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

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### January 9 Shift 2 - Maths

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

1. a. A - B = [-1,2]
2. b. B - A = R-(-2,5)
3. c. A⋃B = R-(2,5)
4. d. A∩B = (-2,-1)

Solution:

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, ∞)}

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 :

1. a. 965/210
2. b. 945/210
3. c. 945/211
4. d. 965/211

Solution:

1. Total ways to distribute 10 balls in 4 boxes is = 410

Total ways of placing exactly 2 and 3 balls in any two of these boxes is = 4C2×10C5×(5!/(2!3!))×2×25

P(E) = 945/210

3. If x = 2 sin θ - sin 2 θ and y = 2 cos θ - cos 2 θ, θ ∈ [0, 2π], then d2y/dx2 at θ =π is:

1. a. -3/8
2. b. 3/4
3. c. 3/2
4. d. -3/4

Solution:

1. dx/d θ = 2 cos θ -2 cos2 θ

dy/d θ = -2 sin θ + 2 sin2 θ dy/dx = cot (3 θ/2)

d2y/dx2 = -(3/2) cosec2(3 θ/2)(d θ/dx)

$\frac{d^{2}y}{dx^{2}}= \left ( -\frac{3}{2}cosec^{2} \: \frac{3\theta}{2} \right )\frac{1}{(2\cos \theta -2\cos 2\theta )}$

($\frac{d^{2}y}{dx^{2}})_{\theta =\pi }= \frac{3}{8}$

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 :

1. a. 2/5
2. b. 5
3. c. 1
4. d. 1/5

Solution:

1. 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

5. In the expansion of $\left ( \frac{x}{\cos \theta }+\frac{1}{x\sin \theta } \right )^{16}$ , if l1 is the least value of the term independent of x when (π/8) ≤ θ ≤ (π/4) and l2 is the least value of the term independent of x when (π/16) ≤ θ ≤ (π/8), then the ratio l2 : l1 is equal to :

1. a. 16:1
2. b. 8:1
3. c. 1:8
4. d. 1:16

Solution:

1. $T_{r+1} =\: ^{16}C_{r}\left ( \frac{x}{\cos \theta } \right )^{16-r}\left ( \frac{1}{x\sin\theta } \right )^{r}$

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

⇒ r = 8

T9 = $\: ^{16}C_{8}\left ( \frac{1}{\sin \theta \cos \theta } \right )^{8}=\: ^{16}C_{8}2^{8}\left ( \frac{1}{\sin 2\theta } \right )^{8}$

l1 = 16C8 28 at θ = π/4

l2 = $^{16}C_{8}\frac{2^{8}}{\left ( \frac{1}{\sqrt{2}} \right )^{8}}= ^{16}C_{8}2^{12} at\: \theta =\frac{\pi }{8}$

l2/l1 = 16:1

6. Let a,b ∈R, a ≠ 0, such that the equation, ax2-2bx+5 = 0 has a repeated root α, which is also a root of the equation x2-2bx-10 = 0. If β is the root of this equation, then α22 is equal to:

1. a. 24
2. b. 25
3. c. 26
4. d. 28

Solution:

1. ax2-2bx+5 = 0 has both roots as α

2α = 2b/a

α= b/a

And α2 = 5/a

b2 = 5a (a≠0) ..(i)

α+β = 2b

And αβ = -10

α= b/a is also a root of x2 -2bx-10 = 0

b2 -2ab2-10a2 = 0

b2 = 5a

5a-10a2 -10a2 = 0

a = 1/4

b2 = 5/4

α2 = 20

β2 = 5

α22 = 25

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

F(x) = $\int_{1}^{x}t^{2}g(t)dt$ ,where $g(t)=\int_{1}^{t}f(u)du$

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

1. a. a point of inflection.
2. b. a point of local maxima
3. c. a point of local minima
4. d. not a critical point

Solution:

1. F'(x) = x2g(x)

Put x = 1

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

Now F’’(x) = 2xg(x) + g’(x)x2

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.

8. Let [t] denotes the greatest integer ≤ t and $\lim_{x\to0} x\left [ \frac{4}{x} \right ]= A$ . Then the function, f(x)=[x2] sin πx is discontinuous, when x is equal to

1. a. √(A+1)
2. b. √A
3. c. √(A+5)
4. d. √(A+21)

Solution:

1. f(x)=[x2]sin π x

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

f(x) is discontinuous at points where [x2] is discontinuous i.e. x2∈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 $\lim_{x\to0} x\left [ \frac{4}{x} \right ]= A$ (indeterminate form (0×∞))

$4-\lim_{x\to0} (\frac{4}{x}) = A$

A = 4

√(A+5) = 3

√(A+1) = √5

√(A+21) =5

√A = 2

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

9. Let a-2b+c = 1. If f(x) = $\begin{vmatrix} x+a & x+2&x+1 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}$, then

1. a. f(-50) = 501
2. b. f(-50) = -1
3. c. f(50) = 1
4. d. f(-50) = -501

Solution:

1. Given f(x) = $\begin{vmatrix} x+a & x+2&x+1 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}$

a -2b+c = 1

Applying R1 🡪 R1 -2R2+ R3

f(x) = $\begin{vmatrix} a -2b+c& 0&0 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}$

Using a-2b+c = 1

f(x) = (x+3)2 -(x+2)(x+4)

f(x) = 1

f(50) = 1

f(-50) = 1

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 :

1. a. (√3/4) - (1/3)
2. b. (1/3) + (√3/4)
3. c. (1/2) + (√3/4)
4. d. (1/2) - (√ 3/4)

Solution:

1. 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 = $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}(f(x)-g(x))dx$

= $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}(1-x-(x-\frac{1}{2})^{2})dx$

= $[x-\frac{x^{2}}{2}-\frac{1}{3}(x-\frac{1}{2})^{3}]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}$

= (√3/4) - (1/3)

11. The following system of linear equations

7x+6y-2z = 0

3x+4y+2z = 0

x-2y-6z = 0, has

1. a. infinitely many solutions, (x,y,z) satisfying y = 2z
2. b. infinitely many solutions (x,y,z) satisfying x = 2z
3. c. no solution
4. d. only the trivial solution

Solution:

1. 7x+6y−2z = 0

3x+4y+2z = 0

x−2y−6z = 0

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

$\begin{vmatrix} 7 & 6&-2 \\ 3&4 & 2\\ 1 & -2 & -6 \end{vmatrix}=0$

⇒ 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.

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

1. a. F, T
2. b. T, F
3. c. F, F
4. d. T, T

Solution:

1. 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.

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:

1. a. (1/2)(√5/3)
2. b. (1/2)√(11/3)
3. c. √(5/6)
4. d. (1/3)√(11/3)

Solution:

1. If 2b = 4/√3

b = 2/√3

Comparing y = (-x/6)+(8/6) with y = mx ±√(a2m2+b2)

m = -1/6 and a2m2 + b2 = 16 / 9

(a2/36) + (4/3) = 16/9

(a2/36) = (16/9) - (4/3)

a2 = 16

e = √(1-b2/a2)

e = √(11/12)

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

1. a. √7
2. b. √(17/2)
3. c. √10
4. d. √8

Solution:

1. |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

15. If $x = \sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta$ and $y = \sum_{n=0}^{\infty }\cos ^{2n}\theta$ where 0 < θ < π/4, then:

1. a. y(1+x) = 1
2. b. x(1-y) = 1
3. c. y(1-x) = 1
4. d. x(1+y) = 1

Solution:

1. y = 1+cos2 θ +cos4 θ + …

$y =\frac{1}{1-\cos ^{2}\theta }$

1/y = sin2 θ

x = 1-tan2 θ + tan4 θ-…

$x =\frac{1}{1-(-\tan ^{2}\theta) } = \cos ^{2}\theta$

x +(1/y) = 1

y(1-x) = 1

16. If $\frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}$ , y(1) = 1, then a value of x satisfying y(x)= e is:

1. a. √3e
2. b. (1/2)√3e
3. c. √2e
4. d. e/√2

Solution:

1. Let y = vx

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

v+x(dv/dx) = vx2/(x2(1+v2) = v/(1+v2)

x(dv/dx) = -v3/(1+v2)

(1/x)dx = (-1/v3)-(1/v)dv

log x = (1/2v2)-log v + log c

log x = (x2/2y2 )-log y + log x + log c

log c + (x2/2y2) - log y = 0

y(1) = 1

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

log c = −1/2

y(x) = e

-(1/2) + (x2/2e2)-1 = 0

x2/e2 = 3

x = ±√3e

17. If one end of focal chord AB of the parabola y2 = 8x is at A(1/2,-2) , then the equation of tangent to it at B is

1. a. x+2y+8 = 0
2. b. 2x-y-24 = 0
3. c. x-2y+8 = 0
4. d. 2x+y-24 = 0

Solution:

1. Let PQ be the focal chord of the parabola y2 = 8x

P(t1) = (2t12, 4t1) and Q(t2) = (2t22, 4t2)

t1t2 = -1

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

Let (1/2, -2) = (2t22, 4t2)

Other end of focal chord will have parameter t1= 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

18. Let an be the nth term of a G.P. of positive terms. If $\sum_{n=1}^{100}a_{2n+1}= 200$ and $\sum_{n=1}^{100}a_{2n}= 100$

1. a. 300
2. b. 175
3. c. 225
4. d. 150

Solution:

1. an is a positive term of GP.

Let GP be a, ar, ar2,.....

$\sum_{n=1}^{100}a_{2n+1}= 200$ = a3+a5+a7+…+a201

200 = ar2 + ar4 + …+ar201

200 = $\frac{ar^{2}(r^{200}-1)}{r^{2}-1}$ …(i)

Also, $\sum_{n=1}^{100}a_{2n}= 100$

100 = a2 + a4+ …+a200

100 = ar + ar3 + …ar199

100 = $\frac{ar(r^{200}-1)}{r^{2}-1}$ ..(ii)

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

And $\sum_{n=1}^{100}a_{2n+1} + \sum_{n=1}^{100}a_{2n}= 300$

a2 + a3+a4 ….+a200+a201 = 300

ar + ar2+ar3+…ar200 = 300

r(a+ar+ar2+…+ar199) = 300

2(a1+a2+a3+…+a200) = 300

$\sum_{n=1}^{200}a_{n} = 150$

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

 X 1 2 3 4 5 P(X) K2 2K K 2K 5K2

Then P(X>2) is equal to:

1. a. 7/12
2. b. 23/36
3. c. 1/36
4. d. 1/6

Solution:

1. We know that $\sum_{X=1}^{5}P(X) = 1$

K2 +2K+K+2K+5K2 = 1

6K2+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 + 5K2

= (1/6) + 2(1/6) +5(1/6)2

= 23/36

20. If $\int \frac{d\theta }{\cos ^{2}\theta (\tan 2\theta +\sec 2\theta )}= \lambda \tan \theta +2\: log_{e}\left | f(\theta ) \right |+C$ where C is constant if integration, then the ordered pair (λ, f(θ)) is equal to:

1. a. (-1, 1-tan θ)
2. b. (-1, 1+tan θ)
3. c. (1, 1+tan θ)
4. d. (1, 1-tan θ)

Solution:

1. Let $I=\int \frac{d \theta}{\cos ^{2} \theta(\sec 2 \theta+\tan 2 \theta)}$

$I=\int \frac{\sec ^{2} \theta d \theta}{\left(\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}\right)+\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)}$

$I=\int \frac{\left(1-\tan ^{2} \theta\right)\left(\sec ^{2} \theta\right) d \theta}{(1+\tan \theta)^{2}}$

Let tan θ= k

sec2 θ = dk

$I=\int \frac{\left(1-k^{2}\right)}{(1+k)^{2}} d k=\int \frac{(1-k)}{(1+k)} d k$

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

21. Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be three vectors such that $\vec{a}=\sqrt{3}$ , $\vec{b}=5$ , $\vec{b}.\vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$ is π/3. If $\vec{a}$ is perpendicular to $\vec{b}\times \vec{c}$ , then $\left | \vec{a}\times (\vec{b}\times \vec{c} )\right |$ is equal to

Solution:

1. $\left | \vec{a}\times (\vec{b}\times \vec{c} )\right | = \left | \vec{a} \right |\left | \vec{b} \times \vec{c}\right |\sin \theta$ where θ is the angle between $\vec{a}$ and $\vec{b}\times \vec{c}$. $\left | \vec{c} \right | = 4$

$\left |\vec{a}\times(\vec{b}\times \vec{c}) \right | = 30$

22. If Cr = 25Cr and C0+5⋅C1+9⋅C2 +⋯+101. C25 =225 .k then k is equal to

Solution:

1. S = 25C0 + 525C1 + 925C2 + ….+9725C24 +10125C25 = 225k ..(i)

Reverse and apply property nCr = nCn-r in all coefficients

S = 10125C0 + 9725C1+…+525C24+25C25 …(ii)

2S = 102[25C0+25C1+…+25C5]

S = 51×225

k = 51

23. If the curves x2 -6x +y2 +8 = 0 and x2-8y+y2+16-k = 0, (k>0) touch each other at a point, then the largest value of k is

Solution:

1. Two circles touch each other if C1C2 = ǀr1± r2ǀ

√k +1 = 5 or ǀ√k-1ǀ = 5

k = 16 or 36

Maximum value of k is 36.

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

Solution:

1. First common term is 23

Common difference = LCM(7,4) = 28

23+(n−1)28 ≤ 407

n−1 ≤ 13.71

n = 14

25. If the distance between the plane, 23x-10y-2z+48 = 0 and the plane containing the lines $\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$ and $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }$ , λ∈ R is equal to $\frac{k}{\sqrt{633}}$ , then k is equal to:

Solution:

1. 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)

$\frac{k}{\sqrt{633}}=\left|\frac{-46-10+5+48}{\sqrt{633}}\right|$

k = 3

### JEE Main 2020 Maths Paper January 9 Shift 2               