**Question 1.**Let f: R -> R be given by f(x) = (x – 1) (x – 2) (x – 5). Define F(x) =

a) F has a local minimum at x = 1

b) F has a local maximum at x = 2

c) F(x) ≠ 0 for all x ϵ (0, 5)

d) F has two local maxima and one local minimum in (0, ∞)

F(x) = (x – 1) (x – 2) (x – 5)

At x = 1 and x = 5, F'(x) changes from – to +

Therefore, F(x) has two local minima points at x = 1 and x = 5.

F(x) has one local maxima point at x = 2.

**Answer:** a,b,c

**Question 2.**

Then the possible value(s) of a is/are:

a) 8

b) –9

c) –6

d) 7

**Answer:** a, b

**Question 3:** Three lines

are given. For which point(s) Q and L_{2} can we find a point P on L_{1} and a point R on L_{3} so that P, Q and R are collinear?

a)

b)

c)

d)

P(λ, 0, 0), Q(0, μ, 1), R(1, 1, r)

Given:

λ/[ λ-1] = – μ/-1 = -1/-r

μ cannot take the values 0 and 1.

**Answer:** c, d

**Question 4:**Let F: R -> R be a function. We say that f has

Then which of the following options is/are correct?

a) f(x) = x|x| has PROPERTY 2

b) F(x) = x^{2/3} has PROPERTY 1

c) f(x) = sin x has PROPERTY 2

d) f(x) = |x| has PROPERTY 1

**Answer:** b, d

**Question 5:**For non-negative integers n, let

Assuming cos^{-1}x takes value in [0, π], which of the following options is/are correct?

a) If α = tan (cos^{-1}f(6)), then α^{2} + 2α – 1 = 0

b) lim_(n-> ∞) f(n) = 1/2

c) f(4) = √3/2

d) sin (7 cos^{-1}f(5)) = 0

(a) α = tan(cos^{-1} f(6)) = tan(cos^{-1}(cos π/8) = tan (π/8)

α^{2} + 2 α – 1 = tan^{2} (π/8) + 2 tan (π/8) – 1

tan 2(π/8) = [2 tan (π/8)]/[1- tan^{2}(π/8)]

This implies, 1 = 2α/[1- α^{2}]

α^{2} + 2 α – 1 = 0

Option (a) is correct.

(b)

Option (b) is not correct.

(c) f(4) = cos(π/6) = √3/2

Option (c) is correct.

(d) sin(7 cos^{-1} f(5)) = sin(7 cos^{-1} (cos π/7)) = sin(7 x π/7) = 0

Option (d) is correct.

**Answer:** a, c, d

**Question 6:**

Where P_{K}^{T }denotes the transpose of the matrix P_{K}. Then which of the following options is/are correct?

a) X – 30I is an invertible matrix

b) The sum of diagonal entries of X is 18

c) If

d) X is a symmetric matrix

P_{1} = P_{1}^{T} = P_{1}^{-1}

P_{2} = P_{2}^{T} = P_{2}^{-1}

…….

P_{6} = P_{6}^{T} = P_{6}^{-1}

And Let

Here A^{T} = A -> A is symmetric matrix

Now, X^{T} = (P_{1}AP_{1}^{T} + ……+ P_{6}AP_{6}^{T})^{T}

= P_{1}AP_{1}^{T} + ……+ P_{6}AP_{6}^{T}

= X

Therefore, X is symmetric matrix.

Again,

Let B =

XB = P_{1}AP_{1}^{T}B + P_{2}AP_{2}^{T}B +……..+ P_{6}AP_{6}^{T}B

= P_{1}AB + P_{2}AB+ ……+ P_{6}AB

= (P_{1} + P_{2 }+ ……+ P_{6})

=

=> α = 30

Since

=>(X – 30I)B = 0 has a nontrivial solution B =

=>X – 30I = 0

X = P_{1}AP_{1}^{T} + …..+ P_{6}AP_{6}^{T}

Trace(X) = tr(P_{1}AP_{1}^{T}) + …..+ tr(P_{6}AP_{6}^{T})

= (2 + 0 + 1) + ……(2 + 0 + 1) = 3 + 3 + ….6 times = 18

**Answer:** b, c, d

**Question 7:**Let x ϵ R and let

And R = PQP^{-1}. Then which of the following options is/are correct?

a) For x = 1, there exists a unit vector

b) There exists a real number x such that PQ = QP

c) det R =

d) For x = 0, if

R = PQP^{-1}

|R| = |P||Q||P^{-1}|

=>det Q = 2(24) – x(0) + x(-4x) = 48 – 4x^{2}

-4 + a + 2b/3 = 0 and -2a + 4b/3 = 0 => a = 2 and b = 3

Therefore, a + b = 5

PQ = QP

=>x + 2 + x = 2 + 2x ->No value exists.

**Question 8:**Let f(x) = sinπx/x

^{2}, x > 0

Let x_{1} < x_{2} < x_{3} <…..< x_{n} < …. be all the points of local maximum of f.

and y_{1} < y_{2} < y_{3} <…..< y_{n} < …. be all the points of local minimum of f.

Then which of the following options is/are correct?

a) |x_{n} – y_{n}| > 1 ;for every n

b) x_{1} < y_{1}

c) x_{n} Є (2n, 2n + 1/2) ;for every n

d) x_{n+1} – x_{n} > 2 ;for every n

*for students, Draw a graph and conclude.

**Question 9:**The value of

**Question 10**: Let |X| denote the number of elements in set X. Let S = {1,2,3,4,5,6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A,B) such that 1 < |B| < |A|, equals.

The number of ordered pairs of (A, B) are

^{6}C_{1}(^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} +…..+^{6}C_{6} )+ ^{6}C_{2} (^{6}C_{3} + ^{6}C_{4} + …+ ^{6}C_{6)} + ^{6}C_{3} (^{6}C_{4} + ^{6}C_{5} +^{6}C_{6}) + ^{6}C_{4} (^{6}C_{5} + ^{6}C_{6}) + ^{6}C_{5} . ^{6}C_{6}

= (^{6}C_{1} . ^{6}C_{2 }+ ^{6}C_{1} . ^{6}C_{3 }+…..+ ^{6}C_{1} . ^{6}C_{6 })+ (^{6}C_{2} . ^{6}C_{3 }+ ^{6}C_{2} . ^{6}C_{4 }+…..+ ^{6}C_{2} . ^{6}C_{6}) + (^{6}C_{3} . ^{6}C_{4 }+ ^{6}C_{3} . ^{6}C_{5 }+ ^{6}C_{3 . }^{6}C_{6}) + ^{6}C_{4 . }^{6}C_{5 }**+ **^{6}C_{4. }^{6}C_{6} + ^{6}C_{5 . }^{6}C_{6}

= (^{12}C_{5} – ^{6}C_{1}) + (^{12}C_{4} – ^{6}C_{2}) + (^{12}C_{3} – ^{6}C_{3}) + (^{12}C_{2} – ^{6}C_{4}) + (^{12}C_{1} – ^{6}C_{5})

= (^{12}C_{1 }+ ^{12}C_{2 }+ ^{12}C_{3 }+ ^{12}C_{4 }+ ^{12}C_{5 }) – ( ^{6}C_{1 }+ ^{6}C_{2 }+ ^{6}C_{3 }+ ^{6}C_{4 }+ ^{6}C_{5})

= 1585 – 62 = 1523

**Question 11:**Five person A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is

Maximum number of hats used of same colour are 2.

They cannot be 3 otherwise atleast 2 hats of same colour are consecutive.

Now the hats used are consider as B B G G R

Which can be selected in 3 ways.

It can be R G G B B or R R G B B or G G R R B

The number of ways of distributing blue hat (single one) in 5 persons equal to 5.

Now either position B and D are filled by green hats and C and E are filled by Red hats or B & D are filled by Red hats and C & E are filled by Green hats.

So, only 2 ways are possible.

Hence, number of ways = 3×5×2 = 30 ways.

**Question 12:**Suppose

holds for some positive integer n. Then

**Question 13:**The value of the integral

Let tan θ = t^{2} =>sec 2θ dθ = dt

2I/3 =

Hint: Split 2t/((t+1)^{4} = 1/(t+1)^{3} – 1/(t+1)^{4}

Solving above integration, we get

I = 1/2

**Question 14:**Let

**Question 15:**Answer the following by appropriately matching the lists based on the information given in the paragraph

Let f(x) = sin(π cosx) and g(x) = cos(2π sinx) be two functions defined for x > 0. Define the following sets whose element are written in the increasing order:

X = {x: f(x) = 0}, Y = {x: f'(x) = 0}

Z = {x: g(x) = 0}, W = {x: g'(x) = 0}

List –I contains the sets X,Y,Z and W. List – II contains some information regarding these sets.

Which of the following is the only correct combination?

a) (II), (R), (S)

b) (I), (P), (R)

c) (II), (Q), (T)

d) (I), (Q), (U)

Note: For answer, check next solution.

**Question 16:**Answer the following by appropriately matching the lists based on the information given in the paragraph.

Let f(x) = sin(π cosx) and g(x) = cos(2π sinx) be two functions defined for x > 0. Define the following sets whose element are written in the increasing order:

X = {x: f(x) = 0}, Y = {x: f'(x) = 0}

Z = {x: g(x) = 0}, W = {x: g'(x) = 0}

List –I contains the sets X,Y,Z and W. List – II contains some information regarding these sets.

Which of the following is the only correct combination?

a) (IV), (Q), (T)

b) (IV), (P), (R), (S)

c) (III), (R), (U)

d) (III), (P), (Q), (U)

f(x) = 0

=> sin(π cos x) = 0

This implies, π cos x = nπ

Cos x = n

=>Cos x = -1, 0, 1

x = {nπ , (2nπ) π /2}

=> x = {nπ/2, n Є I}

f’(x) = 0

=> cos(π cos x)(- π sin x) = 0

=> π cos x = (2n + 1) π/2 or x = nπ

=> cos x = n + (1/2) or x = nπ

=> cos x = ±(1/2) or x = nπ

Therefore, y = {2nπ ± π/3, 2nπ ± 2π/3, n π }

g(x) = 0 => cos (2 π sin x) = 0

=> 2 π sin x = (2n + 1) π/2

=> sin x = (2n+1)/4 = ±(1/4), ±(3/4)

z = {nπ ± sin^{-1}(1/4), nπ ± sin^{-1}(3/4), n Є I}

g’(x) = 0 => -sin (2 π sin x) 2 π cos x = 0

=> 2 π sin x = n π or x = (2n + 1) π/2

=> sin x = n/2 = 0, ±(1/2), ±1 or x = (2n + 1) π/2

And, w = {nπ, (2n + 1) π/2, nπ ± π/6, n Є I}

**Question 17**: Answer the following by appropriately matching the lists based on the information given in the paragraph.

Let the circles C_{1} : x^{2} + y^{2} = 9 and C_{2} : (x – 3)^{ 2} + (y – 4)^{ 2} = 16, intersect at the points X and Y. Suppose that another circle C_{3} : (x – h)^{ 2} + (y – k)^{ 2} = r^{2} satisfies the following conditions:

(i) centre of C_{3} is collinear with the centres of C_{1} and C_{2}

(ii) C_{1} and C_{2} both lie inside C_{3}, and

(iii) C_{3} touches C_{1} at M and C_{2} at N

Let the line through X and Y intersect C_{3} at Z and W, and let a common tangent of C_{1} and C_{3} be a tangent to the parabola x^{2} = 8αy.

There are some expressions given in the List – I whose values are given in List – II below:

Which of the following is the only INCORRECT combination?

a) (IV), (S)

b) (IV), (U)

c) (III), (R)

d) (I), (P)

(i) C_{1} , C_{2} and C_{3 } are collinear.

3k = 4h….(a)

Diameter of C_{3} = 12, so radius = 6 …(b)

C_{3} touches C_{1}

C_{1}(0, 0) and C_{3}(h, k)

|C_{1}C_{3}| = |r – 3|

So, h^{2} + k^{2} = 9…(c)

From (a) and (c)

h = ±9/5 and k = ±12/5

[Neglect -ve values]=> 2h + k = 6

(ii) Equation of line ZW

C_{1} = C_{2}

=> 3x + 4y = 9

=> Distance of ZW from (0, 0)

(ii) Area of ΔmZN = (1/2). Nm. (1/2 . ZW) = 72√6/5

Area of ΔZmW = (1/2). ZW . (om + op) = (1/2) . 24√6/5 . (3 + 9/5) = 288√6/25

Therefore, [Area of ΔmZN]/ [Area of ΔZmW] = 5/4

(iv) Slope of tangent to C_{1} at m = -3/4

Equation of Tangent, y = mx – 2√(1+m^{2})

y = -3x/4 – 3√[1+9/16] = -3x/4 – 15/4

x = -4y/3 – 5 …(i)

Tangent to x^{2} = 4 (2α) y is x = m’y = 2α/m’ …(ii)

On comparing (i) and (ii), we get

m’ = -4/3 and 2α/m’ = -5 => α = 10/3

**Question 18:**Answer the following by appropriately matching the lists based on the information given in the paragraph.

Let the circles C_{1} : x^{2} + y^{2} = 9 and C_{2} : (x – 3)^{ 2} + (y – 4)^{ 2} = 16, intersect at the points X and Y. Suppose that another circle C_{3} : (x – h)^{ 2} + (y – k)^{ 2} = r^{2} satisfies the following conditions:

(i) centre of C_{3} is collinear with the centres of C_{1} and C_{2}

(ii) C_{1} and C_{2} both lie inside C_{3}, and

(iii) C_{3} touches C_{1} at M and C_{2} at N

Let the line through X and Y intersect C_{3} at Z and W, and let a common tangent of C_{1} and C_{3} be a tangent to the parabola x^{2} = 8αy.

There are some expressions given in the List – I whose values are given in List – II below:

Which of the following is the only INCORRECT combination?

a) (II), (T)

b) (I), (S)

c) (I), (U)

d) (II), (Q)

*Note: Refer below solution.

## Video Lessons – Paper 2 Maths

## JEE Advanced 2019 Maths Paper 2 Solutions

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