PAPER-1 (B.E./B.TECH.) JEE MAIN 2019...JEE MAIN 8 APRIL 2019 SHIFT-1 MATHS 1.The mean and variance...

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

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 8th April 2019 (Shift-1)

Time: 09:30 A.M. to 12:30 P.M.

Durations: 3 Hours | Max. Marks: 360

Subject: Mathematics

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JEE MAIN 8 APRIL 2019 SHIFT-1 MATHS

1. The mean and variance for seven numbers are 8 and 16 respectively. If the five observation are 2,4,10,12,14

and the product of other two numbers is

(A) 48

(B) 49

(C) 50

(D) 51

Solution: (A)

Let the remaining numbers are 𝑥 and .

Mean �� =∑𝑥𝑖

𝑁=

2+4+10+12+14+𝑥+𝑦

7

=42+𝑥+𝑦

7= 8 ⇒ 𝑥 + 𝑦 = 14 …(1)

Variance 𝜎2 =∑𝑥𝑖

2

𝑁− (��)2 =

4+16+100+144+196+𝑥2+𝑦2

7− 64

=460 + 𝑥2 + 𝑦2

7− 64 = 16

⇒ 𝑥2 + 𝑦2 = 100 ….(2)

From (1) and (2), we get

𝑥 = 6, y = 8 and 𝑥𝑦 = 48

2. Evaluate ∫sin

5𝑥

2

sin𝑥

2

𝑑𝑥

(A) 𝑥 + 𝑠𝑖𝑛2𝑥 + 𝐶

(B) 2𝑥 + 𝑠𝑖𝑛𝑥 + 2𝑥 + 𝐶

(C) 𝑥 + 2 𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛2𝑥 + 𝐶

(D) 𝑋 + 2 𝑠𝑖𝑛𝑥 + 2𝑠𝑖𝑛𝑥 + 𝐶

Solution: (C)

I = ∫sin

5𝑥

2

sin𝑥

2

𝑑𝑥 = ∫sin(2𝑥 +

𝑥

2)

sin𝑥

2

𝑑𝑥


= ∫sin2𝑥. cos

𝑥2

+ cos2𝑥. sin𝑥2

sin𝑥2

𝑑𝑥

= ∫ (2sin𝑥cos𝑥. cos

𝑥2

sin𝑥2

+ cos2𝑥) 𝑑𝑥

= ∫ (4cos2𝑥

2cos𝑥 + cos2𝑥)𝑑𝑥

= ∫ (2(1 + cos𝑥)cos𝑥 + cos2𝑥)𝑑𝑥

= ∫ (2cos𝑥 + 2cos2𝑥 + 1)𝑑𝑥

= 2sin𝑥 + sin2𝑥 + 𝑥 + 𝐶

3. Given triangle 𝑂𝐴𝑃 where O(0, 0) and 𝐴(0,1). Then the locus of the third vertex 𝑃 if the perimeter of the

triangle is 4, is

(A) 𝑥2

8+

(𝑦−1)2

9=

1

4

(B) 𝑥2

9+

(𝑦−1/2)2

8=

1

4

(C) 𝑥2

8+

(𝑦−1/2)2

9= 1

(D) 𝑥2

8+

(𝑦−1/2)2

9=

1

4

Solution: (D)

OA + PO + PA = 4 ⇒ PO + PA = 3 ⇒ Locus of P is ellipse with foci at O & A and major axis (2b) = 3


Distance between foci = 2𝑏𝑒 = 1 ⇒ e =1

2

⇒ minor axis 2a = 2b√1 − 𝑒2 = 3.2√2

3= 2√2

⇒ Equation of Locus of 𝑃 is:

𝑥2

8+

(𝑦 −12

)2

9=

1

4

4. If the quadratic 𝑥2 − 2𝑥 + 21 = 0 has two distinct roots 𝛼 and 𝛽 such that (𝛼

𝛽)

𝑛

= 1 then the minimum

value of 𝑛 𝑁 is

(A) 2

(B) 3

(C) 4

(D) 6

Solution: (C)

𝑥2 − 2𝑥 + 2 = 0 ⇒ (𝑥 − 1)2 + 1 = 0

⇒ The roots 𝛼, 𝛽 = 1 ± 𝑖

(𝛼

𝛽)

𝑛

= (1 + 𝑖

1 − 𝑖)

𝑛

= (𝑖(1 − 𝑖)

1 − 𝑖)

𝑛

= (𝑖)𝑛

⇒ The minimum value of 𝑛 𝑁 = 4

Since (𝑖)4 = 1

5. If the solution of the equation (𝑥2 + 1)2 𝑑𝑦

𝑑𝑥+ 2𝑥(𝑥2 + 1)𝑦 = 1 satisfies 𝑦(0) = 0 and √𝑎 𝑦(1) =

𝜋

32 then

the value of 𝑎 is

(A) 1

16

(B) 1

4


(C) 2

(D) √2

Solution: (A)

(𝑥2 + 1)2 𝑑𝑦

𝑑𝑥+ 2𝑥(𝑥2 + 1)𝑦 = 1

⇒𝑑𝑦

𝑑𝑥+

2𝑥

𝑥2 + 1𝑦 =

1

(𝑥2 + 1)2

Which is a linear differential equation with integrating factor

𝐼. 𝐹. = 𝑒∫

2𝑥

𝑥2+1𝑑𝑥

= 𝑒ln(𝑥2+1) = 𝑥2 + 1

⇒ The solution of the given differential equation will be

𝑦. (𝑥2 + 1) = ∫ (𝑥2 + 1).1

(𝑥2 + 1)2𝑑𝑥 = ∫

𝑑𝑥

𝑥2 + 1= tan−1𝑥 + c

Now 𝑦(0) = 0 ⇒ C = 0 and 𝑦 =1

(𝑥2+1)tan−1𝑥

√𝑎𝑦(1) =𝜋

32⇒ √𝑎 =

𝜋/32

𝜋/8=

1

4

⇒ 𝑎 =1

16

6. Find the projection of the vector 3�� + 2𝑗 + �� to the vector which is perpendicular to both �� + 𝑗 + �� and �� +

2𝑗 + 3��

(A) 1

√6

(B) 2

√6

(C) 0

(D) 1

√3

Solution: (C)

A vector perpendicular to �� + 𝑗 + �� a and �� + 2𝑗 + 3�� is


|i j k1 1 11 2 3

| = �� − 2𝑗 + ��

⇒ The projection of the given vector 3�� + 2𝑗 + �� on the above vector is

(3�� + 2𝑗 + ��). (�� − 2𝑗 + ��)

|�� − 2𝑗 + ��|

=3 − 4 + 1

√6= 0

7. The shortest distance between the curve 𝑦2 = 𝑥– 2 and the line 𝑦 = 𝑥 is

(A) 3

2√2

(B) 7

4√2

(C) 5

4√2

(D) 11

4√2

Solution: (B)

𝑦2 = 𝑥– 2

Differentiating w.r.t. 𝑥 we get

2𝑦𝑦′ = 1 ⇒ y′ = 1

2𝑦

For the shortest distance, the tangent at point 𝑃 will be parallel to the given line


⇒ 𝑦′|𝑝 =1

2𝑦1= 1 ⇒ 𝑦1 =

1

2

⇒ 𝑥1 = 2 + (1

2)

2

=9

4 (∵ 𝑦1

2 = 𝑥1 − 2)

The shortest distance between the given curve & the line

= The perpendicular distance of point 𝑃 from the line

= |

94

−12

√12 + 12| = |

74

√2| =

7

4√2

8. Given 𝑓(𝑥): [0, 2]𝑅 and 𝑓′′(𝑥) > 0 𝑥 [0,2], if another function 𝜙(𝑥) = 𝑓(𝑥) + 𝑓(2– 𝑥) 𝑥 (0,2),

then

(A) 𝜙(𝑥) is decreasing in (0,1) and increasing in (1,2)

(B) 𝜙(𝑥) is increasing in (0,1) and decreasing in (1,2)

(C) 𝜙(𝑥) is decreasing in (0,2)

(D) 𝜙(𝑥) is increasing in (0,2)

Solution: (A)

𝑓(𝑥): [0,2]𝑅 and 𝑓′′(𝑥) > 0 for 𝑥 [0,2]

⇒ f ′(𝑥) is increasing 𝑥 [0,2]

Now, 𝜙 (𝑥) = 𝑓(𝑥) + 𝑓(2 − 𝑥)

⇒ 𝜙′(𝑥) = 𝑓′(𝑥) − 𝑓′(2 − 𝑥)

For 𝑥 [0,1), x < 2 – x ⇒ 𝑓′(𝑥) < 𝑓′(2 − 𝑥) ⇒ 𝜙′(𝑥) < 0

For𝑥 (1, 2], x > 2 – x ⇒ 𝑓′(𝑥) < 𝑓′(2 − 𝑥) ⇒ 𝜙′(𝑥) > 0

So 𝜙(𝑥) is decreasing in (0, 1) and increasing in (1, 2).

9. If 𝑓(𝑥) =2−𝑥 cos𝑥

2+𝑥 cos𝑥 and 𝑔(𝑥) = loge 𝑥 then ∫ 𝑔(𝑓(𝑥))𝑑𝑥 =

𝜋/4

−𝜋/4

(A) loge 2


(B) loge 1

(C)loge 3

(D) loge 4

Solution: (B)

𝑓(𝑥) =2−𝑥 cos𝑥

2+𝑥 cos𝑥 , 𝑔(𝑥) = loge 𝑥

⇒ 𝑔(𝑓(𝑥)) = log (2 − 𝑥 cos𝑥

2 + 𝑥 cos𝑥)

⇒ 𝑔(𝑓(−𝑥)) = loge (2−(−𝑥) cos (−𝑥)

2+(−𝑥) cos (−𝑥))= log𝑒 (

2+𝑥 cos𝑥

2−𝑥 cos𝑥)

=– 𝑔(𝑓(𝑥))

⇒ 𝑔(𝑓(𝑥)) is an odd function

⇒ ∫ 𝑔(𝑓(𝑥))𝑑𝑥 =𝜋/4

−𝜋/4 0 = loge 1 (By using property of definite integration)

10. Using all the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 to form 9 digit numbers in which all the odd digits are placed at the

even positions. Total number of such numbers is –

(A) 150

(B) 170

(C) 160

(D) 180

Solution: (D)

Odd digits:1, 1, 3

Even digits: 2, 2, 2, 2, 4, 4

The number of ways of placing odd digits at even position = 𝐶3 ×3!

2!= 4 × 3 = 124

The number of ways of placing even digits =6!

4! 2!= 15


∴ Total number of ways = 12 × 15 = 180

11. The area of the region bounded by 𝑦 ≤ 𝑥2 + 3𝑥, 0 ≤ 𝑦 ≤ 4 and 0 ≤ 𝑥 ≤ 3 is

(A) 8

(B) 59

6

(C) 49

6

(D) 16

Solution: (B)

The area of the region 𝑂𝐷𝐶 is

= ∫[4 − (𝑥2 + 3𝑥)]𝑑𝑥 = [4𝑥 −𝑥3

3−

3𝑥2

2]

0

11

0

= 4 −1

3−

3

2=

13

6

So, the area of t he given region 𝑂𝐴𝐵𝐶 is

= Area of 𝑂𝐴𝐵𝐷 – Area of 𝑂𝐶𝐷

= 3 × 4 −13

6

=59

6


12. If two events 𝐴 and 𝐵 are such that 𝐴 ⊂ 𝐵, then which of the following is correct?

(A) 𝑃 (𝐴

𝐵) ≥ 𝑃(𝐴)

(B) 𝑃 (𝐴

𝐵) ≤ 𝑃(𝐴)

(C) 𝑃(𝐴) > 𝑃(𝐵)

(D) None of these

Solution: (A)

A

B

𝑃 (𝐴

𝐵) =

𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

𝑃(𝐴)

𝑃(𝐵) [𝐴 ⊂ 𝐵 ⇒ 𝐴 ∩ 𝐵 = 𝐴]

Now, since 𝑃(𝐵) ≤ 1

⇒ 𝑃 (𝐴

𝐵) =

𝑃(𝐴)

𝑃(𝐵)≥ 𝑃(𝐴)

13. The sum of squares of lengths of intercepts made on the chord 𝑥 + 𝑦 = 𝑛, 𝑛 𝑍 in the circle 𝑥2 + 𝑦2 = 16

is

(A) 420

(B) 384

(C) 484

(D) 648

Solution: (C)


The length of perpendicular from centre (0, 0) to the line 𝑥 + 𝑦 = 𝑛 is |0+0−𝑛

√12+𝐼2| =

𝑛

√2

⇒ Length of intercept = 2√42 − (𝑛

√2)

2

= √64 − 2𝑛2

⇒ Possible values of n are 0, ±1, ±2, … . ±5

⇒ Sum of square of lengths of intercepts

= ∑ (64 − 2𝑛2)

5

𝑛=−5

= 64 × 11 − 2 ∑ 𝑛2

5

𝑛=−5

= 704 − 4 ∑ 𝑛2

5

𝑛=1

= 704 − 220 = 484

14. lim𝑥→0

sin2𝑥

√2−√1+cos𝑥

(A) 4√2

(B) √2

(C) 1

√2

(D) 6√2

Solution: (A)

lim𝑥→0

sin2𝑥

√2 − √1 + cos𝑥

On Rationalization:

lim𝑥→0

sin2𝑥

√2−√1+cos𝑥×

√2+√1+cos

√2+√1+cos


= lim𝑥→0

(2sin

𝑥2

cos𝑥2

)2

1 − cos𝑥× (√2 + √1 + cos𝑥)

= lim𝑥→0

4sin2 𝑥

2cos2 𝑥

2

2sin2 𝑥2

× (√2 + √1 + cos𝑥)

= lim𝑥→0

2cos2𝑥

2(√2 + √1 + cos𝑥) = 2 × 2√2

= 4√2

15. Area between 𝑦 = 𝑥3 and 𝑦 = |𝑥|

(A) 1

(B) 1

2

(C) 1

4

(D) 2

Solution: (C)

Intersection points of 𝑦 = 𝑥3 and 𝑦 = |𝑥| are 0 & 1

∴ 𝐴 = ∫(𝑥 − 𝑥3)𝑑𝑥

1

0

= (𝑥2

2−

𝑥4

4)

0

1

=1

2−

1

4


=1

4

16. If 𝛼 = cos−1 3

5 and 𝛽 = tan−1 1

3, then 𝛼 − 𝛽 is equal to

(A) cos−1 (5

13)

(B) tan−1 (9

13)

(C) cos−1 (12

13)

(D) tan−1(3)

Solution: (B)

𝛼 = cos−13

5= tan−1

4

3

∴ 𝑎 − 𝛽 = tan−14

3− tan−1

1

3

= tan−1

43

−13

1 +43 ×

13

= tan−11

139

= tan−19

13

17. If sin(𝛼 + 𝛽) =3

5 & 𝑠𝑖𝑛(𝛼 − 𝛽) =

5

13 then tan2𝛼 =

(A) 56/33

(B) 57/22

(C) 0

(D) 2

Solution: (A)


sin(𝛼 + 𝛽) =3

5⇒ tan(𝛼 + 𝛽) =

3

4

sin(𝛼 − 𝛽) =5

13⇒ tan(𝛼 + 𝛽) =

5

12

Now, tan2𝛼 = tan((𝛼 + 𝛽) + (𝛼 − 𝛽))

=tan(𝛼 + 𝛽) + tan(𝛼 − 𝛽)

1 − tan(𝛼 + 𝛽)tan(𝛼 − 𝛽)

=

34 +

512

1 −34

×5

12

=56

33

18. 2.20 𝐶0 + 5.20 𝐶1 + 8.20 𝐶1+. . . . . . . +62.20 𝐶20 =

(A) 224

(B) 225

(C) 226

(D) 220

Solution: (B)

Let 𝑆 = 2.20 𝐶0 + 5.20 𝐶1 + 8.20 𝐶2+. . . . . . . +62.20 𝐶20 …(i)

& 𝑆 = 62.20 𝐶0 + 59.20 𝐶1 + 56.20 𝐶2+. . . . . . . +2.20 𝐶20 …(ii) (writing in reverse order & 𝑛𝐶𝑟 =𝑛 𝐶𝑛−𝑟)

Adding (i) & (ii)

2𝑆 = 64(20𝐶0+20𝐶1+20𝐶2+. . . . . +20𝐶20)

⇒ S = 32.220 = 225

19. If 𝑓(𝑥) = 𝑙𝑛 (1−𝑥

1+𝑥) then for |x|< 1, 𝑓 (

2𝑥

1+𝑥2)=

(A) 𝑓(𝑥2)

(B) (𝑓(𝑥))2

(C) – 2𝑓(𝑥)


(D) 2𝑓(𝑥)

Solution: (D)

𝑓(𝑥) = 𝑙𝑛 (1 − 𝑥

1 + 𝑥)

⇒ 𝑓 (22

1 + 𝑥2) = ln (

1 −22

1 + 𝑥2

1 +2𝑥

1 + 𝑥2

)

= ln (1 + 𝑥2 − 2𝑥

1 + 𝑥2 + 2𝑥)

= ln ((1 − 𝑥

1 + 𝑥)

2

)

= 2ln |1 − 𝑥

1 + 𝑥|

= 2ln (1−𝑥

1+𝑥) (∵ |𝑥| < 1)

= 2𝑓(𝑥)

20. The perpendicular distance of point (2, −1, 4) from the line 𝑥+3

10=

𝑦−2

−7=

𝑧

1 lies between

(A) (1,2)

(B) (2,3)

(C) (3,4)

(D) (4,5)

Solution: (C)

Let 𝑃(2, −1,4)

Now any point Q on line 𝑥+3

10=

𝑦−2

−7=

𝑧

1= 𝜆; is

𝑄 = (10𝜆 − 3, −7𝜆 + 2 , 𝜆)

DR’s of PQ = (10𝜆 − 3 − 2, −7𝜆 + 2 + 1, 𝜆 − 4)

= (10𝜆 − 5, 7𝜆 + 3, 𝜆 − 4)

∵ 𝑃𝑄 is perpendicular to given line


∴ 10(10𝜆 + 5) − 7(−7𝜆 + 3) + (𝜆 − 4) = 0

⇒ 150 𝜆 = 75 ⇒ 𝜆 =1

2

∴ 𝑃𝑄 = √(10𝜆 − 5)2 + (−7𝜆 + 3)2 + (𝜆 − 4)2 = √0 +1

4+

49

4= √

50

4

Which lies in (3,4)

21. If |√𝑥 − 2| + √𝑥(√𝑥 − 4) = 2 then find the sum of roots of equation is

(A) 12

(B) 16

(C) 10

(D) 4

Solution: (B)

Given |√𝑥 − 2| + √𝑥(√𝑥 − 4) = 2

⇒ |√𝑥 − 2| + 𝑥 − 4√𝑥 = 2

⇒ |√𝑥 − 2| + (√𝑥)2

− 4√𝑥 + 4 = 6 (Adding 4 on both sides)

⇒ (√𝑥 − 2)2

+ |√𝑥 − 2| = 6

⇒ 𝑡2 + 𝑡 − 6 = 0 (𝑙𝑒𝑡 |√𝑥 − 2| = 𝑡)

⇒ 𝑡 = 2 (or) 𝑡−= −3

⇒ |√𝑥 − 2| = 2

⇒ √𝑥 = 4 or √𝑥 = 0

⇒ 𝑥 = 0 or 𝑥 = 16

22. Let 𝑆1 be the set of minima and 𝑆2 is set of maxima for the curve 𝑦 = 9𝑥4 + 12𝑥3 − 36𝑥2 − 25

(A) 𝑆1 = {−2, −3}; 𝑆2 = {0}


(B) 𝑆1 = {−2}; 𝑆2 = {−1}

(C)𝑆1 = {−2,1}; 𝑆2 = {0}

(D) None

Solution: (C)

𝑦 = 9𝑥4 + 12𝑥3 − 36𝑥2 − 25

𝑑𝑦

𝑑𝑥= 36𝑥3 + 36𝑥2 − 72𝑥

= 36𝑥(𝑥2 + 𝑥 − 2)

= 36𝑥(𝑥 + 2)(𝑥 − 1)

Sign convention for 𝑑𝑦

𝑑𝑥

∴ 𝑆1 = {−2,1} & 𝑆2 = {0}

23. Let 2𝑦 = (cot−1 (√3 cos 𝑥−sin 𝑥

cos 𝑥+√3 sin 𝑥))

2

then 𝑑𝑦

𝑑𝑥 is equal to

(A) 𝑥 +𝜋

2

(B) 2𝑥 −𝜋

2

(C) 𝑥 +𝜋

6

(D) None

Solution: (C)

2𝑦 = (cot−1(√3 cos 𝑥−𝑠𝑖𝑛𝑥

cos 𝑥+√3 sin 𝑥)

2


= (cot−1 (√3 − 𝑡𝑎𝑛𝑥

1 + √3 tan 𝑥))

2

= (cot−1 tan (𝜋

3− 𝑥))

2

= (cot−1 cot (𝜋

2− (

𝜋

3− 𝑥)) )

2

2𝑦 = (𝜋

6+ 𝑥)

2

⇒ 2𝑑𝑦

𝑑𝑥= 2 (

𝜋

6+ 𝑥) . 1 ⇒

𝑑𝑦

𝑑𝑥=

𝜋

6+ 𝑥

24. Find the value of c for which the following equations have non trival solutions:

𝑐𝑥 − 𝑦 − 𝑧 = 0

−𝑐𝑥 + 𝑦 − 𝑐𝑧 = 0

𝑥 + 𝑦 − 𝑐𝑧 = 0L

(A) 1

(B) 0

(C) −1

(D) 2

Solution: (C)

∵ System of equation has non trival solution

∴ ∆ = |𝑐 −1 −1

−𝑐 1 −𝑐1 1 −𝑐

| = 0

⇒ 𝑐(−𝑐 + 𝑐) + 1(𝑐2 + 𝑐) − 1(−𝑐 − 1) = 0

⇒ 𝑐2 + 2𝑐 + 1 = 0

⇒ (𝑐 + 1)2 = 0 ⇒ 𝑐 = −1

25. The sum of coefficient of even power of 𝑥 in (𝑥 + √𝑥3 − 1)6

(𝑥 − √𝑥3 − 1)6

(A) 24


(B) 29

(C) 18

(D) 20

Solution: (A)

𝑠 = (𝑥 + √𝑥3 − 1)6

+ (𝑥 − √𝑥3 − 1)6

= 2 ( 6𝐶0 𝑥6 + 6𝐶2𝑥2(𝑥3 − 1)+ 6𝐶4 𝑥4(𝑥3 − 1)2 + 6𝐶6 (𝑥3 − 1)3)

= 2 ( 6𝐶0 𝑥6 + 6𝐶2 (𝑥5 − 𝑥2)+ 6𝐶4 𝑥4(𝑥6 − 2𝑥3 + 1) +6𝐶6 (𝑥9 − 3𝑥6 + 3𝑥3 − 1))

∴ Sum of coefficient of even power of 𝑥

= 2 ( 6𝐶0− 6𝐶2+ 6𝐶4 + 𝐶4 − 3− 6𝐶6− 6𝐶6)

= 2 (1 − 15 + 15 + 15 − 3 − 1) = 24

26. If a plane passes through intersection of planes 2𝑥 − 𝑦 − 4 = 0 and 𝑦 + 2𝑧 − 4 = 0 and also passes through

the point (1,1,0) then the equation of plane is

(A) 𝑥 − 𝑦 − 𝑧 = 0

(B) 𝑥 − 𝑦 = 0

(C) 𝑥 + 2𝑦 − 𝑧 = 0

(D) 𝑥 − 2𝑦 + 𝑧 = 0

Solution: (A)

Equation of plane passing through intersection of 2𝑥 − 𝑦 − 4 = 0 and 𝑦 + 2𝑧 − 4 = 0 is

⇒ (2𝑥 − 𝑦 − 4) + 𝜆 (𝑦 + 2𝑧 − 4) = 0

⇒ 2𝑥 + 𝑦 (𝜆 − 1) + 2𝜆𝑧 − 4 − 𝜆 = 0

∵ it passes through (1,1,0)

∴ 2 × 1 + 𝜆 − 1 + 0 − 4 − 4𝜆 = 0

⇒ −3𝜆 − 3 = 0 ⇒ 𝜆 = −1

∴ Equation of plane 2𝑥 − 2𝑦 − 2𝑧 = 0

⇒ 𝑥 − 𝑦 − 𝑧 = 0


27. Let 𝐴 = [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

] and 𝐴32 = [

0 −11 0

] then 𝛼 may be

(A) 𝜋

32

(B) 𝜋

64

(C) 𝜋

128

(D) 0

Solution: (B)

∵ 𝐴 = [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

]

⇒ 𝐴2 = [𝑐𝑜𝑠2𝛼 −𝑠𝑖𝑛2𝛼𝑠𝑖𝑛2𝛼 𝑐𝑜𝑠2𝛼

] ⇒ 𝐴3 = [

𝑐𝑜𝑠3𝛼 −𝑠𝑖𝑛3𝛼𝑠𝑖𝑛3𝛼 𝑐𝑜𝑠3𝛼

]

Similarly 𝐴32 = [𝑐𝑜𝑠32𝛼 −𝑠𝑖𝑛32𝛼𝑠𝑖𝑛32𝛼 𝑐𝑜𝑠32𝛼

]

Given 𝐴3 = [0 −11 0

] ⇒ [𝑐𝑜𝑠32𝛼 −𝑠𝑖𝑛32𝛼𝑠𝑖𝑛32𝛼 𝑐𝑜𝑠32𝛼

] = [

0 −11 0

]

On comparing 𝑐𝑜𝑠32𝛼 = 0 & 𝑠𝑖𝑛32𝛼 = 1

⇒ 𝛼 =𝜋

64

28. Given equation of a line is 3𝑥 + 5𝑦 = 15 and a point 𝑃 on this line is equidistant from 𝑥 and 𝑦 − 𝑎𝑥𝑖𝑠 in

which quadrant point 𝑃 lies.

(A) I

(B) III

(C) IV

(D) None of these

Solution: (A)

Let (𝑎, 𝑎) be a point on the line 3𝑥 + 5𝑦 = 15


⇒ 3𝑎 + 5𝑎 = 15

⇒ 𝑎 =15

8

∴ Required point is (15

8 ,

15

8) which lies in I quadrant

PAPER-1 (B.E./B.TECH.) JEE MAIN 2019...JEE MAIN 8 APRIL 2019 SHIFT-1 MATHS 1.The mean and variance...

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Transcript of PAPER-1 (B.E./B.TECH.) JEE MAIN 2019...JEE MAIN 8 APRIL 2019 SHIFT-1 MATHS 1.The mean and variance...