**1**. Let 𝐴 and 𝐵 be two events such that the probability that exactly one of them occurs is 2/5 and the probability that 𝐴 or 𝐵 occurs is 1/2, then the probability of both of them occur together is

a. 0.10

b. 0.20

c. 0.01

d. 0.02

P(𝐴 or 𝐵) = 𝑃(𝐴 ∪ 𝐵) = 1/2

P(exactly one of 𝐴 or 𝐵)= 2/5

⇒ P(𝐴) − 𝑃(𝐴 ∩ 𝐵) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 2/5

⇒ P(𝐴 ∪ 𝐵) − 𝑃(𝐴 ∩ 𝐵) = 2/5

⇒ P(𝐴 ∩ 𝐵) = 1/10

**Answer: (𝑎) **

**2. Let 𝑆 be the set of all real roots of the equation, 3**

^{x}(3^{x}− 1) + 2 = |3^{x}− 1| + |3^{x}− 2|. Then S:a. is a singleton.

b. is an empty set.

c. contains at least four elements

d. contains exactly two elements.

3^{𝑥} (3^{𝑥} − 1) + 2 = |3 ^{𝑥} − 1| + |3^{𝑥} − 2|

Let 3^{𝑥} = 𝑡

t(𝑡 − 1) + 2 = |𝑡 − 1| + |𝑡 − 2|

⇒ 𝑡 ^{2} − 𝑡 + 2 = |𝑡 − 1| + |𝑡 − 2|

We plot 𝑡 ^{2} − 𝑡 + 2 and |𝑡 − 1| + |𝑡 − 2|

As 3 ^{𝑥} is always positive, therefore only positive values of 𝑡 will be the solution.

Therefore, we have only one solution.

**Answer: (𝑎) **

**3. The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:**

a. 4.01

b. 3.99

c. 3.98

d. 4.02

Mean = 10 ⇒ ∑ 𝑥_{𝑖}/20 = 10 ⇒ ∑ 𝑥_{𝑖} = 200

Variance = 4 ⇒ ∑ 𝑥_{𝑖} ^{2}/20 − 100 = 4 ⇒ ∑ 𝑥_{𝑖} ^{2} = 2080

New mean = (200−9+11) / 20 = 202/20 = 10.1

New variance = (2080−81+121)/ 20 − (10.1) ^{2}

= 106 − 102.01 = 3.99

**Answer: (𝑏) **

**4.**Let 𝑎⃗ = 𝑖̂− 2𝑗̂+ 𝑘̂, 𝑏⃗ = 𝑖̂− 𝑗̂+ 𝑘̂ be two vectors. If 𝑐⃗ is a vector such that 𝑏⃗× 𝑐⃗= 𝑏⃗× 𝑎⃗and 𝑐⃗. 𝑎⃗= 0 then 𝑐⃗. 𝑏⃗ is equal to:

a. 1/2

b. – 3/2

c. – 1/2

d. −1

𝑎⃗ = 𝑖̂ − 2𝑗̂ + 𝑘̂

𝑏⃗ = 𝑖̂ − 𝑗̂ + 𝑘̂

𝑏⃗× 𝑐⃗= 𝑏⃗× 𝑎⃗

⇒ 𝑎⃗ × ( 𝑏⃗ × 𝑐⃗ ) = 𝑎⃗× (𝑏⃗× 𝑎⃗)

⇒ (𝑎⃗. 𝑐⃗ ) 𝑏⃗ − (⃗𝑎⃗. 𝑏⃗)𝑐⃗= (𝑎⃗. 𝑎⃗)𝑏⃗ − (𝑎⃗. 𝑏⃗)𝑎⃗

⇒ − (⃗𝑎⃗. 𝑏⃗)𝑐⃗= (𝑎⃗. 𝑎⃗)⃗𝑏⃗ − (𝑎⃗. 𝑏⃗)𝑎⃗

⇒ −4𝑐⃗= 6(𝑖̂− 𝑗̂+ 𝑘̂) − 4(𝑖̂− 2𝑗̂+ 𝑘̂)

⇒ 𝑐⃗ = − 1/2 (𝑖̂ + 𝑗̂ + 𝑘̂)

∴ 𝑏⃗. 𝑐⃗= − ½

**Answer: (𝑐) **

**5.**Let 𝑓: (1,3) → 𝑅 be a function defined by 𝑓(𝑥) = 𝑥[𝑥] /(𝑥

^{2}+1) , where [𝑥] denotes the greatest integer ≤ 𝑥. Then the range of 𝑓 is:

a. ( 2/5 , 3/5 ] ∪ ( 3/4 , 4/5 )

b. ( 2/5 , 4/5 ]

c. ( 3/5 , 4/5 )

d. ( 2/5 , 1/2 ) ∪ ( 3/5 , 4/5 ]

f(𝑥) = 𝑥[𝑥]/(𝑥^{2}+1)

**Answer: (𝑑) **

**6.**If 𝛼 and 𝛽 be the coefficients of 𝑥

^{4}and 𝑥

^{2 }respectively in the expansion of

a. 𝛼 + 𝛽 = −30

b. 𝛼 − 𝛽 = −132

c. 𝛼 + 𝛽 = 60

d. 𝛼 − 𝛽 = 60

= 2[^{6}𝐶_{0}𝑥 ^{6} + ^{6}𝐶_{2}𝑥 ^{4} (𝑥^{2} − 1) + ^{6}𝐶_{4}𝑥 ^{2}(𝑥 ^{2} − 1) ^{2} + ^{6}𝐶_{6} (𝑥^{2} − 1)^{3}]

= 2[32𝑥 ^{6} − 48𝑥 ^{4} + 18𝑥 ^{2} − 1]

⇒ 𝛼 = −96, 𝛽 = 36

⇒ 𝛼 − 𝛽 = −132

**Answer: (𝑏) **

**7.**If a hyperbola passes through the point (10, 16) and it has vertices at (±6, 0), then the equation of the normal at 𝑃 is:

a. 3𝑥 + 4𝑦 = 94

b. 𝑥 + 2𝑦 = 42

c. 2𝑥 + 5𝑦 = 100

d. 𝑥 + 3𝑦 = 58

Vertex of hyperbola is (±𝑎, 0) ≡ (±6, 0) ⇒ 𝑎 = 6

Equation of hyperbola is (𝑥 ^{2}/𝑎^{2}) – (𝑦 ^{2} / 𝑏^{2}) = 1

⇒ (𝑥 ^{2}/36) – (𝑦 ^{2}/𝑏 ^{2}) = 1

As 𝑃 (10,16) lies on the parabola.

(100/36) – (256/ 𝑏^{2}) = 1

⇒ 64/36 = 256/𝑏^{2}

⇒ 𝑏^{2} = 144

Equation of hyperbola becomes (𝑥 ^{2}/36) – (𝑦 ^{2}/144) = 1

Equation of normal is (𝑎 ^{2}𝑥/𝑥_{1}) + (𝑏 ^{2}𝑦 /𝑦_{1}) = 𝑎 ^{2} + 𝑏 ^{2}

⇒ (36𝑥/10) + (144𝑦/16) = 180

⇒ (𝑥/50) + (𝑦 /20) = 1

⇒ 2𝑥 + 5𝑦 = 100

**Answer: (𝑐) **

**8.**

**is equal to:**

a. 0

b. 1/10

c. – 1/10

d. − 1/5

Applying L’Hospital’s Rule:

=

**Answer: (a) **

**9**. If a line, 𝑦 = 𝑚𝑥 + 𝑐 is a tangent to the circle, (𝑥 − 3)

^{2}+ 𝑦

^{2}= 1 and it is perpendicular to a line 𝐿

_{1}, where 𝐿

_{1}is the tangent to the circle, 𝑥

^{2}+ 𝑦

^{2}= 1 at the point ( 1 /√2 , 1/√2 ); then:

a. 𝑐

^{2}+ 7𝑐 + 6 = 0

b. 𝑐

^{2}− 6𝑐 + 7 = 0

c. 𝑐

^{2}− 7𝑐 + 6 = 0

d. 𝑐

^{2}+ 6𝑐 + 7 = 0

For circle, 𝑥 ^{2} + 𝑦 ^{2} = 1

2𝑥 + 2𝑦𝑦 ′ = 0 ⇒ 𝑦 ′ = − 𝑥/y

Slope of tangent to 𝑥 ^{2} + 𝑦 ^{2} = 1 at ( 1/√2 , 1/√2 ) = −1

⇒Slope of tangent to (𝑥 − 3) ^{2} + 𝑦 ^{2} = 1 is 1 ⇒ 𝑚 = 1

Tangent to (𝑥 − 3) ^{2} + 𝑦 ^{2} = 1 is 𝑦 = 𝑥 + 𝑐

Perpendicular distance of tangent 𝑦 = 𝑥 + 𝑐 from centre (3, 0) is equal to radius = 1

| (3 + 𝑐)/√2 | = 1

⇒ 𝑐 + 3 = ±√2

⇒ 𝑐^{2} + 6𝑐 + 9 = 2

⇒ 𝑐^{2} + 6𝑐 + 7 = 0

**Answer: (𝑑) **

**10.**Let 𝛼 = (−1+𝑖√3)/2 . If

a. 𝑥

^{2}+ 101𝑥 + 100 = 0

b. 𝑥

^{2}+ 102𝑥 + 101 = 0

c. 𝑥

^{2}− 102𝑥 + 101 = 0

d. 𝑥

^{2 }− 101𝑥 + 100 = 0

𝛼 = (−1+𝑖√3)/2 = 𝜔

⇒ 𝑎 = (1 + 𝛼)[1 + 𝛼 ^{2} + 𝛼 ^{4} + ⋯ . +𝛼 ^{200}]

⇒ 𝑎 = (1 + 𝛼) [ (1 − (𝛼 ^{2} ) ^{101})/(1 − 𝛼 ^{2}) ]

⇒ 𝑎 = [ (1 − (𝜔 ^{2} ) ^{101} )/(1 – 𝜔) ] = [ (1 – 𝜔)/( 1 – 𝜔) ] = 1

^{3}+ 𝛼

^{6}+ ⋯ . . + 𝛼

^{300}

⇒ 𝑏 = 1 + 𝜔 ^{3} + 𝜔 ^{6}+. . . . +𝜔^{300}

⇒ 𝑏 = 101

Required equation is 𝑥 ^{2} −102𝑥 + 101 = 0

**Answer: (𝑐) **

**11.**The mirror image of the point (1, 2, 3) in a plane is (− 7/3, − 4/3 , − 1/ 3 ). Which of the following points lies on this plane?

a. (1, −1, 1)

b. (−1, −1, 1)

c. (1, 1, 1)

d. (−1, −1, −1)

Image of point P(1, 2, 3) w.r.t. a plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 is 𝑄 (− 7/3 , − 4/3 , − 1/3 )

Direction ratios of 𝑃𝑄: − 10/3, − 10/3 , − 10/3 = 1, 1, 1

Direction ratios of normal to plane is 1, 1, 1

Mid-point of 𝑃𝑄 lies on the plane

∴ The mid-point of 𝑃𝑄 = (− 2/3 , 1/3 , 4/3 )

∴ Equation of plane is 𝑥 + (2/3) + 𝑦 – (1/3) + 𝑧 – (4/3) = 0

⇒ 𝑥 + 𝑦 + 𝑧 = 1

(1, −1, 1) satisfies the equation of the plane.

**Answer: (𝑎) **

**12.**The length of the perpendicular from the origin, on the normal to the curve, 𝑥

^{2}+ 2𝑥𝑦 − 3𝑦

^{2}= 0 at the point (2, 2) is:

a. 2

b. 2√2

c. 4√2

d. √2

Given curve: 𝑥 ^{2} + 2𝑥𝑦 − 3𝑦 ^{2} = 0

⇒ 𝑥 ^{2} + 3𝑥𝑦 − 𝑥𝑦 − 3𝑦 ^{2} = 0

⇒ (𝑥 + 3𝑦)(𝑥 − 𝑦) = 0

Equating we get, 𝑥 + 3𝑦 = 0 or 𝑥 − 𝑦 = 0

(2, 2) lies on 𝑥 − 𝑦 = 0

∴ Equation of normal will be 𝑥 + 𝑦 = 𝜆

It passes through (2, 2)

∴ 𝜆 = 4

𝐿 ∶ 𝑥 + 𝑦 = 4

Distance of 𝐿 from the origin = | −4/√2 | = 2√2

**Answer: (b) **

**13**. Which of the following statements is a tautology?

a. ~(𝑝 ∧ ~𝑞) → (𝑝 ∨ 𝑞)

b. (~𝑝 ∨ ~𝑞) → (𝑝 ∧ 𝑞)

c. 𝑝 ∨ (~𝑞) → (𝑝 ∧ 𝑞)

d. ~(𝑝 ∨ ~𝑞) → (𝑝 ∨ 𝑞)

~(𝑝 ∨ ~𝑞) → (𝑝 ∨ 𝑞)

= (𝑝 ∨ ~𝑞) ∨ (𝑝 ∨ 𝑞)

= (𝑝 ∨ 𝑝) ∨ (𝑞 ∨ ~𝑞)

= 𝑝 ∨ 𝑇

= 𝑇

**Answer: (𝑑) **

**14. If\(\begin{array}{l}I = \int_{1}^{2}\frac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}}\end{array} \), then:**

a. 1/6 < I

^{2}< 1/2

b. 1/8 < I

^{2}< 1/4

c. 1/9 < I

^{2}< 1/8

d. 1/16 < I

^{2}< 1/9

Let

𝑓 ′ (𝑥) = (−(6𝑥 ^{2} − 18𝑥 + 12))/ 2(2𝑥 ^{3} − 9𝑥 ^{2} + 12𝑥 + 4) ^{3/2} = (−3(𝑥 − 1)(𝑥 − 2))/ (2𝑥 ^{3} − 9𝑥 ^{2} + 12𝑥 + 4) ^{3/2} ⇒ 𝑓_{𝑚𝑖𝑛} = 𝑓(1) and 𝑓_{𝑚𝑎𝑥} = 𝑓(2)

= 1/√9 = 1/3

1/3 < I < 1/√8

⇒ 1/9 < I ^{2} < 1/8

**Answer: (𝑐) **

**15.**If

^{−1}is equal to:

a. 6I − 𝐴

b. 𝐴 − 6I

c. 4I − 𝐴

d. 𝐴 − 4I

If

⇒ 10𝐴 ^{−1} = 𝐴 − 6I

**Answer: (𝑏) **

**16. The area (in sq. units) of the region {(𝑥, 𝑦) ∈ 𝑹 : 𝑥**

^{2}≤ 𝑦 ≤ 3 − 2𝑥}, is:a. 31/3

b. 32/3

c. 29/3

d. 34/3

We have 𝑥 ^{2} ≤ 𝑦 ≤ −2𝑥 + 3

For point of intersection of two curves

𝑥 ^{2} + 2𝑥 − 3 = 0

⇒ 𝑥 = −3, 1

⇒

= 32/3 sq. units

**Answer: (𝑏) **

**17. Let 𝑆 be the set of all functions 𝑓:[0,1] → 𝐑, which are continuous on [0, 1] and differentiable on (0, 1). Then for every 𝑓 in 𝑆, there exists a 𝑐 ∈ (0,1), depending on 𝑓, such that:**

a. (f(1)−𝑓(𝑐))/ (1−𝑐) = 𝑓′(𝑐)

b. |f(𝑐) − 𝑓(1)| < |𝑓′(𝑐)|

c. |f(𝑐) + 𝑓(1)| < (1 + c)|𝑓 ′ (𝑐)|

d. |f(𝑐) − 𝑓(1)| < (1 − 𝑐)|𝑓′(𝑐)|

𝑆 is set of all functions.

If we consider a constant function, then option 2, 3 and 4 are incorrect.

For option 1:

(f(1) − 𝑓(𝑐))/ (1 – 𝑐) = 𝑓′(𝑐)

This may not be true for f(𝑥) = 𝑥 ^{2}

None of the options are correct.

**Answer: (Bonus) **

**18**. The differential equation of the family of curves, 𝑥

^{2}= 4b(𝑦 + 𝑏), 𝑏 ∈ 𝑹, is:

a. 𝑥𝑦′′ = 𝑦′

b. x(𝑦′)

^{2}= 𝑥 + 2𝑦𝑦′

c. x(𝑦 ′ )

^{2}= 𝑥 − 2𝑦𝑦′

d. x(𝑦 ′ )

^{2}= 2𝑦𝑦 ′ − 𝑥

𝑥 ^{2} = 4(𝑦 + 𝑏) … (1)

Differentiating both the sides w.r.t. 𝑥, we get

⇒ 2𝑥 = 4𝑏𝑦 ′

⇒ 𝑏 = 𝑥/2𝑦′

Putting the value of 𝑏 in (1), we get

⇒ 𝑥 ^{2} = (2𝑥/𝑦′) (𝑦 + (𝑥 /2𝑦′))

⇒ 𝑥 ^{2} = (2𝑥𝑦 /𝑦′ )+ (𝑥 ^{2}/ 𝑦′^{2})

⇒ x(𝑦 ′) ^{2} = 2𝑦𝑦 ′ + x

**Answer: (b)**

**19.**The system of linear equations

𝜆𝑥 + 2𝑦 + 2𝑧 = 5

2𝜆𝑥 + 3𝑦 + 5𝑧 = 8

4𝑥 + 𝜆𝑦 + 6𝑧 = 10 has:

a. no solution when 𝜆 = 2

b. infinitely many solutions when 𝜆 = 2

c. no solution when 𝜆 = 8

d. a unique solution when 𝜆 = −8

= 18𝜆 − 5𝜆 ^{2} − 24𝜆 + 40 + 4𝜆 ^{2} − 24

⟹ 𝐷 = −𝜆 ^{2} − 6𝜆 + 16

Now, 𝐷 = 0

⇒ 𝜆 ^{2} + 6𝜆 − 16 = 0 ⇒ 𝜆 = −8 or 2

For 𝜆 = 2

= 40 + 4 − 28 ≠ 0

∴ Equations have no solution for 𝜆 = 2.

**Answer: (a) **

**20.**If the 10𝑡ℎ term of an A.P. is 1/20 and its 20

^{𝑡ℎ}term is 1/10 , then the sum of its first 200 terms is:

a.

b. 100

c. 50

d.

𝑇_{10} = 1/20

𝑇_{20} = 1/10

𝑇_{20} − 𝑇_{10} = 10𝑑

⇒ (1/20) = 10𝑑

𝑑 = 1/200

∴ 𝑎 = 1/200

𝑆_{200} = (200/2) [2 (1/200) + 199 ( 1/200)]

=

**Answer: (𝑑) **

**21.**Let a line 𝑦 = 𝑚𝑥 (𝑚 > 0) intersect the parabola, 𝑦

^{2}= 𝑥 at a point 𝑃, other than the origin. Let the tangent to it at 𝑃 meet the 𝑥 −axis at the point 𝑄. If area (Δ𝑂𝑃𝑄) = 4 sq. units, then 𝑚 is equal to __________ .

Let the co-ordinates of 𝑃 be (𝑡 ^{2}, 𝑡)

Equation of tangent at (𝑡 ^{2}, 𝑡) is 𝑦 − 𝑡 = (1/2𝑡) (𝑥 − 𝑡 ^{2} )

Therefore, co-ordinates of 𝑄 will be (−𝑡 ^{2}, 0)

Area of Δ𝑂𝑃𝑄 = 4

⇒ 𝑡 ^{3} = ±8

⇒ 𝑡 = ±2

⇒ 𝑡 = 2 as 𝑡 > 0

𝑚 = 1/𝑡 = ½

**Answer: (0.5)**

**22.**Let f(𝑥) be a polynomial of degree 3 such that 𝑓(−1) = 10, 𝑓(1) = −6, 𝑓(𝑥) has a critical point at 𝑥 = −1 and 𝑓′(𝑥) has a critical point at 𝑥 = 1. Then the local minima at 𝑥 =______

Let the polynomial be

f(𝑥) = 𝑎𝑥^{3} + 𝑏𝑥 ^{2} + 𝑐𝑥 + 𝑑

⇒ 𝑓 ′ (𝑥) = 3𝑎𝑥 ^{2} + 2𝑏𝑥 + 𝑐

⇒ 𝑓 ′′(𝑥) = 6𝑎𝑥 + 2𝑏

𝑓 ′′(1) = 0 ⇒ 6𝑎 + 2𝑏 = 0 ⇒ 𝑏 = −3𝑎

𝑓 ′ (−1) = 0 ⇒ 3𝑎 − 2𝑏 + 𝑐 = 0

⇒ 𝑐 = − 9a

f(−1) = 10 ⇒ −𝑎 + 𝑏 − 𝑐 + 𝑑 = 10

⇒ −𝑎 − 3𝑎 + 9𝑎 + 𝑑 = 10

𝑑 = −5𝑎 + 10

f(1) = -6

⇒ 𝑎 + 𝑏 + 𝑐 + 𝑑 = -6

⇒ 𝑎 − 3𝑎 − 9𝑎 − 5𝑎 + 10 = -6

⇒ 𝑎 = 1

∴ 𝑓 ′ (𝑥) = (3𝑥 ^{2} – 6𝑥 – 9

= 3𝑥^{2} − 2𝑥 − 3

For 𝑓 ′ (𝑥) = 0 ⇒ 𝑥 ^{2} − 2𝑥 − 3 = 0 ⇒ 𝑥 = 3, −1

Minima exists at 𝑥 = 3

**Answer: (3) **

**23.**

⇒ tan 𝛼 = 1/7

⇒ (√2 sin𝛽) /√2 = 1 √10

⇒ sin𝛽 = 1 √10

⇒ tan 𝛽 = 1/3

tan 2𝛽 = (2 tan 𝛽)/ (1 − tan^{2} 𝛽) = (2/3)/(1 – 1/9) = 3/4

tan(𝛼 + 2𝛽) = (tan 𝛼 + tan 2𝛽)/( 1 − tan 𝛼 tan 2𝛽)

= (1/7 + ¾)/(1 – (1/7) × (3/4)) = 1

**Answer: (1) **

**24.**The number of 4 letter words (with or without meaning) that can be made from the eleven letters of the word “EXAMINATION” is _________ .

Word “EXAMINATION” consists of 2𝐴, 2𝐼, 2𝑁, 𝐸, 𝑋, 𝑀, 𝑇,

Case I: All different letters are selected

Number of words formed = ^{8}𝐶 _{4} × 4! = 1680

Case II: 2 letters are same and 2 are different Number of words formed = ^{3}𝐶_{1} × ^{7}𝐶 _{2} × (4!/ 2!) = 756

Case III: 2 pair of letters are same

Number of words formed = ^{3}𝐶 _{2} × 4!/(2!×2!) = 18

Total number of words formed = 1680 + 756 + 18 = 2454

**Answer: (2454)**

**25.**The sum,

= 504

**Answer: (504) **

## Video Lessons – January 8 Shift 2 Maths

## JEE Main 2020 Maths Paper With Solutions Jan 8 Shift 2

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