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Solved Guess Paper – 2

II PUC Mathematics

One Mark Questions

1. Give an example of a relation which is Reflexive and symmetric but not transitive

Ans. R = { (1, 1) , (2, 2), (3, 3), (2, 3),(3, 2), (1, 2),(2,1 )}

(1, 2) R, (2, 3) R.

But (1, 3) R .

2. Find the value of cot(tan-1

x + cot-1

x)

Ans. cot (tan-1

x + cot -1

x)

-1 -1π= cot [ tan x +cot x =1]

2

= 0

3. Define a scalar matrix.

Ans. Scalar matrix :

It is a diagonal matrix whose diagonal elements are equal .

4. x 8 2 8If = find the value of x .

8 x 8 2

Ans.

2

2

x 8 2 8=

8 x 8 2

x = 64 = 4-64

x = 4- 64 + 64

2

.

x = 4

x = ±4

5. Differentiate 3 ex3 with respect to x .

Ans. 3

3

x

x

y = e

Diff w.r.t x

dy d= (e )

dx dx



3

3

x 2

2 x

=e × 3x

dy= 3x .e

dx

6. 1- x

Evaluate dxx

Ans.

1/2+1

1/3

3/23/2

1- x 1 xdx = - dx

x x x

1 x. x 1= .dx - dx = - x )dx

x x x

1= 2 - dx - x . dx

dx

x= 2 x - x . dx = 2 x - + C.

1 / 2 + 1

x 2= 2 x - + C = 2 x - x + C

3 / 2 3

7. Find the vector component of the vector with initial point ( 21) and terminal point ( -5,

7) .

Ans. Let p Q be terminal & initial points p (-5, 7)& q(2,1)

QP = (-j5,-2)i +(7 -1)j

= -7i ,6j are the vector component.

8. What is the equation of the plane that cuts the coordinate axes at(a, 0 , 0 ) , (0 , b, 0)

and (0 , 0 ,c).

Ans. x b 3

+ + = 1a b c

9. Define term corner point in the L. P.P.

Ans. Corner point

The point within or on the boundary of the feasible region represent feasible

solution of the constraints . They are called corner points.



10. If F is an event of a sample spaces „s‟ of an experiment then find P(S/F).

Ans. p(s/ f) = 1

PART – B

Two Mark Question

11. Verify whether the operation defined on Q by a b = ab

4 in associative or not.

Ans.

Let a , b, cab

a*bθ =, 4

a* b*c = bc abc

( ( ) =4

a*16

ab abc(a*b)*c = *c =

4 16

a*(b *c) = (a*b)*c

Hence* is associative,

12. Find the value of -1 -1

tan 3 - sec c - 2

Ans.

-1 -1 -1 -1tan 3 - sec (-2) sec (-2) = π - sec 2

π π -π- π - = π

3 3 3

π 2 -π- =

3 3 3

13. Write the simplest form of tan -1

1-cos x, 0 < x < π

1 +cos x

Ans.



-1

2

2

-1

1-cos xtan ,o < x < π

1 + cos x

x1 - cos x = 2 sin

2

x1 +cos x = 2cos

2

2tan

2

-1sin (x / 2) x= tan tan

2cos 2(x / 2) 2

x=

2

14. Let A( 1 , 3 ), B(0,0) and c(k , 0 ) be the vertices of triangle ABC of area 3 sq. units .

find „k‟ using determinant method .

Ans.

A ( 1, 3) , B(0 , 0) & C(k, 0)

1 1

2 2

3 3

z

x y 11

Δ = x y 1 = 32

x y 1

1 3 11

0 0 1 = 32

k 0 1

Expandalong R

[0 + 0) -1(0 - 3K)] = 6

3K = 6

K = 2

15. Prove that greatest integer function define by f(x) = [x] ,0 < x< 3 is not

differentiable at x = 1.

Ans. f(x) = [x], 0 < x< 3

At x = 1

F(1) = [ 1] = 1

LHS F1(1) =

h 0x 0

1 + h - 1f(1 + h) - f(1) 0 -1 -1Lt = Lt = = '

h h h o

As(1 + h) < 1,[1 + b] = 0

LHS f'does not exist.



RHL f‟(1) =

h 0

f(1 + h) - (f (1) (1 + h) - (1)RHS f'(1) = Lt

x h

1-1 0= = 0

h n

LHL f '(1) RHL f'(1)

f(x) ia not differential at x = 1

16. dy 4find , If x =4t and y =

dx t

Ans.

2

2

4x = 4t y =

t

x = 4t y = 4 / t

Diff w.r.t t Diff w.r.t t

dy dy / dt -4 / t -1= = =

dx dx / dt 4 t

17. Find the intervals in which the function f given by f(x) = x2 +2x – 5 is strictly

increasing.

Ans. f(x) = x2 +2x -5

F‟(x) = 2x +2

For increasing f‟ (x) > 0 2x +2 > 0

x > - 1

f(x) is strictly increasing on ( - 1 , )

18. Evaluate : sin 3x cos 4x dx

Ans.

sin 3x .cos 4x. dx



1 1= 2sin 3x cos 4x . dx = 2cos 4x . sin 3x . dx

2 2

1= (sin 7x -sin x ) dx

2

1 -cos 7n= - (-cos x) + c

2 7

1 cos x 7 x= cos x - +c

2 14

19. Evaluate log x dx

Ans.

log x dx

= log x . 1dx

1= log x . x - x. . dxx

= x log x - x +c

= x(log x-1)+ c

20. From the differential equation of family of curves x y+ = 1,

a b eliminating the constants

“ a” and „b‟ .

Ans.

1

2

2

2

x y+ = 1

a b

1 1Diff . w .r.t x + y = 0

a b

Again diffw.r.t x

1o + y = 0

b

y = 0

y = 0 is therequired differential eqn.

21. Show that the vectors 2i - 3j + 4k and b = i - j + k

Ans.



If a = λb then a .&b are collinear

a = 2i - 3 j + 4k & b = -4i + 6j . 8k

12i - 3 j + 4k = - (-4i + 6j - 8k)

2

a = λb

Hence the two vector are collinear.

22. Find the angle between the vectors a = 2 + j- k and b = i - j+ k

Ans.

a = i + j - k , b = i - j + k

a .b =1 + (1-) + (-1)

= 1-1 -1 = -1

-1

-1

-1

| a | = 3 , |b | = 3

a .bθ = cos

| a| | b |

1θ = cos

3 . 3

1θ = cos -

3

23. Find the distance the parallel lines r = i + 2j - 4k + λ(2i + 3j+ 6k)

andr =3i + 3j- 5k + M(2i + 3j+ 6k)



Ans.

21

12

12

a = i + 2j - 4k & a = 3i + 3j - 5k

a - a = 22 + j - k

i j k

we have b = 22 + 3j + 6k Þ b ×(a - a ) = 2 3 6

2 1 -1

= -a2 + 14j + 4k

|b| = 4 + 9 + 36 = 7

Hencethedistance between two parallel lines

1 1 293d = | -92 + 14j + 4k |= 293 =

7 7 7

24. A fair die is rolled. consider the events E = { 1 , 3 , 5 } , F = { 2 , 3} and G ={ 2 , 3

, 4 , 5 } find i)p(E/F) ii) P (E / G)

Ans E = { 1 , 3 , 5 } , f = { 2 , 3} , G = { 2 , 3 , 4 , 5 }

3P(E) =

6

, 2P(F) =

6 , 4

P(G) =6

21 2P(E F) = , P(F) = , P(G) =

6 6 6

1P(E I F) 16(a)P(E / F) =, = =

2P(F) 26

2P(E I F) 2 16(b) P(E / G) = = =

4P(G) 4 26



PART- C

Three Mark Question

25. Determine whether the relation R is the set A = { 1 , 2 , 3 , 4 , 5, 6} as R = { x , y) y is

divisible by x } is reflexive, symmetric and transitive .

Ans. A= { 1 , 2, 3, 4, 5, 6 }

R = { (x, y ) : y is divisible by x )

i)Since (x, x) R x

= 1 Ax

ii) if x ,y A , ( x,y) R (y, x) R

R is not symmetric

iii) x , y ,z A , (x , y ) R , (y , z ) R ( x y ) R

Thus R is transitive .

Hence R is reflexive transitive not symmetric

26. Find the value of x , if

-1 -1x -1 x +1 πtan + tan =

x - 2 x + 2 4

Ans.

-1 -1x -1 x +1 πtan + tan =

x + 2 x + 2 4

-1

2 2

-1

x -1 x + 1+

πx + 2 x + 2= tan =(x -1)(x + 2) 4

1 -(x - 2)(x + 2)

(x -1)(x - 2) + (x + 1)(x - 2)

x - 2= tan

2 2

2 2

x - 4 - (x -1)

x - 2

2 3

2

2

π=

4

x + 2x - x - 2 + x - 2x + 1x - 2 π= tan

x2 - 4 - x + 1 4

2x 4= 1

-3



2

2

2

2x - 4 = -3

2x = 1

1x =

2

1x = ±

2

27. For what vale of x : [ 1, 2, 1 ]

1 2 0 0

2 0 1 2 = 0

1 0 2 x

Ans.

1 2 0 0

[1.2.1] 2 0 1 2 = 0

0 0 2 x

0

[1 + 4 + 1, 2 + 0 + 0, 0 + 2 + 2] 2 = 0

x

0

[ 6,2,4] 2 = 0

x

0 + 4 + 4x = 0

4 = - 4x

x = -1

28. If yx +x

y = a

b , find

dy

dx .

Ans.

yx +xy = ab

U=yx

V=xy

U + v = ab

Diff w.r.t x



x

x

x

du dv+ = 0 (1)

dx dx

consider u = y

log u = log y

log u = x log y

Diff w.r.t x

1 dv 1 dy= (1)log y + x

u dx y dx

du x dy= u log y +

dx y dx

du x dy= y log y +

dx y dx

y

y

consider u = x

logu = log x

logu = ylog x

Diff w.r.t x

1 dv dy 1= log x + y

v dx dx x

y

x y

dv dy y= v log x +

dx dx x

dv dy y= x log x +

dx dx x

x dy dy y(1) y log y + + x logx + = 0

y dx dx x

x x y y

x y x y-1

x-1 4 x y-1

x y-1

x-1 y

x dy dy yy logy + y + x logx + x = 0

y dx dx x

x dy dyy + x log x = -y logy - x y

y dx dx

dyxy + x logx = - y logy - x y

dx

dy -y logy - x=

dx xy + x logx



29. Verify mean value theorem for the function f(x) = x3 -5x

2-3x, in the interval [ a , b ]

where a = 1 and b = 3 .find all e (1,3)for which f1(c) = 0.

Ans. The given function f(x) = x3 -5x

2 -3x is continuous in (1, 3) & diff in (1, 3)

At a = 1 f(1) = 13 - 5 (1

2) -3 (1) = - 7

B = 3 , f(3) = 33

– 5(32)-3(3) = - 27

2

2

f(b) - f(a)f'(c) =

b - a

-27 - (-7)=

3 -1

f'(1) = -10

3c -10c + 7 = 0

3c -1c - 3c + 7 = 0

((3c - 7) -1(3c - 7) = 0

(c -1)(3c - 7) = 0

7c = 1,c =

3

7(1,3)

3

Hence mean value theorem is verified.

30. Find two positive numbers s whose sum is 16 and the sum of whose cubes is minimum

Ans.

3 3 3 3

x + y = 16 y =16 - x

S = x + y s = x + (16 - x)

ds= 0

dx

2 2 2 2

2

2

2

x - (16 - x) = 0x - (256 + x - 32x) = 0

x = 8

d y= 6x + 6(16 - x) = 96 > 0

dx

sum is minimum when x = 8

sum is minimum when numbers are 8 & 8

31. x

Find dx(x +1)(x + 2)



xdx

(x + 1)(x + 2)

-1 -2dx dx

-1 + 2 -2 + 1+x + 1 x + 2

-1 +2= dx + - dx

x + 2 x + 2

= log (x + 1) + 2 log (x + 2) + c

32. 2

Evaluate x +1)log x dx

Ans.

2

3 3

3 3

3 3

(x + 1) log x . dx

x 1 x= logx + x - + x dx

3 x 3

x + 3x x= log x - + 1 dx

3 3

x + 3x x= log x - - x +c

3 9

33. Find the area of the region bounded by the curves y= x2 and the line y = 4 .

Ans. Since the given curve represented by the equation y = x2 is a parabola

symmetrical about y – axis only therefore the required area of the region ABOA is

given by

4

0

4

0

area of the region BONB

- 2 x dy = 2 bounded by curve y - axis and

the lines y = 0and y = 4

= 2 y dy

43

2

0

2= 2× y

3

23

=3

34. Prove that the equation 2

2 2x dy= x - 2y + xy

dx is a homogenous differential equation .

Ans.



2 2 2

2 2

dyx = x = 2y + xy

dx

x x

2

3

2

dy y y= x 1 - 2

dx x x

dy y yÞ = 1 - 2 + = gx (1)

dx x x

Consider(1) V

2dv+ x = 1 - 2v + v

dx

Diff w.r.t.x

dy dv= v + x

dx dx

Eqn (1) v2dv

+ x = 1 - 2v + vdx

2

2

2

2 2

dvx = 1 - 2v

dx

dy dx=

1 - 2v x

dv dy=

1 x(2 - v )2

dv

12[( ) - v ]2

2 2

Integrate B.S

dv dx=

1 x2[( ) - V ]2

1 1

2 2' 2

1+ v

2log = log x + c

1- v

2



1 + v 2

2 2log

4 1 - 2 v

2

= log x +c

1 1 + 2 vlog = log x + c

2 2 1- 2 v

y1 + 21 xlog = log x + c

2 2 y1- 2

x

y1 + 21 1 x + 2 yxlog = log x + c log = log x + c

y2 2 2 2 x - 2 y1 - 2x

35. Find a vector perpendicular to each of the vectors a = 2i + j+ 3k b = 3i + 5j -2k,

Which has magnitude 10 units .

Ans. Vector = (magnitude ) (unit vector )

a×b= 10

| a×b|

-17i + 13j + 7k= 10

507

36. Show that the points A ( `1, -4 , -3 ) , B(3, 2, -5) C (-3 , 8 , -5) and D(-3 , 2, 1 )are

coplanar.

Ans.

OA = (1,4, -3)

OB =(3, 2, - 5)

OC = (-3,2,1)

ThenAB = (4, -2, -2) = a

AC =( -6

, 6 , (-2,4, -2) = b

AD = (-2 , - 2 ,4 ) = c



4 -2 -2

we have a . (b ×c ) = -2 4 -2

-2 -2 4

= 4(16 - 4) + 2(-8 - 4) - 2(4 + 8)

= 4(12) - 2(12) - 2(12)

a . (b ×c) = 0

Hence the points are coplanar.

37. Find the Cartesian and vector equation of the line that passes through the points (3, -

2,

-5 )and (3, -2 , 6)

Ans.

x - 3 y + 2 z + 5Cartesian equations is : = =

3 - 3 0 1

x - 0 y + 2 z + 5= =

0 0 1

a = 3i - 2j - 5k & b = oi + oj + k

Equation is r =a + λb

r = (3i - 2j - 5k) + λk

38. A die in thrown , If E is the event “ the number appearing in a multiple of 3” and F

be the event “The number appearing is even then find whether E and f are

independent ?

Ans. we have S = {1 , 2, 3, 4, 5, 6}

E = {3,6} F = { 2, 4, 6 }

2 1we have P(E)= =

6 3

3 1P(F) = =

2 2

1EI F = {6} P(EIF) =

6

Consider ,

P(EIF) = (P(E) . P(F)

1 1 1= ×

6 3 2

1 1=

6 6

Clearly , P(EIF) = P(E).P(F)

Thus the E & F are independent



PART - D

Five Mark Questions

39. If f : AA defined by 4x + 3f(x) =

6x - 4

where A = R – {2 / 3} , show that f is invertible

and f - 1

= f .

Ans.

We have f of (x) =f[f(x)]

4x + 3= f + 3

6x - 4

4x + 34 + 3

6x - 4=

4x + 36 + - 4

6x - 4

16x + 12 + 3(6x - 4)

6x - 4=

24x + 12 + 4(6x - 4)

6x - 4

16x + 12=

+ 18x -12

24x + 18- 24

x + 16

34x=

34

fof (x)

f' = f [ Inverse of f is f itself ]

40.

1 2 3 2 3 4 2 3 -1

If A = -4 5 6 B = 5 -3 0 and c = 4 5 6 prove that A(B + C) = AB + AC

7 8 0 4 5 -3 -1 2 3

Ans.

1 2 3 2 3 4 2 3 -1

A = -4 5 6 B = 5 -3 0 C = 4 5 6

7 8 0 4 5 -3 -1 2 3

A(B + C) = AB + BC

LHS : A(B + C)

4 6 3

B + C = 9 2 6

3 7 0



1 2 3 4 6 3

A(B + C) = -4 5 6 9 2 6

7 8 0 6 7 0

4 + 18 + 9 6 + 4 + 21 3 + 12 + 0

= -16 + 45 + 18 -24 + 10 + 42 -12 + 30 + 0

28 + 72 + 0 42 + 16 + 0 21 + 48 + 0

31 31 15

= 47 28 18 (1)

100 58 69

RHS = AB + AC

1 2 3 2 3 4

ABÞ A -4 5 6 × B = 5 -3 0

7 8 0 4 5 -3

2 + 10 + 12 3 - 6 + 15 4 + 0 - 9

= -8 + 25 + 24 -12 -15 + 30 -18 + 0 + 18

15 + 40 + 0 21 - 24 + 0 28 + 0 + 0

24 12 -5

= 41 3 -34

54 -3 28

1 2 3 2 3 -1

AC = A -4 5 6 ×C = 4 5 6

7 8 0 -1 2 3

2 + 8 - 3 3 + 10 + 6 -1 + 12 + 9

= -8 + 20 - 6 -12 + 25 + 12 4 + 30 + 18

14 + 32 + 0 21 + 4 + 0 -7 + 48 + 0

7 19 20

= 6 25 52

46 61 41

31 31 15

AB + AC = 47 28 18 (2)

100 58 69

From (1) & (2) L.H.S = R.H.S

A(B + C) = AB + AC



41. Solve the following system of equations by matrix method x+2y +3z =2, 2x +3y +z

= 1, and x-y –z = -2.

Ans.

x+2y +3z = 2

2x +3y +z = -1

X - y-z= -2

1 2 3 x 2

A = 2 3 1 X = y B = -1

1 -1 -1 z -2

| A| = ad j

-1

A

= 1(-3 + 1) - 2(-2 -1) + 3(-2 - 3)

= - (-2) - 2(-3) + 3(-5)

= -2 + 6 -15

= -11

+(-3 + 1) -(-2 -1) +(-2 - 3)

Co.factor = -(-2 + 3) +(-1 - 3) -(-1 - 2)

+(2 - 9) -(2 - 6) +(3 - 4)

-2 3 -5

= -1 -4 3

-7 5 -1

-2 -1 -7

Adj A = 3 -4 5

-5 3 -1

1A = a

| A |

-1

dj A

-2 -1 -71

A = - 3 -4 511

-5 3 -1



-1X = A B

-2 -1 -7 2-1

= 3 -4 5 -111

-5 3 -1 -2

-4 + 1 + 14-1

= 6 + 4 -1011

-10 - 3 + 2

11-1

= 011

-11

-1

= 0

1

x = -1 , y = 0 , z =1

42. If (tan-1

x )2 = y , prove that (x

2 +1)

2 y +2x (x

2+1)y1 = 2

Ans.

-1 2

-1

-1

2

1 2 2

2

2 1 2

2 2 2

1

(tan x) = y

y = (tan x) 2

Difff w.r.t x

dy 1= 2 tan x ×

dx 1 + x

12xy + (1 + x )y = 2×

1 + x

π2x(1 + y )y + (1 + x )2y - θ = 2

2

(x + 1)2y + 2x(1 + x )y = 2

43. A practical moving along the curve 6y = x2 + z ,find the points on the curve at which

the y – coordinate is changing 8 times as fats as the x – coordinate .

Ans.

2

2

dy dx= 8 (1)

dt dt

we have 6y = x + 2

dy dx6 = 2x (2)

dt dt



dx

6 8dt

dx= 2x

dt

48 = 2x

x = 24

2

2

when x = 24 6y = x + 2

6y = 24 + 2

6y = 576 + 2

6y = 578

578y =

6

289y =

3

289Thus the required point is 24,

3

44. Find he integral of 2 2 2

1 dxwith respect to "x" and evaluate

a - x 5 - 4x - x

Ans.

2 2

-1

2 2 2 2 2

2 2

2 2

1dx

a - x

Put x = asin θ

dx=a cos θ

dθ

dx = a cos θ .dθ

x = a Sin θ

x= Sin θ

a

θ = sin (x / a)

a - x = a - a sin θ

= a (1 - sin θ)

= a cos θ



2 2

= acos θ

1 a cosdx =

a - x

θ

a cos

-1 -1

2 2

dθθ

= 1 dθ

= θ + c

x dx x= sin + c =sin + c

a aa = x

45. Find the area of the circle x2 + b

2 = a

2 by the method of integration and hence find the

area of the circle x2 + b

2= 2 .

Ans.

2 2 2

2 2

2 2 2

2 2 2

2

b

a

b2 2

a

92

2 2 -1

0

2

2 2 -1

2

-1

2

2

x + y = a

Put y = 0 x = a

x = ±a

y + x - a

y = a - x

y = 2 - x

Area of a circle = 4 Area of quadrant OAB

= 4 y dx

= 4 a - x - dx

x a x= 4 a - x + sin

2 2 a

9 a a= 4 a - a + sin -0

2 2 a

a= 4 0 - sin 1

2

a π= 4. .

2 2

A = πA Sq. uni

2 2

2 2 2

ts

Area of thecircle x + y = 2

Area of x + y =aA = 2π sq units

is πa2 8q units



46. Find the general solution of the differential equation 2dy+ y cotx = 2x + x .cot x

dx .

Ans.

2

2 2

2

cotxdx log sin x

2

2

dy+ y cot x = 2x + x - cot x

dx

soln : P = cot x = 2x + x cotx

IF = e = e = sin x

The G S in

b(sin x ) = sin x (2x + x cot x )dx + c

y sin x = 2 xsin x dx + x cos x dx + x

= 2 x(-cos x ) - (-cos x)dx] + x2sin x - 2x sinx dx + c

= 2 - xcosx + 2 sin x + x si

n x - [ - 2x cos x + 2 sin x ] + c

= -2x cos x + 2 sin2

x + x sinx + 2x cos x - 2 sin

2

x +c

y sin x = x sin x + c

47. Derive the formula to find the shortest distance between the two skew lines

1 1 22r = a + λ b and r = a + b

in the vector form .

Ans.

Let l1 & l2 be two skew lines of eqn 1 2 21r = a + λ b &r = a +μb

consider a point

S on the line l1 with position vector 21a & a point with position vector a .

Le tPQ

isashortestdistanceof2 lines l1 &l2 & then it is perpendicular to 2 2b and b

ThenPQ

is an unit vector .

1 2

1 2

n = b × b(1)

| b ×b |

PQ = nd (2)

Let θbe the angle b / w PQ &ST

PQ = ST | cosθ | (3)

PQ .STConsider cos θ =

| PQ . ST |



n dcos θ =

2 1(a - a )

d

1 2 1 2

1 2

(ST)

(b ×b ) (a - a )cosθ =

|(b ×b ) | ST |

Hence the required distance is

d = PQ = ST | cosθ |

PQ = ST

1 2 2 1

21

(b ×b )(a - a )

| (b ×b ) | ST

1 2 2 1

21

(b ×b )(a - a )PQ =

| (b ×b ) |

48. A fair coin in tossed 8 times find the probability of at most 5 heads.

Ans. The repeated toss of coin are Bernoulli trials .

Let x denote the number of heads in a experiment of 8 trials . clearly n = 8 , p

=1/2 , q =1/2

P( = x ) n Cn p

n q

n-n

p( 5 ) = p ( = 0 ) + p ( = 1) + p ( x = z )+p( x = 3) +p(x=4)+p(x = 5)

= 8c0 (1/2)

0(1/2)

8+

8c1(1/2)

1 (1/2)

7+

8c2(1/2)

2(1/2)

6+

8c3(1/2)

3(1/2)

5+

8c4(1/2)

4(1/2)

4+

8c5(1/2)

3(1/2)

5

8

8

8! 8! 8! 8! 8!= 1 / 2 1 + + + + +

11!7! 6!2! 5!3! 4!4! 3!5!

8= 1 / 2 1 + 8 +

× 7

2

8× 7× 6+

3× 2

8+

× 7× 6 × 5

4 3 2

8× 7× 6+

3× 2

8

8

= 1 / 2 1 + 8 + 28 + 56 + 70 + 56

= 219 1 / 2



PART – E

Ten Mark Questions.

49. a) A manufacture produces nuts and bolts. It takes 1 hour of work on machine A

and 3 hours on machine B to produce a package of nuts .It takes 3 horse on

machine A and 1 hour on machine B to produce a package of bolts. He earns a

profit of Rs 17 . 50 per package on nuts and Rs 7.00per package on bolts. How

many package of each should be produced each day 50 as to maximize his profit

if he operate his machines for at the most 12 hours a day ?

b) prove that 2

2

2

1 a a

1 b b = (a - b)(b - c)(c - a)

1 c c

Ans.

a) Let x & y be a no of packets of nuts & bolts.

l ET z = 17.5 x+ 7y be an objective function

Constraints : x+ 3y 12

3x +y 12

X , y 0

Consider x+ 3y 12 if x = 0 , y 4 (0, 4)

y= 0 , x = 12 (12, 0)

3x + y 12 if x = 0 , y = 12 (0, 12)

y = 0,x = 4(4,0)

The region AOCB is the feasible region

Corner point Z = 17 . 5x+7y

(0 , 0 ) 0

(0 , 4) 28

(3, 3) 73.5

(4, 0) 70

The maximum value of Z is 73.5 at (3, 3)



2

2

2

2 2 3 3 3 1

2

1 a a

Δ = 1 b b

1 c c

R R - R , R R - R

1 a a

Δ = 0 b - c (b - c)(b + c)

0 c - a (c - a)(c + a)

2

3 3 2

2

1 a a

Δ = (b - c)(c - a) 0 1 b + c

0 1 c + a

R R - R

1 a a

Δ = (b - c)(c - a) 0 1 b + c

0 0 1

Expand along first column

= (b-c) (c-a)(a-b) 1 1 b + c

0 1

= (a-b)(b-c)(c-a)

50. a) Prove that

b b

a a

π/3

π/6

f(x)dx = f(a + b - x)dx and hence

dxEvaluate

1 + tan x

B) Find all the points of discontinuity of f defined by f(x) = | x| - | x +1|

Ans.

a)

b b

a a

b

a

f(x)dx = f(a + b - x)dx

RHS : f(a + b - x)dx

Put a + b - x = t

diff w.r.t.x

dt-1 = dx =- dt

dx



a

b

a

b

b

a

b

a

f(t) (-dt)

= -f(t)dt

a + b - x = t

Atx = aa + b - a = t

t = b

Atx = b,a + b - b = t

t = b

f(t)dt

f(t)dx LHS

Hence proved

π/3

π/6

π/3 π/3

π/6 π/6

π/3 π/3

π/6 π/6

π/3

π/6

π/3π/3

π/6 π/3

dx*) (1)

1 + tan x

dx dx= (2)

1 + cos xπ1 + tan - x2

Now

dx dx(1) + (2) = +

sin x cos x1 + 1 +

cos x sin x

sin xcosx= + dx

cos x + sin x sin x + cos x

π= 1 dx = x] = π / 3 - π / 6

6

b) The given function is f(x) = |x| - |x+1|

The two functions “g” and ”h” are defined as

G(x) = | x| and h(x)=|x+1|

Then f=g-h

The continuously f g and h is examined first

g(x)= |x| can be written as

-x if x < 0g(x) =

x if 0

Clearly g is defined for all real numbers let “c” be a real number.

Case I.



If c < 0 , g(x) = - x is a polynomial function then f is continuous.

Case II

If c > 0 , g(x) = x is a polynomial function then f is continuous .

Case III

If c =0 Then g( c) = g(0)= 0

x 0- x 0+lim g(x) = lim g(x)= g(0)

g is continous at x = 0

We have , h(x)=| x+1| is written as h(x) =

-(x +1) if x < -1

x +1 if x ³ -1

Case I:

If c< - 1 , then h(x) = -(x+1) is a polynomial function f is continuous

If c > -1 , then h(x) = x+ 1 is a polynomial function f is continuous

If c =1, then h( c) = h (-1) = -1+1 = 0

x -1 + x -1+lim h(x) = lim h(x) =h(-1)

h is continous at x = -1

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