PAPER-1 (B.E./B.TECH.) JEE MAIN 2019 · दഉ༘द༢Մ and दഈ༘Ո༟Մ ⇒दᐌद༘Յᐍᐌद༗Յᐍ༢Մand द༟Ն, ༘Ն द∈ᐌ༘Յ, ՄᐍᐌՅ, ∞ᐍand द༟Ն,

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

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 9th April 2019 (Shift-2)

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

Durations: 3 Hours | Max. Marks: 360

Subject: Mathematics

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JEE MAIN 8 APRIL 2019 SHIFT-2 MATHAMETICS

1. The area of smaller circle out of two circles which touch the parabola 𝑦2 = 4𝑥 at (1, 2) and also touch the x-axis, is

(A) 4𝜋(2 − √2)

(B) 8𝜋(3 − 2√2)

(C) 4𝜋(2 + √2)

(D) 2𝜋(1 + √2)

Solution: (B)

Equation of tangent to

Parabola at (1, 2) is

𝑦 ⋅ 2 = 2(𝑥 + 1) ⇒ 𝑥 − 𝑦 + 1 = 0 …(1)

Equation of family of circle touching the parabola at (1, 2) is

(𝑥 − 1)2 + (𝑦 − 2)2 + 𝜆(𝑥 − 𝑦 + 1) = 0 …(2)

Since circle touches x-axis. Putting 𝑦 = 0

𝑥2 + (𝜆 − 2)𝑥 + (𝜆 + 5) = 0

Now 𝐷 = 0 ⇒ (𝜆 − 2)2 − 4 ⋅ 1 ⋅ (𝜆 + 5) = 0

⇒ 𝜆 = 4 ± 4√2 for smaller circle 𝜆 = 4 − 4√2.

Radius of smaller circle = 4 − 2√2

Area of circle = 𝜋(4 − 2√2)2

= 8𝜋(3 − 2√2)

2. If 𝑓(𝑥) = {𝑎|𝜋 − 𝑥| + 1 𝑥 ≤ 5𝑏|𝑥 − 𝜋| + 3 𝑥 > 5

is continuous ∀ 𝑥 ∈ 𝑅, then 𝑎 − 𝑏 equal to

(A) 3

4+𝜋


(B)

1

5−𝜋

(C) 2

5−𝜋

(D) 1

5+𝜋

Solution: (C)

For continuity at 𝑥 = 5

lim𝑥→5+

𝑓(𝑥) = lim𝑥→5−

𝑓(𝑥) = 𝑓(5)

⇒ lim𝑥→5+

𝑏|𝑥 − 𝜋| + 3 = lim𝑥→5−

𝑎|𝑥 − 𝜋| + 1 = 𝑎(5 − 𝜋) + 1

⇒ 𝑏(5 − 𝜋) + 3 = 𝑎(5 − 𝜋) + 1 = 𝑎(5 − 𝜋) + 1

⇒ 𝑏 − 𝑎 =2

𝜋 − 5

3. If the system of linear equations 2𝑥 + 3𝑦 − 𝑧 = 0, 2𝑥 + 𝑘𝑦 − 3𝑧 = 0 and 2𝑥 − 𝑦 + 𝑧 = 0 has non triangle

solution then 𝑥

𝑦+

𝑦

𝑧+

𝑧

𝑥+ 𝑘 will be

(A) 1

(B) −2

(C) 3

(D) 4

Solution: (C)

Given homogeneous system is

2𝑥 + 3𝑦 − 𝑧 = 0 …(1)

2𝑥 + 𝑘𝑦 − 3𝑧 = 0 …(2)

2𝑥 − 𝑦 + 𝑧 = 0 …(3)

It will have non trivial solutions if

|2 3 −12 𝑘 −3𝑧 −1 1

| = 0 ⇒ 𝑘 = 7

From equations (1) and (3), we get 4𝑥 + 2𝑦 = 0 ⇒ 𝑦 = −2𝑥 and 𝑧 = 𝑦 − 2𝑥 = −4𝑥

So, 𝑥

𝑦+

𝑦

𝑧+

𝑧

𝑥+ 𝑘


=𝑥

−2𝑥+

−2𝑥

−4𝑥+

−4𝑥

𝑥+ 7

= −1

2+

1

2− 4 + 7 = 3

4. If 𝑓(𝑥) = [𝑥] − [𝑥

4] where [. ]𝐺𝐼𝐹, then

(A) lim𝑥→4+

𝑓(𝑥) and lim𝑥→4−

𝑓(𝑥) both does not exist

(B) Function 𝑓(𝑥) is continuous at 𝑥 = 4

(C) lim𝑥→4+

𝑓(𝑥) and lim𝑥→4−

𝑓(𝑥) exist but not equal

(D) lim𝑥→4+

𝑓(𝑥) exist but lim𝑥→4−

𝑓(𝑥) does not exist

Solution: ()

𝑅𝐻𝐿 = lim𝑥→4+

[𝑥] − [𝑥

4] = 4 − 1 = 13

𝐿𝐻𝐿 = lim𝑥→4−

[𝑥] − [𝑥

4] = 3 − 0 = 3

𝑓(𝑥) = 4 − 1 = 3

Hence given function is continuous at 𝑥 = 4

5. There are two towers of 5 𝑚 and 10 𝑚. The line joining there tops makes 15𝑜 with ground then find the

distance between them.

(A) 5(√3 − 1)

(B) 10(1 + √3)

(C) 5(2 + √3)

(D) 2(2 + √3)

Solution: (C)


In 𝛥𝐵𝐷𝑀

tan15𝑜 =5

𝐾

⇒ 𝐾 =5

tan15𝑜

=5(√3 + 1)

√3 − 1×

√3 + 1

√3 + 1

=5(3 + 1 + 2√3)

3 − 1

= 5(2 + √3)

6. Balls are arranged in the form of an equilateral triangle with 𝑛 rows such that 𝑟𝑡ℎ row has 𝑟 balls in it. 2𝑓 99

more balls are added to existing balls then a square of (𝑛 − 2) balls in each row can be formed. Number of

balls in equilateral triangle is

(A) 290

(B) 190

(C) 200

(D) 175

Solution: (B)

From the given condition we can write

𝑛(𝑛 + 1)

2+ 99 = (𝑛 − 2)2

⇒ 𝑛(𝑛 + 1) + 198 = 2(𝑛 − 2)2

⇒ 𝑛2 − 9𝑛 − 190 = 0

⇒ 𝑛 = 19


Hence number of balls sin equilateral triangle =

19⋅20

2= 190

7. If the lines 𝑥 + (𝑎 − 1)𝑦 = 1 and 2𝑥 + 𝑎2𝑦 = 1 where 𝑎 ∈ 𝑅 − {0, 1} are perpendicular to each other, then

distance of their point intersection from origin is

(A) √5

2

(B) √2

5

(C) 2

√5

(D) √2

5

Solution: (B)

Since given lines are perpendicular, hence

1 ⋅ 2 + (𝑎 − 1) ⋅ 𝑎2 = 0

⇒ 𝑎3 − 𝑎2 + 2 = 0

⇒ (𝑎 + 1) ⋅ (𝑎2 − 2𝑎 + 2) = 0

⇒ 𝑎 = −1

Lines are 𝑥 − 2𝑦 = 1 and 2𝑥 + 𝑦 = 1.

Point of intersection of two lines is (3

5,

−1

5)

Hence distance from origin = √9

25+

1

25

= √2

5

8. If tangent of 𝑦2 = 𝑥 at (𝛼, 𝛽) where 𝛽 > 0 is also a tangent of ellipse 𝑥2 + 2𝑦2 = 1, then 𝛼 equals to

(A) √2 + 1

(B) 2√2 − 1

(C) 2 + √2

(D) 2√2 + 1


Solution: (A)

Since (𝛼, 𝛽) lies on the parabola ⇒ 𝛼 = 𝛽2

⇒ (𝛼, 𝛽) = (𝛽2, 𝛽)

Tangent to parabola 𝑦 ⋅ 𝛽 =1

2(𝑥 + 𝛽2)

⇒ 𝑦 =𝑥

2𝛽+

𝛽

2 … (1)

Tangent to parabola is also a tangent to ellipse 𝑥2

4= 1 ⋅ (

1

4𝛽2) +

1

2

⇒ 𝛽2 = √2 + 1 ⇒ 𝛼 = √2 + 1.

9. The area enclosed by 𝑦2

2≤ 𝑥 ≤ 𝑦 + 4 equals to

(A) 18

(B) 53

3

(C) 16

(D) 14

Solution: (A)

Required are

= ∫ (𝑦 + 4 −𝑦2

2) 𝑑𝑦

4

−2

= (𝑦2

2+ 4𝑦 −

𝑦3

6)|

−2

4

= (8 + 16 −32

3) − (2 − 8 +

4

3)


= 18

10. The domain of the function 𝑓(𝑥) =1

(𝑥2−4)+ 𝑙𝑛(𝑥3 − 𝑥) is

(A) (−∞, − 1) ∪ (0, 1)

(B) (−1, 0) ∪ (1, ∞)

(C) (−∞, − 1) ∪ (1, ∞) − {2, − 2}

(D) (−1, 0) ∪ (1, 2) ∪ (2, ∞)

Solution: (D)

Given function is defined only when

𝑥3 − 𝑥 > 0 and 𝑥2 − 4 ≠ 0

⇒ 𝑥(𝑥 − 1)(𝑥 + 1) > 0 and 𝑥 ≠ 2, − 2

𝑥 ∈ (−1, 0) ∪ (1, ∞) and 𝑥 ≠ 2, − 2

Hence domain is (−1, 0) ∪ (1, 2) ∪ (2, ∞)

11. Consider the common tangent(s) to the circles 𝑥2 + 𝑦2 = 4 and 𝑥2 + 𝑦2 + 6𝑥 + 8𝑦 − 24 = 0. Then which

of the following point lines on the common tangent

(A) (6, −2)

(B) (3, 7)

(C) (4, 6)

(D) (2, 4)

Solution: ()


Let the circle 𝐶1: 𝑥2 + 𝑦2 = 4 center 𝐶1: (0, 0), 𝑟1 = 2

Circle 𝐶2: 𝑥2 + 𝑦2 + 6𝑥 + 8𝑦 − 24 = 0 center 𝐶2(−3, − 4)𝑟2 = 7

Since 𝐶1𝐶2 − |𝑟1 − 𝑟2|, hence both the circles touches each other internally.

Equation of common tangent is

𝑆1 − 𝑆2 = 0

⇒ 6𝑥 + 8𝑦 − 20 = 0

⇒ 3𝑥 + 4𝑦 − 10 = 0

Clearly point (6, −2) lies on the common tangent.

12. Consider a circle 𝐶1 with diameter 3𝑦 = 𝑥 + 7 and points 𝐴(−8, 5) and 𝐵(6, 5) lies on it. Rectangle 𝐴𝐵𝐶𝐷

which is inscribe inside the circle is completed, then area of this rectangle is

(A) 80

(B) 60

(C) 84

(D) 42

Solution: (C)


Since 𝐴𝐵 is horizontal line coordinates of 𝐶 and 𝐷 can be taken as 𝐶(6, 𝛼), 𝐷(−8, 𝛽)

Now mid point of 𝐴𝐶 and 𝐵𝐷 must lie an diameter

⇒ 𝐵𝐶 = 6 and 𝐴𝐵 = 14

Area 14 × 6 = 84.

13. An inverted cone is filled with water at the rate of 5 𝑐𝑚3 𝑚𝑚⁄ has semi vertical angle 𝜃 = tan−1 (1

2). Then

the rate of charge of height of water when height of water is 10 𝑐𝑚 is

(A) 1

5𝜋𝑐𝑚/𝑚𝑚

(B) 1


(C) 2


(D) 3


Solution: (A)

Let at any time 𝑡, height of water level is ℎ and radius of cone height filled with water be 𝑟.


Hence 𝑉 =

1

3𝜋𝑟2ℎ

𝑉 =1

3𝜋

ℎ3

4

⇒𝑑𝑣

𝑑𝑡=

𝜋

12⋅ 3ℎ2

𝑑ℎ

𝑑𝑡

⇒ 5 =𝜋

4⋅ 100

𝑑ℎ

𝑑𝑡

⇒𝑑ℎ

𝑑𝑡=

1

5𝜋𝑐𝑚 𝑚𝑚⁄

14. If 25% people in a city read news paper 𝐴, 20% read 𝐵 and 8% read both 𝐴 and 𝐵. Also 30% of those who

read 𝐴 but not 𝐵 look into advertisement, 40% of those who read 𝐵 but not 𝐴 look into advertisement 50% of

those who read both 𝐴 and 𝐵 look in advertisement, then how many people look into advertisement.

(A) 13.3%

(B) 13.9%

(C) 13%

(D) 12.9%

Solution: (B)

To simply, lets consider total percent as be 100. Then from venn diagram

𝐴 = 25, 𝐵 = 20 𝐴 ∩ 𝐵 = 8

30% of 𝐴 ∩ �̅� = 5 ⋅ 1%

40% of �̅� ∩ 𝐵 = 4 ⋅ 8%

50% of 𝐴 ∩ 𝐵 = 4%

Total persons look into advertisement = 13 ⋅ 9%


15. Two points 𝐵 and 𝐶 lies on the line

𝑥+2

3=

𝑦−1

0=

𝑧

4 at a distance of 5 units with each other. If 𝐴 has

coordinates (1, − 1, 2), then area of 𝛥𝐴𝐵𝐶 is

(A) √43

(B) 2√34

(C) √34

(D) 3√32

Solution: (C)

For 𝐷

Clearly 𝛥𝐴𝐵𝐶 =1

2⋅ 𝐴𝐷 ⋅ 𝐵𝐶

Let point 𝐷(3𝑟 − 2, 1, 4𝑟)

DR’s of 𝐴𝐷 3𝑟 − 3, 2, 4𝑟 − 2

Since 𝐴𝐷 is perpendicular to given line

⇒ 3 ⋅ (3𝑟 − 3) + 0 ⋅ 2 + 4 ⋅ (4𝑟 − 2) = 0

⇒ 𝑟 =17

25⇒ 𝐷 (

1

25, 1,

68

25)

⇒ 𝐴𝐷 =2

5√34

⇒ Area of 𝛥𝐴𝐵𝐶 =1

2⋅

2

5⋅ √34 ⋅ 5 = √34

16. If a line makes an angle 𝜋

3 and

𝜋

4 with the x-axis and y-axis respectively then the angle made by this line with

the z-axis is

(A) 𝜋

4


(B)

𝜋

12

(C) 5𝜋

12

(D) 2𝜋

3

Solution: (D)

As we know is a line is inclined at an angle 𝛼, 𝛽 and 𝛾 with the positive direction of x, y and z-axis, then

cos 𝛼, cos 𝛽 and cos 𝛾 are called direction cosines of the line and also cos2 𝛼 + cos2 𝛽 + cos2 𝛾 = 1.

Hence from given condition

cos2𝜋

3+ cos2

𝜋

4+ cos2 𝛾 = 1

⇒ cos2 𝛾 =1

4

⇒ cos 𝛾 = ±1

2

⇒ 𝛾 =𝜋

3,2𝜋

3

Hence (𝐷) is the correct option.

17. The mean and median of 10, 22, 26, 29, 34, 𝑥, 42, 67, 70, 𝑦 is increasing order are 42 and 35 respectively.

Then the value of 𝑦

𝑥 is

(A) 7

3

(B) 8

3

(C) 3

(D) 7

2

Solution: (A)

Since number of observations are even (10)


Median =

34+𝑥

2⇒

34+𝑥

2= 35 ⇒ 𝑥 = 36 …(i)

Mean =10+22+26+29+34+𝑥+42+67+70+𝑦

10= 42

⇒ 𝑥 + 𝑦 = 120 ...(ii)

from (i) and (ii)

𝑥 = 36, 𝑦 = 84

⇒𝑦

𝑥=

84

36=

7

3.

18. The coefficients of three consecutive terms in binomial expansion (1 + 𝑥)𝑛 are in the ratio 2: 15: 70, then

the average of these coefficients is

(A) 974

(B) 227

(C) 232

(D) 347

Solution: (C)

From the given condition

𝑛𝐶𝑟: 𝑛𝐶𝑟+1: 𝑛𝐶𝑟+2 = 2: 15: 70

⇒ 𝑛𝐶𝑟

𝑛𝐶𝑟+1=

2

15 and

𝑛𝐶𝑟+1

𝑛𝐶𝑟+2=

2

70

⇒ 17𝑟 = 2𝑛 − 15 and 17𝑟 = 3𝑛 − 31

⇒ 𝑛 = 16 and 𝑟 = 1

Hence average = 𝑛𝐶𝑟+ 𝑛𝐶𝑟+1+ 𝑛𝐶𝑟+2

3

= 16𝐶1 + 16𝐶2 + 16𝐶3

3

= 232


19. The value of 𝑐, so that for all real 𝑥, the vectors 𝑐𝑥�̂� − 6𝑗̂ + 3�̂� and 𝑥�̂� + 2𝑗̂ + 2𝑐𝑥�̂� make an obtuse angle, is

(A) 𝑐 < 0

(B) 𝑜 < 𝑐 <4

3

(C) −4

3< 𝑐 < 0

(D) 𝑐 > 0

Solution: (C)

Since angle between the vectors is obtuse, hence (𝑐𝑥�̂� − 6𝑗̂ + 3�̂�) ⋅ (𝑥�̂� + 2𝑗̂ + 2𝑐𝑥�̂�) < 0 ∀𝑥

⇒ 𝑐𝑥2 − 12 + 6𝑐𝑥 < 0 ∀𝑥

⇒ 𝑐𝑥2 + 6𝑐𝑥 − 12 < 0 ∀𝑥

⇒ 𝑐 < 0 and 36𝑐2 − 4 ⋅ 𝑐 ⋅ (−12) < 0

⇒ 𝑐 < 0 and −4

3< 𝑐 < 0

⇒ 𝑐 ∈ (−4

3, 0)

Hence (𝐶) is the correct

20. The sum of the first three terms of an arithmetic progression (A.P.) is 33 and their product is 1155. The

11th term of the A.P. is

(A) −25

(B) 25

(C) −36

(D) 36

Solution: (A)

Let the first 3 terms of A.P. are 𝑎 − 𝑑, 𝑎, 𝑎 + 𝑑


⇒ sum = 3𝑎 = 33 ⇒ 𝑎 = 11

And product (𝑎 − 𝑑) 𝑎 (𝑎 + 𝑑) = 𝑎(𝑎2 − 𝑑2) = 1155

⇒ 112 − 𝑑2 =1155

11= 105

⇒ 𝑑 = ±4

For 𝑑 = 4, the A.P. is

7, 11, 15, ….

So 𝑇11 = 7 + 10.4 = 47

For 𝑑 = −4, the A.P. is

15, 11, 7, …

So, 𝑇11 = 15 + 10(−4) = −25

21. If a curve satisfy cos 𝑥 ⋅𝑑𝑦

𝑑𝑥− 𝑦 sec 𝑥 = 6𝑥 and 𝑦 (

𝜋

3) = 0, then the value of 𝑦 (

𝜋

6) is

(A) −𝜋2

√3

(B) 𝜋2

2√3

(C) −𝜋2

2√3

(D) 𝜋2

√3

Solution: (C)

The given D.E. can be written as

𝑑𝑦

𝑑𝑥− 𝑦 tan 𝑥 = 6𝑥 sec 𝑥 …(i)

Integrating factor = 𝑒− ∫ tan 𝑥 𝑑𝑥 = cos 𝑥

Solution is


𝑦 ⋅ cos 𝑥 = ∫ 6𝑥 sec 𝑥 ⋅ cos 𝑥 𝑑𝑥

⇒ 𝑦 ⋅ cos 𝑥 = 3𝑥2 + 𝑐 …(ii)

Given by 𝑦 (𝜋

3) = 0, hence 𝑐 =

−𝜋2

3

Curve is 𝑦 cos 𝑥 = 3𝑥2 −𝜋2

3

Put 𝑥 =𝜋

6, we get

𝑦 (𝜋

6) ⋅

√3

2= 3 ⋅

𝜋2

36−

𝜋2

3⇒ 𝑦 (

𝜋

6) = −

𝜋2

2√3

22. value of 𝑝 → (𝑞𝑣𝑟) is false, then the statements 𝑝, 𝑞, 𝑟 respectively are

(A) 𝑇 𝑇 𝑇

(B) 𝐹 𝐹 𝐹

(C) 𝑇 𝐹 𝐹

(D) 𝐹 𝑇 𝑇

Solution: (C)

Simply by truth value table we can solve.

𝑝 𝑞 𝑟 (𝑞𝑣𝑟) 𝑝 → (𝑞𝑣𝑟) 𝑇 𝑇 𝑇 𝑇 𝑇 𝑇 𝑇 𝐹 𝑇 𝑇 𝑇 𝐹 𝑇 𝑇 𝑇 𝑇 𝐹 𝐹 𝐹 𝐹 𝐹 𝑇 𝑇 𝑇 𝑇 𝐹 𝑇 𝐹 𝑇 𝑇 𝐹 𝐹 𝑇 𝑇 𝑇 𝐹 𝐹 𝐹 𝐹 𝑇

If the compound event 𝑝 → (𝑞𝑣𝑟) is false then 𝑝 is 𝑇, 𝑞 is 𝐹 and 𝑟 is 𝐹.

23. If𝑓(2) = 6, then lim𝑥→2

∫2𝑡𝑑𝑡

𝑥−2

𝑓(𝑥)

6 equals to


(A) 6𝑓′(2)

(B) 12 𝑓′(2)

(C) 2𝑓′(2)

(D) none of these

Solution (B)

The gives limit can be written as

lim𝑥→2

∫ 2𝑡𝑑𝑡𝑓(𝑥)

6

(𝑥−2) applying L' Hospotal's rule

= lim𝑥→2

2𝑓(𝑥)⋅𝑓′(𝑥)

1

= 2𝑓(2) ⋅ 𝑓′(2)

= 12𝑓′(2)

24. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + ⋯ upto 11 terms is

(A) 985

(B) 946

(C) 991

(D) 989

Solution (B)

1 + 2 × 3 + 3 × 5 + 4 × 7 + 5 × 9 + ⋯

𝑇𝑟 = 𝑟(2𝑟 − 1)

Sum of 11 terms

𝑆11 = ∑ 𝑟

11

𝑟=1

(2𝑟 − 1)

= 2 ∑ 𝑟2 − ∑ 𝑟

= 2 ⋅11 ⋅ (11 + 1)(22 + 1)

6−

11(11 + 1)

2

= 44 × 23 − 66


= 1012 − 66 = 946

25. The value of sin 10𝑜 sin 30𝑜 sin 50𝑜 sin 70𝑜 is

(A) 1

8

(B) 1

8 √2

(C) 1

16

(D) 3

16

Solution (C)

sin 10𝑜 sin 30𝑜 sin 50𝑜 sin 70𝑜

=1

4sin 10𝑜(2 sin 50𝑜 sin 70𝑜)

=1

4sin 10𝑜(cos 20𝑜 − cos 120𝑜)

=1

8(2sin 10𝑜 cos 20𝑜 + sin 10𝑜)

=1

8(sin 30𝑜 − sin 10𝑜 + sin 10𝑜)

=1

16

26. ∫ 𝑥1

0cot−1(1 − 𝑥2 + 𝑥4)𝑑𝑥 equals to

(A) 𝜋

2=

1

2𝑙𝑛 2

(B) 𝜋

4− 𝑙𝑛 2

(C) 𝜋

4−

1

2𝑙𝑛 2

(D) 𝜋

2− 𝑙𝑛 2

Solution: (C)

Given integral 𝐼 = ∫ 𝑥1

0cot−1(1 − 𝑥2 + 𝑥4)𝑑𝑥 𝑥2 = 𝑡, 2𝑥𝑑𝑥 = 𝑑𝑡


⇒ 𝐼 =

1

2∫ cot−1(1 − 𝑡 + 𝑡2)𝑑𝑡

1

0

⇒ 𝐼 =1

2∫ tan−1 (

𝑡−(𝑡−1)

1+𝑡(𝑡−1)) 𝑑𝑡

1

0

⇒ 𝐼 =1

2[∫ tan−1 𝑡 𝑑𝑡 − ∫ tan−1(𝑡 − 1)𝑑𝑡

1

0

1

0]

⇒ 𝐼 =1

2[2 ∫ tan−1 𝑡 𝑑𝑡

1

0]

⇒ 𝐼 = ∫ 1 ⋅ tan−1 𝑡 𝑑𝑡1

0

⇒ 𝐼 = tan−1𝑡 ⋅ 𝑡|01 − ∫

1

1+𝑡2 ⋅ 𝑡 𝑑𝑡1

0

𝐼 =𝜋

4−

1

2𝑙𝑛(1 + 𝑡2)|0

1

𝐼 =𝜋

4−

1

2𝑙𝑛 2

27. If (𝑚2 + 1)𝑥2 − 3𝑥 + (𝑚2 + 1)2 = 0 and value of 𝑚 is chosen such that sum of roots is maximum then

the difference of cube of roots for that value of 𝑚 is

(A) 9√5

(B) 3√5

(C) 8√5

(D) 6√5

Solution: (C)

Let roots are 𝛼 and 𝛽

∴ 𝛼 + 𝛽 =3

𝑚2+1 ∵ 𝛼 + 𝛽 is maximum ∴ 𝑚 = 0

∴ equation is 𝑥2 − 3𝑥 + 1 = 0

𝛼 + 𝛽 = 3, 𝛼𝛽 = 1

Now, 𝛼3 − 𝛽3 = (𝛼 − 𝛽)(𝛼2 + 𝛽2 + 𝛼𝛽)

= √(𝛼 + 𝛽)2 − 4𝛼𝛽((𝛼 + 𝛽)2 − 𝛼𝛽)

= √9 − 4(9 − 1)

= 8√5

PAPER-1 (B.E./B.TECH.) JEE MAIN 2019 · दഉ༘द༢Մ and दഈ༘Ո༟Մ ⇒दᐌद༘Յᐍᐌद༗Յᐍ༢Մand द༟Ն, ༘Ն द∈ᐌ༘Յ, ՄᐍᐌՅ, ∞ᐍand द༟Ն,

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