Download Solved Mathematics Question Paper (Question + Solution) held on 2nd Sept of JEE Mains 2020

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session

FINAL JEE–MAIN EXAMINATION – SEPTEMBER, 2020 (Held On Wednesday 02nd SEPTEMBER, 2020)

TEST PAPER WITH SOLUTION

MATHEMATICS 1.

TIME : 3 PM to 6 PM

The area (in sq. units) of an equilateral triangle

1

inscribed in the parabola y2 = 8x, with one of

n

its vertices on the vertex of this parabola, is : (1) 64 3

(2) 256 3

(3) 192 3

(4) 128 3

2

Sol.

3

Official Ans. by NTA (3)

4

(2t2,4t) A y2=8x

Number of blue lines = Number of sides = n

Sol.

O (0,0)

30°

EN

Number of red lines = number of diagonals

nC

B 4t 2 Þ t =2 3 2 = 2t t

3.

AB = 8t = 16 3

Area = 256.3·

Let n > 2 be an integer. Suppose that there are

A

2.

3 = 192 3 4

n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are

(1) 199

(2) 101

(3) 201

(4) 200

n(n - 1) - n = 99 n 2

If the equation cos4q+sin4q + l = 0 has real solutions for q, then l lies in the interval : é 3 5ù (1) ê - , - ú ë 2 4û

æ 1 1ù (2) ç - , - ú è 2 4û

æ 5 ö (3) ç - , -1 ÷ 4 è ø

1ù é (4) ê -1, - ú 2û ë

Official Ans. by NTA (4) Sol. l = – (sin4q + cos4q) l = – (sin2q + cos2q)2 – 2sin2qcos2q

connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is :-

– n = 99 n Þ

n -1 - 1 = 99 Þ n = 201 2

LL

tan 30° =

2

= nC2 – n

l=

sin 2 2q -1 2

sin 2 2q é 1 ù Î ê0, ú 2 ë 2û

Official Ans. by NTA (3)

1ù é l Î ê-1, - ú 2û ë 1

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 4.

Let f(x) be a quadratic polynomial such that

6.

f(–1) + f(2) = 0. If one of the roots of f(x) = 0

Let a, b, cÎR be all non-zero and satisfy a3 + b3 + c3 = 2. If the matrix

is 3, then its other root lies in : (1) (–3, –1)

(2) (1, 3)

(3) (–1, 0)

(4) (0, 1)

æa b cö ç ÷ A =çb c a÷ ç c a b÷ è ø

Official Ans. by NTA (3) satisfies ATA = I, then a value of abc can be :

Sol. f(x) = a(x – 3) (x – a) f(2) = a(a – 2)

(1)

2 3

(2) -

1 3

f(–1) = 4a(1 + a) f(–1) + f(2) = 0 Þ a(a – 2 + 4 + 4a) = 0

(3) 3

(4)

a ¹ 0 Þ 5a = – 2

Official Ans. by NTA (4)

1 3

a= -

EN

Sol. ATA = I

2 = – 0.4 5

Þ a2 + b2 + c2 = 1

and ab + bc + ca = 0

a Î (–1, 0)

Let f : R ® R be a function which satisfies f(x + y) = f(x) + f(y)"x,yÎR. If f(1) = 2 and g(n) =

(n -1)

å f(k), n Î N k=1

Þ a+b+c=±1

LL

5.

Now, (a + b + c)2 = 1

then the value of n, for

which g(n) = 20, is :

(2) 9

A

(1) 5 (3) 20

(4) 4

So, a3 + b3 + c3 – 3abc

= (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = ± 1 (1 – 0) = ± 1 Þ 3 abc = 2 ± 1 = 3, 1 Þ abc = 1,

Official Ans. by NTA (1) Sol. f(x + y) = f(x) + f(y) Þ f(n) = nf(1)

7.

Let f : (–1, ¥) ® R be defined by f(0) = 1 and f(x) =

f(n) = 2n

1 3

1 log e (1 + x), x ¹ 0 . Then the function f: x

(1) decreases in (–1, ¥) n -1

g(n) =

k=1

2

æ (n - 1)n ö = n(n – 1) 2 ÷ø

å 2n = 2 çè

(2) decreases in (–1, 0) and increases in (0, ¥) (3) increases in (–1, ¥)

g(n) = 20 Þ n(n – 1) = 20

(4) increases in (–1, 0) and decreases in (0, ¥)

n=5

Official Ans. by NTA (1)

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session x - l n(1 + x) Sol. f '(x) = 1 + x 2 x x - (1 + x) l n(1 + x) = x 2 (1 + x)

9.

(3 + 2

Suppose h(x) = x – (1 + x) ln(1 + x) Þ h'(x) = 1 – ln(1 + x) – 1 = – ln(1 + x) Sol.

h'(x) < 0, " x Î (0, ¥) h(0) = 0 Þ h'(x) < 0 " x Î (–1, ¥)

)

2

) (

54 = 3 - 6 i

10.

A

Þ a1 + a11 = 0

where d is common difference Þ a1 = -5d

-54

)

1/ 2

)

2

(

+ 3 - 2 -54

) (

)

1/ 2

)

= 6, –6, 2 6i, – 2 6i,

11 =0 2

Þ a1 + a1 + 10d = 0

can be :

)

(

Official Ans. by NTA (2) Sol. a1 + a2 + a3+.....+a11 = 0 Þ (a1 + a11) ×

1/2

-54 = 3 + 2 ´ 3 ´ 6 i

EN

121 (4) 10

(3 + 2

= ± 3+ 6 i ± 3- 6 i

LL

72 (3) 5

)

(4) - 6

6

(3 + 2

is 0 (a 1 ¹ 0), then the sum of the A.P., a1, a3, a5,...,a23 is ka1, where k is equal to : 72 5

(

– 3 - 2 -54

(3)

(3 - 2

If the sum of first 11 terms of an A.P., a1 a2, a3,...

(2) -

1/2

(2) 6

(

Þ f(x) is a decreasing function for all xÎ(–1, ¥)

121 10

)

(1) -2 6

= 3+ 6 i

Þ f'(x) < 0 " xÎ(–1, ¥)

(1)

-54

Official Ans. by NTA (1)

h'(x) > 0, " x Î (–1, 0)

8.

The imaginary part of

1/x

æ æp öö lim ç tan ç + x ÷ ÷ x®0 øø è è4

is equal to :

(1) 2

(2) e

(3) 1

(4) e2

Official Ans. by NTA (4) 1/x

Sol.

ì æp öü lim í tan ç + x ÷ ý x®0 øþ î è4

=

e

1ì æp ö ü lim ítan ç + x ÷ -1ý x®0 x î è4 ø þ

a1 + a3 + a5 +......+a23 12 = (a1 + a 23 ) ´ = (a1 + a1 + 22d) ´ 6 2

æ 1+ tan x -1+ tan x ö ÷ x(1 - tan x) ø

ç ®0 è = e xlim

2 tan x

æ æ -a1 ö ö = ç 2a1 + 22 ç ÷÷ ´ 6 è 5 øø è

= -

= e xlim ®0 x(1- tan x) = e2

72 -72 a1 Þ K = 5 5 3

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 11.

The equation of the normal to the curve y = (1+x)2y + cos2(sin–1x) at x = 0 is :

12.

p For some qÎæç 0, ö÷ , if the eccentricity of the è 2ø

(1) y = 4x + 2

(2) x + 4y = 8

hyperbola, x2–y 2sec2q = 10 is

(3) y + 4x = 2

(4) 2y + x = 4

eccentricity of the ellipse, x2sec2q + y2 = 5, then the length of the latus rectum of the ellipse, is:

Official Ans. by NTA (2) Sol. Given equation of curve y = (1 + x)2y + cos2(sin–1x)

5 times the

30

(2)

4 5 3

(3) 2 6

(4)

2 5 3

(1)

at x = 0 y = (1 + 0)2y + cos2(sin–10) y=1+1 Official Ans. by NTA (2) y=2

So we have to find the normal at (0, 2) 2 y ln(1+ x) + cos2 cos -1 1 - x 2 Now y = e

y = e 2y ln(1+ x ) +

(

1 - x2

y = e2yln(1+x) + (1–x2)

)

)

equation of hyperbola Þ x2 – y2sec2q = 10

EN

(

æ pö Sol. Given qÎç 0, ÷ è 2ø

Þ

2

…(1)

LL

Now differentiate w.r.t. x

x2 y2 =1 10 10 cos2 q

Hence eccentricity of hyperbola

(e H ) = 1 +

é ù æ 1 ö y ' = e 2y ln(1+x) ê 2y × ç + ln(1 + x).2y' ú - 2x ÷ è1+ x ø ë û

10 cos2 q 10

…(1)

ìï b 2 üï íe = 1 + 2 ý a ïþ îï

Put x = 0 & y = 2

A

é ù æ 1 ö y ' = e 2´2 l n1 ê 2 ´ 2 ç + ln(1 + 0).2y 'ú - 2 ´ 0 ÷ è1+ 0 ø ë û

y' = e0[4 + 0] – 0

Now equation of ellipse Þ x2sec2q + y2 = 5

y' = 4 = slope of tangent to the curve 1 so slope of normal to the curve = - {m1m2=–1} 4

Hence eccenticity of ellipse

Hence equation of normal at (0, 2) is 1 y - 2 = - (x - 0) 4

Þ 4y – 8 = –x Þ x + 4y = 8

4

a 2 ïü ïì e = 1 í ý b 2 ïþ ïî

x2 y2 + =1 Þ 5cos2 q 5

(e E ) = 1 -

5cos2 q 5

(e E ) = 1 - cos 2 q = | sin q |= sin q

…(2)

ì æ p öü íQ qÎ ç 0, ÷ ý è 2 øþ î

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session given Þ eH =

Sol. Hence normal is ^r to both the lines so normal vector to the plane is r ˆ ´ (2iˆ + 3ˆj - k) ˆ n = (iˆ - 2 ˆj + 2k)

5 ee

Hence 1 + cos2q = 5sin2q 1 + cos2q = 5(1 – cos 2q) 1 + cos2q = 5 – 5cos 2q

ˆi ˆj kˆ r ˆ + 4) n = 1 -2 2 = ˆi(2 - 6) - ˆj( -1 - 4) + k(3 2 3 -1

6cos2q = 4 cos2q =

2 3

…(3)

r n = -4iˆ + 5ˆj + 7kˆ

Now length of latus rectum of ellipse

Now equation of plane passing through 2a 10 cos q 20 4 5 = = = b 3 5 3 5 2

=

(3,1,1) is Þ –4(x–3) + 5(y – 1) + 7(z – 1) = 0 Þ –4x + 12 + 5y – 5 + 7z – 7 = 0

Which of the following is a tautology ?

EN

13.

2

(1) (~p) Ù (pÚq)®q

(2) (q®p)Ú~(p®q)

Þ –4x + 5y + 7z = 0

(3) (p®q)Ù(q®p)

(4) (~q)Ú(pÙq)®q

Plane is also passing through (a, –3, 5) so this

Official Ans. by NTA (1) Sol. Option (1) is ~pÙ(pÚq) ® q

point satisfies the equation of plane so put in equation (1)

–4a + 5 × (–3) + 7 × (5) = 0

LL

º (~pÙp)Ú(~pÙq) ® q º CÚ(~pÙq) ® q º (~pÙq)®q º ~(~pÙq)Úq º (pÚ~q)Úq

15.

A

º (pÚq)Ú(~qÚq) º (pÚq) Ú t

14.

Þ –4a –15 + 35 = 0

Þ a= 5 Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1ÇE2ÇE3) = 0.

Then P( E C2 Ç E 3C /E1) is equal to :

so ~pÙ(pÚq) ® q is a tautology

(1) P( E 3C ) – P(E2)

A plane passing through the point (3, 1,1) contains two lines whose direction ratios are 1,

(3) P( E 3C ) – P( E C2 )

–2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (a, –3, 5), then a is equal to: (1) –10

(2) 5

(3) 10

(4) –5

Official Ans. by NTA (2)

…(1)

(2) P( E C2 ) + P(E3) (4) P(E3) – P( E C2 )

Official Ans. by NTA (1) Sol. Given E1, E2, E3 are pairwise indepedent events so P(E1ÇE2) = P(E1).P(E2) and P(E2ÇE3) = P(E2).P(E3) and P(E3ÇE1) = P(E3).P(E1) & P(E1ÇE2ÇE3) = 0

5

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session By solving these equation

æ E 2 Ç E3 ö P ëé E1 Ç (E 2 Ç E 3 ) ûù ÷= Now P ç P(E1 ) è E1 ø

=

we get x =

P(E1 ) - [ P(E1 Ç E2 ) + P(E1 Ç E3 ) - P(E1 Ç E2 Ç E3 )]

-11l 7l ; y = l; z = 2 2

Also given, x2 + y2 + z2 = 1

P(E1) 2

=

P(E1 ) - P(E1 ).P(E 2 ) - P(E1 )P(E 3 ) - 0 P(E1 )

Þ l=±

= 1 – P(E2) – P(E3) = [1 – P(E3)] – P(E2)

1 121 49 +1+ 4 4

= P(E C3 ) - P(E 2 )

so, there are 2 values of l.

Let A = {X = (x, y, z)T : PX = 0 and

\ so, there are 2 solution set of (x,y,z).

EN

16.

2

æ -11l ö æ 7l ö + (l )2 + ç ÷ = 1 Þ ç ÷ è 2 ø è 2 ø

17.

é1 2 1ù x 2 + y 2 + z 2 = 1} where P = ê -2 3 -4 ú , ê ú ëê 1 9 -1úû

then the set A :

LL

(1) is a singleton

Consider a region R = {(x, y) ÎR2 : x2£y £2x}.

If a line y = a divides the area of region R into two equal parts, then which of the following is true?

(1) a3–6a2 + 16 = 0

(2) 3a2 – 8a + 8 = 0

(3) a3 – 6a3/2–16 = 0

(4) 3a2 –8a3/2+8 = 0

Official Ans. by NTA (4)

(2) contains exactly two elements

(3) contains more than two elements

y=x

A

6

C A

y=a

O(0,0)

* y ³ x2 Þ upper region of y = x2 y £ 2x Þ lower region of y = 2x According to ques, area of OABC = 2 area of OAC

x + 2y + z = 0

infinite many solutions,

B

a

é1 2 1ù Sol. Given P = ê -2 3 -4 ú , Here |P| = 0 & also ê ú êë 1 9 1 úû

ü ï Þ -2x + 3y - 4z = 0 ý Q D = 0, so system have ï x + 9y - z = 0 þ

(2,4)

4

Official Ans. by NTA (2)

é 1 2 1 ù éxù Þ ê -2 3 -4 ú ê y ú = 0 ê úê ú êë 1 9 1 úû êë z úû

y=2x

Sol.

(4) is an empty set

given PX = 0

2

4

Þ

a

yö yö æ æ ò0 çè y - 2 ÷ø dy = 2 ò0 çè y - 2 ÷ø .dy

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session Þ

4 1 é2 ù = 2 ê a 3/2 - .a 2 ú 3 4 û ë3

Þ -

Þ 3a2 - 8a3/ 2 + 8 = 0 18.

Þ

If a curve y = f(x), passing through the point

ì Put x = 1& y = 2 ü 2x = lnx + C í ý y î then we get C = -1þ

-2x = l n(x) - 1 y

(1,2), is the solution of the differential equation, Þ y=

æ1ö 2x2dy= (2xy + y2)dx, then f ç ÷ is equal to : è2ø 1 (1) 1 - log 2 e

EN 19.

{x4+(k + 6)a}+..... where a ¹ 0 and x ¹ 1. If

Þ t+x

dt t2 = t+ dx 2

dt t = dx 2

Þ 2ò

dt = t2

ò

x10 - x + 45a(x - 1) , then k is equal to : x -1

(1) –5

(2) 1

(3) –3

(4) 3

Official Ans. by NTA (3) Sol. S = [x + ka + 0] + [x2 + ka + 2a] + [x3 + ka + 4a] + [x4 + ka + 6a] +......9 terms Þ S = (x + x2 + x3 + x4+.....9 terms) + (ka + ka + ka + ka +.......9 terms) + (0 + 2a + 4a + 6a + .......9 terms)

é x9 - 1 ù = S x ê ú + 9ka + 72a Þ ë x -1 û

2

Þ x

S=

LL

dt 2x 2 t + x 2 t 2 = dx 2x 2

A

Þ t+x

Let S be the sum of the first 9 terms of the series: {x + ka} + {x2 + (k + 2)a} + {x3+(k+4)a}+

dy 2xy + y 2 = {Homogeneous D.E.} dx 2x 2

ì let y = xt ü ï ï í dy dt ý ïî Þ dx = t + x dx ïþ

2x 1 - log e x

1 æ1ö so, f ç ÷ = è 2 ø 1 + loge 2

(4) 1+ loge2

Official Ans. by NTA (2) Sol. 2x2dy = (2xy + y2) dx Þ

Þ f(x) =

1 (2) 1 + log 2 e

-1 (3) 1 + log 2 e

2x 1- lnx

dx x

yü ì æ 1ö Þ 2 ç - ÷ = l n(x) + C í Put t = ý xþ è tø î

(x10 - x) + (9k + 72)a(x - 1) ÞS= (x - 1) Compare with given sum, then we get, (9k + 72) = 45 Þ k = -3

7

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 20.

The set of all possible values of q in the interval (0, p) for which the points (1, 2) and (sin q,

Sol. Let a be the first term and d be the common difference of the given A.P. Where d > 0

cosq) lie on the same side of the line x + y =

X= a+

1 is : æ pö (1) ç 0, ÷ è 4ø

æ 3p ö (2) ç 0, ÷ è 4 ø

æ p 3p ö (3) ç , ÷ è4 4 ø

æ pö (4) ç 0, ÷ è 2ø

= a + 5d S(X - x i )2 Þ varience = 11

Þ 90 × 11 = (25d 2 + 16d2 + 9d2 + 4d2) × 2

Official Ans. by NTA (4) Sol. Given that both points (1, 2) & (sinq, cosq) lie on same side of the line x + y – 1 = 0 n q, B(si

) co sq

Þ d = ±3 Þ d = 3 22.

x+y–1=0

2)

6 4 ü 1 ì3 If y = å k cos - í cos kx - sin kx ý , 5 î5 þ k =1

then

dy at x = 0 is________. dx

EN

A (1,

0 + d + 2d + ... + 10d 11

Official Ans. by NTA (91)

3 4 p 0<a< Sol. Put cos a = ,sin a = 5 5 2

æ Put (1,2) in ö æ Put (sin q, cos q in ö So, ç ÷ ç ÷>0 è given line ø è given line ø

Þ (1 + 2 – 1) (sin q + cos q – 1) > 0 Þ sin q + cos q > 1 ¸ by 2

1 1 1 sin q + cos q > 2 2 2

A

Þ

pö 1 æ Þ sin ç q + ÷ > 4ø 2 è

Þ

3p 4

p 4

p 2

If the variance of the terms in an increasing A.P., b 1 , b 2 , b 3 ,....b 11 is 90, then the common difference of this A.P. is_______.

Official Ans. by NTA (3.00)

3 4 cos kx - sin kx 5 5

= cos a . cos kx – sin a . sin kx = cos(a + kx) As we have to find derivate at x = 0 We have cos–1 (cos(a + kx)) = (a + kx) Þ y=

p p 3p < q+ < 4 4 4

Þ 0<q< 21.

}

LL

{

Now

6

å (a + kx) k =1

dy = Þ dx at x = 0

23.

6

åk = k= x

6 ´ 7 ´ 13 = 91 6

Let the position vectors of points 'A' and 'B' be ˆi + ˆj + kˆ and 2iˆ + ˆj + 3kˆ , respectively. A point

'P' divides the line segment AB internally in the ratio l : 1 (l > 0). If O is the origin and uuur uuur uuur uuur OB × OP - 3 | OA ´ OP |2 = 6 , then l is equal to______.

Official Ans. by NTA (0.8)

8

Final JEE - Main Exam September, 2020/02-09-2020/Evening Session l A(i + j + k)

n

B(2i + j + 3k)

Using section formula we get

OP =

2l + 1 ˆ l + 1 ˆ 3l + 1 ˆ i+ j+ k l +1 l +1 l +1

Now OB × OP =

=

4 l + 2 + l + 1 + 9l + 3 l +1

24.

æ 1ö For a positive integer n, ç1 + ÷ is expanded è xø

in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2 : 5 : 12, then n is equal to________.

Official Ans. by NTA (118) Sol. nCr–1 : nCr : nCr+1 = 2:5:12 n

Now

14l + 6 l +1

Cr -1 2 = Cr 5

n

Þ 7r = 2n + 2

…(1)

n n

Þ 17r = 5n – 12

Þ n = 118

| OA ´ OP |2 =

(2l + 1)2 + l 2 + l 2 (l + 1) 2

6l 2 + 1 = (l + 1)2

Let [t] denote the greatest integer less than or equal to t. Then the value of

Þ l = 0,

8 = 0.8 10

2

1

| 2x - [3x] | dx

5 < 3x < 6 ; 2

Now

ò | 2x - [3x] | dx 1

14l + 6 (6l 2 + 1) 3 ´ =6 Þ l +1 (l + 1) 2 Þ 10l2 – 8l = 0

ò

is_______.

Official Ans. by NTA (1.0) Sol. 3 < 3x < 6 Take cases when 3 < 3x < 4, 4 < 3x < 5,

LL

2l + 1 ˆ -l ˆ -l ˆ i+ j+ k l +1 l +1 l +1

…(2)

On solving (1) & (2)

25.

=

Cr 5 = C r +1 12

EN

ˆi ˆj kˆ OA ´ OP = 1 1 1 2l + 1 3l + 1 1 l +1 l +1

A

Sol.

1

4/3

=

ò 1

=

5/3

(3 - 2x)dx + ò (4 - 2x) dx + 4/3

2

ò (5 - 2x)dx

5/3

2 3 4 + + =1 9 9 9

Þ l = 0.8

9

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