The top 2,45,00 students in JEE Main are qualified to appear for JEE Advanced. The JEE Advanced includes two question papers – Paper 1 and Paper 2. Both the papers together will have a total of 54 questions. These questions are from Chemistry, Physics and Maths. The questions asked in JEE Advanced are based on the Class 11 and Class 12 syllabus.

**Question 1: Consider a triangle Δ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is;**

a. x^{2} + y^{2} – 3x + y = 0

b. x^{2} + y^{2} + x + 3y = 0

c. x^{2} + y^{2} + 2y – 1 = 0

d. x^{2} + y^{2} + x + y = 0

Answer: b

As we know mirror image of the orthocenter lies on the circumcircle.

Image of (1, 1) in x-axis is (1, -1)

Image of (1, 1) in x + y + 1 = 0 is (-2, -2).

∴ The required circle will be passing through both (1, -1) and (-2, -2).

∴ Only x^{2} + y^{2} + x + 3y = 0 satisfy both.

**Question 2: The area of the region {(x, y): 0 ≤ x ≤ 9/4, 0 ≤ y ≤ 1, x ≥ 3y, x + y ≥ 2}**

a. 11/32

b. 35/96

c. 37/96

d. 13/32

Answer: a

A rough sketch of the required region is;

∴ Required area is area of ΔACD + Area of ΔABC

i.e (¼) + (3/32) = 11/32 sq.units.

**Question 3: Consider three sets E _{1} = {1, 2, 3}, F_{1} = {1, 3, 4} and G_{1} = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E_{1}, and let S_{1} denote the set of these chosen elements. Let E_{2} = E_{1} – S_{1 }and F_{2} = F_{1} ⋃ S_{1}. Now two elements are chosen at random, without replacement, from the set F_{2} and let S_{2} dentoe the set of these chosen elements.**

**Let G _{2} = G_{1} ⋃ S_{2}. Finally, two elements are chosen at random, without replacement from the set G_{2} and let S_{3} denote the set of these chosen elements. Let E_{3} = E_{2} ⋃ S_{3}. Given that E_{1} = E_{3}, let p be the conditional probability of the event S_{1} = {1, 2}. Then the value of p is;**

a. 1/5

b. 3/5

c. 1/2

d. 2/5

Answer: a

We will follow the tree diagram,

P(E_{1} = E_{3}) = ⅓ [(½ × 1/10) + (½ × 0) + (½ × 1/10) + (½ × ⅙) + (⅔ × 1/10) + (⅓ × 0)]

= ⅓ (¼)

Required probability = ⅓ (½ × 1/10)/(⅓ × ¼ )

= 1/5

**Question 4: Let θ _{1}, θ_{2}, …., θ_{10} be positive valued angles (in radian) such that θ_{1}+ θ_{2}+ ….+ θ_{10} = 2π. Define the complex numbers z_{1} = **

**\(\begin{array}{l}e^{i\theta _{1}}\end{array} \)**

**, z\(\begin{array}{l}z_{k-1}e^{i\theta _{k}}\end{array} \) for k = 2, 3, …, 10, where i = √-1. Consider the statement P and Q given below:**

_{k}=**P: |z _{2 }– z_{1}| + |z_{3 }– z_{2}| + …. +|z_{10 }– z_{9}| + |z_{1 }– z_{10}| ≤ 2π**

**Q: |z _{2}^{2} – z_{1}^{2}| + |z_{3}^{2} – z_{2}^{2}| + …. +|z_{10}^{2} – z_{9}^{2}| + |z_{1}^{2} – z_{10}^{2}| ≤ 4π**

**Then, **

a. P is TRUE and Q is FALSE

b. Q is TRUE and P is FALSE

c. Both P and Q are TRUE

d. Both P and Q are FALSE

Answer: c

|z_{2} – z_{1}| = length of line AB ≤ length of arc AB

|z_{3} – z_{2}| = length of line BC ≤ length of arc BC

∴ Sum of length of these 10 lines ≤ Sum of length of arcs (i.e. 2π)

(As θ_{1}+ θ_{2}+ ….+ θ_{10} = 2π)

∴ |z_{2 }– z_{1}| + |z_{3 }– z_{2}| + …. + |z_{1 }– z_{10}| ≤ 2π

And |z_{k}^{2} – z^{2}_{k-1}| = |z_{k }– z_{k-1}| |z_{k }+ z_{k-1}|

As we know |z_{k }+ z_{k-1}| ≤ |z_{k}| + |z_{k-1}| ≤ 2

|z_{2}^{2} – z_{1}^{2}| + |z_{3}^{2} – z_{2}^{2}| + …. + |z_{1}^{2} – z_{10}^{2}| ≤ 2(|z_{2 }– z_{1}| + |z_{3 }– z_{2}| + …. + |z_{1 }– z_{10}|) ≤ 2(2π)

∴ Both P and Q are true.

**Question Stem for Question Nos. 5 and 6**

**Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, …, 100}. Let p _{1} be the probability that the maximum of chosen numbers is at least 81 and p_{2} be the probability that the minimum of chosen numbers is at most 40.**

**Question 5: The value of (625/4)p _{1} is**

Answer: 76.25

For p_{1}, we need to remove the cases when all three numbers are less than or equal to 80.

So p_{1} = 1 – (80/100)^{3}

= 61/125

So (625/4)p_{1} = (625/4)× 61/125

= 305/4

= 76.25

**Question 6: The value of (125/4)p _{2} is**

Answer: 24.50

For p_{2}, we need to remove the cases when all three numbers are greater than 40.

So p_{2} = 1 – (60/100)^{3}

= 98/125

So (125/4)p_{2} = (125/4)×98/125

= 24.50

**Question Stem for Question Nos. 7 and 8**

**Let α, β and γ be real numbers such that the system of linear equations**

**x + 2y + 3z = α**

**4x + 5y + 6z = β**

**7x + 8y + 9z = γ – 1 is consistent. **

**Let |M| represent the determinant of the matrix.**

**M = **

**\(\begin{array}{l}\begin{bmatrix} \alpha & 2 &\gamma \\ \beta & 1&0 \\ -1 & 0& 1 \end{bmatrix}\end{array} \)**

**Let P be the plane containing all those (α, β, γ) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.**

**Question 7: The value of |M| is **

Answer: 1

**Question 8: The value of D is**

Answer: 1.50

**Solution for Q.7 and Q.8**

**∆** =

Given system of equation will be consistent even if α = β = γ-1 = 0, i.e. equations will form homogeneous system.

So α = 0, β = 0, γ = 1

M =

= -1×(-1)

= 1

As given equations are consistent

x + 2y + 3z – α = 0 …P_{1}

4x + 5y + 6z – β = 0 …P_{2}

7x + 8y + 9z – (γ – 1) = 0 …P_{3}

For some scalar λ and μ

μP_{1} + λP_{2} = P_{3}

μ(x + 2y + 3z – α) + λ(4x + 5y + 6z – β) = 7x + 8y + 9z – (γ – 1)

comparing coefficients

μ+ 4λ = 7, 2μ + 5λ = 8, 3μ + 6λ = 9

λ = 2 and μ = -1 satisfy all these conditions

comparing constant terms,

αμ – βλ = – (γ – 1)

α – 2β + γ = 1

So equation of plane is

x – 2y + z = 1

distance from (0, 1, 0) = |(-2-1)/√6| = 3/√6

D = (3/√6)^{2} = 9/6 = 3/2 = 1.50

**Question Stem for Question Nos. 9 and 10**

**Consider the lines L _{1} and L_{2} defined by**

**L _{1} : x√2 + y – 1 = 0 and L_{2 }: x√2 – y + 1 = 0**

**For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L _{1} and the distance of P from L_{2} is λ^{2}. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is √270.**

**Let the perpendicular bisector of RS meet C at two distinct points R’ and S’. Let D be the square of the distance between R’ and S’.**

**Question 9: The value of λ ^{2} is**

Answer: 9

**Question 10: The value of D is**

Answer: 77.14

**Solution for Q.9 and Q.10**

C: |(x√2 + y – 1)/√3| |(x√2 – y + 1)/√3| = λ^{2}

⇒ C: |2x^{2} – (y-1)^{2}| = 3λ^{2}

C cuts y-1 = 2x at R(x_{1}, y_{1}) and S(x_{2}, y_{2})

So |2x^{2} – 4x^{2}| = 3λ^{2}

⇒ x = ±√(3/2) |λ|

So |x_{1 }– x_{2}| = √6|λ| and |y_{1} – y_{2}| = 2| x_{1} – x_{2}| = 2√6|λ|

Since RS^{2} = (x_{1} – x_{2})^{2} + (y_{1}-y_{2})^{2}

⇒ 270 = 30λ^{2}

⇒ λ^{2} = 270/30 = 9

Since slope of RS = 2 and midpoint of RS is ((x_{1}+x_{2})/2, (y_{1}+y_{2})/2) ≡ (0, 1)

So R’S’ ≡ y-1 = -½ x

Solving y-1 = -½ x with ‘C’ we get x^{2} = (12/7) λ^{2}

⇒ |x_{1}-x_{2}| = 2√(12/7) |λ| and |y_{1}-y_{2}| = ½ |x_{1}-x_{2}| = √(12/7) |λ|

Hence D = (R’S’)^{2} = (x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} = (12/7)9×5 ≈ 77.14

**Question 11: For any 3 × 3 matrix M, let |M| denote the determinant of M. Let E = **

**\(\begin{array}{l}\begin{bmatrix} 1 & 2 & 3\\ 2& 3 &4 \\ 8 & 13 & 18 \end{bmatrix}\end{array} \)**

**,**

**P = **

**\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 0\\ 0& 0 &1 \\ 0 & 1 & 0 \end{bmatrix}\end{array} \)**

**,**

**And F = **

**\(\begin{array}{l}\begin{bmatrix} 1 & 3 & 2\\ 8& 18 &13 \\ 2 & 4 & 3 \end{bmatrix}\end{array} \)**

**If Q is a nonsingular matrix of order 3 × 3, then which of the following statements is(are) TRUE?**

a. F = PEP and P^{2} =

b. |EQ + PFQ

^{-1}| = |EQ| +|PFQ

^{-1}|

c. |(EF)

^{3}| > |EF|

^{2 }

d. Sum of the diagonal entries of P

^{-1}EP + F is equal to the sum of diagonal entries of E + P

^{-1}FP

Answer: (a, b, d)

P is formed from I by exchanging second and third rows or by exchanging second and third columns.

So, PA is a matrix formed from A by changing the second and third rows.

Similarly, AP is a matrix formed from A by changing the second and third columns.

Hence, Tr(PAP) = Tr(A) …(1)

(a) Clearly P.P =

And PE =

⇒ PEP =

⇒ PEP = F ⇒ PFP = E …(2)

(b) Since |E| = |F| = 0

So, |EQ + PFQ^{–1} | = |PFPQ + PFQ^{–1}| = | P| | F| | PQ + Q^{–1} | = 0

Also, |EQ|+|PFQ^{–1 }| = 0

(c) From (2); PFP = E and |P| = -1

So, |F| = |E|

Also, |E| = 0 = |F|

So, |EF|^{3 }= 0 = |EF|^{2}

(d) since P^{2} = I ⇒ P^{ –1} = P

So, Tr (P^{-1}EP + F) = Tr(PEP + F) = Tr(2F )

Also Tr (E + P^{ –1}FP) = Tr(E + PFP) = Tr(2E )

Given that Tr(E) = Tr(F)

⇒ Tr(2E) = Tr(2F)

**Question 12: Let F: R→ R be defined by f(x) = (x ^{2}-3x-6)/(x^{2}+2x+4). Then which of the following statements is (are) TRUE?**

a. f is decreasing in the interval (-2, -1)

b. f is increasing in the interval (1, 2)

c. f is onto

d. Range of f is [-3/2, 2]

Answer: (a, b)

f(x) = (x^{2}-3x-6)/(x^{2}+2x+4)

⇒ f’(x) = 5x(x+4)/(x^{2}+2x+4)^{2}

⇒ f(x) has local maxima at x = -4 and minima at x = 0

Range of f(x) is [-3/2, 11/6]

**Question 13: Let E, F and G be three events having probabilities P(E) = 1/8, P(F) = ⅙ and P(G) = ¼, and P(E⋂F⋂G) = 1/10. For any event H, if H ^{c} denotes its complement, then which of the following statements is(are) TRUE?**

a. P(E⋂F⋂G^{c}) ≤ 1/40

b. P(E^{c}⋂F⋂G) ≤ 1/15

c. P(E⋃F⋃G) ≤ 13/24

d. P(E^{c}⋂F^{c}⋂G^{c}) ≤ 5/12

Answer: (a, b, c)

Let P(E⋂F) = x, P(F⋂G) = y and P(E⋂G) = z

Clearly x, y, z ≥ 1/10

Since x + z ≤ 27/120

⇒ x, z ≤ 15/120

x+y ≤ 32/120

⇒ x, y ≤ 20/120

And y + z ≤ 42/120 ⇒ y, z ≤ 30/120

Now P(E⋂F⋂G^{c}) = x – 12/120 ≤ 3/120 = 1/40

P(E^{c}⋂F⋂G) = y – 12/120 ≤ 80/120 = 1/15

P(E⋃F⋃G) ≤ P(E) + P(F) + P(G) = 13/24

And P(E^{c}⋂F^{c}⋂G^{c}) = 1 – P(E⋃F⋃G) ≥ 11/24 ≥ 5/12

**Question 14: For any 3 × 3 matrix M, let |M| denote the determinant of M. Let I be the 3 × 3 identify matrix. Let E and F be two 3 × 3 matrices such that (I – EF) is invertible. If G = (I – EF) ^{–1}, then which of the following statements is(are) TRUE?**

a. |FE| = |I – FE| |FGE|

b. (I – FE) (I + FGE) = I

c. EFG = GEF

d. (I – FE) (I – FGE) = I

Answer: (a, b, c)

I – EF = G^{–1}

G – GEF = I …(1)

And G – EFG = I …(2)

Clearly GEF = EFG (option C is correct)

Also (I – FE)(I + FGE) = I – FE + FGE – FE + FGE

= I – FE + FGE – F(G – I)E

= I – FE + FGE – FGE + FE

= I (option B is correct and D is incorrect)

Now, (I – FE)(I – FGE) = I – FE – FGE + F(G – I)E

= I – 2FE

(I – FE)(- FGE) = – FE

|I – FE||FGE| = |FE|

**Question 15: For any positive integer n, let S _{n}: (0, ∞) → R be defined by S_{n}(x) = ∑_{k=1}^{n} cot^{-1}((1 + k(k+1)x^{2})/x) where for any x ∈R , cot^{–1}(x) ∈ (0, π) and tan^{–1}( x)∈ (-π/2, π/2). Then which of the following statements is (are) TRUE?**

a. S_{10}(x) = π/2 – tan^{-1}(1 + 11x^{2})/10x, for all x >0

b. lim _{n→∞ }cot(S_{n}(x)) = x, for all x >0

c. The equation S_{3}(x) = π/4 has a root in (0, ∞)

d. tan(S_{n}(x)) ≤1/2, for all n≥1 and x > 0

Answer: (a, b)

S_{n}(x) =

=

= tan^{-1}(n+1)x – tan^{-1}x

= tan^{-1}(nx/(1+(n+1)x^{2}))

Now,

(a) S_{10}(x) = tan^{-1}(10x/(1+11x^{2})) = π/2 – tan^{-1}((1+11x^{2})/10x)

(b) lim_{n→ ∞}cot (S_{n}(x)) = cot (tan^{-1}x/x^{2}) = x

(c) S_{3}(x) = π/4 => 3x/(1+4x^{2}) = 1

⇒ 4x^{2}-3x+1=0 has no real root.

(d) for x = 1, tan(S_{n}(x)) = n/(n+2) which is greater than 1/2 for n≥3 so this option is incorrect.

**Question 16: For any complex number w = c + id, let arg(w) ∈ (-π, π], where i = √-1 . Let α and β be real numbers such that for all complex numbers z = x + iy satisfying arg(z+α)/(z+β) = π/4, the ordered pair (x, y) lies on the circle x ^{2} + y^{2} + 5x – 3y + 4 = 0. Then which of the following statements is (are) TRUE?**

a. α = -1

b. αβ = 4

c. αβ = -4

d. β = 4

Answer:(b, d)

Circle x^{2} + y^{2} + 5x – 3y + 4 = 0 cuts the real axis (x-axis) at (-4, 0), (-1, 0)

Clearly α = 1 and β = 4

**Question 17: For x ∈ R, the number of real roots of the equation 3x ^{2} – 4|x^{2 }– 1| + x – 1 = 0 is**

Answer: 4

3x^{2} – 4|x^{2 }– 1| + x – 1 = 0

Let x ∈ [-1, 1]

⇒ 3x^{2} – 4(-x^{2}+1) + x – 1 = 0

⇒ 3x^{2} + 4x^{2} – 4 + x – 1 = 0

⇒ 7x^{2} + x – 5 = 0

⇒ x = -1±√(1+140)/2

Both values are acceptable.

Let x ∈ (-∞, -1) ⋃ (1, ∞)

x^{2}-4(x^{2}-1)+x-1 = 0

⇒ x^{2}-x-3 = 0

⇒ x = 1±√(1+12)/2

Again both are acceptable.

Hence, total number of solution = 4.

**Question 18: In a triangle ABC, let AB = √23, and BC = 3 and CA = 4. Then the value of (cot A + cot C)/cot B is**;

Answer: 2

With standard notations

Given c = √23, a = 3 and b = 4

Now (cot A + cot C)/cot B = (cos A/sin A + cos C/sin C)/(cos B/sin B)

= ((b^{2}+c^{2}-a^{2})/2bc sin A + (a^{2}+b^{2}-c^{2})/2ab sin C)/((c^{2}+a^{2}-b^{2})/2ac sin B)

= ((b^{2}+c^{2}-a^{2})/4∆ + (a^{2}+b^{2}-c^{2})/4∆)/((c^{2}+a^{2}-b^{2})/4∆)

= 2b^{2}/(a^{2}+c^{2}-b^{2})

= 2

**Question 19: Let **

**\(\begin{array}{l}\vec{u}, \vec{v}\: and \: \vec{w}\end{array} \)**

**be vectors in three-dimensional space, where\(\begin{array}{l}\vec{u}\: and \: \vec{v}\end{array} \) are unit vectors which are not perpendicular to each other and \(\begin{array}{l}\vec{u} . \vec{w}= 1\end{array} \), \(\begin{array}{l}\vec{v} . \vec{w}= 1\end{array} \) and \(\begin{array}{l}\vec{w} . \vec{w}= 4\end{array} \). If the volume of the parallelepiped, whose adjacent sides are represented by the vectors u, v and w is √2 , then the value of \(\begin{array}{l}\left | 3\vec{u} +5\vec{v}\right |\end{array} \) is;**

Answer: 7

Given

a.so,

Let

4K^{2} – 2K = 0

⇒

Or

Therefore

= 49

⇒