JEE Main 2020 Paper With Solutions Maths Shift 1 - Download PDF

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

Required area = area of the circle – area bounded by given line and parabola

JEE Main 2020 Paper With Solutions Maths Shift 1

Required area = πr² –

\begin{array}{l} \int_{0}^{1} (y - y^{2}) d y \end{array}

Area =

\begin{array}{l} 2 π - (\frac{y^{2}}{2} - \frac{y^{3}}{3})_{0}^{1} \end{array}

= 2𝜋 – (1/6)

= (1/6)(12𝜋 – 1)

Answer:(b)

Question 2. Total number of six-digit numbers in which only and all the five digits 1,3,5,7 and 9 appear, is
a) 5⁶
b) (½)(6!)
c) 6!
d) (5/2) 6!

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 =

\begin{array}{l} \frac{6!}{2!} \end{array}

Total numbers = ⁵C₅×⁵C₁×(6!/2!)

= (5/2) 6!

Answer: (d)

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
a) 1/8
b) 3/16
c) -1/8
d) -3/16

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}

\begin{array}{l} P (\bar{3} \cap \bar{4} \cap \bar{5}) = 1 - (\frac{5}{32} + \frac{2}{32} + \frac{1}{32}) = \frac{24}{32} \end{array}

ƩXP(X) =

\begin{array}{l} (- 1 \times \frac{24}{32}) + (3 \times \frac{5}{32}) + (4 \times \frac{2}{32}) + (5 \times \frac{1}{32}) \end{array}

= 1/8

Answer: (a)

Question 4. If Re (z-1)/(2z + i) = 1, where z = x+iy, then the point (x,y) lies on a
a) circle whose centre is at (-1/2, -3/2)
b) straight line whose slope is 3/2.
c) circle whose diameter is √5/2
d) straight line whose slope is -2/3.

z = x + iy

\begin{array}{l} \frac{x + i y - 1}{2 x + 2 i y + i} = \frac{(x - 1) + i y}{2 x + i (2 y + 1)} (\frac{2 x - i (2 y + 1)}{2 x - i (2 y + 1)}) \end{array}

\begin{array}{l} \frac{2 x (x - 1) + y (2 y + 1)}{4 x^{2} + (2 y + 1)^{2}} = 1 \end{array}

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

Answer: (c)

Question 5. If f(a + b + 1 – x) = f(x) ∀x, where a and b are fixed positive real numbers, then

\begin{array}{l} \frac{1}{(a + b)} \int_{a}^{b} x (f (x) + f (x + 1)) d x \end{array}

is equal to
a)

\begin{array}{l} \int_{a - 1}^{b - 1} f (x) d x \end{array}

b)

\begin{array}{l} \int_{a + 1}^{b + 1} f (x + 1) d x \end{array}

c)

\begin{array}{l} \int_{a - 1}^{b - 1} f (x + 1) d x \end{array}

d)

\begin{array}{l} \int_{a + 1}^{b + 1} f (x) d x \end{array}

f(a + b + 1 – x) = f(x) …(i)

f(a + b – x) = f(x+1) …(ii)

\begin{array}{l} I = \frac{1}{(a + b)} \int_{a}^{b} x (f (x) + f (x + 1)) d x \end{array}

…(iii)

From (i) and (ii)

\begin{array}{l} I = \frac{1}{(a + b)} \int_{a}^{b} (a + b - x) (f (x + 1) + f (x)) d x \end{array}

…(iv)

Adding (iii) and (iv)

\begin{array}{l} 2 I = \int_{a}^{b} (f (x) + f (x + 1)) d x \end{array}

JEE Main 2020 Maths Paper With Solutions Shift 1

Answer: (c)

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
a) 2√3
b) √3
c) 3/√2
d) 3√2

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

Answer: (d)

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

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.

Answer: (a)

Question 8. The greatest positive integer 𝑘, for which 49^𝑘 + 1 is a factor of the sum 49¹²⁵ + 49¹²⁴ + ⋯ + 49² + 49 + 1, is
a) 32
b) 60
c) 65
d) 63

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

Answer: (d)

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

JEE Main 2020 Paper with Solutions Maths Paper Han 7th Shift 2 Solved Questions

Question 10. If y(α) =

\begin{array}{l} \sqrt{2 (\frac{\tan α + \cot α}{1 + \tan^{2} α}) + \frac{1}{\sin^{2} α}} \end{array}

where α ∈ (3π/4, π) then dy/dα at α = 5π/6 is
a) -1/4
b) 4/3
c) 4
d) -4

y(α) =

\begin{array}{l} \sqrt{2 (\frac{\tan α + \cot α}{1 + \tan^{2} α}) + \frac{1}{\sin^{2} α}} \end{array}

\begin{array}{l} y (α) = \sqrt{2 \frac{1}{\sin α \cos α \times \frac{1}{\cos^{2} α}} + \frac{1}{\sin^{2} α}} \end{array}

\begin{array}{l} y (α) = \sqrt{2 \cot α + c o s e c^{2} α} \end{array}

y(α) = √(1+cotα )²

y(α) = -1 – cotα

dy/dα = 0 + cosec²α

dy/dα = cosec² 5π/6

dy/dα = 4

Answer: (c)

Question 11. If y = mx + 4 is a tangent to both the parabolas, y²= 4x and x²= 2by, then b is equal to
a) -64
b) 128
c) -128
d) -32

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

Answer: (c)

Question 12. Let 𝛼 be a root of the equation x²+ x+ 1= 0 and the matrix A =

\begin{array}{l} \frac{1}{\sqrt{3}} [\begin{array}{c} 1 & 1 & 1 \\ 1 & α & α^{2} \\ 1 & α^{2} & α^{4} \end{array}] \end{array}

then the matrix A³¹ is equal to
a) A
b) A²
c) A³
d) I₃

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

= ω or ω²

\begin{array}{l} A = \frac{1}{\sqrt{3}} [\begin{array}{c} 1 & 1 & 1 \\ 1 & α & α^{2} \\ 1 & α^{2} & α^{4} \end{array}] \end{array}

=

\begin{array}{l} \frac{1}{\sqrt{3}} [\begin{array}{c} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}] \end{array}

A² =

\begin{array}{l} \frac{1}{3} [\begin{array}{c} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}] \end{array}

\begin{array}{l} A^{2} = \frac{1}{3} [\begin{array}{c} 3 & 0 & 0 \\ 0 & 0 & 3 \\ 0 & 3 & 0 \end{array}] \end{array}

\begin{array}{l} A^{2} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}] \end{array}

\begin{array}{l} A^{4} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}] [\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}] \end{array}

\begin{array}{l} A^{4} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}] = I \end{array}

A⁴ = I

A²⁸ = 𝐼

Therefore, we get

A³¹ = A²⁸ A³

A³¹ = IA³

A³¹ = A³

Answer: (c)

Question 13. If g(x) = x² + x – 1 and (gof)(x) = 4x² – 10x + 5, then f(5/4) is equal to
a) -3/2
b) -1/2
c) 1/2
d) 3/2

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

Answer: (b)

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
a) 5√2
b) 10√2
c) 10
d) 5

(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)

\begin{array}{l} {(\frac{\frac{\sqrt{2} λ}{k + 1}}{1 - \frac{k - 1}{k + 1}})}^{2} = 50 \end{array}

2λ²/4 = 50

λ² = 100

λ = 10, λ = -10

Answer: (c)

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

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

\begin{array}{l} \vec{A B} = (2, 0, 1) \end{array}

\begin{array}{l} \vec{A C} = (3, - 1, 1) \end{array}

\begin{array}{l} \vec{n} = \vec{A B} \times \vec{A C} \end{array}

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)

Answer: (b)

Question 16. Let x^k + y^k = a^k, (a,k>0) and (dy/dx) + (y/x)^1/3 = 0, then k is
a) 1/3
b) 3/2
c) 2/3
d) 4/3

x^k + y^k = a^k

kx^k-1 + ky^k-1 (dy/dx) = 0

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

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

1-k = 1/3

k = 2/3

Answer: (c)

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:
a) [-6, 20]
b) (-∞, 20]
c) (-∞, 11]
d) [-3, 11]

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

Answer (b)

Question 18. If y = y(x) is the solution of the differential equation e^y(dy/dx)-1 = e^x such that y(0) = 0, then y(1) is equal to
a) log_e2
b) 2e
c) 2 + log_e 2
d) 1+log_e 2

𝑒^y(𝑦^′ − 1) = 𝑒^𝑥

⇒ dy/dx = 𝑒^𝑥−𝑦 + 1

Let x-y = t

1 – dy/dx = dt/dx

So, we can write

⇒ 1 − dt/dx = 𝑒^𝑡 + 1

⇒ −𝑒^−𝑡 𝑑𝑡 = 𝑑𝑥

⇒ 𝑒^−𝑡 = 𝑥 + 𝑐

⇒ 𝑒^𝑦−𝑥 = 𝑥 + 𝑐

1 = 0 + 𝑐

⇒ 𝑒^𝑦−𝑥 = 𝑥 + 1

at 𝑥 = 1

⇒ 𝑒^𝑦−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
a) 16
b) 27
c) 7
d) 21/2

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.

Answer: (a)

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

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

\begin{array}{l} | \begin{array}{c} 2 & 2 a & a \\ 2 & 3 b & b \\ 2 & 4 c & c \end{array} | = 0 \end{array}

R₂ R₂ – R₁

R₃ R₃ – R₁

\begin{array}{l} | \begin{array}{c} 2 & 2 a & a \\ 0 & 3 b - 2 a & b - a \\ 0 & 4 c - 2 a & c - a \end{array} | = 0 \end{array}

⇒ (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.

\begin{array}{l} lim_{x \to 2} \frac{3^{x} + 3^{3 - x} - 12}{3^{- \frac{x}{2}} - 3^{1 - x}} \end{array}

is equal to

JEE Main 2020 Paper With Solution Maths Shift 1

JEE Main 2020 Paper With Solution Maths Shift 1

Put 3^x/2= t

JEE Main Maths 2020 Solved Paper For Shift 1

Answer: (36)

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

For 𝑛 natural number variance is given by

\begin{array}{l} σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\frac{\sum x_{i}}{n})^{2} \end{array}

\begin{array}{l} \frac{\sum x_{i}^{2}}{n} = \frac{1^{2} + 2^{2} + . . n^{2}}{n} = \frac{n (n + 1) (2 n + 1)}{6 n} \end{array}

\begin{array}{l} \frac{\sum x_{i}}{n} = \frac{1 + 2 + . . n}{n} = \frac{n (n + 1)}{2 n} \end{array}

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

Answer: (18)

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

(1 + x + x² + x³ … . +x^2𝑛)(1 − x + x² − x³ … . + x^2𝑛) is 61, then 𝑛 is equal to

Let (1 + x + x² + ⋯ + x^2𝑛)(1 − x + x² − ⋯ + x^2𝑛) = a𝜊 + a1𝑥 + a2𝑥² + ⋯

Put x = 1

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

Put x = −1

2n + 1 = a_o – a₁ + a₂ – a₃ +. . . . . . . . . . . . . (2)

Add (1) and (2)

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

2n + 1 = 61

n = 30

Answer: (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:

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

Answer: (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

P is the centroid which is =

\begin{array}{l} (\frac{1 + 6 + \frac{3}{2}}{3}, \frac{0 + 2 + 6}{3}) \end{array}

P = (17/6, 8/3)

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

PQ =

\begin{array}{l} \sqrt{4^{2} + 3^{2}} = 5 \end{array}

Answer (5)

Video Lessons – January 7 Shift 1 Maths

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JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

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JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1

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