paper-1 (b.e./b.tech.) jee main 2019€¦ · jee main 8 april 2019 shift-2 mathematics page no. 1...

PAPER-1 (B.E./B.TECH.)

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 8th April 2019 (Shift-2)

Time: 02:30 P.M. to 05:30 P.M.

Durations: 3 Hours | Max. Marks: 360

Subject: Mathematics

www.embibe.com

JEE MAIN 8 APRIL 2019 SHIFT-2 Mathematics

Page No. 1

1. The number of times a fair coin should be tossed so that P (at least one head) is greater than 90%, is

(A) 3

(B) 4

(C) 5

(D) 6

Solution: (B)

Let a fair coin is tossed ‘n’ times then 𝑃 (at least one head)

= 1– 𝑃(All tail)

= 1– (1

2)

𝑛

>90

100 (given)

⇒ (1

2)

𝑛

< 1 −90

100=

1

10

⇒ 2𝑛 > 10

⇒ 𝑛 ≥ 4

2. A right circular cylinder of maximum value is cut from a sphere of radius 3m. The height of the cylinder is

(A) 2√3

(B) 3√2

(C) 3

(D) 3√3

Solution: (A)


Page No. 2

From the diagram, we get radius of cylinder = √𝑅2 − (ℎ

2)

2

⇒ Volume of cylinder 𝑉 = 𝜋 (𝑅2 −ℎ2

4) ℎ

For maximum volume: 𝑑𝑣

𝑑ℎ= 0

⇒ 𝜋 (𝑅2 −3ℎ2

4) = 0

⇒ ℎ2 =4𝑅2

3=

4 × 9

3= 12

⇒ ℎ = 2√3

3. Given 𝑓(1) = 1 and 𝑓′(1) = 3, then the derivative of 𝑓 (𝑓(𝑓(𝑥))) + (𝑓(𝑥))2

𝑎𝑡 𝑥 = 1 is

(A) 43

(B) 33

(C) 38

(D) 41

Solution: (B)

𝑦 = 𝑓 (𝑓(𝑓(𝑥))) + (𝑓(𝑥))2

⇒𝑑𝑦

𝑑𝑥= 𝑓′ (𝑓(𝑓(𝑥))) × 𝑓′(𝑓(𝑥)) × 𝑓′(𝑥) + 2𝑓(𝑥) × 𝑓′(𝑥)


Page No. 3

⇒ (dy

𝑑𝑥)

𝑥=1= 𝑓′ (𝑓(𝑓(1))) × 𝑓′(𝑓(1)) × 𝑓′(1) + 2𝑓(1) × 𝑓′(1)

= 𝑓′(𝑓(1)) × 𝑓′(1) × 3 + 2 × 1 × 3

= 𝑓′(1) × 3 × 3 + 6

= 3 × 3 × 3 + 6

= 27 + 6

= 33

4. How many four digits number greater than 4321 can be formed with the digit 0, 1, 2, 3, 4 and 5 if repetition

is allowed?

(A) 360

(B) 306

(C) 310

(D) 288

Solution: (C)

∴ Total Numbers = 216 + 72 + 184 + 4 = 310


Page No. 4

5. Evaluate: ∑𝑘

2𝑘

20

𝑘=1

(A) 2 –1

213

(B) 2 –11

220

(C) 2–11

219

(D) 2 –1

220

Solution: (C)

𝑆 = ∑𝑘

2𝑘

20

𝑘=1

=1

2+

2

22+

3

23+

4

24+ ⋯ . . +

20

220

1

2 𝑆 = ∑

𝑘

2𝑘+1

20

𝑘=1

=1

22+

2

23+

3

24+ ⋯ +

19

220+

20

22

By subtraction,

1

2 𝑆 = (

1

2+

1

22+

1

23+ … … . +

1

220) –

20

221

=

1

2(1−(

1

2)

20)

1−1

2

−20

221 (Using sum of 𝑛 terms of G.P.)

= 1–1

220 –20

221 = 1 –11

220

⇒ 𝑆 = 2 −11

219

6. If |𝐴| = |1 1 22 6 𝑐4 62 𝑐2

| and |𝐴| ∈ [2, 16]. Given 2, 𝑏, 𝑐 are in AP, then range of 𝑐 is

(A) [2, 4]

(B) [2 + 21 3⁄ , 4]

(C) [3, + 2 + 21 3⁄ ]

(D) [4, 6]


Page No. 5

Solution: (D)

|1 1 12 𝑏 𝑐4 𝑏2 𝑐2

| = |1 0 02 𝑏 − 2 𝑐 − 24 𝑏2 − 4 𝑐2 − 4

| = (𝑏 − 2)(𝑐 − 2)(𝑐 − 𝑏)

Let common difference of 𝐴𝑃 = 𝑑

∴ 𝑏 = 2 + 𝑑, 𝑐 = 2 + 2𝑑

|𝐴| = 𝑑 × 2𝑑 × 𝑑 = 2𝑑3 ∈ [2, 16]

⇒ 𝑑3 ∈ [1, 8] ⇒ 𝑑 ∈ [1, 2] ⇒ 2𝑑 ∈ [2, 4]

⇒ 2 + 2𝑑 ∈ [4, 6]

∴ 𝑐 ∈ [4, 6]

7. If the points 𝑃(3, 2, 1), 𝑄(2, −1, 3) and 𝑅(4, 𝑦, 𝑧) lie on the same line, then distance of R from origin is

(A) √37

(B) √33

(C) √42

(D) 6

Solution: (C)

Direction Rates of 𝑃𝑄: 1, 3, −2

Direction rates of 𝑃𝑅: 1, 𝑦 − 2, 𝑧 − 1

Since 𝑃, 𝑄, 𝑅 are collinear

⇒1

1=

𝑦 − 2

3=

𝑧 − 1

−2

⇒ 𝑦 = 5 and 𝑧 = −1 ⇒ 𝑅(4, 5, −1)

⇒ 𝑂𝑅 = √42

8. 𝑧 =√3

2+

𝑖

2, then the value of (1 + 𝑖𝑧 + 𝑧5 + 𝑖𝑧8)9 is equal to

(A) −1


Page No. 6

(B) 1 (C) (1 + 2𝑖𝑧)9 (D) 0 Solution: (A)

𝑧 =√3

2+

𝑖

2 ∴ 𝑖𝑧 = −

1

2+

√3

2𝑖 = 𝜔

∴ (1 + 𝑖𝑧 + 𝑧5 + 𝑖𝑧8)9 = (1 + 𝜔 +𝜔5

𝑖5+ 𝑖 ×

𝜔8

𝑖8)

9

= (1 + 𝜔 − 𝑖𝜔2 + 𝑖𝜔2)9 = (−𝜔2)9 = −𝜔18 = −1

9. ∫𝑑𝑥

𝑥3(1+𝑥6)2 3⁄ =

(A) (1+𝑥6)

1 3⁄

2𝑥2 + 𝑐

(B) (1+𝑥6)

1 3⁄

𝑥2 + 𝑐

(C) −(1+𝑥6)

1 3⁄

2𝑥2 + 𝑐

(D) −(1+𝑥6)

1 3⁄

𝑥2 + 𝑐

Solution: (C)

∫𝑑𝑥

𝑥3(1 + 𝑥6)2 3⁄ = ∫

𝑑𝑥

𝑥7 (1

𝑥6 + 1)2 3⁄

Put 𝑡 =1

𝑥6 + 1

𝑑𝑡 = −6

𝑥7𝑑𝑥 ⇒

𝑑𝑥

𝑥7= −

𝑑𝑡

6

∴ −1

6 ∫

1

𝑡2 3⁄ 𝑑𝑡 =1

6×

𝑡1 3⁄

1

3

+ 𝑐

= −1

2(

1

𝑥6+ 1)

1

3

+ 𝑐

= −(1 + 𝑥6)1 3⁄

2𝑥2+ 𝑐

10. The marks of a student in 6 tests are 41, 45, 54, 57, 43 and 𝑥. If the mean marks for these 6 tests is 48, then the standard deviation of these 6 tests is

(A) 10

√2

(B) 10

√3

(C) 10

3

(D) 2

3

Solution: (B) Given mean of 6 observations is 48


Page No. 7

⇒41 + 45 + 54 + 57 + 43 + 𝑥

6= 48

⇒ 𝑥 = 48

Now variance 𝜎2 =∑(𝑥−�̅�)2

𝑛

=72 + 32 + 62 + 92 + 52 + 02

6

=100

3

Hence standard deviation =10

√3.

11. A tangent is drawn to the parabola 𝑦2 = 4𝑥 at the point where it is cut by the circle 𝑥2 + 𝑦2 = 5, then

which of the following point lies on tangent

(A) (−1, 4)

(B) (2, 7)

(C) (3, 4)

(D) (4, 3)

Solution: (C)

Let’s first find points of intersection of two curves.

𝑥2 + 4𝑥 = 5

⇒ 𝑥 = 1, − 5 (neglect 𝑥 = −5)

Hence point of intersections are (1, 2) and (1, − 2).


Page No. 8

Tangent to the parabola at (1, 2)

𝑦 ⋅ 2 = 2(𝑥 + 1)

⇒ 𝑦 = 𝑥 + 1 clearly point (3, 4) satisfy the given equation of tangent.

12. If 𝑓′(3) + 𝑓′(2) = 0, then lim𝑥→0

(1+𝑓(3+𝑥)−𝑓(3)

1+𝑓(2−𝑥)−𝑓(2))

1

𝑥 is equal to

(A) 𝑒𝜋

(B) 𝑒2

(C) 1

(D) 𝑒1

2

Solution: (C)

This limit is 1∞ form

∴ lim𝑥→0

(1 + 𝑓(3 + 𝑥) − 𝑓(3)

1 + 𝑓(2 − 𝑥) − 𝑓(2))

1

𝑥

= 𝑒lim𝑥→0

1

𝑥(

1+𝑓(3+𝑥)−𝑓(3)

1+𝑓(2−𝑥)−𝑓(2)−1)

= 𝑒lim𝑥→0

1

𝑥(

𝑓(3+𝑥)−𝑓(3)−𝑓(2−𝑥)+𝑓(2)

1+𝑓(2−𝑥)−𝑓(2))

(0

0 form)

= 𝑒lim𝑥→0

𝑓′(3+𝑥)+𝑓′(2−𝑥)

𝑥(−𝑓(2−𝑥))+1+𝑓(2−𝑥)−𝑓(2)

= 𝑒0 = 1

13. If sides of triangle are in AP and the longest angle is double smallest angle then find ratio of sides

(A) 3: 5: 6

(B) 4: 5: 6

(C) 2: 3: 5

(D) 3: 4: 5

Solution: (B)

Let 𝑎 < 𝑏 < 𝑐 are three sides and 𝜃 is the smallest angle. Hence three angles are 𝜃, 𝜋 − 3𝜃, 2𝜃


Page No. 9

Given, 2𝑏 = 𝑎 + 𝑐 ⇒ 2 sin 𝐵 = sin 𝐴 + sin 𝐶 (using sine rule)

⇒ 2sin 3𝜃 = sin 𝜃 + sin 2𝜃

⇒ 2(3sin 𝜃 − 4sin3𝜃) = sin 𝜃(1 + 2cos 𝜃)

⇒ 6 − 8(1 − cos2𝜃) = 1 + 2cos 𝜃

⇒ cos 𝜃 =3

4, −

1

2 (−

1

2 is reserved since 𝜃 is acute)

∴ 𝑎: 𝑏: 𝑐 = sin𝐴: sin 𝐵: sin 𝐶 = sin 𝜃: sin 3𝜃: sin 2𝜃

= 1: 3 − 4 sin2𝜃: 2cos 𝜃 = 1: 4cos2𝜃 − 1: 2 cos 𝜃

= 4: 5: 6

14. If the equation (1 + 𝑚2)𝑥2 − 2(1 + 3𝑚)𝑥 + (1 + 8𝑚) = 0 has no real solutions, then number of values of

𝑚 is

(A) 5

(B) 3

(C) only 1

(D) Infinite

Solution: (D)

Since the given equation has no real solution.

⇒ 𝐷 < 0

⇒ 4(1 + 3𝑚)2 − 4 ∙ (1 + 𝑚2)(1 + 8𝑚) < 0

⇒ (1 + 9𝑚2 + 6𝑚) − (1 + 8𝑚 + 𝑚2 + 8𝑚3) < 0

⇒ −8𝑚(2𝑚 − 1)2 < 0

Clearly for infinite values of 𝑚 given condition is true.

15. Consider two perpendicular lines 𝐿1 and 𝐿2. If points (ℎ, 𝑘), (1, 2) and (−3, 4) lie on 𝐿1 while the points

(ℎ, 𝑘) and (4, 3) lies on 𝐿2, then the value of ℎ

𝑘 equals to

(A) 1

3


Page No. 10

(B) 3

(C) 2

3

(D) 7

3

Solution: (B)

Slope of line 𝐿1 is 4−2

−3−1=

2

−4=

−1

2

Hence slope of 𝐿2 is 2.

From line 𝐿1𝑘−2

ℎ−1=

−1

2

⇒ ℎ + 2𝑘 = 5…..(1)

From line 𝐿2 .𝑘−3

ℎ−4= 2

⇒ 𝑘 = 2ℎ − 5 ..…(2)

From (1) and (2) ℎ = 3 and 𝑘 = 1

⇒ℎ

𝑘= 3

16. Two poles of height 20 𝑚 and 80 𝑚 are standing on ground at same distance. Find the distance of point of

intersection, of the lines from top of each pole to the base of other, from the ground.

(A) 8

(B) 12

(C) 15

(D) 16

Solution: (D)


Page No. 11

Equation of line 𝑂𝑄 𝑦 =80

𝑎𝑥

Equation of line 𝐴𝐵 𝑥

𝑎+

𝑦

20= 1

Solving for 𝑃(𝑥0, 𝑦0),𝑎𝑦0 80⁄

𝑎+

𝑦0

20= 1

⇒ 𝑦0 (1

80+

1

20) = 1 ⇒ 𝑦0 = 16

17. The difference of length of major axis and minor axis of an ellipse is 10 and are of the focus is (5√3, 0). If

the coordinate axis are the axes of ellipse, then the latus rectum of ellipse is

(A) 9

5

(B) 5

(C) 13

2

(D) 7

Solution: (B)

Given difference of length of major and minor axes is 10

⇒ 2𝑎 − 2𝑏 = 10 ⇒ 𝑎 − 𝑏 = 5 …(1)


Page No. 12

Also focus of ellipse is (5√3, 0)

⇒ 𝑎𝑒 = 5√3 …(2)

Since 𝑏2 = 𝑎2(1 − 𝑒2)

⇒ 𝑏2 = 𝑎2 − 75

⇒ (𝑎 − 𝑏)(𝑎 + 𝑏) = 75

⇒ 𝑎 + 𝑏 = 15 …(3)

From (1) and (3)

𝑎 = 10 and 𝑏 = 5

Hence length of the latus rectum = 2𝑏2

𝑎= 2 ⋅

25

10= 5

18. The area of triangle formed by tangent as well as normal to the circle 𝑥2 + 𝑦2 = 4 at (√3, 1) and x-axis is

(A) 4

√3

(B) 3

(C) 2√3

5

(D) 2

√3

Solution: (D)

Equation tangent at 𝑃(√3, 1)

𝑥 ⋅ √3 + 𝑦 ⋅ 1 = 4


Page No. 13

x-intercept of this tangent is 4

√3

Hence area of 𝛥 =1

2⋅

4

√3⋅ 1 =

2

√3

19. Let 𝑓(𝑥, 𝑦) = {(𝑥, 𝑦): 𝑦2 ≤ 4𝑥, 0 ≤ 𝑥 ≤ 𝜆} and 𝑆(𝜆) is area such that 𝑆(𝜆)

𝑆(4)=

2

5, then the vale of 𝜆 is equal to

(A) 4 (4

25)

1

3

(B) 4 (2

25)

1

3

(C) 3 (4

25)

1

3

(D) 2 (2

25)

1

3

Solution: (A)

𝑆(𝜆) = 2 ∫ 2√𝑥 𝑑𝑥 = 4 [𝑥

3

2

3

2

]

0

𝜆

𝜆

0

=8

3𝜆

3

2

∴ 𝑆(4) =8

3(4)

3

2

Given 𝑆(𝜆)

𝑆(4)=

2

5⇒

𝜆32

432

=2

5


Page No. 14

⇒ 𝜆 = 4 (4

25)

1

3

20. If 𝑎, 𝑏, 𝑐 are in G.P. and equations 𝑎𝑥2 + 2𝑏𝑥 + 𝑐 = 0 and 𝑑𝑥2 + 2𝑒𝑥 + 𝑓 = 0 have a common root then:

(A) 𝑎

𝑑,

𝑏

𝑒,

𝑐

𝑓 are in A.P.

(B) 𝑑, 𝑒, 𝑓 are in G.P.

(C) 𝑎

𝑑,

𝑏

𝑒,

𝑐

𝑓 are in H.P.

(D) 𝑑, 𝑒, 𝑓 are in A.P.

Solution: (C)

∵ 𝑎, 𝑏, 𝑐 are in G.P. ⇒ 𝑏2 = 𝑎𝑐 …(i)

Equation 𝑎𝑥2 + 2𝑏𝑥 + 𝑐 = 0 → 𝐷 = 4𝑏2 − 4𝑎𝑐 = 0 (∵ 𝑏2 = 𝑎𝑐)

⇒ equation has equal roots ⇒ roots are =−𝑏

𝑎,

−𝑏

𝑎 (∵ sum of roots 𝛼 + 𝛼 =

−2𝑏

𝑎)

∵ both equations have one root in common

∴ common root is −𝑏

𝑎 it will satisfy equation (ii)

∴ 𝑑 (−𝑏

𝑎)

2

+ 2𝑒(−𝑏)𝑎 + 𝑓 = 0 ⇒ 𝑑𝑏2 − 2𝑎𝑒𝑏 + 𝑎2𝑓 = 0

⇒ 𝑑(𝑎𝑐) − 2𝑎𝑒𝑏 + 𝑎2𝑓 = 0 (∵ 𝑏2 = 𝑎𝑐)

⇒ 𝑑𝑐 − 2𝑎𝑒𝑏 + 𝑎𝑓 = 0

⇒𝑑𝑐

𝑎𝑐−

2𝑒𝑏

𝑎𝑐+

𝑎𝑓

𝑎𝑐= 0 (dividing by 𝑎𝑐)

⇒𝑑

𝑎−

2𝑒𝑏

𝑏2+

𝑓

𝑐= −⇒

𝑑

𝑎+

𝑓

𝑐=

2𝑒

𝑏

⇒𝑑

𝑎,

𝑒

𝑏,

𝑓

𝑐 are in A.P. ⇒

𝑎

𝑑,

𝑑

𝑒,

𝑐

𝑓 are in H.P.

21. 𝑓(𝑥) is defined as 𝑓(𝑥) = {

|𝑥| + [𝑥]; −1 <, 𝑥 < 1𝑥 + |𝑥|; 1 ≤ 𝑥 < 2

|𝑥| + [𝑥]; 2 ≤ 𝑥 < 3, then the number of point of discontinuity of 𝑓(𝑥) is


Page No. 15

(A) 1

(B) 2

(C) 0

(D) 3

Solution: (B)

𝑓(𝑥) = {

|𝑥| + [𝑥]; −1 <, 𝑥 < 1𝑥 + |𝑥|; 1 ≤ 𝑥 < 𝑧

|𝑥| + [𝑥]; 2 ≤ 𝑥 < 3

= {

−𝑥 − 1; −1 < 𝑥 ≤ 0𝑥 + 0; 0 ≤ 𝑥 < 1

2𝑥; 1 ≤ 𝑥 < 2𝑥 + 2; 2 ≤ 𝑥 < 3

At 𝑥 = 0, 1, 2 𝑓 is changes its definition

∴ at 𝑥 = 0 L.H.L= −1, R.H.L = 0 ⇒ 𝑓 is discontinuous at 𝑥 = 0

𝑥 = 1 L.H.L= 1, R.H.L = 2 ⇒ 𝑓 is discontinuous at 𝑥 = 1

𝑥 = 2 L.H.L = R.H.L = 𝑓(2) = 4 ⇒ 𝑓 is continuous at 𝑥 = 2

∴ Pt of discontinuity are 0,1.

23. Equation of plane passing through line of intersection of planes 𝑥 + 𝑦 + 𝑧 = 1 and 2𝑥 + 3𝑦 + 𝑧 = 5 and

perpendicular to the plane 𝑥 − 𝑦 − 𝑧 = 0

(A) 𝑟. (𝑗 + �̂�) = 3

(B) 𝑟. (𝑗 − 𝑘) = 3

(C) 𝑟. (𝑗 − �̂�) = 2

(D) None

Solution: (B)

Equation of plane passing through line of intersection of plane 𝑃1 and 𝑃2 is 𝑃1 + 𝜆𝑃2 = 0

∴ equation of required plane is (𝑥 + 𝑦 + 𝑧 − 1) + 𝜆(2𝑥 + 3𝑦 + 𝑧 − 5) = 0

⇒ (2𝜆 + 1)𝑥 + (3𝜆 + 1)𝑦 + (1 + 𝜆)𝑧 − 1 − 5𝜆 = 0 …(i)


Page No. 16

∵ Plane (i) is perpendicular to 𝑥 − 𝑦 − 𝑧 = 0

∴ (2𝜆 + 1)1 + (3𝜆 + 1)(−1) + (1 + 𝜆)(−1) = 0

⇒ −2𝜆 − 1 = 0 ⇒ 𝜆 =−1

2

∴ equation to required plane is −1

2𝑦 +

1

2𝑧 +

3

2= 0

⇒ 𝑦 − 𝑧 − 3 = 0

⇒ 𝑦 − 𝑧 = 3

And vector form is 𝑟 ∙ (𝑗 − �̂�) = 3

24. If the slope of tangent at point (𝑥, 𝑦) of curve 𝑦 = 𝑓(𝑥) is given by 2𝑦

𝑥2. If this curve passes through the

centre of the circle 𝑥2 + 𝑦2 − 2𝑥 − 2𝑦 = 0. Then curve is

(A) 𝑥 ln 𝑦 = 2𝑥 − 1

(B) 𝑥 ln 𝑦 = 2(𝑥 − 1)

(C) 𝑥2 ln 𝑦 = 2𝑥 + 1

(D) 𝑥2 ln 𝑦 = (𝑥 − 1)

Solution: (B)

Slope of tangent 𝑑𝑦

𝑑𝑥=

2𝑦

𝑥2

⇒𝑑𝑦

𝑦=

2

𝑥2𝑑𝑥

Integrating both side we get

ln 𝑦 =−2

𝑥+ 𝑐

∵ it passes through centre (1, 1) of given circle

⇒ 0 = −2 + 𝑐 ⇒ 𝑐 = 2

∴ solution ln 𝑦 = −2

𝑥+ 2

⇒ 𝑥 ln 𝑦 = −2 + 2𝑥

⇒ 𝑥 ln 𝑦 = 2(𝑥 − 1)


Page No. 17

25. Which of the following is not a tautology

(A) 𝑝 → (𝑝 ∨ 𝑞)

(B) (𝑝 ∨ 𝑞) → 𝑝

(C) (𝑝 ∨ ~𝑝)

(D) (𝑝 ∨ 𝑞) → (𝑝 ∧ ~𝑞)

Solution: (D)

A B C D

p q 𝑝𝑞 𝑝∧ 𝑞

𝑝→ (𝑝 ∨ 𝑞)

(𝑝 ∧ 𝑞)→ 𝑝

(𝑝∨ ~𝑝)

(𝑝∧ ~𝑞)

(𝑝 ∨ 𝑞)→ (𝑝 ∧ ~𝑞)

T T T T T T T F F

T F T F T T T T T

F T T F T T T F F

F F F F T T T F T

paper-1 (b.e./b.tech.) jee main 2019€¦ · jee main 8 april 2019 shift-2 mathematics page no. 1...

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