eb??a@=d:;=cb>a?a ilnmjklnhgf · (a)p= 6 ,q=27(b)p=3,q= 6 ; 6 (c) p=6,q= 6 ; 6 (d) p=3 ,q=27 1 4...

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Class – XII MATHEMATICS- 041

SAMPLE QUESTION PAPER 2019-20

Time: 3 Hrs. Maximum Marks: 80

General Instructions:

(i) All the questions are compulsory.(ii) The question paper consists of 36 questions divided into 4 sections A, B,

C, and D.(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises

of 6 questions of 2 marks each. Section C comprises of 6 questions of 4marks each. Section D comprises of 4 questions of 6 marks each.

(iv) There is no overall choice. However, an internal choice has been providedin three questions of 1 mark each, two questions of 2 marks each, twoquestions of 4 marks each, and two questions of 6 marks each. You haveto attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted.

SECTION A

Q1 - Q10 are multiple choice type questions. Select the correct option

1 If A is any square matrix of order 3 × 3 such that |A| = 3, then the value of |adjA| is ? (a) 3 (b) (c) 9 (d) 27

1

2 Suppose P and Q are two different matrices of order 3 × n and n × p , then the order of the matrix P × Q is? (a) 3 × p (b) p × 3 (c) n × n (d) 3 × 3

1

3 If 2ı̂ + 6ȷ̂+ 27k × ı̂+ pȷ̂ + qk = 0⃗ ,then the values of p and q are ? (a) p= 6 ,q=27(b)p=3,q= (c) p=6,q= (d) p=3 ,q=27

1

4 If A and B are two events such that P(A)=0.2 , P(B)=0.4 and P(A ∪ B)=0.5 , then value of P(A/B) is ? (a)0.1 (b)0.25 (c)0.5 (d) 0.08

1

5 The point which does not lie in the half plane 2푥 + 3푦 − 12 ≤ 0 is (a) (1,2) (b) (2,1) (c) (2,3) (d)(−3, 2)

1

6 If sin x + sin y = , then the value of cos x + cos y is ________

(a) (b) (c) (d) π

1

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7 An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colours is

(a) ퟐퟓ(b) ퟏ

ퟏퟓ (c) ퟖ

ퟏퟓ(d) ퟒ

ퟏퟓ

1

8 푑푥√9− 25x

(a) sin + c(b) sin + c

(c) log + c(d) log + c

1

9 What is the distance(in units) between the two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 ? (a) 0(b) 3(c)

√(d) 6

1

10 The equation of the line in vector form passing through the point(−1,3,5) and parallel to line = , z = 2. is (a) r⃗ = −ı̂ + 3ȷ̂+ 5k + λ 2ı̂ + 3ȷ̂+ k .(b) r⃗ = −ı̂ + 3ȷ̂+ 5k + λ(2ı̂+ 3ȷ̂)(c) r⃗ = 2ı̂ + 3ȷ̂ − 2k + λ −ı̂ + 3ȷ̂+ 5k(d) r⃗ = (2ı̂+ 3ȷ̂) + λ −ı̂ + 3ȷ̂+ 5k

1

(Q11 - Q15) Fill in the blanks 11 If f be the greatest integer function defined asf(x) = [x] and g be the modulus

function defined as g(x) = |x| , then the value of g of − is___________ 1

12 If the function푓(푥) = when 푥 ≠ 1

푘 when 푥 = 1 is given to be continuous at

푥 = 1, then the value of 푘 is ____

1

13 If 1 22 1

xy = 5

4 , then value of y is _____. 1

14 If tangent to the curve y + 3x − 7 = 0 at the point (ℎ, 푘) is parallel to line x − y = 4, then value of k is ___?

OR

1

For the curve 푦 = 5푥 − 2푥 ,if 푥 increases at the rate of 2units/sec, then at 푥 = 3the slope of the curve is changing at_________

15 The magnitude of projection of 2ı̂ − ȷ̂+ k on ı̂ − 2ȷ̂+ 2k is_________ OR

1

Vector of magnitude 5 units and in the direction opposite to 2ı̂+ 3ȷ̂ − 6k is___

(Q16 - Q20) Answer the following questions 16 Check whether (l + m + n) is a factor of the

determinantl + m m + n n + l

n l m2 2 2

or not. Give reason.

1

17 Evaluate ∫ (푥 + 1)푑푥.

1

18 Find ∫ 푑푥. 3

1



OR Find ∫(푐표푠 2푥 − 푠푖푛 2푥)푑푥

19 Find ∫ xe( )dx. 1

20 Write the general solution of differential equation = e 1

SECTION – B

21 Express sin√

;where− < 푥 < , in the simplest form.

OR

2

Let R be the relation in the set Z of integers given by R = {(a, b) : 2 divides a – b}.Show that the relation R transitive? Write the

equivalence class [0]. 22 If = ae + be , then show that − − 2y = 0. 2

23 A particle moves along the curve x = 2y . At what point, ordinate increases at the same rate as abscissa increases?

2

24 For three non-zero vectors a⃗, b⃗ and c⃗ , prove that [ 푎⃗ - 푏⃗ 푏⃗ - 푐⃗ 푐⃗ - 푎⃗ ] = 0 .

OR

2

If 푎⃗ + 푏⃗ + 푐⃗ = 0 푎푛푑 |푎⃗| = 3, 푏⃗ = 5, |푐⃗| = 7 , then find the value of 푎⃗ . 푏⃗ + 푏⃗.푐⃗ +푐⃗ . 푎⃗ .

25 Find the acute angle between the lines = = and = = 2

26 A speaks truth in 80% cases and B speaks truth in 90%cases. In what percentage of cases are they likely to agree with each other in stating the same fact?

2

SECTION – C 27 Let 푓: A → B be a function defined as 푓(푥) = , whereA = R − {3}and

B = R − {2}. Is the function f one –one and onto? Is f invertible? If yes, then find its inverse.

4

28 If √1− x + 1 − y = a(x − y) , then prove that =√

. OR

4

If x = a(cos 2θ + 2θ sin 2θ) and y = a(sin 2θ − 2θ cos 2θ) ,

find at θ = .

29 Solve the differential equation x dy − y dx = x + y dx .

4



30 Evaluate ∫ |푥 − 2푥|푑푥. 4

31 Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. Also, find mean of the distribution.

OR

4

There are three coins, one is a two headed coin (having head on both the faces), another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If It shows head. What is probability that it was the two headed coin ?

32 Two tailors A and B earn ₹150 and ₹200 per day respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. Form a L.P.P to minimize the labour cost to produce (stitch) at least 60 shirts and 32 pants and solve it graphically.

4

SECTION D 33 Using the properties of determinants, prove that

(푦+ 푧) 푥 푥푦 (푧 + 푥) 푦푧 푧 (푥 + 푦)

= 2푥푦푧(푥 + 푦 + 푧) .

OR

6

If A =2 3 41 −1 00 1 2

, find A . Hence, solve the system of equations

푥 − 푦 = 3 ; 2푥 + 3푦 + 4푧 = 17;

푦 + 2푧 = 7

34 Using integration, find the area of the region {(x, y) ∶ x + y ≤ 1 , x + y ≥ 1, x ≥ 0, y ≥ 0 }

6

35 A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of semi-circular ends is π ∶ π + 2.

OR

Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

6

36 Find the equation of a plane passing through the points 퐴(2,1,2) and 퐵(4,−2,1) and perpendicular to planer⃗. ı̂ − 2k = 5. Also, find the coordinates of the point, where the line passing through the points (3,4,1) and (5,1,6) crosses the plane thus obtained.

6



Class – XII MATHEMATICS (041)

SQP Marking Scheme (2019-20)

TIME: 3 Hrs. Maximum Marks: 80

SECTION A

1 (c) 9 1

2 (a) 3 × p 1

3 (b)p=3,q= 1

4 (b)0.25 1

5 (c) (2,3) 1

6 (b) 1 7 (c) ퟖ

ퟏퟓ1

8 (b) sin + c 1

9 (a) 0 1 10 (b) 풓⃗ = − ̂ + ퟑ ̂ + ퟓ풌 + 흀(ퟐ ̂ + ퟑ ̂) 1 11 푔 − = 푔(−2) = 2 1

12 2 1

13 푦 = 2 1 14 −3

2

OR

decreasing at rate of 72 units/sec.

1

15

57−2ı̂ − 3ȷ̂+ 6k

2 units

OR

1

16 Apply푅 → 푅 + 푅

l + m + n m + n + l n + l + mn l m2 2 2

= 2(l + m + n)1 1 1n l m1 1 1

; yes (l + m + n) is a factor 1

17 ∫ (푥 + 1)푑푥=∫ (푥 )푑푥 + ∫ 1푑푥 =퐼 + 퐼

= 0 +[푥] (As 퐼 is odd function)

=2+2 = 4

1



18

푓 (푥)푓(푥) 푑푥 = log|푓(푥)| + 푐

Let x + sin x = 푡 So (1 + cos x )푑푥 = 푑푡 퐼 = 3∫ = 3 log|푡| + 푐 = 3 log|(x + sin x)| + 푐 or directly by writing formula

OR

∫ cos 4푥 푑푥= + 푐

1

19 let (1 + x ) = 푡 so 2푥푑푥 = 푑푡 ⟹ 퐼 = ∫푒 푑푡 = 푒 +c = 푒( ) +c 1

20 dydx

= e e

⟹dye

= e dx

⇒ −e + 푐 = e⇒ 퐞퐱 + 퐞 퐲 = 풄

integrating both sides 1

SECTION B

21 = sin√

+√

if − < 푥 <

=sin sin 푥 cos + cos푥 sin if − + < 푥 + < +

=sin sin 푥 + if 0 < 푥 + < i.e. principal values

= 푥 +OR

Let 2 divides 푎 – 푏 and 2 divides (푏 – 푐) : where 푎, 푏, 푐 ∈ 푍 So 2 divides (푎 –푏) + (푏 − 푐) 2 divides (푎 – 푐): Yes relation R is transitive [0] = {0, ± 2, ± 4, ± 6,...}

1

1

1

1

22 y = ae + be ………………(1) = 2ae − be ………….(2)

= 4ae + be ………..(3) putting values on LHS = − − 2y=(4ae + be )− (2ae − be )− 2(ae + be ) =4ae + be − 2ae + be − 2ae − 2be =0

1

1



23

⟹ 2xdxdt

= 2dxdt

⟹ x = 1

x = 2y ………(1) ⟹ 2x = 2 (given = )

from (1) y =

so point is 1,

1

1

24 = a⃗− b⃗ . b⃗ − c⃗ × (c⃗ − a⃗)

= a⃗. b⃗ × c⃗ − b⃗. (c⃗ × a⃗)

푎⃗+ 푏⃗ + 푐⃗ . 푎⃗ + 푏⃗ + 푐⃗ = 0 ⇒ 푎⃗. 푎⃗ + 푎⃗. 푏⃗ + 푎⃗. 푐⃗ + 푏⃗. 푎⃗ + 푏⃗. 푏⃗ + 푏⃗. 푐⃗ + 푐⃗. 푎⃗ + 푐⃗. 푏⃗ + 푐⃗. 푐⃗ = 0.⟹ |푎⃗| + 푏⃗ + |푐⃗| + 2 푎⃗. 푏⃗ + 푏⃗. 푐⃗ + 푐⃗. 푎⃗ = 0⟹ 3 + 5 + 7 + 2 푎⃗. 푏⃗ + 푏⃗. 푐⃗ + 푐⃗. 푎⃗ = 0⟹ 2 푎⃗. 푏⃗ + 푏⃗. 푐⃗ + 푐⃗. 푎⃗ = −(9 + 25 + 49)

⟹ 푎⃗. 푏⃗ + 푏⃗. 푐⃗ + 푐⃗. 푎⃗ = −832

= a⃗ − b⃗ . b⃗ × c⃗− b⃗ × a⃗ − c⃗ × c⃗ + c⃗ × a⃗ = a⃗ − b⃗ . b⃗ × c⃗− b⃗ × a⃗ + c⃗ × a⃗ ……(c⃗ × c⃗ = 0) = a⃗ − b⃗ . b⃗ × c⃗ + a⃗ × b⃗ + c⃗ × a⃗ = a⃗. b⃗ × c⃗ + a⃗. a⃗ × b⃗ +a⃗. (c⃗ × a⃗)− b⃗. b⃗ × c⃗ − b⃗. a⃗ × b⃗ − b⃗. (c⃗ × a⃗) = a⃗. b⃗ × c⃗ + 0+0 − 0− 0 − b⃗. (c⃗ × a⃗)

=0 (STP remains same if vectors a⃗, b⃗ , c⃗ are changed in cyclic order)

OR

1

1

12

12

1

25

cos 휃 =3ı+ 4ȷ + 5k . 4ı̂ − 3ȷ̂+ 5k3ı+ 4ȷ + 5k 4ı̂ − 3ȷ̂ + 5k

⟹ cos 휃 =12 − 12 + 25

√9 + 16 + 25√9 + 16 + 25

⟹ cos 휃 =25

√50√50

⟹ cos 휃 =12

⟹ 휃 =휋3

Vector in the direction of first line 푏⃗ = 3ı+ 4ȷ+ 5k

Vector in the direction of second line 푑⃗ = 4ı̂ − 3ȷ̂ + 5k

Angle 휃 between two lines is given by cos 휃 =⃗. ⃗⃗ ⃗

12

12

1

26 P(A)= = , P(B)= =P(Agree)=P(Both speaking truth or both telling lie)

=푃(퐴퐵 표푟 퐴̅퐵 ) 1



= 푃(퐴)푃(퐵)표푟푃(퐴̅)푃(퐵) = +

= = = = 74% 1

SECTION C

27

⇒2푥 + 3푥 − 3

=2푥 + 3푥 − 3

⇒ (2푥 + 3)(푥 − 3) = (2푥 + 3)(푥 − 3)⇒ (2푥 푥 − 6푥 + 3푥 − 9) = (2푥 푥 − 6푥 + 3푥 − 9)⇒ −6푥 + 3푥 = −6푥 + 3푥⇒ 9푥 = 9푥⇒ 푥 = 푥

푦 =2x + 3x − 3

⇒ 푥푦 − 3푦 = 2푥 + 3⇒ 푥푦 − 2푥 = 3푦 + 3⇒ 푥(푦 − 2) = 3(푦 + 1)

f(x) = f3(푦 + 1)(푦 − 2)

=2 ( )

( ) + 3( )

( ) − 3since f(x) =

2x + 3x − 3

2(3y + 3) + 3(y − 2)3y + 3− 3y + 6

=9푦9

= 푦

Let 푦 = f(x) = ………………………(1) Let 푥 , 푥 ∈ A = R − {3} Let 푓(푥 ) = 푓(푥 )

Now 푓(푥 ) = 푓(푥 ) ⇒ 푥 = 푥so 푓(푥) is one-oneFor onto

⇒ 푥 = ( )( )

………………………(2)equation (2) is defined for all real values of y except 2i.e 푦 ∈ R − {2} which is same as given set 퐵 = R − {2}(co-domain=range) Also 푦 = f(x)

Thus for every y ∈ B,there exists x ∈ A such that f(x) = 푦 Thus function is onto. Since f(x) is one-one and onto so f(x)is invertible. Inverse is given by 푥 = 푓 (푦) = ( )

( )

12

112

1

1

28 1 − x + 1 − y = a(x − y)

1− sin 퐴 + 1 − sin 퐵 = a(sin퐴 − sin퐵) cos퐴 + cos퐵 = a(sin퐴 − sin퐵)

⟹ 2 cos퐴 + 퐵

2cos

퐴 − 퐵2

= 2a cos퐴 + 퐵

2sin

퐴 − 퐵2

⟹ cos퐴 − 퐵

2= a sin

퐴 − 퐵2

Let 푥 = sin퐴 , 푦 = sin퐵 12

1



⟹ cot퐴 − 퐵

2= a

⟹퐴−퐵

2= cot 푎

⟹ 퐴− 퐵 = 2 cot 푎 ⟹ sin 푥 − sin 푦 = 2 cot 푎

⇒1

√1 − x−

11 − y

푑푦푑푥

= 0

⇒ 푑푦푑푥

=1 − y

√1− x

differentiating w.r.t. x

OR

1

12

1

x = a(cos 2θ + 2θ sin 2θ)

⇒dxdθ

= a(−2 sin 2θ + 2 sin 2θ+ 4θcos 2θ)

y = a(sin 2θ − 2θ cos 2θ)

⇒dydθ

= a(2cos 2θ + 4 θ sin 2θ − 2cos 2θ)

⇒푑푦푑푥

=a(4 θ sin 2θ)a(4θcos 2θ)

⇒푑푦푑푥

= sin 2θcos 2θ

= tan 2θ

Differentiating again with respect to x, we get

⇒d ydx

= 2 푠푒푐 2θ.dθdx

⇒d ydx

= 2 푠푒푐 2θ.1

a(4θcos 2θ)

d ydx

= 2 푠푒푐π4

.1

a 4 cos

=8√2휋푎

⇒ = a(4θcos 2θ)……………..(1)

⇒ = a(4 θ sin 2θ)………………(2) using (1)and (2)

1

1

12

12

1

29 x

dydx

− y = x + y

⇒ xdydx

= y + x + y

let y = vx

⇒푑푦푑푥

= 푣 + 푥푑푣푑푥

⇒ =

…………………………………(1)

differentiating with w.r.t. x

put in (1)

1



⇒ 푣 + 푥푑푣푑푥

=푣푥 + √푥 + 푣 푥

x

⇒ 푣 + 푥푑푣푑푥

=푥(푣 + √1 + 푣 )

푥

⇒ 푥푑푣푑푥

= 푣 + 1 + 푣 − 푣

⇒ 푥푑푣푑푥

= 1 + 푣

⇒푑푣

√1 + 푣=푑푥푥

⇒푑푣

√1 + 푣=

푑푥푥

⇒ log 푣 + 1 + 푣 = log푥 + log 푐

⇒ log 푣 + 1 + 푣 = log 푐푥

⇒ 푣 + 1 + 푣 = 푐푥

⇒푦푥

+ 1 +푦푥

= 푐푥

⇒ 푦 + 푥 + 푦 = 푐푥

integrating both sides

12

1

1

30

|푥 − 2푥| =−(푥 − 2푥) 푤ℎ푒푛 1 ≤ 푥 < 2(푥 − 2푥) 푤ℎ푒푛 2 ≤ 푥 ≤ 3

Consider I=∫ |푥 − 2푥|푑푥

I=∫ |푥 − 2푥|푑푥 +∫ |푥 − 2푥|푑푥

I=∫ −(푥 − 2푥) 푑푥 +∫ (푥 − 2푥)푑푥

I=− − 푥 + − 푥

I=− − + +

I= = 2 1

1

1 1

31 Let X denotes the smaller of the two numbers obtained So X can take values 1,2,3,4,5,6 P(X=1 is smaller number) P(X=1)= = =

(Total cases when two numbers can be selected from first 7 numbers are 7 )

P(X=2)= =

P(X=3)= =

P(X=4)= = =

P(X=5)= =

P(X=6)= =

푥 1 2 3 4 5 6

12

2



푝 621

521

421

321

221

121

푝 푥 621

1021

1221

1221

1021

621

Mean =∑푝 푥 = + + + + + = =

OR

12

1

∴ 푃(퐸 ) = 푃(퐸 ) = 푃(퐸 ) =13

= 1 푃 퐴

퐸 = 푃(coin showing head given that it is a biased coin)

=75

100=

34

푃 퐴퐸 = 푃(coin showing head given that it is unbiased coin)

=12

푃(gettingtwo headedcoin when it is known that it shows Head)

푃 퐸퐴 =

푃(퐸 )푃 퐴퐸

푃(퐸 )푃 퐴퐸 + 푃(퐸 )푃 퐴

퐸 + 푃(퐸 )푃 퐴퐸

Let 퐸 = event of selecting a two headed coin 퐸 = event of selecting a biased coin, which shows 75% times Head 퐸 =event of selecting a unbiased coin. A = event that tossed coin shows head.

. 푃 퐴퐸 = 푃(coin showing head given that it is two headed coin)

By Bayes theorem

= ×

× × × =

×=

× =

Required probability=

12

112

1

1

32

푥 ≥ 0 ,푦 ≥ 0

3푥 + 5푦 = 30 푥 + 푦 = 8

Let tailor A works for 푥 days and tailor B works for y days Objective function : To minimize labour cost 푍 = 150푥 + 200푦 (in ₹) Subject to constraints 6푥 + 10푦 ≥ 60 i.e. 3푥 + 5푦 ≥ 30 4푥 + 4푦 ≥ 32 i.e. 푥 + 푦 ≥ 8

consider equations to draw the graph and then we will shade feasible region

12

112



corner points of feasible region are A(10,0),B(5,3) and C(0,8) Value of Z at these corner points

Point 푍 = 150푥 + 200푦 (in ₹) A(10,0) =1500+0=1500 B(5,3) =750+600=1350 (minimum) C(0,8) =0+1600=1600

So minimum value of Z is ₹1350 when tailor A works for 5 days and tailor B works for 3 days.

To check draw 150푥 + 200푦 < 1350 i.e 3푥 + 4푦 < 27 As there is no region common with feasible region so minimum value is ₹1350

1

1

SECTION D

33

LHS=(푦 + 푧) 푥 푥

푦 (푧 + 푥) 푦푧 푧 (푥 + 푦)

퐴푝푝푙푦 퐶 → 퐶 − 퐶 , 퐶 → 퐶 − 퐶

=(푦 + 푧) 푥 − (푦 + 푧) 푥 − (푦 + 푧)

푦 (푧 + 푥) − 푦 0푧 0 (푥 + 푦) − 푧

=(푦 + 푧) (푥 + 푦 + 푧)(푥 − 푦 − 푧) (푥 + 푦 + 푧)(푥 − 푦 − 푧)

푦 (푧 + 푥 + 푦)(푧 + 푥 − 푦) 0푧 0 (푥 + 푦 + 푧)(푥 + 푦 − 푧)

Taking (푥 + 푦 + 푧)common from 퐶 as well as 퐶

=(푥 + 푦 + 푧)(푦 + 푧) (푥 − 푦 − 푧) (푥 − 푦 − 푧)

푦 (푧 + 푥 − 푦) 0푧 0 (푥 + 푦 − 푧)

Apply 푅 → 푅 −푅 − 푅

=(푥 + 푦 + 푧)2yz −2푧 −2푦푦 (푧 + 푥 − 푦) 0푧 0 (푥 + 푦 − 푧)

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Apply 퐶 → 푦 퐶 and 퐶 → 푧 퐶

=( )2yz −2푦푧 −2푦푧푦 (푦푧 + 푦푥 − 푦 ) 0푧 0 (푧푥 + 푧푦 − 푧 )

Apply 퐶 → 퐶 +퐶 and 퐶 → 퐶 +퐶

=( )2yz 0 0푦 (푦푧 + 푦푥) 푦푧 푧 (푧푥 + 푧푦)

expanding along 푅

= ( ) 2푦푧[(푦푧 + 푦푥)(푧푥 + 푧푦)− 푦 푧 ]

=2(푥+ 푦 + 푧) [푥푦푧 + 푥 푦푧 + 푥푦 푧 + 푦 푧 − 푦 푧 ] = 2푥푦푧(푥 + 푦 + 푧) (푥 + 푦 + 푧) = 2푥푦푧(푥 + 푦 + 푧)

OR

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|A| = 2(−2)− 3(2− 0) + 4(1− 0) = −6 ≠ 0 ∴ A exists

Cofactors 퐴 = −2 퐴 = −2 퐴 = 1

퐴 − 2 퐴 = 4 퐴 = −2

퐴 = 4 퐴 = 4 퐴 = −5

퐴푑푗 퐴 =−2 −2 1−2 4 −24 4 −5

퐴푑푗 퐴 =−2 −2 4−2 4 41 −2 −5

퐴 =퐴푑푗 퐴

|퐴| =1−6

−2 −2 4−2 4 41 −2 −5

⇒ 푋 = 퐴 퐵

⇒ 푋 =1−6

−2 −2 4−2 4 41 −2 −5

1737

** A =2 3 41 −1 00 1 2

System of equations can be written as 퐴푋 = 퐵

Where A =2 3 41 −1 00 1 2

,푋 =푥푦푧

,퐵 =1737

Now 퐴푋 = 퐵

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⇒ 푋 =1−6

−34− 6 + 28−34 + 12 + 28

17− 6 − 35

⇒ 푋 =1−6

−126

−24

⇒ 푋 =푥푦푧

=2−14

⇒ 푥 = 2, 푦 = −1, 푧 = 4

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34

x + (1− 푥) = 1 x + x − 2푥 + 1 = 1 2x − 2푥 = 0 2푥(푥 − 1) = 0 푥 = 0 표푟 푥 = 1

1 − 푥 푑푥 − (1 − 푥)푑푥

0 +12

.휋2

− 0 − 1 −12

x + y = 1………………..(1) x + y = 1…………………..(2) solving (1) and(2)

Required area = shaded area ACBDA =area(OACBO)− area(OADBO) =∫ (푦 − 푦 )푑푥

= √ + sin 푥 − 푥 −

− square units

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35 Let 푟 be the radius and ℎbe the height of half cylinder Volume = 휋푟 ℎ = 푉(constant)……………..(1)

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⇒ (2휋푟) +−1푟

4푉휋

+ 2푉 = 0

⇒ (2휋푟) =1푟

4푉 + 2푉휋휋

⇒ 휋푟 = 푉2 + 휋휋

⇒12휋푟 ℎ =

휋 푟휋 + 2

⇒ℎ

2푟=

휋휋 + 2

⇒ ℎ푒푖푔ℎ푡:푑푖푎푚푒푡푒푟 = 휋:휋 + 2

ℎ푒푖푔ℎ푡: 푑푖푎푚푒푡푒푟 = 휋:휋 + 2

Total surface area of half cylinder isS=2 휋푟 + 휋푟ℎ + 2푟ℎ…………………(2) From (1) put the value of ℎ in (2)

S=(휋푟 ) + 휋푟 + 2푟

S=(휋푟 ) + + 2푉

= (2휋푟) + + 2푉 ……………..(3)

퐹표푟 maxima/minima = 0

⇒ V= ………………..(4) 퐹rom (1) and (4)

Differentiating (3) with respect to 푟

= (2휋) + + 2푉 =positive (as all quantities are +ve)so S is minimum when

OR

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Let 2r be the base and h be the height of triangle ,which is inscribed in a circle of radius R

Area of triangle= (푏푎푠푒)(ℎ푒푖푔ℎ푡)

A = (2푟)(ℎ) = 푟ℎ…………(1)

Area being positive quantity, A will be maximum or minimum if 퐴 is

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푝푢푡 푖푛 (2) 푍 = ℎ (2ℎ푅 − ℎ ) ⇒ 푍 = (2ℎ 푅 − ℎ )

⇒ 6ℎ 푅 − 4ℎ = 0⇒ 6푅 = 4ℎ(ℎ ≠ 0)

⇒ ℎ =3푅2

⇒푑 푍푑ℎ

= 12ℎ푅 − 12ℎ

⇒푑 푍푑ℎ

= 123푅2

푅 − 123푅2

= 18푅 − 27푅 = −푣푒

푟 =3푅

4

푟 =√3푅

2

tan휃 =ℎ푟

=√

= √3휃 =휋3

maximum or minimum.

푍 = 퐴 = 푟 ℎ ……………………..(2) 푁ow In triangle OLB 퐵퐿 = 푂퐵 −푂퐿 In ∆OBD Z = A2 = r2 h2 푟 = 푅 − (ℎ − 푅) ⇒ 푟 = 2ℎ푅 − ℎ

⇒ = 6ℎ 푅 − 4ℎ ……………(3)

퐹표푟 maxima/minima = 0

differentiating (3) w.r.t. h

so Z=퐴 is maximum when ℎ =

⇒ 퐴 is maximum when ℎ =

when ℎ = , 푟 = 2ℎ푅 − ℎ = 2푅. −

Triangle ABC is equilateral triangle

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36 Let 푃(푥,푦, 푧) be any point on the plane in which 퐴(2,1,2) and 퐵(4,−2,1)lie.∴ 퐴푃⃗ 푎푛푑 퐴퐵⃗lie on required plane. Also required plane is perpendicular to given plane r⃗. ı̂ − 2k = 5 ∴normal to given plane 푛⃗ = ı̂ − 2k lie on required plane. ⇒ 퐴푃⃗,퐴퐵⃗and푛⃗ are coplanar.Where퐴푃⃗ = (푥 − 2)ı̂+ (y − 1)ȷ̂+ (z − 2)k 퐴퐵⃗ = =2ı̂ − 3ȷ̂ − k ⇒Scaler triple product 퐴푃⃗ 퐴퐵⃗ 푛⃗ = 0

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⇒푥 − 2 푦 − 1 푧 − 2

2 −3 −11 0 −2

= 0

⇒ (푥 − 2)(6− 0)− (푦 − 1)(−4 + 1) + (푧 − 2)(0 + 3) = 0⇒ 6푥 − 12 + 3푦 − 3 + 3푧 − 6 = 0

∴ 2(2휆 + 3) + (−3휆 + 4) + (5휆 + 1) = 7 4휆 + 6− 3휆 + 4 + 5휆 + 1 = 7

6휆 = −4

휆 = −23

⇒ 2푥 + 푦 + 푧 = 7………………….(1) Line passing through points 퐿(3,4,1)and 푀(5,1,6) is ⇒ = = = 휆………….(2) ⇒General point on the line is 푄(2휆 + 3,−3휆+ 4,5휆 + 1)As line (2) crosses plane (1) so point Q should satisfy equation(1)

푄(− + 3,2 + 4,− + 1)= 푄 , 6,−

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eb??a@=d:;=cb>a?a ilnmjklnhgf · (a)p= 6 ,q=27(b)p=3,q= 6 ; 6 (c) p=6,q= 6 ; 6 (d) p=3 ,q=27 1 4...

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