**Question 1.**The area of the region, enclosed by the circle x

^{2}+ y

^{2}= 2 which is not common to the region bounded by the parabola y

^{2}= 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

Required area = πr^{2} –

Area =

= 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) 5

^{6}

b) (½)(6!)

c) 6!

d) (5/2) 6!

Selecting all 5 digits = ^{5}C_{5} = 1 way

Now, we need to select one more digit to make it a 6 digit number = ^{5}C_{1} = 5 ways

Total number of permutations =

Total numbers = ^{5}C_{5}×^{5}C_{1}×(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}

ƩXP(X) =

= 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

2x^{2} + 2y^{2} – 2x + y = 4x^{2} + 4y^{2} + 4y + 1

x^{2} + y^{2} + 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

a)

b)

c)

d)

f(a + b + 1 – x) = f(x) …(i)

f(a + b – x) = f(x+1) …(ii)

From (i) and (ii)

Adding (iii) and (iv)

**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

x^{2}/a^{2} + y^{2}/b^{2} = 1

a>b

Now 2ae = 6

2a/e = 12

ae = 3 and a/e = 6

a^{2} = 18

a^{2}e^{2} = c^{2} = a^{2}-b^{2} = 9

b^{2} = 9

length of latus rectum = 2b^{2}/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 49

^{125}+ 49

^{124}+ ⋯ + 49

^{2}+ 49 + 1, is

a) 32

b) 60

c) 65

d) 63

1 + 49 + 49^{2} + …. + 49^{125} = (49^{126} – 1)/(49 – 1)

= (49^{63} + 1)( 49^{63} – 1)/48

= (49^{63} + 1)((1 + 48)^{63}-1)/48

= (49^{63} + 1)(1 +48)^{63} -1)/48

= (49^{63} + 1)(1 +48 *I*-1)/48: where *I* is an integer

= (49^{63} + 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

**Question 10.**If y(α) =

a) -1/4

b) 4/3

c) 4

d) -4

y(α) =

y(α) = √(1+cotα )^{2}

y(α) = -1 – cotα

dy/dα = 0 + cosec^{2}α

dy/dα = cosec^{2} 5π/6

dy/dα = 4

**Answer:** (c)

**Question 11.**If y = mx + 4 is a tangent to both the parabolas, y

^{2 }= 4x and x

^{2 }= 2by, then b is equal to

a) -64

b) 128

c) -128

d) -32

Any tangent to the parabola y^{2} = 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 x^{2} = 2by

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

Or 2x^{2} – bx – 16b = 0

D = 0

b^{2 }+ 128b = 0

b = 0 (not possible)

b = -128

**Answer:** (c)

**Question 12.**Let 𝛼 be a root of the equation x

^{2 }+ x+ 1= 0 and the matrix A =

^{31}is equal to

a) A

b) A

^{2}

c) A

^{3}

d) I

_{3}

The roots of equation 𝑥^{2} + 𝑥 + 1=0 are complex cube roots of unity.

= ω or ω^{2}

=

A^{2} =

A^{4} = I

A^{28} = 𝐼

Therefore, we get

A^{31} = A^{28} A^{3}

A^{31} = IA^{3}

A^{31} = A^{3}

**Answer: **(c)

**Question 13.**If g(x) = x

^{2}+ x – 1 and (gof)(x) = 4x

^{2}– 10x + 5, then f(5/4) is equal to

a) -3/2

b) -1/2

c) 1/2

d) 3/2

g(x) = x^{2} + x – 1

gof(x) = 4x^{2} – 10x + 5

g(f(x) = 4x^{2} – 10x + 5

f^{2}(x) + f(x) – 1 = 4x^{2} – 10x + 5

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

t^{2} + 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) tan

^{2}𝑥 − √2 𝜆 tan 𝑥 = 1 − 𝑘, where (𝑘 ≠ −1) and are real numbers. If tan

^{2}(α + β) = 50, then value of 𝜆 is

a) 5√2

b) 10√2

c) 10

d) 5

(k+1) tan^{2}x – √2 tan x = 1 – k

tan^{2}(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)

2λ^{2}/4 = 50

λ^{2} = 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)

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

_{e }2

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 + log_{2}2

**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 − 4𝑑^{2})(25 − 𝑑^{2}) = 504

4𝑑^{4} − 125𝑑^{2} + 121 = 0

4𝑑^{4 }− 4𝑑^{2} − 121𝑑^{2 }+ 121 = 0

𝑑^{2 }= 1 𝑜𝑟 𝑑^{2} = 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

R_{2} R_{2} – R_{1}

R_{3} R_{3} – R_{1}

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

⇒ 3𝑏𝑐 − 2𝑎𝑐 − 3𝑎𝑏 + 2𝑎^{2} − [4𝑏𝑐 − 4𝑎𝑐 − 2𝑎𝑏 + 2𝑎^{2}] = 0

⇒ −𝑏𝑐 + 2𝑎𝑐 − 𝑎𝑏 = 0

⇒ 𝑎𝑏 + 𝑏𝑐 = 2𝑎𝑐

1/c + 1/a = 2/b

**Question 21.**

Put 3^{x/2 }= t

**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

σ^{2} = (n^{2}-1)/12 = 10

n = 11

Variance of (2, 4, 6…) = 4×variance of (1,2,3,4…) = 4×(m^{2}-1)/12

= (m^{2}-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 + x^{2} + x^{3} … . +x^{2𝑛})(1 − x + x^{2} − x^{3} … . + x^{2𝑛}) is 61, then 𝑛 is equal to

Let (1 + x + x^{2} + ⋯ + x^{2𝑛})(1 − x + x^{2} − ⋯ + x^{2𝑛}) = a𝜊 + a1𝑥 + a2𝑥^{2} + ⋯

Put x = 1

2n + 1 = a_{o} + a_{1} + a_{2} + a_{3 } + . . . . . . . . . . . . (1)

Put x = −1

2n + 1 = a_{o} – a_{1} + a_{2} – a_{3} +. . . . . . . . . . . . . (2)

Add (1) and (2)

2(2n + 1) = 2(a_{𝜊} + a_{2} + a_{4} + . . . . . . . . . . .

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 =

P = (17/6, 8/3)

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

PQ =

**Answer **(5)

## Video Lessons – January 7 Shift 1 Maths

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

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