c. no solution

d. only the trivial solution

7x+6y−2z = 0

3x+4y+2z = 0

x−2y−6z = 0

As the system of equations are homogeneous the system is consistent.

⇒ Infinite solutions exist (both trivial and non-trivial solutions)

When y = 2z

Let’s take y = 2, z = 1

When (x,2,1)is substituted in the system of equations

⇒7x+10 = 0 ,3x+10 = 0 , x-10 = 0 (which is not possible)

∴ y = 2z ⇒ Infinitely many solutions does not exist.

For x = 2z, lets take x = 2, z = 1, y = y

Substitute (2,y,1)in system of equations ⇒ y = -2

∴For each pair of (x,z), we get a value of y.

Therefore, for x = 2z infinitely many solutions exists.

**Answer: (b) **

**12.**If p -> (p ∧~ q) is false. Then the truth values of p and q are respectively

a. F, T

b. T, F

c. F, F

d. T, T

Given p ->(p ∧~ q)

Truth table:

p |
q |
~q |
(p ∧~ q) |
p->(p ∧~ q) |

T |
T |
F |
F |
F |

T |
F |
T |
T |
T |

F |
T |
F |
F |
T |

F |
F |
T |
F |
T |

p→(p ∧~ q) is false when p is true and q is true.

**Answer: (d)**

**13.**The length of minor axis (along y-axis) of an ellipse of the standard form is 4/√3. If this ellipse touches the line x + 6y = 8, then its eccentricity is:

a. (1/2)(√5/3)

b. (1/2)√(11/3)

c. √(5/6)

d. (1/3)√(11/3)

If 2b = 4/√3

b = 2/√3

Comparing y = (-x/6)+(8/6) with y = mx ±√(a^{2}m^{2}+b^{2})

m = -1/6 and a^{2}m^{2} + b^{2 }= 16 / 9

(a^{2}/36) + (4/3) = 16/9

(a^{2}/36) = (16/9) – (4/3)

a^{2} = 16

e = √(1-b^{2}/a^{2})

e = √(11/12)

**Answer: (b)**

**14.**If z be a complex number satisfying |Re(z)|+|Im(z)| = 4, then |𝑧| cannot be:

a. √7

b. √(17/2)

c. √10

d. √8

|Re(z)|+|Im(z)| = 4

Let z = x+iy

⇒|x|+|y| = 4

z lies on the rhombus.

Maximum value of |z| = 4 when z = 4, -4, 4i, -4i

Minimum value of |z| = 2√2 when z = 2±2i, ±2+2i

|z|∈[2√2 ,4]

|z|∈[√8 ,√16]

|z| ≠ √7

**Answer: (a)**

**15.**If

a. y(1+x) = 1

b. x(1-y) = 1

c. y(1-x) = 1

d. x(1+y) = 1

y = 1+cos^{2} θ +cos^{4} θ + …

1/y = sin^{2} θ

x = 1-tan^{2} θ + tan^{4} θ-…

x +(1/y) = 1

y(1-x) = 1

**Answer: **(c)

**16.**If

a. √3e

b. (1/2)√3e

c. √2e

d. e/√2

Let y = vx

dy/dx = v+x(dv/dx)

v+x(dv/dx) = vx^{2}/(x^{2}(1+v^{2}) = v/(1+v^{2})

x(dv/dx) = -v^{3}/(1+v^{2})

(1/x)dx = (-1/v^{3})-(1/v)dv

log x = (1/2v^{2})-log v + log c

log x = (x^{2}/2y^{2} )-log y + log x + log c

log c + (x^{2}/2y^{2}) – log y = 0

y(1) = 1

log C + (1/2)-0 = 0

log c = −1/2

y(x) = e

-(1/2) + (x^{2}/2e^{2})-1 = 0

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

x = ±√3e

**Answer: (a)**

**17.**If one end of focal chord AB of the parabola y

^{2 }= 8x is at A(1/2,-2) , then the equation of tangent to it at B is

a. x+2y+8 = 0

b. 2x-y-24 = 0

c. x-2y+8 = 0

d. 2x+y-24 = 0

Let PQ be the focal chord of the parabola y^{2} = 8x

P(t_{1}) = (2t_{1}^{2}, 4t_{1}) and Q(t_{2}) = (2t_{2}^{2}, 4t_{2})

t_{1}t_{2} = -1

(1/2, -2) is one of the ends of the focal chord of the parabola

Let (1/2, -2) = (2t_{2}^{2}, 4t_{2})

Other end of focal chord will have parameter t_{1}= 2

The co-ordinate of the other end of the focal chord will be (8,8).

The equation of the tangent will be given as 8y = 4(x+8)

⇒2y−x = 8

**Answer:(c)**

**18.**Let a

_{n}be the n

^{th }term of a G.P. of positive terms. If

a. 300

b. 175

c. 225

d. 150

a_{n} is a positive term of GP.

Let GP be a, ar, ar^{2},…..

_{3}+a

_{5}+a

_{7}+…+a

_{201}

200 = ar^{2} + ar^{4} + …+ar^{201}

200 =

Also,

100 = a_{2} + a_{4}+ …+a_{200}

100 = ar + ar^{3} + …ar^{199}

100 =

From (i) and (ii), r = 2

And

a_{2} + a_{3}+a_{4} ….+a_{200}+a_{201} = 300

ar + ar^{2}+ar^{3}+…ar^{200} = 300

r(a+ar+ar^{2}+…+ar^{199}) = 300

2(a_{1}+a_{2}+a_{3}+…+a_{200}) = 300

**Answer: (d)**

**19.**A random variable X has the following probability distribution:

X |
1 |
2 |
3 |
4 |
5 |

P(X) |
K |
2K |
K |
2K |
5K |

Then P(X>2) is equal to:

a. 7/12

b. 23/36

c. 1/36

d. 1/6

We know that

K^{2} +2K+K+2K+5K^{2} = 1

6K^{2}+5K-1 = 0

6K (K+1)-(K+1) = 0

(K+1)(6K-1) = 0

K = -1 or K = 1/6

K cannot be negative.

P(X>2) = P(X = 3) + P(X = 4)+ P(X = 5)

= K+2K + 5K^{2}

= (1/6) + 2(1/6) +5(1/6)^{2}

= 23/36

**Answer: **(b)

**20.**If

a. (-1, 1-tan θ)

b. (-1, 1+tan θ)

c. (1, 1+tan θ)

d. (1, 1-tan θ)

Let

Let tan θ= k

sec^{2} θ = dk

I = (2/(1+k))-1)dk

I = 2 lnǀ1+kǀ -k+c

I = 2 lnǀ1+tan θǀ -tan θ +c

Given I = tan θ+ 2 log f(θ) + c

λ= -1, f(θ) = ǀ1+ tan θǀ

**Answer: (b)**

**21.**Let

**Answer : (30) **

**22.**If C

_{r}=

^{25}C

_{r}and C

_{0}+5⋅C

_{1}+9⋅C

_{2}+⋯+101. C

_{25}=2

^{25}.k then k is equal to

S = ^{25}C_{0} + 5^{25}C_{1} + 9^{25}C_{2} + ….+97^{25}C_{24} +101^{25}C_{25} = 2^{25}k ..(i)

Reverse and apply property ^{n}C_{r} = ^{n}C_{n-r} in all coefficients

S = 101^{25}C_{0} + 97^{25}C_{1}+…+5^{25}C_{24}+^{25}C_{25} …(ii)

Adding (i) and (ii)

2S = 102[^{25}C_{0}+^{25}C_{1}+…+^{25}C_{5}]

S = 51×2^{25}

k = 51

**Answer: (51) **

**23.**If the curves x

^{2}-6x +y

^{2}+8 = 0 and x

^{2}-8y+y

^{2}+16-k = 0, (k>0) touch each other at a point, then the largest value of k is

Two circles touch each other if C_{1}C_{2} = ǀr_{1}± r_{2}ǀ

√k +1 = 5 or ǀ√k-1ǀ = 5

k = 16 or 36

Maximum value of k is 36.

**Answer: (36) **

**24.**The number of terms common to the A.P.’s 3,7,11,…407 and 2,9,16,…709 is .

First common term is 23

Common difference = LCM(7,4) = 28

23+(n−1)28 ≤ 407

n−1 ≤ 13.71

n = 14

**Answer: (14)**

**25.**If the distance between the plane, 23x-10y-2z+48 = 0 and the plane containing the lines

We find the point of intersection of the two lines, and the distance of given plane from the two lines is the distance of plane from the point of intersection.

(2p-1, 4p+3, 3p-1) = (2q-3, 6q-2, λq+1)

p = -1/2 and q = 1/2

λ= -7

Point of intersection is (-2,1,-5/2)

k = 3

**Answer: **(3)

## Video Lessons – January 9 Shift 2 Maths

## JEE Main 2020 Maths Paper With Solutions Jan 9 Shift 2

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