 # JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

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### January 7 Shift 1 - Maths

Question 1. The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = 𝑥 and the straight line 𝑦 = 𝑥, is

1. a) (1/3)(12𝜋 − 1)
2. b) (1/6)(12𝜋 − 1)
3. c) (1/3)(6𝜋 − 1)
4. d) (1/6)(24𝜋 − 1)

Solution:

1. Required area = area of the circle – area bounded by given line and parabola Required area = πr2 - $\int_{0}^{1}(y-y^{2})dy$

Area = $2\pi -(\frac{y^{2}}{2}-\frac{y^{3}}{3})_{0}^{1}$

= 2𝜋 - (1/6)

= (1/6)(12𝜋 - 1)

Question 2. Total number of six-digit numbers in which only and all the five digits 1,3,5,7 and 9 appear, is

1. a) 56
2. b) (½)(6!)
3. c) 6!
4. d) (5/2) 6!

Solution:

1. Selecting all 5 digits = 5C5 = 1 way

Now, we need to select one more digit to make it a 6 digit number = 5C1 = 5 ways

Total number of permutations = $\frac{6!}{2!}$

Total numbers = 5C5×5C1×(6!/2!)

= (5/2) 6!

Question 3. An unbiased coin is tossed 5 times. Suppose that a variable 𝑋 is assigned the value 𝑘 when 𝑘 consecutive heads are obtained for 𝑘 = 3, 4, 5, otherwise 𝑋 takes the value -1. The expected value of 𝑋, is

1. a) 1/8
2. b) 3/16
3. c) -1/8
4. d) -3/16

Solution:

1. k = no. of consecutive heads

P(k = 3) = 5/32 {HHHTH, HHHTT, THHHT, HTHHH, TTHHH}

P(k = 4) = 2/32 {HHHHT, THHHH}

P(k = 5) = 1/32 {HHHHH}

$P(\bar{3}\cap \bar{4}\cap \bar{5})= 1-(\frac{5}{32}+\frac{2}{32}+\frac{1}{32}) = \frac{24}{32}$

ƩXP(X) = $(-1\times \frac{24}{32})+(3\times \frac{5}{32})+(4\times \frac{2}{32})+(5\times \frac{1}{32})$

= 1/8

Question 4. If Re (z-1)/(2z + i) = 1, where z = x+iy, then the point (x,y) lies on a

1. a) circle whose centre is at (-1/2, -3/2)
2. b) straight line whose slope is 3/2.
3. c) circle whose diameter is √5/2
4. d) straight line whose slope is -2/3.

Solution:

1. z = x + iy

$\frac{x+iy-1}{2x+2iy+i} = \frac{(x-1)+iy}{2x+i(2y+1)}\left ( \frac{2x-i(2y+1)}{2x-i(2y+1)}\right )$

$\frac{2x(x-1)+y(2y+1)}{4x^{2}+(2y+1)^{2}}= 1$

2x2 + 2y2 - 2x + y = 4x2 + 4y2 + 4y + 1

x2 + y2 + x + (3/2)y + (1/2) = 0

Circle’s centre will be (-1/2, -3/4)

Radius = √[(1/4) + (9/16) - (1/2)] = √5/4

Diameter = √5/2

Question 5. If f(a + b + 1 - x) = f(x) ∀x, where a and b are fixed positive real numbers, then $\frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx$ is equal to

1. a) $\int_{a-1}^{b-1}f(x)dx$
2. b) $\int_{a+1}^{b+1}f(x+1)dx$
3. c) $\int_{a-1}^{b-1}f(x+1)dx$
4. d) $\int_{a+1}^{b+1}f(x)dx$

Solution:

1. f(a + b + 1 - x) = f(x) …(i)

f(a + b - x) = f(x+1) …(ii)

$I = \frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx$ …(iii)

From (i) and (ii)

$I = \frac{1}{(a+b)}\int_{a}^{b}(a+b-x)(f(x+1)+f(x))dx$ …(iv)

$2I = \int_{a}^{b}(f(x)+f(x+1))dx$ Question 6. If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

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

Solution:

1. Let the equation of ellipse be

x2/a2 + y2/b2 = 1

a>b

Now 2ae = 6

2a/e = 12

ae = 3 and a/e = 6

a2 = 18

a2e2 = c2 = a2-b2 = 9

b2 = 9

length of latus rectum = 2b2/a = 2×9/√18 = 3√2

Question 7. The logical statement (p⇒q) ∧ (q ⇒∼ p) is equivalent to

1. a) ∼ p
2. b) p
3. c) q
4. d) ∼ q

Solution:

1.  𝒑 𝒒 𝒑 ⇒ 𝒒 ∼ 𝒑 𝒒 ⇒∼ 𝒑 (𝒑 ⇒ 𝒒) ∧ (𝒒 ⇒∼ 𝒑) T T T F F F T F F F T F F T T T T T F F T T T T

Clearly (𝑝 ⇒ 𝑞) ∧ (𝑞 ⇒∼ 𝑝) is equivalent to ∼ 𝑝.

Question 8. The greatest positive integer 𝑘, for which 49𝑘 + 1 is a factor of the sum 49125 + 49124 + ⋯ + 492 + 49 + 1, is

1. a) 32
2. b) 60
3. c) 65
4. d) 63

Solution:

1. 1 + 49 + 492 + …. + 49125 = (49126 - 1)/(49 - 1)

= (4963 + 1)( 4963 - 1)/48

= (4963 + 1)((1 + 48)63-1)/48

= (4963 + 1)(1 +48)63 -1)/48

= (4963 + 1)(1 +48 I-1)/48: where I is an integer

= (4963 + 1)I

Greatest positive integer is k = 63

Question 9. A vector 𝑎⃗ = 𝛼𝑖̂ + 2𝑗̂ + 𝛽𝑘̂ (𝛼, 𝛽 ∈ 𝑹) lies in the plane of the vectors, ⃗𝑏 = 𝑖̂ + 𝑗̂ and 𝑐⃗ = 𝑖̂ − 𝑗̂ + 4𝑘̂. If 𝑎⃗ bisects the angle between 𝑏⃗ and 𝑐⃗, then

1. a) 𝑎⃗. 𝑖̂ + 3 = 0
2. b) 𝑎⃗. 𝑘̂ + 4 = 0
3. c) 𝑎⃗. 𝑖̂ + 1 = 0
4. d) 𝑎⃗. 𝑘̂ + 2 = 0

Solution:

1. Question 10. If y(α) = $\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}$ where α ∈ (3π/4, π) then dy/dα at α = 5π/6 is

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

Solution:

1. y(α) = $\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}$

$y(\alpha )= \sqrt{2\frac{1}{\sin \alpha \cos \alpha \times \frac{1}{\cos ^{2}\alpha }}+\frac{1}{\sin ^{2}\alpha }}$

$y(\alpha )= \sqrt{2\cot \alpha +cosec^{2} \alpha }$

y(α) = √(1+cotα )2

y(α) = -1 - cotα

dy/dα = 0 + cosec2α

dy/dα = cosec2 5π/6

dy/dα = 4

Question 11. If y = mx + 4 is a tangent to both the parabolas, y2 = 4x and x2 = 2by, then b is equal to

1. a) -64
2. b) 128
3. c) -128
4. d) -32

Solution:

1. Any tangent to the parabola y2 = 4x is y = mx + a/m

Comparing it with y = mx + 4, we get 1/m = 4

So m = ¼.

Equation of tangent becomes y = (x/4) + 4

y = (x/4) + 4 is a tangent to x2 = 2by

x2 = 2b{(x/4)+4}

Or 2x2 - bx - 16b = 0

D = 0

b2 + 128b = 0

b = 0 (not possible)

b = -128

Question 12. Let 𝛼 be a root of the equation x2 + x+ 1= 0 and the matrix A = $\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}$ then the matrix A31 is equal to

1. a) A
2. b) A2
3. c) A3
4. d) I3

Solution:

1. The roots of equation 𝑥2 + 𝑥 + 1=0 are complex cube roots of unity.

= ω or ω2

$A= \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}$

= $\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}$

A2 = $\frac{1}{{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}$

$A^{2}=\frac{1}{3}\begin{bmatrix} 3 & 0 & 0\\ 0 & 0 & 3\\ 0 & 3 & 0 \end{bmatrix}$

$A^{2}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$

$A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$

$A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}= I$

A4 = I

A28 = 𝐼

Therefore, we get

A31 = A28 A3

A31 = IA3

A31 = A3

Question 13. If g(x) = x2 + x - 1 and (gof)(x) = 4x2 - 10x + 5, then f(5/4) is equal to

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

Solution:

1. g(x) = x2 + x - 1

gof(x) = 4x2 - 10x + 5

g(f(x) = 4x2 - 10x + 5

f2(x) + f(x) - 1 = 4x2 - 10x + 5

Putting x = 5/4 and f(5/4) = t

t2 + t + ¼ = 0

t = -1/2 or f(5/4) = -1/2

Question 14. Let 𝛼 and 𝛽 are two real roots of the equation (𝑘 + 1) tan2 𝑥 − √2 𝜆 tan 𝑥 = 1 − 𝑘, where (𝑘 ≠ −1) and are real numbers. If tan2(α + β) = 50, then value of 𝜆 is

1. a) 5√2
2. b) 10√2
3. c) 10
4. d) 5

Solution:

1. (k+1) tan2x - √2 tan x = 1 - k

tan2(α + β) = 50

∵ tan α and tan β are the roots of the given equation

Now

tan α + tanβ = √2/(k+1),

tanα tan β= (k -1)/(k+1)

$\left ( \frac{\frac{\sqrt{2 }\lambda}{k+1}}{1-\frac{k-1}{k+1}} \right )^{2} = 50$

2/4 = 50

λ2 = 100

λ = 10, λ = -10

Question 15. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and 𝑅 be any point (2, 1, 6). Then the image of 𝑅 in the plane 𝑃 is:

1. a) (6, 5, 2)
2. b) (6, 5, −2)
3. c) (4,3, 2)
4. d) (3,4, −2)

Solution:

1. Points A(2, 1, 0), B(4, 1, 1) C(5, 0, 1)

$\vec{AB } = (2,0,1)$

$\vec{AC } = (3,-1,1)$

$\vec{n } = \vec{AB}\times \vec{AC }$

Equation of the plane is x + y -2z = 3….(1)

Let the image of point (2, 1, 6) is (l, m, n)

(l-2)/1 = (m-1)/1 = (n-6)/-2 = -2(-12)/6 = 4

⇒ l = 6, m = 5, n = −2

Hence the image of R in the plane P is (6, 5, −2)

Question 16. Let xk + yk = ak, (a,k>0) and (dy/dx) + (y/x)1/3 = 0, then k is

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

Solution:

1. xk + yk = ak

kxk-1 + kyk-1 (dy/dx) = 0

dy/dx = -(x/y)k-1

dy/dx + (y/x)1-k = 0

1-k = 1/3

k = 2/3

Question 17. Let the function, f:[-7, 0] →R be continuous on [−7, 0] and differentiable on (−7, 0). If f(-7)= −3 and 𝑓(𝑥) ≤ 2, for all 𝑥 ∈ (−7, 0), then for all such functions 𝑓, 𝑓(−1) + 𝑓(0) lies in the interval:

1. a) [-6, 20]
2. b) (-∞, 20]
3. c) (-∞, 11]
4. d) [-3, 11]

Solution:

1. f(-7) = -3 and f’(x) ≤ 2

Applying LMVT in [-7,0], we get

(f(-7) - f(0))/-7 = f’(c) ≤ 2

(-3-f(0))/-7 ≤ 2

f(0) + 3 ≤ 14

f(0) ≤ 11

Applying LMVT in [-7, -1], we get

(f(-7) - f(-1))/(-7 +1) = f’(c) ≤ 2

-3 - f(-1))/-6 = f’(c) ≤ 2

f(-1) + 3 = ≤ 12

f(-1) ≤ 9

Therefore f(-1)+ f(0) ≤ 20

Question 18. If y = y(x) is the solution of the differential equation ey(dy/dx)-1 = ex such that y(0) = 0, then y(1) is equal to

1. a) loge 2
2. b) 2e
3. c) 2 + loge 2
4. d) 1+loge 2

Solution:

1. 𝑒y(𝑦 − 1) = 𝑒𝑥

⇒ dy/dx = 𝑒𝑥−𝑦 + 1

Let x-y = t

1 - dy/dx = dt/dx

So, we can write

⇒ 1 − dt/dx = 𝑒𝑡 + 1

⇒ −𝑒−𝑡 𝑑𝑡 = 𝑑𝑥

⇒ 𝑒−𝑡 = 𝑥 + 𝑐

⇒ 𝑒𝑦−𝑥 = 𝑥 + 𝑐

1 = 0 + 𝑐

⇒ 𝑒𝑦−𝑥 = 𝑥 + 1

at 𝑥 = 1

⇒ 𝑒𝑦−1 = 2

⇒ 𝑦 = 1 + log22

Question 19. Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is -1/2, then the greatest number amongst them is

1. a) 16
2. b) 27
3. c) 7
4. d) 21/2

Solution:

1. Let 5 numbers be 𝑎 − 2𝑑, 𝑎 − 𝑑, 𝑎, 𝑎 + 𝑑, 𝑎 + 2𝑑

5𝑎 = 25

𝑎 = 5

(𝑎 − 2𝑑)(𝑎 − 𝑑) a (𝑎 + 𝑑)(𝑎 + 2𝑑) = 2520

(25 − 4𝑑2)(25 − 𝑑2) = 504

4𝑑4 − 125𝑑2 + 121 = 0

4𝑑4 − 4𝑑2 − 121𝑑2 + 121 = 0

𝑑2 = 1 𝑜𝑟 𝑑2 = 121/4

d = 11/2

For d = 11/2, a + 2d is the greatest term, a + 2d = 5 + 11 = 16.

Question 20. If the system of linear equations

2x + 2ay + az = 0

2x + 3by + bz = 0

2x + 4cy + cz = 0,

where a, b, c ∈ R are non-zero and distinct; has non-zero solution, then

1. a) a + b + c = 0
2. b) a, b, c are in A.P
3. c) 1/a, 1/b, 1/c are in A.P
4. d) a, b, c are in G.P

Solution:

1. $\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c & c \end{vmatrix}=0$

R2 R2 - R1

R3 R3 - R1

$\begin{vmatrix} 2 & 2a &a \\ 0& 3b-2a &b-a \\ 0& 4c-2a & c-a \end{vmatrix}=0$

⇒ (3b − 2𝑎)(𝑐 − 𝑎) − (4𝑐 − 2𝑎)(𝑏 − 𝑎) = 0

⇒ 3𝑏𝑐 − 2𝑎𝑐 − 3𝑎𝑏 + 2𝑎2 − [4𝑏𝑐 − 4𝑎𝑐 − 2𝑎𝑏 + 2𝑎2] = 0

⇒ −𝑏𝑐 + 2𝑎𝑐 − 𝑎𝑏 = 0

⇒ 𝑎𝑏 + 𝑏𝑐 = 2𝑎𝑐

1/c + 1/a = 2/b

Question 21. $\lim_{x\to 2}\frac{3^{x}+3^{3-x}-12}{3^{-\frac{x}{2}}-3^{1-x}}$ is equal to

Solution:

1. Put 3x/2 = t Question 22.If variance of first 𝑛 natural numbers is 10 and variance of first 𝑚 even natural numbers is 16, 𝑚 + 𝑛 is equal to

Solution:

1. For 𝑛 natural number variance is given by

$\sigma ^{2} = \frac{\sum x_{i}^{2}}{n}- (\frac{\sum x_{i}}{n})^{2}$

$\frac{\sum x_{i}^{2}}{n}= \frac{1^{2}+2^{2}+..n^{2}}{n} = \frac{n(n+1)(2n+1)}{6n}$

$\frac{\sum x_{i}}{n}= \frac{1+2+..n}{n} = \frac{n(n+1)}{2n}$

σ2 = (n2-1)/12 = 10

n = 11

Variance of (2, 4, 6…) = 4×variance of (1,2,3,4…) = 4×(m2-1)/12

= (m2-1)/3

= 16

m = 7

Therefore, n + m = 11 + 7 = 18

Question 23. If the sum of the coefficients of all even powers of 𝑥 in the product

(1 + x + x2 + x3 … . +x2𝑛)(1 − x + x2 − x3 … . + x2𝑛) is 61, then 𝑛 is equal to

Solution:

1. Let (1 + x + x2 + ⋯ + x2𝑛)(1 − x + x2 − ⋯ + x2𝑛) = a𝜊 + a1𝑥 + a2𝑥2 + ⋯

Put x = 1

2n + 1 = ao + a1 + a2 + a3 + . . . . . . . . . . . . (1)

Put x = −1

2n + 1 = ao - a1 + a2 - a3 +. . . . . . . . . . . . . (2)

2(2n + 1) = 2(a𝜊 + a2 + a4 + . . . . . . . . . . .

2n + 1 = 61

n = 30

Question 24. Let S be the set of points where the function, (𝑥) = |2 − |𝑥 − 3||, 𝑥 ∈ 𝐑, is not differentiable. Then, the value of ∑𝑥∈S f(f(x)) is equal to

Solution:

1. Solution:

There will be three points 𝑥 = 1, 3, 5 at which f(x) is non-differentiable.

So f(f(1)) + f(f(3)) + f(f(5))

= f(0) + f(2) + f(0)

= 1+1+1

= 3

Question 25. Let A(1,0), B(6,2), C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangle APC, APB and BPC have equal areas, then the length of the line the segment PQ, where Q is the point ( -7/6, -1/3), is

Solution:

1. P is the centroid which is = $\left ( \frac{1+6+\frac{3}{2}}{3} , \frac{0+2+6}{3}\right )$

P = (17/6, 8/3)

Q = (-7/6, -1/3)

PQ = $\sqrt{4^{2}+3^{2}} = 5$

### JEE Main 2020 Maths Paper January 7 Shift 1             