Namma Kalvi 11th Maths Public Exam 2019 Model Question ... · 2 +1 Std - Mathematics Sura s Model...

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11thSTD.

Time : 2.30 Hours Mathematics Marks : 90

Sura’s Model Question Paper 1

Section - INote : (i) Answer all the questions. [20 × 1 = 20] (ii) Choose the correct or most suitable answer

from the given four alternatives. Write the option code and the corresponding answer.

1. If A = {(x, y) : y = sin x, x ∈ R} and B = {(x, y) : y = cos x, x ∈ R} then A ∩ B contains(1) no element (2) infinitelymanyelements(3) onlyoneelement(4) cannot be determined.

2. The range of the function 1

1 2sin− x is

(1) (–∞, –1) ∪ 1

3,∞

(2) −

1

1

3,

(3) −

11

3, (4) (–∞, –1] ∪ 1

3,∞

3. The value of log 2512 is

(1) 16 (2) 18 (3) 9 (4) 12

4. The number of roots of (x + 3)4 + (x + 5)4 = 16 is(1) 4 (2) 2 (3) 3 (4) 0

5. 1

cos 80º–

3sin80º

=

(1) 2 (2) 3 (3) 2 (4) 4

6. If sin a + cos a = b, then sin 2a is equal to(1) b2– 1, if b ≤ 2 (2) b2 – 1, if b > 2 (3) b2 – 1, if b ≥ 1 (4) b2 – 1, if b ≥ 2

7. The number of 5 digit numbers all digits of which are odd is(1) 25 (2) 55 (3) 56 (4) 625

8. The number of permutations of n different things taking r at a time when 3 particular things are to be included is(1) n – 3Pr–3 (2) n – 3Pr (3) nPr–3 (4) r! n – 3Cr–3

9. If nC10 > nCr for all possible r, then a value of n is(1) 10 (2) 21 (3) 19 (4) 20.

10. Choose the correct statement(1) Matrix addition is not association (2) Matrix addition is not commutative(3) Matrix multiplication is association(4) Matrix multiplication is commutative

11. Find the odd one out of the following :

(1) 0 2

2 0−

(2) 0

7

2

7

2 0

−

(3) 0 3 2

3 2 0

.

.−

(4) 0 1

1 0

12. Match List - I with List II.

List I List II

1. log tan x dx( )∫0

2

π

(a)16

35

2. x x dx110

0

1

−( )∫ (b)120

46

3. sin7

0

2

xdx

π

∫ (c)1

132

4. x e dxx5 4

0

−∞

∫ (d) 0

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Namma Kalvi

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2 +1 Std - Mathematics Sura’s Model Question Paper - 2018-19

The Correct match is (1) 1 – d 2 – c 3 – a 4 – b (2) 1 – 4 2 – c 3 – b 4 – a (3) 1 – b 2 – a 3 – c 4 – a (4) 1 – b 2 – c 3 – a 4 – d

13. Two vertices of a triangle have position vectors

3 4 4i j k∧ ∧ ∧+ − and 2 3 4i j k

∧ ∧ ∧+ + . If the position vector

of the centroid is i j k∧ ∧ ∧+ +2 3 , then the position vector

of the third vertex is

(1) − − +∧ ∧ ∧

2 9i j k (2) − − −∧ ∧ ∧

2 6i j k

(3) 2 6i j k∧ ∧ ∧− + (4) − + +

∧ ∧ ∧2 6i j k

14. Choose the incorrect pair :

1. sin x x ∈2. cos x x ∈3. log x x > 04. e–x x > 0

15. lim3x

x→

=

(1) 2 (2) 3 (3) does not exist (4) 0

16. The function y = | |3 43 4

x

x

−−

is discontinuous at x =

(1) 0 (2) 3

4

(3) 4

3 (d) 1

17. For the curve x ydy

dx+ = 1, at 1

414

,

is

(1) 1

2 (2) 1

(3) −1 (d) 2

18. x

xdx

+

−∫ 2

12is

(1) x x x2 21 2 1− − + −log + c

(2) sin log− − + −1 2

2 1x x x + c

(3) 2 12 1

log sinx x x+ − − − + c

(4) x x x2 21 2 1− + + −log + c

19. Assertion (A) : In rolling die, getting number

Reason (R) :Inadiecontainsonlynumbers1,2,3,4, 5, 6

(1) Both (A) and (R) are true and (R) is the correct rexplanation of (A)

(2) Both (A) and (R) are true but (R) is not the correct explantion of (A)

(3) (A) is true (R) is false(4) (A) is false (R) is true

20. If A and B are two events such that A ⊂ Band P(B) ≠ 0, then which of the following is correct?

(1) P (A/B) = P A

P B

( )( )

(2) P(A/B) < P(A)

(3) P(A/B)≥P(A) (4) P(A/B)>P(B)

Section - II (i) AnsweranySEVEN questions. (ii) Question number 30 iscompulsory. 7 × 2 = 1421. Show that the relation R on the set A = {1, 2, 3} given

byR={(1,1)(2,2)(3,3)(1,2)(2,3)}isreflexivebutneithersymmetricnortransitive.

22. A model rocket is launched from the ground. The height ‘h’reachedbytherocketaftert seconds from liftoffisgivenbyh(t) = –5t2 + 100t, 0 ≤ t ≤ 20. At what time the rocket is 495 feet above the ground?

23. Findthevaluesofsin(−1110°)24. AKabaddicoachhas14playersreadytoplay.How

manydifferentteamsof7playerscouldthecoachputon the court?

25. Find a negative value of miftheCo-efficientofx2 in the expansion of (1 + x)m, |x| < 1 is 6.

26. Find the value or values of m for which m ( )i j k∧ ∧ ∧

+ + is a unit vector.

27. Solve: lim2x→

−−

x

x

4 162

28. Solve: y = sin x + cos x

29. Integrate x43 30. Five mangoes and 4 apples are in a box. If two fruits

arechosenatrandom,findtheprobabilitythat(i)oneis a mango and the other is an apple (ii) both are of thesamevariety.

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+1 Std - Mathematics Sura’s Model Question Paper - 2018-19 3

Section - III (i) Answerany SEVEN questions. (ii) Question number 40iscompulsory. 7 × 3 = 2131. A plumber can be paid according to the following

schemes:Inthefirstschemehewillbepaidrupees500plusrupees70perhour,andinthesecondschemehe will paid rupees 120 per hour. If he works x hours, then for what value of xdoesthefirstschemegivebetter wages?

32. If in two circles, arcs of same length subtend angles 60°and75°atthecenter,findtheratiooftheirradii?

33. Find the sum of all 4-digit numbers that can be formed using digits 1, 2, 3, 4, and 5 repetitions not allowed?

34. If a, b, c are in A.P, b, c, d are in G. P, c, d, eareinH.P., then show that a, c, e are in G. P.

35. Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1,–1) is equal to 20.

36. Construct the matrix A =[ aij]3 × 3, where aij = i – j. StatewhetherAissymmetricorskew-symmetric.

37. Find the vectors of magnitude 10 3 that are perpendicular to the plane which contains

i j k i j k∧ ∧ ∧ ∧ ∧ ∧

+ + + +2 and 3 4 ,

38. If f(x) = 2 33 2

x x

x x

++

sinsin

, x ¹ 0 is continuous at x = 0,

thenfindf(0).39. Find the derivatives from the left and from the right

at x =1(iftheyexist)ofthefollowingfunctions.Arethe functions differentiable at x = 1, f(x) = 1 2− x

40. If f ″(x) = 12x – 6 and f (1) = 30, f ′(1)=5findf (x)

Section - IVAnswer all questions. 7 × 5 = 3541. The formula for converting from Fahrenheit to

Celsius temperatures is y x= −5

9

160

9. Find the

inverse of this function and determine whether the inverse is also a function.

(OR)

Prove that 3 isanirrationalnumber.(Hint:Followthe method that we have used to prove 2 ∉ Q.)If

cos(α−β)+cos(β−γ)+cos(γ−α)=−32

then provethatcosα+cosβ+cosγ=sinα+sinβ+sinγ=0.

(OR) By the principle of mathematical induction, prove

that for n ≥1

1 2 3 1

23 3 3 3

2

+ + + + =+

.. ( )n

n n

42. Find the sum up to the 17th term of the series 11

1 21 3

1 2 31 3 5

3 3 3 3 3 3

+ ++

+ + ++ +

+�

(OR) Arayoflightcomingfromthepoint(1,2)isreflected

at a point A on the x-axis and it passes through the point (5,3). Find the co-ordinates of the point A.

43. Solve 4 4 44 4 44 4 4

− + ++ − ++ + −

x x x

x x x

x x x

= 0.

(OR) Show that the following vectors are coplanar

(i) i j k i j k j k∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧

− + − + − − +2 3 2 3 4 2, ,

(ii) 5 6 7 7 8 9 3 20 5i j k i j k i j k∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧

+ + − + + +, ,

44. Examinethecontinuityofthefollowing:(i)x + sin x (ii) x2 cos x (iii) ex tan x (iv) e2x + x2 (v) x.ln x

(OR)

Find the derivative with tan−1sin

1 + cosx

x

with

respect to tan–1 cos

1 + sinx

x

.

45. 1

3 4x x+ − − (OR) AfirmmanufacturesPVCpipesinthreeplantsviz,

X,YandZ.ThedailyproductionvolumesfromthethreefirmsX,YandZarerespectively2000units,3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s totalproduction,(i) findtheprobabilitythattheselectedpipeisa

defective one.(ii) if the selected pipe is a defective, then what is

theprobabilitythatitwasproducedbyplantY?

46. Evaluate x x

x xdx

2

2

5 33 2

+ ++ +∫

(OR)

47. Solve: x x x+ + + = +5 21 6 40

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ANSWERS

1. (2) infinitelymanyelements

2. (4) (–∞, –1] ∪ 1

3,∞

3. (2) 184. (1) 45. (4) 4

6. (1) b2– 1, if b ≤ 2

7. (2) 55

8. (4) r! n – 3Cr–3

9. (4) 20

10. (3) Matrix multiplication is association

11. Hint : (a), (b), (a) are skew symmetric and (d)symmetric

(4) 0 1

1 0

12. (4) 1 – b 2 – c 3 – a 4 – d

13. (1) − − +∧ ∧ ∧

2 9i j k

14. Hint : e–x canbedefinedin (4) e–x ; x > 0

15. (3) does not exist

16. (3) 4

3

17. (3) –1

18. (4) x x x2 21 2 1− + + −log + c]

19. (4) (A) is false (R) is true

20. (3) P(A/B)≥P(A)

Section - II (i) AnsweranySEVEN questions. (ii) Question number 30 iscompulsory. 7 × 2 = 1421. Since A = {1, 2, 3} (1, 1) (2, 2) (3, 3) ∈ R. ⇒ R is reflexive. Also (1, 2) ∈ R but (2, 1) ∉ R ⇒ Risnotsymmetric. And (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R ⇒ R is not transitive. Risreflexivebutneithersymmetricnortransitive.

22. h(t) = – 5t2 + 100t, 0 ≤ t ≤ 20 Let the time be ‘t’ sec, when the rocket is 495 feet

above the ground \ 0 < h(t) < 495 ⇒ 0 < –5t2 + 100t < 495 ⇒ 0 < – 5t2 + 100t – 495 < 0

[subtracting 495] ⇒ –5t2 + 100t – 495 = 0 ⇒ t2 – 20t + 99 = 0

[Dividedby–5] ⇒ (t – 11) (t – 9) = 0 ⇒ t = 11 or 9. \ At 11 or 9 sec, the rocket is 495 feet above the

ground.

23. sin(−1110°)=−sin(1110°)since(sinθisanoddfunction)=−sin(1080+30°)

=−sin(3×360+30°)=−sin30°=−1

2

24. Here7playersmustbeselectedfrom14players.This can be done in 14C7ways.

Hence,numberofdifferentteamsofplayers

= =14

7

14

7 7C

!

! !

=

× × × × × × ×× × × × × ×

14 13 12 11 10 9 8 7

7 7 6 5 4 3 2 1

!

!

= × × × ×13 11 2 3 4

= 3432.

25. 1 11

2

2+( ) = + +−( ) +x mx

m mxm

!. .

[Binomial theorem for rational index]

\Co-efficientof xm m

21

2=

−( )

Given m m

m m−( ) = ⇒ − =

1

26 12

2

⇒ m2 – m – 12 = 0 ⇒ (m−4)(m + 3) = 0⇒ m=4or−3.\ Negative value of mis−3.

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26. Let a→

= m( i j k∧ ∧ ∧+ + )

| a→

| = m 1 1 12 2 2+ + = m 3

To make a→

as a unit vector, | a→

| = +1

\m 3 = ± 1 ⇒ m = ± 1

3.

27. limx

xx→

−−2

416

2 = lim

x→2

xx

4 42

2

−−

∵ lim .x a

n nnx a

x an a

→

−−−

=

1

= 4(2)4–1 = 4(2)3 = 4(8) = 32

28. Given y = sin x + cos x

dydx

= ddx

x ddx

xsin cos ( ) + ( )

= cos x – sin x.

29. x43 = x dx43ò = x dx41 3( )∫/

= x dx4 3/ò = x c

4

31

14

3

+

+ +

= x c

7

3

7

3

+ = 3

7

7

3x c+

30. S = { 5 mangoes and 4 apples} ⇒ n (S) = 9C2

[ 2 fruits are taken from 9 fruits]

(i) Let A be the event of getting one mango and one apple

∴ n(A) = 5C1 × 4C1 = 5 × 4 = 20

∴ P(A) = 20 20

9 8

2 1

20 2

9 8

5

992C

= ××

= ××

=

(ii) Let B be the event of getting 2 fruits from the samevariety.

∴ n(B) = 5C2 + 4C2 = 5 4

2 1

4 3

2 1

20

2

12

2

××

+ ××

= +

= 10 + 6 = 16

∴ P(B) = nn

B

S C

( )( )

= = ××

= ××

=16 16

9 8

2 1

16 2

9 8

4

992

Section - III (i) Answerany SEVEN questions. (ii) Question number 40iscompulsory. 7 × 3 = 21

31. Let the number of hours to complete the job is x. Wagesfromthefirstscheme=`(500+70x) Wages from the second scheme = `120x Given 500+70x > 120x ⇒ 500 > 120x–70x ⇒ 500 > 50x

⇒ 500

50> x > x ⇒

50

5

10

1

> x

⇒ 10 > x

⇒ x < 10

\ Number of hours should be less than ten hours.

32. Let r1, r2 be the radii of two circles at the centres of which arcs of equal lengths subtend angles of 60° and75°respectively.

∴θ = 60º

= 60 ×∴ = ° = × =θπ π

160 60

180 3 radians

⇒ θ = ⇒ = ⇒ =θ θ11

1 1lr

l r

⇒ l = ⇒ =l rπ3

1 ...(1)

⇒ θ2 = 75º

= 75×θπ π π

275 75

180

15

36

5

12= ° = × = = radian

∴θ2 = ∴ = ⇒ =θ θ22

2 2lr

l r

⇒ l = ⇒ = ×l r5

62

π ...(2)

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From (1) and (2), π π3

5

121 2r r= = π π

3

5

121 2r r=

⇒ ⇒ = × =rr1

2

5

12

3 5

4

ππ

= ⇒ = × =rr1

2

5

12

3 5

4

ππ

⇒ r1 : r2 = 5 : 4

33. The number of 4-digit numbers that can be formed using the 5 digits is 5P4 = 120Letusfindthesumofthedigitsintheunitplace.

Byfilling1intheunitplace,remaining3placescanbefilledwithremaining4digitsis4P3=24ways.

Similarly,eachofthedigits2,3,4,5appear24timesin unit place.

\ Sum of all these 120 numbers are (4P3 × 1) (4P3 × 2) + (4P3 × 4) + (4P3 × 5)

= 4P3 (1 + 2 + 3 + 4 + 5)

= 4P3 × 15 = 24 × 15 = 360

Similarlysumofthedigitsintheten’splaceis3600

Sum of the digits in the hundreds place is 36000

Sum of the digits in the thousands place is 360000

Hence,sumofall the4–digitnumbersformedbyusing the digits 1, 2, 3, 4 and 5 is= 360 + 3600 + 36000 + 360000= 399960.

34. Since a, b, c are in A. P.

b a c= +2 ...(1)

and b, c, d are in G. P ⇒ c bd= ...(2)

Also c, d, eareinH.P ⇒ d cec e

=+

2 ...(3)

From (2), c2 = bd = a c ce

c e+

+

2

2 [using (1) and

(3)]

= a c ce

c e+( )+

⇒ ca c ec e

=+( )+

⇒ c ce ae ce2 + = + ⇒ c2 = ae ⇒ a, c, e are in G. P.

35. Let P(h, k) be the point on the locus and A(3,5) B (1,–1) be the given points.

Bythegivencondition,PA2 + PB2 = 20.

P (h, k)A(3,5)

B(1,−1)

1 2 3 4 50

xx¢

y¢

y

Given PA2 + PB2 = 20 ⇒ (h – 3)2 + (k – 5)2 + (h – 1)2 + (k + 1)2 = 20 h2 – 6h + 9 + k2 –10k + 25 + h2 – 2h + 1 + k2 + 2k + 1 = 20 ⇒ 2h2 + 2k2 – 8h – 8k + 36 = 20 ⇒ 2h2 + 2k2 – 8h – 8k + 36 – 20 = 0 ⇒ 2h2 + 2k2 – 8h – 8k + 16 = 0 Dividingby2weget, h2 + k2 – 4h – 4k + 8 = 0 ∴ Locus of (h,k) is x2 + y2 – 4x – 4y + 8 = 0

36. Given aij = i – j Let A = [aij]3×3 ∴ a11 = 1 – 1 = 0 a21 = 2 – 1 = 1 a31 = 3 – 1 = 2 a12 = 1 – 2 = – 1 a22 = 2 – 2 = 0 a32 = 3 – 2 = 1 a13 = 1 – 3 = – 2 a23 = 2 – 3 = –1 a33 = 3 – 3 = 0

⇒ A = 0 1 2

1 0 1

2 1 0

− −−

⇒ ∴ AT = 0 1 2

1 0 1

2 1 0

−− −

= – 0 1 2

1 0 1

2 1 0

− −−

= – A

Since AT=–A,Aisaskew-symmetricmatrix.

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37. Let a→

= i j k∧ ∧ ∧+ +2

and b→

= i j k∧ ∧ ∧+ +3 4

A unit vector which is perpendicular to the vector a→

and b→

is a b

a b

→ →

→ →×

×| |

a→

× b→

=

i j k∧ ∧ ∧

1 2 1

1 3 4

= i∧

(8 – 3) – j∧

(4 – 1) + k∧

(3 – 2)

= 5 i∧

– 3 j∧

+ k∧

| a→

× b→

| = 5 3 12 2 2+ −( ) +

= 25 9 1+ + = 35

\ A unit vector which is perpendicular to the vector

a→

and b→

is 5 3

35

i j k∧ ∧ ∧− +

Hence, a vector of magnitude 10 3 , which is

perpendicular to the vectors a→

and b→

is

± − +

∧ ∧ ∧10 3

355 3i j k

38. Given f(x) is continuous at x = 0

⇒ f(0) = limx

f x→

( )0

⇒ f(0) = limsin

sinx

x xx x→

++0

2 3

3 2

⇒ f(0) = lim

sin

sinx

xx

xx

→

+

+

0

2 3

3 2

⇒ f(0) = 2 3 1

3 2 1

+ ( )+ ( ) =

55

= 1

f(0) = 1

39. Given f(x) = 12− x

f (1) = 1 1 0− =

\f ′(1–) = limx

xx→ −

− −−1

21 0

1 [ f(1) = 0]

= limx

xx→ −

−−1

21

1

= limx

x xx→ −

+ −− −( )1

1 1

1

.

= limx

x xx x→ −

+ −− −( ) −1

1 1

1 1

..

= lim limx x

xx

xx→ − → −

+−

= − +−1 1

1

1

1

1= –∞

[ Negative sign comes inside the square root]

\ f ′(1+) = limx

f x fx→ +

( ) − ( )−1

1

1

= lim limx x

xx

xx→ + → +

− −−

= −−1

2

1

21 0

1

1

1

= lim limx x

x xx

x x

x x→ + → +

+ −− −( ) =

+ −

− − −1 1

1 1

1

1 1

1 1

. .

.

= limx

xx→ +

− +−

=1

1

10

From (1) and (2), f ′(1–) and f ′(1+) are not equal.

\ f ′(x) is not differentiable at x = 1.

40. Given f ″(x) = 12x – 6, f (1) = 30

f ′(1) = 5

Integrating on both sides we get,

f x dx''( )∫ = 12 6x dx−( )∫

⇒ f ′(x) = 12 6x dx dx− ∫∫

= 12

26

2x x c− +

f ′(x) = 6x2 – 6x + c ... (1)

since f ′(1) = 5, we get

⇒ f ′(1) = 6(1)2 – 6(1) + c

⇒ 5 = 6 – 6 + c

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c = 5

\f ′(x) = 6x2 – 6x + c [From (1)]

Integrating again with respect to ‘x’ we get

′∫ f x dx( ) = 6 6 52x x dx− +( )∫ = 6 6 5

2x dx x dx dx− + ∫∫∫

⇒ f (x) = 6

3

6

25

3 2.x x x k− + +

f (x) = 2 3 53 2x x x k− + + ... (2)

since f (1) = 30 we get

f (1) = 2 1 3 1 5 13 2

( ) ( ) ( )− + + k

⇒ 30 = 2 – 3 + 5 + k

⇒ 30 = 4 + k

\ k = 26

Substituting k = 26 in (2) we get,

f (x) = 2 3 5 263 2x x x− + +

Section - IVAnswer all questions. 7 × 5 = 35

41. Let f(x) = 5 160

9

x −

Given y = 5

9

160

9

x −

⇒ y = 5 160

9

x −

Then 9y = 5x – 160 ⇒ 5x = 9y + 160

⇒ x = 9 160

5

y +

Let g(y) = 9 160

5

y +

Now gof(x) = g[f (x) = g x5 160

9

−

= =

−

+=

− += =

95 160

9160

5

5 160 160

5

5

5

xx x x

and fog(y) = f(g(y)) = f y9 160

5

+

= =

+

−=

+ −=

59 160

5160

9

9 160 160

9

yy y

Thus gof = Ix and fog = Iy.

This implies that f and g are bijections and inverses to each other.

f –1(y) = 9 160

5

y +

Replacing ybyx, we get f –1(x) = 9 160

5

x +=

9

5

x + 32

(OR)

Suppose 3 is a rational number.

Then 3 can be written as 3 =mn

Where m and n are rational numbers with no common factors other than 1.Squaring both sides we get,

3 3

2

2

2 2= ⇒ =mn

n m

multiplyingby2weget, 6n2 = 2m2 ⇒ 3(2n2) =2m2 ...(1)Since 2n2isdivisibleby2,m2 is also an even number⇒ m must be even ⇒ m = 2k for some natural number k⇒ 3n2 = (2k)2

⇒ 3n2 = 4k2 [From (1)]⇒ n is also an even number.Thus both m and n are even numbers having a common factor 2.This contra dicts our initial assumption that m and n do not have a common factor.

Hence 3 cannot be a rational number.

⇒ 3 is an irrational number.Henceproved.

42. Given cos (α−β) + cos (β−γ) + cos (γ−α)=−3

2

⇒ cos αcosβ+sinαsinβ+cosβcosγ

+sinβsinγ+cosγcosα+sinαsinγ=−3

2

⇒ 2 [cos α cos β + cos βcos γ+cos γcos α+sin αsin β+sin βsinγ+sin αsin γ]=−3

⇒ (2 cos αcosβ+2cosβcosγ+2cosγcosα)+(2sinαsinβ+2sinβsinγ+2sinγsinα) + 3 = 0

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⇒ (2 cos αcosβ+2cosβcosγ+2cosγcosα)+(2sinαsinβ+2sinβsinγ+2sinγsinα)+ (cos2α+sin2α)+(cos2β+sin2β) + (cos2γ+sin2γ)=0

⇒ (cos2α+cos2β+cos2γ+2cosαcosβ+2cosβcosγ+2cosγcosα)+(sin2α+sin2β+sin2γ+2sinαsinβ+2sinβsinγ+2sinγsinα)=0

⇒ (cosα+cosβ+cosγ)2+(sinα+sinβ+sinγ)2 = 0 ⇒ cosα+cosβ+cosγ=0 andsinα+sinβ+sinγ=0.

(OR)

Step 1: Put n = 1

⇒ 13 = ⇒ =+

=

⇒ =

11 1 1

2

1 2

2

1 1

3

2 2( ) ( )

⇒ 1 = 1\ p(n) is true.

Step 2: Let us assume that p(K) is true.

13 + 23 + 33 + ... + K3 = 1 2 31

2

3 3 3 3

2

+ + + + =+

� KK K( ) ...(1)

Step 3: To prove that p(K + 1) is true.

i. e to P. T. 13 + 23 + ... + K3 + (K+1)3 = 1 2 11 2

2

3 3 3 3

2

+ + + + + =+ +

� K KK K

( )( )( )

LHS=13 + 23 + … + K3 + (K + 1)3

= KK( )+

1

2

2

+ (K + 1)3 [Using (1)]

= ( )KK

K+ + +

1

41

2

2

= (K +1)K + 4K + 4

4

22

= (K +1) (K + 2)

4=

(K +1)(K + 2)

2= RHS

2 2

p(K + 1) is true

Bytheprincipleofmathematicalinduction,p(n) is true for all values of n.

43. Let Tn be the nth term of the given series

T Snnn

n n

n n= + + +

+ + + + −=

+

+ −

1 2

1 3 5 2 1

1

2

21 2 1

3 3 3

2

...

... ( )

( )

( )

∵ nnn a l

n n n n

= +

= +

2

1

4 22

2 2

( )

( )( )

= + × = +

= + +

n nn

n

n n

2 2

2

2

2

1

4

1 1

4

1

42 1

( ) ( )

( )

Let Sn denote the sum of n terms of the given series.

Then S Tn kk

n

k

n

k

n

k

n

k k

k k

n

= = + +

= + +

=

=

===

∑

∑∑∑1

2

2

111

1

42 1

1

42 1

1

4

( )

(nn n n n+ + + + +

1 2 1

6

2 1

2

)( ) ( )(n )

= + + + + +[ ]

= + + + + +

1

241 2 1 6 1 6

1

242 1 6 6 6

2 2

n n n n n n

n n n n n n

( )( ) ( )( )

( )( )

= + + + + +

= + + = +

21

242 2 6 12

1

242 9 13

242 9

3 2 2

3 2 2

n n n n n n

n n n nnn

S nn + 13

NowwehavetofindS17

∴ = + + = + +[ ]

= =

S17

217

242 17 9 17 13

17

24578 153 13

17

24744 17

( ) ( )

( ) (( ) .31 527=

S17=527

(OR)

Let P (1,2) and B(5,3) are the given points.Bythepropertyofreflector∠XAB = ∠OAP = θLet m1 be the slope of the x - axis, m2 and m3 be the slopes of the lines AP and AB.TofindXAB,Clearlym1 = 0 Since it represents slope of x - axis.

O

θ θA (x, 0)

P (1, 2)

B (5, 3)

y

x

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m2 = 2 0

1

2

1

−−

=−x x

and m y yx x

=−−

2 1

2 1

m3 = 3 0

5

3

5

−−

=−x x

tan θ = m m

m m1 3

1 31

−+ .

tan θ = 0

3

5

1 03

5

3

5

−−

+ −

=−

x

x x ... (1)

To find ∠OAP,

tan (–θ) = m m

m mx

xx

1 2

1 21

02

1

1 02

1

2

1

−+

=−

−

+−

=−

[Since OAP is in the clockwise direction]

⇒ – tan θ = 2

1− x [ tan θ (– θ) = tan θ]

⇒ tan θ = −−2

1 x ... (2)

From (1) and (2),

3

5 − x =

−−2

1 x ⇒ 3 – 3x = – 10 + 2x

⇒ 3 + 10 = 2x + 3x

⇒ 13 = 5x

⇒ x = 13

5

∴ The required co-ordinates of A is 13

50,

44. Let A = 4 4 44 4 44 4 4

− + ++ − ++ + −

x x xx x xx x x

... (1)

Putting x = 0 in (1) we get,

A = 4 4 44 4 44 4 4

= 0 [ R1 º R2 º R3]

⇒x2 is a factor of (1)

Putting x = – 12 we get,

A = 4 12 4 12 4 12

4 12 4 12 4 12

4 12 4 12 4 12

+ − −− + −− − +

= 16 8 8

8 16 8

8 8 16

− −− −− −

= 0 8 8

0 16 8

0 8 16

− −−

− = 0

[ApplyingC1 ® C1 + C2 + C3]

∴ (x + 12) is also a factor of (1).

SincetheleadingdiagonalofAisofdegree3,only3factors and a constant k are available

∴ A = 4 4 44 4 44 4 4

− + ++ − ++ + −

x x xx x xx x x

= k (x2) (x + 12)

Putting x = 1, we get

3 5 5

5 3 5

5 5 3

= k (1)2 (1 + 12)

⇒ 3(9 – 25) – 5(15 – 25) + 5 (25 – 15) = 13k[ Expanding along R1]

⇒ 3(–16) – 5(–10) + 5(10) = 13k

⇒ – 48 + 50 + 50 = 13k

⇒ 52 = 13k

⇒ k = 4

∴ 4 4 44 4 44 4 4

− + ++ − ++ + −

x x xx x xx x x

= 4 (x2) (x + 12)

⇒ 4(x2) (x + 12) = 0

⇒ x = 0 or x = – 12

(OR)

Let a→

= i j k∧ ∧ ∧− +2 3 , b

→= − + −

∧ ∧ ∧2 3 4i j k ,

c→

= − +∧ ∧j k2

Let a→

= s b t c→ →

+

⇒ i j k∧ ∧ ∧− +2 3 = s( − + −

∧ ∧ ∧2 3 4i j k ) + t( − +

∧ ∧j k2 )

⇒ i j k∧ ∧ ∧− +2 3 = (–2s) i

∧ + (3s – t) j

∧ + (–4s + 2t) k

∧

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Equating the like components both sides, we get – 2s = 1 ... (1) 3s – t = – 2 ... (2) – 4s + 2t = 3 ... (3)

From (1), s = – 1

2

Substituting s = – 1

2 in (2) we get,

31

2

−

– t = – 2 ⇒ –

3

2– t = – 2

– t = – 2 + 3

2

– t = − + = −4 3

2

1

2

t = 1

2

Substituting s = – 1

2, t =

1

2 in (3) we get,

− −

+

41

22

1

2 = + 3

⇒ 2 + 1 = 3⇒ 3 = 3whichsatisfiesthe(3)equation.Thus, one vector is a linear combination of other two vectors.Hence,thegivenvectorsareco-planar.

(ii) Let a→

= 5 6 7i j k∧ ∧ ∧+ +

b→

= 7 8 9i j k∧ ∧ ∧− +

c→

= 3 20 5i j k∧ ∧ ∧+ +

Let a→

= s b t c→ →

+

⇒5 6 7i j k∧ ∧ ∧+ + = s( 7 8 9i j k

∧ ∧ ∧− + )

+ t( 3 20 5i j k∧ ∧ ∧+ + )

⇒ 5 6 7i j k∧ ∧ ∧+ + = (7s + 3t) i

∧ + (–8s + 20t) j

∧

+ (9s + 5t) k∧

Equating the like components, both sides we get. 5 = 7s + 3t ... (1) – 8s + 20t = 6 ... (2) 9s + 5t = 7 ...(3)

(1) × 20 ⇒ 140s + 60t = 100 (+) (–) (–) (2) × 3 ⇒ – 24s + 60t = 18

164s = 82

⇒s = 82

164 =

1

2

Substituting s = 1

2 in (1) we get,

71

2

+ 3t = 5

⇒ 3t = 5 – 7

2 =

10 7

2

− =

3

2

⇒ t = 3

2 3× =

1

2

Substituting s = 1

2, t =

1

2 in (3) we get,

91

25

1

2

+

= 7

⇒9

2

5

2+ = 7

⇒14

2 = 7

⇒ 7 = 7 which satisfies the (3)equation.

Thus, one vector is a linear combination of other two vectors.

Hence,thegivenvectorsareco-planar.

45. (i) Let f (x) = x + sin x Since the algebraic function x is continuous

for all x ∈ and the circular function sin x is continuous for all x Î .

f (x) = x + sin x is continuous for all x ∈ .(ii) Let f(x) = x2 cos x The algebraic function x2 and the circular

function cos x is continuous for all x ∈ . f(x) = x2 cos x is continuous for all x Î .

(iii) Let f(x) = ex tan x We know the exponential function ex is

continuous for all x ∈ .

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But tan x is not continuous at odd multiples of π2

.

\ f(x) = ex tan x is continuous for all x ∈ –

(2n + 1) π2

, n ∈ Z.

(iv) Let f(x) = e2x + x2 Since the exponential function e2x and algebraic

function (x2) is continuous for all x ∈. f(x) = e2x + x2 is continuous for all x ∈ .

(v) Let f(x) = x. In(x) The algebraic function x is continuous for all

x∈,butthelogarithmiccurveexistsonlyforpositive values of x

∴f(x) is continuous in (0, ∞).

(OR)

u = tan–12

2 2

22

2

sin cos

cos

x x

x

[ sin 2θ = 2 sin θ cos θ and 1 + cos 2θ = 2 cos2θ]

u = tan–1 tanx2

⇒ u =

x2

∴dudx

= 1

2

Given v = tan–1 cos

sin

xx1+

v = tan–1 sin

cos

π

π2

12

+

− +

x

x

sin cos sin cosπ π2 2

+

= = − +

x x x xand

= tan–12

4 2 4 2

24 2

2

sin cos

sin

π π

π

+

+

+

x x

x

[ sin θ = 2sin θ/2 cos θ/2]

= tan–1 cot tan tanπ π π4 2 2 4 2

1+

= − +

−x x

= π4 2

− x

∴dvdx

= –1

2

∴dudv

=

dudxdvdx

=

1

21

2−

= 1

2

2

1× −

∴dudv

= –1

46. Let I = 1

3 4x xdx

+ − −∫ Rationalising the denominator we get

I = x xx x x x

dx+ + −

+ − −( ) + + −( )∫3 4

3 4 3 4

= x x

x xdx

+ + −( )+( )− −( )∫

3 4

3 4

a b a b a b+( ) −( )= −

2 2

= x x

dx+ + −( )

∫3 4

7

= 1

73 4x dx x dx+ + −

∫∫

= 1

7

3

1

21

4

1

21

1

21

1

21x x

c+( )

++−( )

+

++ +

=

1

7

3

3

2

4

3

2

3

2

3

2x xc

+( )+−( )

+

= 1

73

2

3 43

2

3

2

×+( ) + −( )

+x x c

= 2

213 4

3

2

3

2x x c+( ) + −( )

+

(OR)

Lettheeventsbedefinedasfollows:A1:ProductionofPVCpipefromPlantX

A2:ProductionofPVCpipefromplantY

A3:ProductionofPVCpipefromplantZ

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B : Selecting a defective pipe

Given P (A1) = 2000

2000 3000 5000

2000

10000

2

10+ += =

= 2000

2000 3000 5000

2000

10000

2

10+ += =

P (A2) = 3000

10000

3

10=

P(A3) = 5000

10000

5

10=

P (B/ A1) = 3

100

P (B/ A2) = 4

100

P (B/ A3) = 2

100

(i) ∴ P(B) = P(A1). P(B/A1) + P (A2) . P(B/A2) + P (A3) . P(B/A3)

= 2

10

3

100

3

10

4

100

5

10

2

100× + × + ×

= 6

1000

12

1000

10

1000

28

1000

7

250+ + = =

(ii)ByBayes theorem, probability of selectingdefective pipe from plant Y

P (A2/ B) = P(A P(B / A

P(A P(B / A P(A P(B / A P(A P(B / A

2 2

1 1 2 2 3 3

). )

). ) ). ) ). )+ +

=

3

10

4

100

12

10007

250

×=

P(B) =

12

1000

250

7

3

7

3

4

× =

∴P(A2/ B) = 3

7

47. Let I = x xx x

dx2

2

5 3

3 2

+ ++ +∫

Bylongdivisionmethod,

x x x x

x x

x

22

2

1

3 2 5 3

3 2

2 1

+ + + +( ) ( ) ( )

+ +

+

− − −

∴I = 12 1

3 22

+ ++ +

∫

xx x

dx

I = x +12 1

3 22

dx xx x

dx+ ++ +∫∫

( )

= x + I1 ....(1)

Consider I1 = 2 1

3 22

xx x

dx++ +∫

Now 2x + 1 = A . ddx

(x2 + 3x + 2) + B

⇒ 2x + 1 = A (2x + 3) + B Equating the like terms both sides, we get 2 = 2A ⇒A = 1 and 1 = 3A + B ⇒1 = 3 + B ⇒B = –2 ∴ 2x + 1 = 1 (2x + 3) – 2

I = 2 1

3 2

2 3

3 22

3 22 2 2

xx x

dx xx x

dx dxx x

++ +

= ++ +

−+ +∫∫ ∫

= log |x2 + 3x + 2| – 2dx

x x23

9

4

9

42+ + − +

∫

= log |x2 + 3x + 2| – 2 dx

x +

−

∫3

2

1

2

2 2

I1 = log |x2 + 3x + 2| – 2

1

21

2

3

2

1

23

2

1

2

××

+ −

+ ++log

x

xc

I1 = log |x2 + 3x + 2| – 21

2log

xx

c++

+ ...(2)

Substituting (2) in (1) we get,

I = x + log |x2 + 3x + 2| – 21

2log

xx

c++

+

(OR)

Given x x x+ + + = +5 21 6 40

Squaring both sides we get,

x x x+ + +( ) = +( )5 21 6 402 2

= x x x+ + +( ) = +( )5 21 6 402 2

⇒ x + 5 + x + 21 + 2 ( )( )x x+ +5 21 = 6x + 40

⇒ 2x + 26 + 2 ( )( )x x+ +5 21 = 6x + 40

⇒ 2 ( )( )x x+ +5 21 = 6x + 40 – 2x – 26

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⇒ 2 ( )( )x x+ +5 21 = 4x + 14

⇒ ( )( )x x+ +5 21 = 2x+7 Squaring again we get,

(x + 5) (x + 21) = (2x+7)2

⇒ x2 + 21x + 5x + 105 = 4x2 + 49 + 28x

⇒ x2 + 26x + 105 = 4x2 + 49 + 28x

⇒ 3x2 + 2x – 56 = 0

x =

− ± − −2 4 4 3 56

6

( )( ) ∵ x b b ac

aa b c

= − ± −

= = = −

24

2

3 2 56, ,

x = − ± +2 4 672

6

⇒ x = − ±2 26

6 ⇒ x = 4,

−14

3

Case (i) When x = 4,

4 5 4 21+ + + = 6 4 40( ) +

9 25+ = 64

3 + 5 = 8 8 = 8 which is true ⇒ x = 4 is a root.

Case (ii) When x = −14

3

− + + − +14

35

14

321 = 6

14

340

−

+

1

3

49

3+

= 12 which is not true.

\ x =−14

3 is not a root.

\Theonlyrootis4.

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11thSTD.





1. Let R be the universal relation on a set X with more than one element. Then R is(1) notreflexive (2) not symmetric (3) transitive (4) none of the above

2. The range of the function f(x) = |⎣x⎦ – x|, x ∈ R is(1) [0,1] (2) [0,∞) (3) [0,1) (4) (0,1)

3. The solution of 5x –1<24and5x+1>–24is(1) (4,5) (2) (–5,–4)(3) (–5, 5) (4) (–5, 4)

4. Find the odd one out of the following(1) x3 + 3x2+2x+1 (2) (x2+2x + 1)(x + 4) (3) x2+ 5x + 6 (4) (x+2)(x + 3) (x + 4)

5. cos1º+cos2º+cos3º+....+cos179º=(1) 0 (2) 1(3) –1 (4) 89

6. Assertion (A) : cos x =−1

2 and 0 < x<2π, then the

solutions are x = 2

3

π, 4

3

π .

Reason (R) : cos is negative in thefirst and fourthquadrantonly

(1) Both A and (R) are true and (R) is the correct explanationof(A)

(2) BothAandRaretruebut(R)isnotthecorrectexplantionofA

(3) A is true R is false(4) A is false R is true

7. The product of r consecutive positive integers is divisibleby(1) r! (2) r! + 1 (3) (r + 1) (4) none of these

8. In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is(1) 110 (2) 10C3 (3) 120 (4) 116

9. The equation of the locus of the point whose distance from y-axisishalfthedistancefromoriginis(1) x2 + 3y2=0 (2) x2 – 3y2 = 0 (3) 3x2 + y2 = 0 (4) 3x 2 – y2 = 0

10. θ is acute angle between the lines x2– xy – 6y2 = 0,

then 2 cos 3sin4sin 5 cos

θ θθ θ

++

is

(1) 1 (2) − 1

9

(3) 5

9 (4)

1

9


List I List II

1. a bb a

(a) Identity

2. 00b

b−

(b) Singularmatrix

3. a ab b

(c) Skew-Symmetric

4. 1 0

0 1

(d) Symmetric

The Correct match is (1) 1–d 2–c 3–b 4–a (2) 1–c 2–d 3–b 4–a (3) 1–b 2–a 3–d 4–c (4) 1–b 2–d 3–a 4–c

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12. If ⋅ denotes the greatest integer less than or equal totherealnumberunderconsiderationand–1≤x < 0, 0≤y <1,1≤z <2,thenthevalueofthedeterminant

x y z

x y z

x y z

+ + +

11

1is

(1) z (2) y

(3) x (4) x +1

13. The value of AB BC DA CD→ → → →

+ + + is

(1) AD→

(2) CA→

(3) 0→

(4) −→AD

14. The vector in the direction of the vector i j k∧ ∧ ∧− +2 2

thathasmagnitude9is(1) i j k

∧ ∧ ∧− +2 2 (2) i j k

∧ ∧ ∧− +2 2

3

(3) 3 2 2( )i j k∧ ∧ ∧− + (4) 9 2 2( )i j k

∧ ∧ ∧− +

15. limcos

x

xx→

−0

1 2

(1) 0 (2) 1

(3) 2 (4) doesnotexist

16. Choose the correct statement(1) Derivative of odd functionis odd(2) Derivativeofevenfunctioniseven(3) Inverse of odd function is even (4) Inverse function of sin x is sin–1x

17. e x dxx−∫ 4 cos is

(1) e x−4

17[4 cos x – sin x] + c

(2) e x−4

17[– 4 cos x + sin x] + c

(3) e x−4

17[4 cos x + sin x] + c

(4) e x−4

17 [– 4 cos x – sin x] + c

18. 23 5x dx+∫ is

(1) 3 2

2

3 5x+( )log

+ c (2)2

2 3 5

3 5x

x

+

+( )log+ c

(3) 2

2 3

3 5x+

log+ c (4)

2

3 2

3 5x+

log+ c

19. Choose the incorrect pair :1. A and B disjoint P(A∩B) = P(A) + P(B)2. A and B independent P(A∩B) = P(A) P(B)3. A and B disjoint P(A∩B) = 04. A and B independent P(A/B) = P(B/A)

20. Aspeaks truth in75%casesandBspeaks truth in80% cases. Probability that they contradict eachother in a statement is

(1) 7

20 (2)

13

20

(3) 3

5 (4)

2

5

Section - II (i) AnsweranySEVEN questions. (ii) Question number 30 iscompulsory. 7 × 2 = 1421. Find the largest possible domain of the real valued

function f (x) = 4

9

2

2

−

−

x

x.

22. Solve. 12 2x − a

23. Ina∆ABCifa = 3, b = 5 and c=7,findcosAandcos B.

24. Write the nth term of the following sequences. 2,2,4,4, 6, 6, ...

25. Showthat2x2 + 3xy–2y2 + 3x + y + 1 = 0 represents a pair of perpendicular lines.

26. IfAisasquarematrixand|A|=2,findthevalueof|AAT|.

27. If a→

and b→

are two vectors such that a b a b→ → → →

= = =10, 15 and 75 2. , find the anglebetween a

→ and b

→.

28. Evaluate limx

x

x→

+ −( )+1

2

21 1

1.

29. y = xcosx. Find dydx

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30. Solve. sin x

x


31. Supposethat120studentsarestudyingin4sectionsof eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. DefinearelationfromAtoBas“x related to y if the student x belongs to the section y”. Is this relation a function? What can you say about the inverserelation?Explainyouranswer.

32. Factorize: x4+1(Hint:Trycompletingthesquare.)33. Ina∆ABC,provethata cos A + b cos B + c cos C =

2a sin B sin C.34. Find 10013 approximately.(twodecimalplaces).35. A rod of length l slides with its ends on two

perpendicular lines. Find the locus of its mid – point.36. Find the angle A of the triangle whose vertices are

A(0,–1,2),B(3,1,4)andC(5,7,1).37. For what value of a is this function f (x) =

xx

x

x

41

11

1

−−

≠

=

,

,

if

ifα continuous at x = 1?

38. y = xlogx + (logx)x ;finddydx

.

39. Solve 1

1 2 2x x−( ) +( )

40. SupposethechancesofhittingatargetbyapersonXis3timesin4shots,byYis4timesin5shots,andbyZis2timesin3shots.Theyfiresimultaneouslyexactly one time.What is the probability that thetargetisdamagedbyexactly2hits?

Section - IVAnswer all questions. 7 × 5 = 3541. LetA= {1, 2, 3, 4} andB= {a, b, c, d}.Give a

function from A → B for each of the following :(i) neither one- to -one and nor onto. (ii) not one-to-one but onto.(iii) one-to-one but not onto. (iv) one-to-one and onto.

(OR)

Solve : log8 x + log4 x + log2x = 11.

42. Show that cot .7 12

2 3 4 6°

= + + +

(OR) Acommitteeof7peopleshastobeformedfrom8

menand4women. Inhowmanywayscan thisbedone when the committee consists of(i) exactly3women?(ii) at least 3 women? (iii) at most 3 women?

43. Find the value of n if the sum to n terms of the series 3 75 243 3+ + + ... . is 435

(OR) Find the length of the perpendicular and the co-

ordinates of the foot of the perpendicular form (–10, –2)tothelinex + y–2=0.

44. If (k,2),(2,4)and(3,2)areverticesofthetriangleof area 4 square units then determine the value of k.

(OR)

Find the angle between the vectors 2 i j k∧ ∧ ∧+ − ˆ and

i j k∧ ∧ ∧+ +2 using vector product.

45. A function fisdefinedasfollows:

f (x) =

0 0

0 1

4 2 1 3

4 3

2

, ;

, ;

, ;

,

for

for

for

for

x

x x

x x x

x x

<

≤ <

− + − ≤ <

− ≥

Is the function continuous?(OR)

If y = (cos –1x)2 prove that (1 – x2) d y

dxx

dy

dx

2

2 −

–2=0.Hencefindy2 when x = 0

46. (i) 2 1

9 4 2

x

x x

+

+ −

(OR) Theprobabilitythatacarbeingfilledwithpetrolwill

alsoneedanoilchangeis0.30;theprobabilitythatitneedsanewoilfilteris0.40;andtheprobabilitythatboththeoilandfilterneedchangingis0.15.(i) Iftheoilhadtobechanged,whatistheprobability

thatanewoilfilterisneeded?

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(ii) Ifanewoilfilterisneeded,whatistheprobabilitythat the oil has to be changed?

47. Prove that 1 11

+

1 1

2+

1 13

+

. . . 1 1+

n

=

(n + 1) for all n ∈Nbytheprincipleofmathematicalinduction.

(OR)

Evaluate x dx

x x

3

4 23 2+ +∫

ANSWERS

1. (3) transitive2. (3) [0,1)3. (3) (–5,5)4. Hint:(3)isofdegree2andallothersaredegree3 (3) x2 + 5x + 65. (1) 06. Hint : cos is negative in the second and third quadrant (3) (A) is true (R) is false

7. (4) 1168. (4) 3x2 – y2 = 0

9. (3) 5

9

10. (2)

1

2

1

2

2 1

−

11. (1) 1–d 2–c 3–b 4–a

12. (3) 0→

13. (3) 3 2 2( )i j k∧ ∧ ∧− +

14. (4) doesnotexist15. (3) 316. (2) Inversefunctionofsinx is sin–1 x

17. (4) 2

3 2

3 5x+

log+ c

18. (1) 5

13

19. Hint : A and B are independent then

P(A∩B) = P(A) P(B) (4) A and B independent P(A/B) = P (B/A)

Section - II

(i) AnsweranySEVEN questions. (ii) Question number 30 iscompulsory. 7 × 2 = 14

20. Givenf (x) = 4

9

2

2

−

−

x

x.

When x = 2,f (x) = 0

When x = –2,f (x) = 0

For all the other values, we get negative value in the square root which is not possible.

∴Domain = ϕ

21. 12 2x − a

= 1

( )( )x x− +a a

Let = 1

( )( )x x− +a a = A B

x a x a−+

+

⇒ 1 = A (x + a) + B (x – a) ...(1) Putting x = a in (1) we get,

1 = A(2a) ⇒ A = 1

2a Putting x = – a in (1) we get, 1 = B(–2a) ⇒ B = – 1

2a

∴ 12 2x − a

= 1

2

1

2 1

2

1

2

a

x a

a

x a a x a a x a−−

+=

−−

+

( ) ( )

22. cos A = b c abc

2 2 2

2+ − [Bycosineformula]

= 25 49 9

2 5 7

65

70

13

14

+ − = =( )( )

cos B = c a b

ca

2 2 2

2

49 9 25

2 7 3

33

42

11

14

+ − = + − = =( )( )

23. Givensequenceis2,2,4,4,6,6,... Theoddtermare2,4,6,...andeventerms

arealso2,4,6...

∴ =

+

an nn nn

1 if is odd

if is even

24. Givenequationofpairoflinesis 2x2 + 3xy –2y2 +

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3x + y + 1 = 0 The condition to represent pair of perpendicular lines is a + b = 0 Here a=2,2h = 3, b=–2,2g=3,2f = 1 and c = 1

h = 3

2, ⇒a + b=2–2=0

25. GivenAisasquarematrixand |A| = 2

∴ |AAT| = |A| |AT| = |A| . |A| [ |A|T = |A|]

=2(2)=4 Byproperty1

26. Given| a→

| = 10, | b→

| = 15 and a→

. b→=75 2

Let q be the angle between the vectors a→

and b→

.

∴ cos q = a b

a b

→ →

→ →⋅

= 75 2

10 15

5

1

( ) = 22

= 1

2 = cos π

4.

q = π4

27. limx

xx→

+ −( )+1

2

2

1 1

1 =

1 0

1 12

++

= 1

2

28. Taking logarithm,

log y = log xcos x ⇒ log y = cos x. log x

1

y.dydx

= cos x. ddx

(log x) + log x. ddx

(cos x)

1

y.dydx

= cos x × 1x

+ logx (–sinx)

= cos x

x – sin x. log x

⇒ dydx

= y [cos x

x – sin x. log x]

⇒ dydx

= xcos x [cos x

x – sin x. log x]

29. Let I = sin x

xdxò

Put x u=

⇒1

22

xdx du dx

xdu= ⇒ =

∴ I= 2 2 2sin . sin cosu du u du u c= =− +∫∫ = − +2cos x c


30. Givenn(A)=120,n(B) = 4

1

2

3

4

.

.

.

.

120

P

Q

R

S

A B

xRy is the student x belongs to the section y. ThisrelationisafunctionsinceeverystudentofsetA

will be mapped on to some section in B. ∴ f is a function from A → B. The inverse relation is f –1: B → A. The inverse relation is not a function since one

section will have more than one student.

31. Givenequationisx4 + 1 x4 + 1 = (x2)2 + 12 [ a2 + b2 = (a + b)2–2ab

where a = x2, b = 1]

= (x2 + 1)2–2x2(1)

= (x2 + 1)2 – ( 2 x)2

= (x2 + 1 + 2 x) (x2 + 1 – 2 x)

[ a2 – b2 = (a + b) (a – b)] = (x2 + 2 x + 1) (x2 – 2 x + 1) = 0

32. a b c k

sin sin sinA B C= = =

⇒ a = k sin A,

b = k sin B and c = k sin C ...(1)

LHS = a cos A + b cos B + c cos C = k sin A cos A + k sin B cos B

+ k sin C cos C [Using (1)] = k

2[2sinAcosA+2sinBcosB

+2sinCcosC]

Let the four sections be P, Q, R and S.

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= k2[sin2A+sin2B+sin2C]

= k2

[4 sin A sin B sin C]

[UsingExercise3.7Qn.no.1(i)] = 2(k sin A) sin B sin C = 2a sin B sin C [ k sin A = a] = RHS. Hence proved.

33. Given 1001 1000 1 1000 11

1000

10 11

1000

31

3

1

3

1

3

31

3

= +( ) = ( ) +

= +

×

1

3

Since 1

10001< , binomialexpansionisvalid.

∴ = +( )

=+ +

−

1001 10 1 001

101 001

1

3

1

3

1

31

2001

31

3.

(. )!

(. )) ...2 +

app

= + +

−

+

10 1001

3

1

3

2

3

000001

2

1 0

. .

.

app

=10 00033000001

9−

.app

= 10 [1.00033 – .00000011] app

=10[1.000329]=10[1.00033]

10001

3( ) ≅ 10.0033

34. Let the two perpendicular lines be the co-ordinate axes. LetAB be a rod of length l. Let the co-ordinates of A and B be (a, 0) and (0, b).

Let p(x1, y1) be the mid point of AB.

A (a,0) O

B (0,b)

y

xa

b

l

x′

P (x1, y

1)

Then, x1 = a + 0

2 and y1 = 0

2

+ b (using mid-point formula)

⇒ x a y b1 1

2 2= = and ...(1)

In ∆OAB we have OA2 + OB2 = AB2

⇒ a2 + b2 = l2

⇒(2x1)2+(2y1)

2 = l2 [∵ From (1), a=2x1 and b=2y1]

⇒ 4x12 + 4y1

2 = l2

Hence, the locus of (x1, y1) is 4x2 + 4y2 = l2

35. Let OA→

= − −∧ ∧j k2 , OB

→ = 3 4i j k

∧ ∧ ∧+ + and

OC→

= 5 7i j k∧ ∧ ∧+ +

AB→

= OB→

– OA→

= ( 3 4i j k∧ ∧ ∧+ + ) – ( − −

∧ ∧j k2 )

= 3 2 6i j k∧ ∧ ∧+ +

∴ | AB→

| = 3 2 62 2 2+ + = 9 4 36+ + = 49 =7

AC→

= OC→

– OA→

= ( 5 7i j k∧ ∧ ∧+ + ) – ( − −

∧ ∧j k2 )

= 5 8 3i j k∧ ∧ ∧+ +

| AC→

| = 5 8 32 2 2+ + = 25 64 9+ +

= 98 = 49 2× =7 2

Now, AB→

. AC→

= ( 3 2 6i j k∧ ∧ ∧+ + ) . ( 5 8 3i j k

∧ ∧ ∧+ + )

= 15+16+18=49

∴ cos A = AB AC

AB AC

→ →

→ →⋅

| || |

= 49

7 7 2× =

1

2 = cos

π4

⇒ A = π4

36. Givenf(x) =

xx

x

x

41

11

1

−−

≠

=

,

,

if

ifα

limx

f x→ −

( )1

= limx

xx→ −

−−1

41

1 = lim

x

x x

x→ −

+( ) −( )−( )1

2 21 1

1

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= limx

x x x

x→ −

+( ) +( ) −( )−( )1

21 1 1

1

= limx

x x→ −

+( ) +( )1

21 1 =(2)(2)=4

limx

f x→ +

( )1

= limx

xx→ +

−−1

41

1 = lim

xx x

→ ++( ) +( )

1

21 1 = 4

∴ limx

f x→ −

( )1

= limx

f x→ +

( )1

= 4

Since f(x) is continuous at x = 1, one sided limits = value of the function. ∴ f(1) = a = 4⇒ a = 4

37. y = xlogx + (logx)x

Let y = u + v Þ dydx

= dudx

dvdx

+ ...(1)

Now u = xlog x

Taking logarithm,

log u = log xlogx = logx. logx = (logx)2

1u

dudx

= 2logx × 1x

= 2log x

x

⇒ 1u

dudx

= u 2log xx

= xlogx 2log xx

...(2)

Also v = (log x)x

Taking logarithm,

log v = log ((log x)x) = x. log (log x)

1v

dvdx = x.

ddx

(log (logx))

+ log log(x). ddx

(1)

= xx x

.log

1 1× + log (logx)

= 1

log x + log (logx)

⇒dvdx

= v1

loglog log

xx+ ( )

= (logx)x 1

loglog (log )

xx+

...(3)

Substituting(2)and(3)in(1)weget,

dydx

= xlogx 2log xx

+ (log x)x 1

loglog log

xx+ ( )

38. 1

1 2 2x x−( ) +( )

Let I = 1

1 22x x

dx−( ) +( )∫

Consider 1

1 22x x−( ) +( )

= A B C

x x x−+++

+( )1 2 22

[Bypartialfraction]

⇒ 1 = A (x +2)2 + B (x −1)(x +2)+C(x −1)...(1)

Putting x=−2weget,

1 = C(−2−1)⇒1=C(−3)⇒C=−1

3

Putting x=1weget,1=(A)(1+2)2

⇒ 1 = A(9)⇒ A = 1

9

Putting x = 0 in (1) we get,

1 = 4A+B(−2)+C(−1)⇒1=4A−2B−C

⇒ 1 = 41

92

1

3

− − −

B ⇒1=

4

9–2B+

1

3

⇒2B=4

9

1

31

4 3 9

9

2

9+ − =

+ −=−

⇒ B = −1

9

∴ 1

1 22x x−( ) +( )

=1

9 1

1

9 2

1

3 22x x x−( )

−+( )−

+( )

∴I = dxx x−( ) +( )∫

1 22

=1

9

1

1

1

9 2

1

3 22x

dx dxx

dxx−

−+−

+( )∫∫∫

= 1

91

1

92

1

32

2log logx x x dx− − − − +( )−∫

= 1

91

1

92

1

3

2

2 1

2 1

log logx xx

c− − − −+( )− +

+− +

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=1

91

1

92

1

3

2

1

1

log logx xx

c− − − −+( )−

+−

=1

91

1

92

1

3 2log logx x

xc− − − +

+( )+

39. P(targetishitbyX)=P (X) = 3

4

∴ P ( X ) = 1 – 3

4

1

4=

P (targetishitbyY)= P(Y)=4

5

⇒ P ( Y ) = 1 – 4

5

1

5=

P (targetishitbyZ)= P(Z)=2

3

⇒ P ( Z ) = 1 – 2

3

1

3=

P(targetishitbyexactly2hits) = P X Y Z P X Y Z P X Y Z∩ ∩( ) + ∩ ∩( ) + ∩ ∩( ) = P X P Y P Z P X P Y P Z P X P Y P Z( ) ( ) ( ) + ( ) ( ) ( ) + ( ) ( ) ( ). . . . . .

= 3

4

4

5

1

3

3

4

1

5

2

3

1

4

4

5

2

3× × + × × + × ×

= 12

60

6

60

8

60+ +

= 12 6 8

60

26

60

+ + =

= 13

30


40. (i) not one-one and not onto.

1

2

3

4

a

b

c

d

A B

Let f = {(1, b)(2,b) (3, c) (4, c)}

Different elements in A does not have different images in B

∴ f is not one- one Now, Co-domain = {a, b, c, d},Range={b, c} Co-domain ≠ range ∴ f is not onto. Hence f is neither one - one and

nor onto.(ii) not one-to-one but onto. GivenA={1,2,3,4},andB={a, b, c, d}

Let f=A→B. Thefunctiondoesnotexistfornotone-onebut

onto. Since f = A → B, f is onto ⇒ f must be one one since n(A) = n(B)

(iii) one-to-one but not onto. The function does not exist for one-to-one

but not onto. Since f=A→B, f is one-one ⇒ f must be

onto [∴ n(A) = n(B)](iv) one-to-one and onto.

1

2

3

4

a

b

c

d

A B

Let f:A→Bdefinedby f = {(1, a)(2,b) (3, c) (4, d)} Here different elements have different

images ∴ f is one-to-one. Also Co-domain = {a, b, c, d}=Range. ∴ f is onto. ∴ f is one-to-one and onto.

(OR)Givenlog8 x + log4 x + log2x = 11

⇒1 1 1

8 4 2log log logx x x

+ + = 11

⇒ 11 1 1

23

22 2

log log logx x x

+ + = 11

⇒1 4 1

2 2 23log 2log logx x x

+ + = 11

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⇒1 1

3

1

21

2logx

+ +

= 11

⇒ 1 2 3 6

62logx

+ +

= 11

⇒1 11

62logx

= 11

⇒1

116

116

2log

×

x= = . = 11 ×

111

6

116

2log

×

x= = . = 6

⇒1

2log x

= 6

⇒ log2x = 6

⇒ 26 = x

⇒ x = 64

41. LHS = cot 71

2

�

= cos

sin

71

2

71

2

°

°

Multiplyingthenumeratoranddenominatorby2sin

71

2

�

= =° °

°=

°− °

=2 71

27

1

2

2 71

2

15

1 15

2 2

22

sin cos

sin

sin

cos

sin cos sin

s

∵ A A A

iin cos2

1 2A A= −

= =− °( )

− − °( ) =− ° °

− °sin

cos

sin cos cos sin

cos c

45 30

1 45 30

45 30 45 30

1 45 oos sin sin30 45 30° +( )

= =⋅ − ⋅

− ⋅ + ⋅

= − − +

1

2

3

2

1

2

1

2

11

2

3

2

1

2

1

2

3 1

2 21

3 1

2 2

=−

×− −

=3 1

2 2

2 2

2 2 3 1 =

3 1

2 2 3 1

−− −

= =−

− −×

+ ++ +

3 1

2 2 3 1

2 2 3 1

2 2 3 1

= =−( ) + +( )

( ) − +( )=

+ + − − −− + +( )

3 1 2 2 3 1

2 2 3 1

2 6 3 3 2 2 3 1

8 1 3 2 32 2

= =+ −−

=+ −−

×++

2 6 2 2 2

4 2 3

6 1 2

2 3

2 3

2 3

=+ − + + −

−2 6 2 2 2 18 3 6

4 3

= 2 6 2 2 2 3 2 3 6+ − + + −

= + + − + = + + +6 3 2 2 2 3 2 6 3 2 2

= = + + + =2 3 4 6 RHS Hence proved.

(OR)(i)exactly3women?Sincethecommitteemusthaveexactly3women,remaining 4 men can be selected from 8 men.This can be done in 8C3ways

8C4 × 4C3 = 8 7 6 5

4 3 2 1

4 3 2

3 2 1

2

× × ×× × ×

× × ×× ×

=280

(ii) at least 3 women?The following are the choices to select at least 3 women

Men (8) Women (4) Combinations(a) 4 3 8C4 × 4C3(b) 3 4 8C3 × 4C4

∴Requirednumberofwaysofformingthecommittee

= 8C4 × 4C3 + 8C3 × 4C4

=

× × ×× × ×

× +× ×× ×

×8 7 6 5

4 3 2 14

8 7 6

3 2 11

∵ 4

3

4

1

4

4

4

1

C C

C

= =

=

=280+56 = 336 (iii) at most 3 women The following are the choices to select at most 3

womenMen (8) Women (4) Combination

a) 4 3 8C4 × 4C3b) 5 2 8C5 × 4C2c) 6 1 8C6 × 4C1d) 7 0 8C7 × 4C0

Hence, requirednumberofwaysofforming the49committee is8C4 × 4C3 + 8C5 × 4C2 + 8C6 × 4C1 + 8C7 × 4C0 = 8C4 × 4C1 + 8C3 × 4C2 + 8C2 × 4C1 + 8C1 × 4C0

=× × ×× × ×

× +× ×× ×

×××

+××

× + ×8 7 6 5

4 3 2 14

8 7 6

3 2 1

4 3

2 1

8 7

2 14 8 1

=280+336+112+8=736.

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42. Givenseriesis 3 75 243+ + + ... and Sn = 435 3.

Givenseriesis1 3 5 3 9 3( ) + ( ) + ( ) + ...

Here a d= =3 4 3,

∴ The given series an arithmetic progression

∴ = + −

= + − =

S

given S

n

n

n a n d

n n

22 1

435 32

2 3 1 4 3 435 3

[ ( ) ]

( ) ∵

⇒

435 32

2 3 4 3 4 3

435 32

4 3 2 3

= + −

= −

n n

n n

⇒

⇒

435 3 23

22 1

435 3 3 22

= ⋅ −[ ]= −( )

n n

n n

⇒ 435 = 2n2 – n

⇒ 2n2 – n – 435 = 0.

⇒(n=15)(2n+29) = 0

⇒n – 15 or n = −29

2 which is not possible

⇒n = 15(OR)

Given equation of line is x + y–2 = 0. ... (1)Anylineperpendiculartox + y–2=0willbeoftheform x – y + k = 0Thislinepassesthrough(–10,–2)

BC

x + y −2 = 0

A (−10, −2)

–870

–1

−30

2

–15

(n – 15)

29

2

29

2

(2n+29)2x2

∴–10+2+k = 0⇒ k = 8∴Equation of AB is x – y+8 = 0...(2)The foot of the perpendicular is the point of intersectionof(1)and(2). (1) ⇒x + y–2 = 0 (2)⇒x – y + 8 = 0 2x + 6 = 0 ⇒ x = –3Substituting x = –3 in (1) we get, –3 + y–2 = 0⇒ y = 5

∴Co-ordinate of the foot of the perpendicular B is (–3,5). ...(2)Now lengthof theperpendicular from (–10, –2) tothe line x + y–2=0is

= ±− − −

+

= ±−

10 2 2

1 1

14

22 2 =

14

2

[Since length of the perpendicular cannot be negative]

= 14

2

2

2

14 2

27 2× = =

43. Let the vertices of the triangle be A(k,2)B(2,4)andC(3,2)

Also area of DABC = 4 sq. units. We know that, area of DABC

= absolute value of 1

2

1

1

1

1 1

2 2

3 3

x yx yx y

⇒ 4 = absolute value of 1

2

2 1

2 4 1

3 2 1

k


2[k(4–2)

–2(2–3)+1(4–12)][ExpandedalongR1]


2[2k+2–8]


2[2k – 6]

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⇒ 4 = ± −( )1

22 4k

Case (i)

when 4 = 1

22 6k −( )

⇒ 8 = 2k – 6⇒ 14 = 2k⇒ k = 7Case (ii)

when 4 = –1

22 6k −( )

⇒ 8 = –2k + 6 ⇒ 8–6 = –2k⇒ 2 = –2k⇒ k = – 1

(OR)

Let a→

= 2 i j k∧ ∧ ∧+ − and b

→= i j k

∧ ∧ ∧+ +2

Let q be the angle between the vectors a→

and b→

a→

× b→

=

i j k∧ ∧ ∧

−2 1 1

1 2 1

= i j k∧ ∧ ∧

+( ) − +( ) + −( )1 2 2 1 4 1

= 3 3 3i j k∧ ∧ ∧− + = 3( i j k

∧ ∧ ∧− + )

| a→

× b→

| = 3 1 1 12 2 2+ + −( ) = 3 3

| a→

| = 2 1 12 2 2+ + −( ) = 6

| b→

| = 1 2 12 2 2+ + = 6

sin q = | |

| || |

a b

a b

→ →

→ →×

= 3 3

6 6=

3 3

6

= 3

2 = sin

π3

q = π3

44. Givenf(x) =

0 0

0 1

4 2 1 3

4 3

2

, ;

, ;

, ;

,

for

for

for

for

x

x x

x x x

x x

<

≤ <

− + − ≤ <

− ≥

(i) At the point x = 0

limx

f x→ −

( )0

= limx→ −

0

0 = 0

limx

f x→ +

( )0

= x = 0

and f (0) = x = 0 lim

xf x

→ −( )

0

= limx

f x→ +

( )0

= f (0) = 0 ∴ f (x) is continuous at x = 0

(ii) At the point x = 1 lim

xf x

→ −( )

1

= x = 1

limx

f x→ +

( )1

= – x2 + 4x–2 = –1+4–2 = 4 – 3 = 1

f(1) = – x2 + 4x–2 = –1+4–2=1 lim

xf x

→ −( )

1

= limx

f x→ +

( )1

= f(1) = 1

∴ f(x) is continuous at x = 1

(iii) At the point x = 3

limx

f x→ −

( )3

= limx

x x→ −

− + −3

24 2

= – (3)2+4(3)–2 = –9+12–2=1

and limx

f x→ +

( )3

= limx

x→ +

− = − =3

4 4 3 1

f (3) = 4 – 3 = 1

∴ limx

f x→ −

( )3

= limx

f x→ +

( )3

= f (3) = 1 ∴ f(x) is continuous at x = 3.

(OR)

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Giveny = (cos–1 x)2

Differentiating with respect to ‘x’ we get,

y′ = 2.cos–1 x. ddx

( cos–1 x)

⇒ y′ = 2.cos–1 x −

−

1

12x

...(1)

⇒y′ 12− x = –2cos–1 x

Squaring both sides we get,

(y′)2 (1 – x2) = 4 (cos–1 x)2

Differentiating again with respect to ‘x’ we get,

(y′)2(–2x)+(1 – x2)2y′.y″=4(2)cos–1 x. ddx

(cos–1 x)

⇒ –2x (y′)2+2(1–x2) y′.y″ = 8 cos–1 x. −

−

1

12x

⇒–2x (y′)2+2(1–x2) y′.y″ = 4 −

−

−

2

1

1

2

cos x

x⇒ –2x (y′)2+2(1–x2) y′.y″ = 4 y′ [From (1)]Dividingthroughoutby2y′we get, – x y′ + (1 – x2) y″ = 2

⇒ (1 – x2) y″– xy′–2 = 0

i.e., (1 – x2) d ydx

x dydx

2

2 − –2= 0

Hence proved.

45. Let I = 2 1

9 42

x

x x

+

+ −∫ dx

Now,2x + 1 = A B.ddx

x x9 42+ −( )+

⇒ 2x+1=A(4−2x) + BEquating the co-efficients of x term and constant terms, we get

2 = −2A⇒A=−1 1 = 4 A + B ⇒1=−4+B⇒ B = 5 ∴2x+1 = −1(4−2x) + 5

∴ I = 2 1

9 4

14 2

9 4

5

9 42 2 2

x dx

x x

x dx

x x

dx

x x

+( )+ −

= −−( )+ −

++ −

∫ ∫ ∫

I = I1 + 5I2 ... (1)

I1 = 4 2

9 42

−

+ −∫

x

x xdx

Let t = 9+4x−x2 ⇒ dt=(4−2x) dx

∴I1 = − =− +∫dt

tt c2

I1 = − + −2 9 42x x

I2 = dx

x x

dx

x x

dx

x x9 4 4 9 4 4 4 92 2 2+ −=

− − −( )=

− − + − −( )∫ ∫∫

= dx

x x

dx

x x

dx

x x9 4 4 9 4 4 4 92 2 2+ −=

− − −( )=

− − + − −( )∫ ∫∫

= dx

x

dx

x− −( ) −

=( ) − −( )

∫ ∫2 13 13 2

2 2 2

= sin− −

1 2

13

x

dx

a xx a

2 2

1

−= ( )

∫ −sin /

Substituting I1 and I2in (1) we get,

I = − + − +−( )+−

2 9 4 52

13

2 1x xx

sin C+ c

= 52

132 9 4

1 2sin− −

− + − +

x x x C+ c

(OR)

Lettheeventsbedefinedasfollows:B:Carbeingfilledwithpetrolwillalsoneedanoil change.⇒ P (B) = 0.30E1 :Carneedsanewoilfilter⇒ P (E1) = 0.40 ∴ P (B∩ E1) = 0.15(i)Iftheoilhadtobechanged,theprobabilitythatanewoilfilterisneeded = P(E1/B)

= E

B

P E B

P(B)

1 1 0 15

0 30

1

2

=

∩( )= =.

. = 0.5

(ii)Ifanewfilterisneeded,theprobabilitythattheoil has to be changed.

= P(B/ E1) = B

E

P B E

P E1

1

1

0 15

0 40

=∩( )

( ) = .

.=0.375

46. Let p(n): be 1 11

+

1

1

2+

1

1

3+

. . . 1 1+

n

= (n + 1)

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Step 1: p(1): 1 11

+

= (1 + 1)

⇒ p(1):2=2⇒ p(1) is true.Step2: Let p(m) be true

1 11

+

11

2+

11

3+

... 1 1+

m

= (m+1)...(2)

Step 3: To prove that p(m + 1) is true

i.e. to prove that 1 11

+

1

1

2+

... 1 1+

m

1 11

++

m

= (m+2)

LHS = 1 11

+

1

1

2+

. . . 1 1+

m

1 11

++

m

= (m + 1) 1 11

++

m = m + 1 + 1

= (m+2)=RHS⇒ p(m + 1) is true.By theprincipleofmathematical induction,p(n) is true for all n ∈ N.

(OR)

Let I = x dxx x

3

4 23 2+ +∫

= x x dx

x x

2

2 2 23 2

⋅+ +∫

( )

Let t = x2 ⇒dt=2x dx ⇒ dt2

= x dx

∴ I = t dt

t ttdt

t t

.2

3 2

1

2 3 22 2+ +

=+ +∫ ∫

Now t = A ddx (t2 + 3t+2)+B

⇒ t = A(2t + 3) + B

Equating the x - term and constant terms we get,

1 = 2A⇒A=1

2

⇒ 0 = 3A + B ⇒0=3

2 + B

⇒B = -3

2 ; ∴t =

1

2(2t+3)-

3

2

∴I = 1

2 3 22

t dtt t

+ +∫

= 1

2

2 3

2 3 2

3

2 3 22 2

tt t

dt dtt t

++ +( ) −

+ +

∫ ∫

⇒ I = 1

43 2 3

39

4

9

42

2

2

log t t dt

t t+ + −

+ + − +

∫

= 1

43 2 3

3

2

1

2

2

2 2log t t dt

t+ + −

+

−

∫

Now I1 = dt

t t23 2+ +∫ =

dt

t t2

2 2

33

2

3

22+ +

−

+

∫

= 3 1

43 2 3

3

2

1

23

2

1

2

2log logt t

t

t+ + −

+ −

+ +

I = 14

log |t2 + 3t+2|–3log tt++

1

2 + c

∴ I = 1

4 log |x4 + 3x2+2|–33

4

1

2

2

2log

xx

c++

+

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11thSTD.





1. The number of constant functions from a set containing m elements to a set containing n elements is

(1) mn (2) m (3) n (4) m + n

2. The shaded region in the adjoining diagram represents.

(1) A\B (2) B\A (3) A∆B (4) A

A B ∪

3. The equation whose roots are numerically equal but opposite in sign to the roots of 3x2 – 5x – 7 = 0 is(1) 3x2 – 5x – 7 = 0 (2) 3x2 + 5x – 7 = 0 (3) 3x2 – 5x + 7 = 0 (4) 3x2 + x – 7

4. If kxx x x x+( ) −( )

=+

+−2 1

2

2

1

1, then the value of k

is

(1) 1 (2) 2 (3) 3 (4) 4

5. Choose the incorrect pair :

1. sin x in IInd quadrant−1

5

2. cos x in Ist quadrant 13. sec x in IInd quadrant –24. tan x in IIIrd quadrant 20

6. If cos x = −12

and 0 < x<2π,thenthesolutionsare

(1) x = π3

, 4

3

π

(2) x = 2

3

π , 4

3

π

(3) x = 2

3

π , 7

6

π

(4) x = 2

3

π , 5

3

π

7. In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is

(1) 125 (2) 124 (3) 64 (4) 63

8. Thecoefficientofx6 in (2 + 2x)10 is

(1) 10C6 (2) 26 (3) 10C6 26 (4) 10C6 2

10


List I List II1. {(1, 1), (2,2), (3,3)(1,2)} (a) equivalence2. {(1,2), (2,1), (2,3), (3,2)} (b) transitive3. {(1,2), (2,3), (1,3)} (c) Symmetric4. {(1,1), (2,2), (3,3), (1,2),

(2,1), (2,3), (1,3)}(d) reflexive

The Correct match is (1) 1 – c 2 – d 3 – b 4 – a (2) 1 – d 2 – c 3 – b 4 – a (3) 1 – b 2 – a 3 – d 4 – c (4) 1 – b 2 – a 3 – b 4 – c

10. The area of the triangle formed by the lines x2 – 4y2 = 0 and x = a is

(1) 2a2 (2) 3

2

2a

(3) 1

2

2a (4) 2

3

2a

11. If A is a matrix of order 3 × 3, then (A2)–1 =

(1) 12

A (2) A–2 (3) (A–1)2 (4) I

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12. Choose the correct statement.(1) One-to-one function have inverse (2) Onto function have inverse (3) bijection function have inverse (4) many - to - one function hae inverse

13. The value of λwhen the vectors a i j k→ → → →

= + +2 λ

and b i j k→ → → →

= + +2 3 are orthogonal is

(1) 0 (2) 1 (3) 3

2 (4) − 5

2

14. Assertion (A) : f (x) =x xx x

+ <− ≥

1 2

2 1 2

,

,then f ′(2) does

not exist.

Reason (R) : f (x) is not continuous at 2. (1) Both A and R are true and R is the correct

explanation of A(2) Both A and R are true but R is not the correct

explantion of A(3) A is true R is false(4) A is false R is true

15. The derivative of f (x) = x |x| at x = –3 is(1) 6 (2) – 6(3) does not exist (4) 0

16. If f (x) is an odd function, then f ′(x) is an ___________ function(1) even (2) odd (3) even or odd (4) neither even nor odd

17. Find the odd one of the following

(1) x2 (2) x4 (3) cos x (4) sin x

18. sin cos

sin cos

8 8

2 21 2x x

x xdx

−−∫ is

(1) 1

22sin x c+ (2) − +1

22sin x c

(3) 1

22cos x c+ (4) − +1

22cos x c

19. A, B, and C try to hit a target simultaneously but independently. Their respective probabilities of

hitting the target are 34

, 12

, 58

. The probability that

the target is hit by A or B but not by C is

(1) 21

64 (2)

7

32

(3) 9

64 (4)

7

8

20. If two events A and B are such that P ( A ) = 3

10 and

P( A B∩ ) = 12

, then P (A∩B) is

(1) 1

2 (2)

1

3

(3) 1

4 (4)

1

5

Section - II (i) Answer any SEVEN questions. (ii) Question number 30 is compulsory. 7 × 2 = 1421. Discuss the following relations for reflexivity,

symmetricity and transitivity: TherelationRdefinedon the set of all positive integers by “mRn if m divides n”.

22. Simplify: ( 1000)2

3−−

23. An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.

24. In how many ways a cricket team of eleven be chosen out of a batch of 15 players if there is no restriction on the selection?

25. Show that the lines are 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0 are parallel lines.

26. Using cofactors of elements of second row, evaluate

| A |, where A = 5 3 82 0 11 2 3

.

27. Find the area of the parallelogram whose two adjacent

sides are determined by the vectors i j k i j k∧ ∧ ∧ ∧ ∧ ∧

+ + − +2 3 and 3 2

i j k i j k∧ ∧ ∧ ∧ ∧ ∧

+ + − +2 3 and 3 2

28. y = sin (ex)finddydx .

29. If f ′(x) = x

x22+ and f(1) =

54,findf(x).

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30. If A and B are two events such that P(A∪B) = 0.7, P(A ∩ B) = 0.2, and P(B) = 0.5, then show that A and B are independent.

Section - III (i) Answer any SEVEN questions. (ii) Question number 40 is compulsory. 7 × 3 = 2131. Let A = {a, b, c}. What is the equivalence relation of

smallest cardinality on A? What is the equivalence relation of largest cardinality on A?

32. Find all the angles between 0° and 360° which satisfy

the equation sin2θ=34

.

33. Provethatif1≤r≤n, then n × (n – 1)Cr – 1 = (n – r + 1)∙nCr – 1

34. If y xx x x= + + + +

2 3 4

2 3 4... then show that

x yy y y= − + − +

2 3 4

2 3 4! ! !...

35. Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B(6, –5).

36. Show that the vectors a i j k b i j k c i j k→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧

= + + = + − = − +2 3 6 6 2 3 and 3 6 2, , ,

a i j k b i j k c i j k→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧

= + + = + − = − +2 3 6 6 2 3 and 3 6 2, , , a r e mutually orthogonal.

37. At the given point x0 discover whether the given function is continuous or discontinuous citing the reasons for your answer:

(i) x0 = 1, f(x) =

xx

x

x

21

11

2 1

−−

≠

=

,

,

(ii) x0 = 3, f(x) =

xx

x

x

29

33

5 3

−−

≠

=

,

,

if

if

38. If y = sin–1xthenfindy″ .

39. Evaluate 112

13x x

dx

+∫

40. One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball isdrawnfromeachbag,findtheprobabilitythat(i)both are white (ii) both are black (iii) one white and one black.


41. From the curve y = x, draw (i) y = –x (ii) y = 2x (iii)

y = x + 1 (iv) y = 12

x + 1 (v) 2x + y + 3 = 0.

(OR)

Draw a graph x≤3y, x≥y.

42. Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5 km apart,findthedistanceoftheintruderfromB.

(OR)

How many strings are there using the letters of the word INTERMEDIATE, if

(i) The vowels and consonants are alternative (ii) All the vowels are together

(iii) Vowels are never together (iv) No two vowels are together.

43. If p – q is small compared to either p or q, then show

that p

q

n p n q

n p n qn �

( ) ( )( ) ( )

+ + −− + +

1 11 1

Hencefind1516

8 .

(OR)

A photocopy store charges ` 1.50 per copy for the first 10 copies and ` 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying. (i) Draw graph of the cost as x goes from 0 to 50

copies.(ii) Find the cost of making 40 copies

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44. Verify that det(AB) = (det A) (det B) for A = 4 3 21 0 72 3 5

−

−

and B = 1 3 32 4 09 7 5

−

.

(OR)

Let a b c→ → →

, , be unit vectors such that a b a c→ → → →

= =. . 0

and the angle between b c→ →

and is =3π . Prove that

a b c→ → →

= ± ×23

( ) .

45. (a) Which of the following functions f has a removable discontinuity at x = x0? If the discontinuity isremovable,findafunctiongthatagreeswithf for x ≠x0 and is continuous on R.

(i) f (x) = x x

xx

2

0

2 8

22

− −+

= −,

(ii) f(x) = xx

x3

0

64

44

++

= −,

(iii) f (x) = 3

9

−−

xx

, x0 = 9

(OR)

(b) If sin y = x sin(a + y), then prove that dy

dx

a y

a=

sin +sin

2 ( ), a ≠ nπ.

(OR)

46. (a) Integrate the following with respect to x:

3 1

2 2 32x

x x

+− +

(OR)

(b) A year is selected at random. What is the probability that(i) it contains 53 Sundays (ii) it is a leap year which contains 53 Sundays

47. Find the equation of the straight line upon which the length of perpendicular from origin is 3 2 units and this perpendicular makes an angle of 75° with the positive direction of x-axis.

(OR)48. For what value of a and b is the function

f xx x c

ax b x c( ) =

≤+ >

2 ,

, is differentiable at x = c.

ANSWERS1. (3) n2. (3) A∆B3. (2) 3x2 + 5x – 7 = 04. (3) 35. Hint : sin x is +ve in second quadrant

sin x in IInd quadrant −1

5

6. (2) x = 2

3

π , 4

3

π

7. (2) 124

8. (2) 1 – d 2 – c 3 – b 4 – a

9. (3) (1,2)

10. (3) 1

2

2a

11. (3) bijection function have inverse12. (4) rectangular13. Hint : Rf ′ (2) = 2 and L f ′ (2) = 1

14. (1) 1

2

15. (1) 6

16. Hint : (1), (2), (3) are even functions and (4) is odd.

(4) sin x

17. (3) −

+ +3

4

3

12cos

cosx x c

18. (2) − 1

22sin x + c

19. (1) 21

64

20. (4) 1

5

Section - II (i) Answer any SEVEN questions. (ii) Question number 30 is compulsory. 7 × 2 = 14

21. Given relation is “mRn if m divides n”. Reflexivity : mRm since m divides m for all

positive integers m. ∴Risreflexive.

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Symmetricity : mRn ⇒ nRm. m divides n ⇒4 divides 2 ≠2 divides 4. ∴ R is not symmetric Transitive : mRn and nRp ⇒ mRp. m divides n and n divides p then m divides p. ∴ R is transitive. ∴Risreflexive,notsymmetricandtransitive.

22. ( 1000)2

3−−

= −

−

− −10

= 10 = =32

2×3 2

1

10

1

100

23. Using example 3.62, the equilateral triangle has the maximumareaforanyfixedperimeter.

∴ Let a be the side of the equilateral triangle. Given Perimeter = 120 m ⇒ a + a + a = 120 m ⇒ 3a = 120 ⇒ a = 40 m. Hence the dimensions of the park are 40 m, 40 m and

40 m.

24. The total number of ways of selecting 11 players out of 15 players.

= 15C11

= 15C15–11 = 15C4 [nCr = nCn–r]

= 15 14 13 12

4 3 2 11365

× × ×× × ×

= .

25. If the equation of two lines are in general form as a1 x + b1 y1 + c = 0 and a2 x + b2 y + c2 = 0

aa

1

2

= bb

1

2

or a1 b2 = a2b1

Given lines are 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0

3

12 =

2

8

⇒ 1

4 =

1

4

Hence the given lines are parallel.

26. Given A = 5 3 8

2 0 1

1 2 3

Co-factor of 2 = A21 = (–1)1+2 3 8

2 3 = – (9 – 16) = 7

Co-factor of 0 = A22 = (–1)2+2 5 8

1 3 = 15 – 8 = 7

Co-factor of 1 = A23 = (–1)2+3 5 3

1 2

= – (10 – 3) = – 7 ∴ |A| = a21 A21 + a22 A22 + a23 A23

= 2(7) + 0(7) + 1(–7) = 14 – 7 = 7

27. Let the adjacent sides of the parallelogram are

a→

= i j k∧ ∧ ∧+ +2 3 and b

→= 3 2i j k

∧ ∧ ∧− +

a→

× b→

=

i j k∧ ∧ ∧

−

1 2 3

3 2 1

= i j k∧ ∧ ∧

+( ) − −( ) + − −( )2 6 1 9 2 6

= 8 8 8i j k

∧ ∧ ∧+ − = 8( )i j k

∧ ∧ ∧+ −

| a→

× b→

| = 8 1 1 12 2 2+ + −( ) = 8 3

∴ Area of the parallelogram = 8 3 sq. units.

28. Given y = sin (ex) Let u = ex ⇒ y = sin u

∴ dudx

= ex and dydu

= cos u

Now, dydx

= dydu

× dudx

= cos u × ex = ex cos (ex) [ u = ex]

29. Given f ′(x) = x

x22+

Integrating both sides with respect to ‘x’ we get,

f x dx'( )∫ = x

xdx

22+

∫

= 1

2 22

2

⋅ +x xlog + c

⇒ f (x) = x2

4 + 2 log |x| + c ... (1)

But f(1) = 5

4

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⇒ 5

4 =

1

4 + 2 log|1| + c

⇒ 5

4 =

1

4 + c [ log |1| = 0]

c = 5

4

1

4− =

44

= 1

∴ f (x) = x x c

2

42+ +log [From (1) ]

30. Given P(A∪B) = 0.7, P(A∩B) = 0.2, P(B) = 0.5 We know P(A∪B) = P (A) + P(B) – P(A∩B) 0.7 = P(A) + 0.5 – 0.2 ⇒ 0.7 = P(A) + 0.3 ⇒ P(A) = 0.7 – 0.3 = 0.4 Now P (A). P(B) = (0.4) (0.5) = 0.20 and P(A∩B) = 0.2 ⇒ P(A∩B) = P (A). P(B) Thus, A and B are independent events.

Section - III (i) Answer any SEVEN questions. (ii) Question number 40 is compulsory. 7 × 3 = 2131. Given A = {a, b,c} (i) Let R = {(a, a) (b, b) (c,c)} Risreflexive R is symmetric and R is transitive ⇒ R is an

equivalence relation. This is the equivalence relation of smallest

cardinality on A. ∴n (R) = 3 (ii) Let R = {(a, a) (a, b) (a,c) (b,a) (b, b) (b, c) (c,

a) (c, b) (c,c)} Risreflexivesince(a, a) (b, b) and (c, c) ∈R R is symmetric since (a, b)∈R ⇒(b, a) ∈R (b, c)∈R ⇒(c, a) ∈R (c, a)∈R ⇒(a, c) ∈R R is also transitive since (a, b) (b, c) ∈R ⇒(a, c) ∈R Hence R is are equivalence relation of largest

cardinality on A. ∴n (R) = 9

32. Given sin2θ=3

4

⇒ sin θ = ±3

2

⇒ sin sinθ θ= =−3

2

3

2or

⇒sinθ=sin60° orsinθ=−sin60°

⇒sinθ=sin60° orθ=180−60°

⇒θ=60° orθ=120°

33. LHS = n × (n – 1)Cr – 1

= ×

−− − − +

n nr n r

( )!

( )!( )!

1

1 1 1

= ×

−− −

n nr n r

( )!

( )!( )!

1

1

= ×

− − +− + − −

n n n rn r r n r

( )!( )

( )( )!( )!

1 1

1 1 [Multiplying and dividing by (n – r + 1)]

=

− +− + −

− = − + − = −n n r

n r rn n n n r n r n!( )

( )!( )![ ( )! ! ( )( )! (

1

1 11 1∵ and rr +1)!]

=− +

− + −− = − + − = −

n n rn r r

n n n n r n r n!( )

( )!( )![ ( )! ! ( )( )! (

1

1 11 1∵ and rr +1)!]

= − + ⋅

− − +( )

!

( )!( )!n r 1

1 1

nr n r

= − + × −( )n r nr1

1C = RHS ∵n n

r n rrC =−

!

!( )! Hence proved.

34. Given y x x x x= + + + +2 3 4

2 3 4...

Multiplying by –1, both sides we get,

− = − − − − −y x x x x2 3 4

2 3 4...

⇒ − = − − = − − − −

y x x x x x

log( ) log( ) ...1 12 3

2 3

∵

⇒e–y = (1 – x)

⇒x = 1 – e–y

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⇒⇒ = − − + − + −

⇒ = − + −

x y y y y

x y y

1 11 2 3 4

1 11 2

2 3 4

2

! ! ! !...

! !++ − +

⇒ = − + − +

y y

x y y y y

3 4

2 3 4

3 4

2 3 4

! !...

! ! !...

⇒

⇒

35. The slope of AB is

m = − −

−5 3

6 2

∵slope = −−

y yx x

2 1

2 1

= − = −8

42

∴ Slope of a line perpendicular to AB = − =1 1

2m

Let P be the mid point of AB.

Then the co-ordinates of P are 2 6

2

3 5

2

+ −

,

= (4, –1)

So, equation of the line passing through (4, –1) with

slope 1

2 is

y x+ = −11

24( ) [ y – y1 = m(x – x1)]

⇒2y + 2 = x – 4 ⇒x – 2y – 4 – 2 = 0 ⇒x – 2y – 6 = 0

36. Given a→

= 2 3 6i j k∧ ∧ ∧+ + , b

→ = 6 2 3i j k

∧ ∧ ∧+ − ,

c→

= 3 6 2i j k∧ ∧ ∧− +

a→

. b→

= ( 2 3 6i j k∧ ∧ ∧+ + ).( 6 2 3i j k

∧ ∧ ∧+ − )

= 2(6) + 3(2) + 6(–3)= 12 + 6 – 18 = 0

b→

. c→

= ( 6 2 3i j k∧ ∧ ∧+ − ).( 3 6 2i j k

∧ ∧ ∧− + )

= 6(3) + 2(–6) – 3(2) = 18 – 12 – 6 = 0

c→

. a→

= ( 3 6 2i j k∧ ∧ ∧− + ).( 2 3 6i j k

∧ ∧ ∧+ + )

= 3(2) – 6(3) + 2(6) = 6 – 18 + 12 = 0

Since a→

. b→

= b→

. c→

= c→

. a→

= 0 the given vectors are mutually orthogonal.

37. (i) Given f(x) =

xx

x

x

21

11

2 1

−−

≠

=

,

,

limx

f x→ −

( )1

= limx

xx→ −

−−1

21

1

= limx

x xx→ −

+( ) −( )−1

1 1

1

= limx

x→ −

+1

1 = 2

limx

f x→ +

( )1

= limx

x→ +

+1

1 = 2

and f(1) = 2 (given)

∴ limx

f x→ +

( )1

= limx

f x→ −

( )1

= f (1) = 2

∴ f(x) is continous at x0 = 1

(ii) Given f(x) =

xx

x

x

29

33

5 3

−−

≠

=

,

,

if

if

limx

f x→ −

( )3

= limx

xx→ −

−−3

29

3

= limx

x xx→ −

+( ) −( )−3

3 3

3

= limx

x→ −

+3

3 = 6

limx

f x→ +

( )3

= limx

xx→ +

−−3

29

3

= limx

x→ +

+( )3

3 = 6

But f (3) = 5

∴ limx

f x→ −

( )3

= limx

f x→ +

( )3

¹ f(3)

∴ f(x) continuous at x0 = 3

38. Given y = sin–1 x

y′ = 1

12− x

= 12

12−( )−

x

∴ y″ = − −( ) −( )− −1

21 1

212

12x

ddx x

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yn = −

−( ) −( )−1

21 22

3

2x x = x x12

32−( )−

y″ = x

x12

32−( )

39. Let I = 1

1

2

1

3x xdx

+∫

Here, the exponents of x are 1

2 and

1

3 and the LCM

of their denominator is 6. So, to remove fractional exponents, put x = t6 ⇒ dx =

6t5 dt.

∴ I = 6

61

5

3 2

5

2

tt t

dt t dtt t+

=+( )∫ ∫ = 6

1

3tt

dt+∫

= 61 1

1

3( )t

tdt+ −

+∫ x t

x t

1

2 6

1

2 3

1

3 6

1

3 2

= =

= =

( )

( )

t

t

= 61 3 1 1

1

3t t tt

dt+( ) − +( ) −+∫

= 61

1

3 1

1

1

1

3tt

t tt t

dt+( )+

− +( )+

−+

∫

= 6 1 31

1

2∫ +( ) − −+

t tt

dt

= 6 2 1 31

1

2∫ + + − −+

t t tt

dt

= 6 11

1

2∫ − + −+

t tt

dt

= 63 2

1

3 2t t t t c− + − +

+log | |

= 2 3 6 6 1

1

3

1

6

1

6⋅ − + − + +x x x x clog | |

40. (i) Bag A has 5 white and 3 black balls (Totally 8 balls)

Bag B contains 4 white and 6 black balls. (Totally 10 balls)

∴P (White ball from I bag) = 5

8

P (White ball from II bag) = 4

10

∴Required probability = 5

8

4

10

20

80

1

4× = =

(ii) P (Black ball from I bag) = 3

8

P (Black ball from II bag) = 6

10

∴Required probability = 3

8

6

10

18

80

9

40× = =

(iii) P (White ball from I bag and black ball from II bag) + P (Black ball from I

bag and white ball from II bag)

= 5

8

6

10

3

8

4

10

30

80

12

80× + × = + =

42

80

21

40=


41. (a) y = – x

1–1–2–3–4–5 2 3 4 5

xx¢

y¢

y

–1

1

2

–2

–3

–4

–5

3

4

5

y = x

y = −

x

Graph of y = – xisthereflectionofthegraphofy = x about the X - axis.

(OR) (b) Given in equalities are x ≤ 3y, x ≥ y

Suppose x = 3y ⇒ x3

= y

If x = y

x 1 2 –1 –2y 1 2 –1 –2

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x 0 3 6 –3y 0 1 2 –1

xx′

y′

yx = y

-4

4

3

2

1

0 3 6 9 −3 −6 −9

x = 3y

−1

−3

−2

000000000000

111111111

42. (a) Let P be the intruder, A and B are the soldiers. Let x be the distance between the intruder and

soldier B.

In ∆ABP, Given ∠PAB = 30º and ∠PBC = 45º In ∆ABP,∠APB = 15° In ∆ABP,usingsineformula,

⇒ 5

15 30sin sin� �=

x = 5

15 30sin sin� �=

x

⇒ x = 5

1530 5

1

2 15sinsin

sin�

��= × =

5

1530 5

1

2 15sinsin

sin�

��= ×

Now, sin 15° = sin (45 – 30)

= sin 45 cos 30 – cos 45 sin 30

30° 45°

45°15°

135°5

90°C

P

B

x

A

= 1

2

3

2

1

2

1

2

3 1

2 2× − × =

−

Substituting this value in (1) we get,

x = x =−

=−

5

2 3 1

2 2

5 2

3 1( ) Km

(OR)(b)

(i) V C V C V C V C V C V C Case(i) Whenthefirstplaceisavowel,numberofpermutation

= 6

2 3

!

! !

Remaining 6 places have consonants

∴ Number of permutation = 6

2

!

!

∴ Number of ways = 6

2 3

!

! !

6

2

!

!

Case(ii) Whenthefirstplaceisaconsonant,

Number of ways = 6

2 3

!

! !

6

2

!

!

∴ Total number of ways = 26

2 3

!

! !

6

2

!

!

= 2 (60) (360) = 43200(ii) 6 consonants out of which 2 are alike can be place

in 6

2

!

! ways and 6 vowels out of which 3E’s are alike

and 2I’s are alike can be arranged in 7 places in

71

3

1

26

P × ×! !

ways

∴ Total number of words = × × ×6

27

1

3

1

26

!

! ! !P

=× × ×

× × ×6 5 4 3

2

7

1

1

3

1

2

! !

! !

= 30 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 151200.

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(iii) Vowels are never together

Total number of arrangements = 12

2 2 3

!

! ! !

By (ii), number of arrangements, when all the vowels

are to gether is = 7

2

6

2 3

!

!

!

! !

∴Number of arrangements, when all the vowels are

never together = 12

2 2 3

7

2

6

2 3

!

! ! !

!

!

!

! !−

= 19958400 – 151200 = 19807200(iv) No two vowels are together is the same as (i) 43200.

43. Let p = q + h

h is numerically very small and so h12 h3 ... may be

neglected.

RHS = n p n qn p n q

n q h n qn q h n

+( ) + −( )−( ) + +( )

= +( ) +( ) + −( )+( ) +( ) + +

1 11 1

1 11 1(( )q

= nq q nh h nq qnq q nh h nq q

+ + + −− + − + +

_=

2 1

2 1

nq n hnq n h

+ +( )+ −( )

= 1

1

2

11

2

+ +

+ −

nn

hq

nn

hq

.

.

= 11

21

1

2

1

+ +

+ −

−nn

hq

nn

hq

. .

= 11

21

1

2+ +

− −

nn

hq

nn

hq

. . =11

2

1

2+ + − −

nn

nn

hq

= 1 + 1

nhq

. ... (1)

LHS = pq

q hq

hq n

hq

n n n

=+

= +

= +

1 1 1

1 11

. ... (2)

From (1) and (2), LHS = RHS

Now 15

168 = 8 1 15 8 1 16

8 1 15 8 1 16

+( )( ) + −( )( )−( )( ) + +( )( )

[n = 8, p = 15 and q = 16]

= 9 15 7 16

7 15 9 16

135 112

105 144

247

249

( )( ) + ( )( )( ) + ( )

= ++

=

15

16

0.99196

(OR)

(i) Since x and y represent the number of copies and the cost of photocopying,

y = 1.50 x, 0 ≤ x ≤ 10 [given] For x > 10, y = 10(1.50)+ (x – 10) ...(1) [ First 10 copies 1 Re and the remaining (x – 10)] = 15 + x – 10 = x + 5 ∴y = x + 5 if x > 10 ∴y = 1.50x, 0 ≤ x ≤ 10 x + 5 if x > 10

20

30

10

cost

of

copie

s

Number of copies

0 10

(10, 15)

(20, 25)

(30, 29)

(40, 31.5)40

20

50

30

60

40

70

50 60 70

y

xx¢

y¢

(ii) Find the cost of making 40 copies Cost of 10 copies 1.50 × 10 = 15.0 Rs. (per copy 1.5 Rs.) after 10 copies 1 × 10 = 10.00 Rs.(Pen copy 1 Rs.) after 10 copies 0.40 × 10 = 4.00 Rs. (per copy 0.40 paise) after 10 copies 0.25 × 10 = 2.50 Rs. (per copy 0.25 paise) cost of making 40 copies = 31.50 Rs.

44. Given A = 4 3 2

1 0 7

2 3 5

−

−

and B = 1 3 3

2 4 0

9 7 5

−

AB = 4 3 2

1 0 7

2 3 5

1 3 3

2 4 0

9 7 5

−

−

−

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= 4 6 18 12 12 14 12 0 10

1 0 63 3 0 49 3 0 35

2 6 45 6 12 35 6 0 25

− − + − + −+ + + + + +− − − − + −

det (AB) = −

− − −

20 10 2

64 52 38

49 17 19

Expanding along R1 we get,

det (AB) = – 2052 38

41 19− − – 10

64 38

49 19− −

+ 264 52

49 17− −

= – 20(–988 + 646) – 10(–1216 + 1862) + 2(–1088 + 2548)

= – 20(–342) – 10(646) + 2(1460)

= – 6840 – 6460 – 292 = 3300 ... (1)

|A| = 4 3 2

1 0 7

2 3 5

−

−

= 4 0 7

3 5− – 3

1 7

2 5− – 2

1 0

2 3

= 4(0 – 21) – 3(–5 –14) – 2(3 + 0)

= –84 + 57 – 6 = –33

|B| = 1 3 3

2 4 0

9 7 5

−

= 14 0

7 5– 3

−2 0

9 5+ 3

−2 4

9 7

= 1(20 + 0) –3 (–10 + 10) + 3 (–14 – 36) = 20 + 30 –150 = – 100

∴ |A| |B| = –33 (–100) = – 3300 ... (2)

From (1) and (2) det (AB) = det (A) det (B)(OR)

Given a→

, b→

and c→

are unit vectors.

⇒ | a→

| = | b→

| = | c→

| = 1

a→

. b→

= a→

. c→

= 0,

and angle between b→

and c→

is π3

a→

. b→

= a→

. c→

= 0

⇒ a→

is ⊥r to both b→

and c→

.

a→

is ⊥r to b→

× c→

⇒ a→

= l ( b→

× c→

) for some scalar l.

∴ | a→

|2 = l2 | b→

× c→

|2

⇒ 1 = l2[| b→

|2 | c→

|2 – ( b→

. c→

)2]

[ | a→

| = 1 and | a→

× b→

|2 = | a→

|2 | b→

|2 – ( a→

. b→

)2]

⇒ 1 = l2 [(1) (1) – | b→

|2| c→

|2 cos2 π3

]

[ angle between b→

and c→

is π3

]

⇒ 1 = l2 [1 – cos2 π3

]

[ | b→

| = | c→

| = 1]

⇒ 1 = l2 [1 – 1

4]

⇒ 1 = l2 3

4

⇒ l2 = 4

3 ⇒ l = ± 2

3

Substituting l = ± 2

3 in (1) we get,

a→

= ± 2

3( b

→× c

→)

45. (i) Given f(x) = x x

x

22 8

2

− −+

f(x) does not exist at x = – 2

∴ It has a removable discontinuity at x = – 2

limx

f x→−

( )2

= limx

x xx→−

− −+2

22 8

2

limx

f x→−

( )2

= limx

x xx→−

−( ) +( )+( )2

4 2

2

= limx

x→−

−2

4 = – 2 – 4 = – 6

∴ The continuous function g(x) can be written as

g(x) =

x xx

x

x

22 8

22

6 2

− −+

≠ −

− = −

if

if

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(ii) Given f(x) = xx

364

4

++

The given function does not exist at x = – 4.

Hence, f(x) has a removable discontinuity at x = – 4.

limx

f x→−

( )4

= limx

xx→−

++4

364

4

= limx

x x x

x→−

+( ) − +( )+( )4

24 4 16

4

= limx

x x→−

− +4

24 16

= (–4)2 – 4 (– 4) + 16 = 16 + 16 + 16 = 48 ∴ The continuous function g(x) can be written as

g(x) =

xx

x

x

364

44

48 4

++

≠ −

= −

if

if

(iii) Given f(x) = 3

9

−−

xx

The curve does not exist at x = 9. Hence f(x) has a removable discontinuity at x = 9.

limx

f x→

( )9

= limx

xx→

−−9

3

9

= limx

x

x x→

−

+( ) −( )9

3

3 3

= limx x→ +( )9

1

3 =

1

3 9+

= 1

3 3+ =

1

6

After removing the discontinuous point, the continuous functions g(x) can be written as

g(x) =

3

99

1

69

−−

≠

=

xx

x

x

if

if

(OR)

Given sin y = x sin (a + y) ... (1)

Differentiating with respect to ‘x’ we get,

cos y dydx

= x cos (a + y) dydx

+ sin (a + y ) (1)

[product rule]

⇒ cos y dydx

= x cos (a + y) dydx + sin (a + y)

⇒ dydx

(cos y – x cos (a + y)) = sin (a + y)

⇒ dydx

= sin

cos cos

sin

cossin

sincos

a yy x a y

a y

y ya y

a y

+( )− +( ) =

+( )−

+( ) ⋅ +( ) =

sin

cos cos

sin

cossin

sincos

a yy x a y

a y

y ya y

a y

+( )− +( ) =

+( )−

+( ) ⋅ +( )

[from (1)]

⇒ dydx

= sin

sin cos sin cos

2 a ya y y y a y

+( )+( ) − +( )

= sin

sin

2 a ya y y

+( )+ −( )

[ sin (A+B) = sin A cos B – cos A sin B]

dydx

= sin

sin

2 a ya+( )

Hence proved.

46. Let I = 3 1

2 2 32

x dxx x+( )− +∫

Now, 3x + 1 = A B.ddx

x x2 2 32− +( )+

⇒ 3x + 1 = A (4x−2)+B

Equating the co - efficients of x-term and constant term we get,

3 = 4A ⇒ A = 3

4

1 = −2A+B⇒ 1

= −23

4

+ B

⇒ 1 = −3

2+B

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⇒ B = 1 + 3

2 =

5

2

∴3x + 1 = 3

4

(4x−2)+

5

2

∴I = 3 1

2 2 32

x dxx x+( )− +∫

= 3

4

4 2

2 2 3

5

2 2 2 32 2

xx x

dx dxx x

−( )− +

+− +∫∫

= 3

42 2 3

5

4

2log x x− + +

5

4 3 22

dxx x− +∫

/

= 3

42 2 3

5

4 1

4

1

4

3

2

2

2

log x x dx

x x− + +

− + − +∫

= 3

42 2 3

5

4 1

2

5

4

2

2log x x dx

x− + +

−

+

∫

= 3

42 2 3

5

4 1

2

5

2

2

2 2log x x dx

x

− + +

−

+

∫ + c

= Adding and Subtracting

Co-efficient of 1

2

1

21

2

x

= −( )

=2

1

4

= 34

2 2 35

4

1

5

2

1 2

5 2

2 1log tan

/

/x x x c− + + ×

−

−+

dxa x a

x a C2 2

11

+= +

∫ −

tan ( / )

= 3

42 2 3

5

2

2 1

5

2 1log tanx x x

− + +−

+

−C

(OR)

(i) A year has 365 days = 52 weeks + 1 day That 1 day may be Sunday, or Monday or

Tuesday or Wednesday or Thursday or Friday or Saturday.

∴ n(S) = 7, n (A) = 1

∴ P(A) = nn

A

S

( )( )

= 1

7

A year is selected at random.

It may be an ordinary year for which the

probability is 3

4or a leap year for which the

probability is 1

4

∴Required probability

= 3

4

1

7

1

4

2

7× + ×

= 3

28

2

28

5

28+ =

(ii) leap year w hich contains 53 Sundays A leap year has 366 days = 52 weeks + 2 days

Those two days may be Sunday and Monday (or) Monday and Tuesday (or) Tuesday and Wednesday (or) Wednesday and Thursday (or) Thursday and Friday (or) Friday and Saturday (or) Saturday and Sunday.

∴ n(S) = 7, n(B) = 2

∴ P(B) = 2

7

∴ Probability that the year selected is a leap year and it contains 53 Sundays.

= 1

4

2

7

1

14× =

47. Let OL be the perpendicular from the origin to the required line.

A O

75º

B

y

xax′

L

Given OL = p = 3 2 and ∠XOL = 75°

∴ p = 3 2 and α = 75°

Equation of the line is normal form is xcosα+ysinα = p ⇒ x cos 75° + y sin 75° = 3 2 ...(1) Now cos 75° = cos (45º + 30°) = cos 45º cos 30º – sin 45º sin 30º

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= 3

2 2

1

2 2

3 1

2 2− = − ...(2)

sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

= 3

2 2

1

2 2

3 1

2 2+ = +

...(3)

Substituting (2) and (3) in (1) we get,

x y3 1

2 2

3 1

2 23 2

−

+ +

= = x y3 1

2 2

3 1

2 23 2

−

+ +

=

⇒ x ( 3 – 1) + y( 3 + 1) = 3 2 × 2 2 = 12

⇒ x ( 3 – 1) + y( 3 + 1) = 12 which is the required equation

(OR)

It is given that f(x) is differentiable at x = c and every differentiable function is continuous.

So, f(x) is continuous at x = c.

∴ limx c

f x→ −

( ) = limx c

f x f c→ +

( ) = ( )

⇒ limx c

x→

2 = limx c

a c b c→

+ +( ) = 2 (ax + b) = c2

⇒ c2 = ac + b ... (1)

Now, f(x) is differentiable at x = c.

∴f ′(c–) = f ′(c+)

⇒ limx c

f x f cx c→ −

( ) − ( )−

= limx c

f x f cx c→ +

( ) − ( )−

⇒ limx c

x cx c→ −

−−

2 2

= limx c

ax b cx c→ +

+( ) −−

2

⇒ limx c

x c x cx c→ −

+( ) −( )−

= limx c

ax b ac bx c→ +

+( ) − +( )−

[Using (1)]

⇒ limx c

x c→ −

+ = limx c

a x cx c→ +

−( )−

⇒ limx c

x c→ −

+ = limx c

a→ +

⇒ 2c = a Substituting a = 2c in (1) we get c2 = c(2c) + b ⇒ c2 – 2c2 = b ⇒ b = –c2. ∴ a = 2c and b = – c2

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