Z1543480043risc Maths Self Assessment Paper-9.pdf

MATHEMATICS

Time : 3 Hours Maximum Marks : 100

ISC Solutions

9

Self Assessment Paper Section ‘A’ 1. (i) Given, A = {2, 3}, B = {4, 5, 6} \  A × B = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} \  n(A × B) = 6 \ No. of subsets of A × B = 26 Þ Total no. of relations from A to B = 26 = 64. 3 1 + 2 tan −1 (ii) LHS = sin 2 3 p p   −1 −1  = sin  sin  + 2 tan  tan    3 6 −1

p p p p = + 2 . = + 3 6 3 3 2p = = RHS. 3 (iii)



x2  2x   7  2 + 2   = 3    y  3 y   −3 

 x 2   4 x   21 Þ  2  + 6 y  =  −9   y      Þ x2 + 4x = 21 and y2 + 6y = –9 2 Þ x + 4x – 21 = 0 and y2 + 6y + 9 = 0 Þ (x + 7)(x – 3) = 0 and (y + 3)2 = 0 Þ x = –7 or 3, y = –3.

(iv) Given, y =

ex sin x

y − tan −1 y  y → 0 y − sin y

0   0 form   

(v) Given: lim

On applying L’ Hospitals rule, we get 1 1− 0  1+ y 2 form  = lim   0 y → 0 1 − cos y  

Again, applying L’ Hospital’s rule, we get 0 − ( −1)(1 + y 2 )−2 ⋅ 2 y y→0 sin y

= lim

2y

= lim

y → 0 (1 + y 2 ) 2

2

= lim

y → 0 (1 + y 2 ) 2

=

2 (1 + 0 )

2

⋅ sin y  y  ⋅  sin y    sin y = 1  lim y→0 y 

⋅ 1 = 2. 

(vi) The given function is f(x) = 5x – 3 At x = 0, f(0) = 5 × 0 – 3 = –3

f ( x ) = lim ( 5x − 3 ) = 5 × 0 − 3 = −3      xlim →0 x →0

f ( x ) = f (0) \   xlim →0

Therefore, f is continuous at x = 0 At        x = –3, f(–3) = 5 × (–3) = –18 lim f ( x ) = lim ( 5x − 3) = 5 ×( −3) − 3 = −18 x →−3 x →−3    \     lim f ( x ) = f ( −3) x →−3

( )

d x x d dy sin x dx e − e dx ( sin x ) \   = dx sin 2 x x sin x e − e x cos x = sin 2 x e x ( sin x − cos x ) = . sin 2 x

Therefore, f is continuous at x = –3. (vii) ∫ e x (tan x + log sec x ) dx = ∫ e x tan x dx + ∫ e x log sec x dx II

I

= ∫ e tan x dx + log (sec x )e x x

  

−∫

d (log sec x ). ∫ e x dx dx

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

2

= ∫ e x tan x dx + log sec x ⋅ e x − ∫ e x tan x dx + c

= ex · log sec x + C. (viii) Let I = ∫

2

x

1

3−x+ x

Þ I = ∫

2

3−x

1

3 − (3 − x ) + 3 − x



[using

b

∫a

2

3−x

1

x+ 3−x

dx ...(i) dx b

f ( x ) = ∫ f ( a + b − x ) dx] a



Þ I = ∫



Adding eq. (i) and eq. (ii), we get 2

x+ 3−x

1

x+ 3−x

2I = ∫

dx ...(ii)

dx

2

= ∫ 1dx 1

= [ x]12 = 2 − 1 = 1

\ I =



i.e.,

2

∫1

1 2 x 3−x+ x

dx =

1 2

(ix) The given differential equation is dy 1 − cos x         = dx 1 + cos x 2 x dy 2 sin 2 x Þ = = tan 2 x dx 2 2 cos2 2 dy  2 x  Þ =  sec − 1 dx  2 

Separating the variables, we get



x   dy =  sec 2 − 1 dx  2 



Now, integrating both sides, we get : dy =  sec 2 x − 1 dx ∫ ∫ 2  x dx − ∫ dx 2 x Þ y = 2 tan − x + C. 2 This is the required solution. (x) Given,

Þ



Probability of selection of first horse =



Probability of selection of second horse =

∫ dy = ∫ sec

2

1 4 1 3



Probability that only one of them will be  1 1 1  1 selected =  1 −  + ×  1 −     4 3 3 4 1 2 1 3 3 2 = × + × = + 4 3 3 4 12 12 5 = . 12 2. Given, R = {(a, b) : |a – b| is multiple of 4}, where a, b  A = {x  z : 0 ≤ x ≤ 12} = {0, 1, 2, … 12}. Reflexivity : For any a  A, we have |a – a| = 0, which is multiple of 4. Þ (a, a)  R Thus, (a, a)  R for all a  A. So, R is reflexive. Symmetry : Let (a, b)  R. Then (a, b)  R Þ |a – b| is multiple of 4 Þ |a – b| = 4l for some l  N Þ |b – a| = 4l for some l  N [ |a – b| = |b – a|] Þ (b – a)  R So, R is symmetric. Transitivity : Let (a, b)  R and (b, c)  R. Then, (a, b)  R and (b, c)  R Þ |a – b| is multiple of 4 and |b – c| is multiple of 4 Þ |a – b| = 4l and |b – c| = 4m for some l, m  N Þ a – b = ± 4l and b – c = ± 4m Þ a – c = ± 4l ± 4m Þ a – c is multiple of 4 Þ |a – c| is multiple of 4 Þ (a, c)  R Thus, (a, b)  R and (b, c)  R Þ (a, c)  R So, R is transitive. Hence, R is an equivalence relation. x 3. We have, f ( x )= Clearly, domain (f) = R. 1+ x 2 In order to find the range of f, we proceed as follows : Let   f(x) = y. Then, x =y            y = f ( x ) ⇒ 1+ x 2 y x2 Þ = y2 ⇒ x = ± 1+ x 2 1 − y2 Since x takes real values. Therefore, 1 – y2 > 0 Þ y2 – 1 < 0 Þ y  (–1, 1) Hence, Range (f) = (–1, 1)

Solutions

3

Clearly, Range (f)  Domain f. Therefore, fof : R → R and fofof : R → R. Now,

( fofof )( x ) = (( fof ) of )( x ) = ( fof ) ( f ( x ))



   x  =ff    1+ x 2    1+ x 2   

x

( fofof )( x ) = fof 

Þ

 x  1+ x 2 Þ ( fofof ) ( x ) = f   x2  1+  1+ x 2

    x =f     1+ 2x 2   

x =



1+ 2x 2 x2 1+ 1+ 2 x 2

=

x 1+ 3x 2





63 3 1 + 2 tan −1 = sin −1 65 5 5 Taking LHS, sin

−1

2

 63  1 −   + sin −1  65 

1 5 2  1 1+    5 2.

    Using   −1 −1 2 1−x  cos x = sin  2x    2 tan −1 x = sin −1  1 + x 2 



= sin −1



= sin −1

i.e., AX = B, where

x  6 1 −2 3        A = 1 4 1  , X =  y  and B = 12  1 −3 2   z   1  1 −2 3

Now, A = 1 4 1 1 −3 2 = 1[8 + 3] + 2[2 – 1] + 3[–3 – 4] = 11 + 2 + 3(–7) = 13 – 21 = –8 ≠ 0 Since, A is a non-singular matrix, so the system of equations has unique solution. Now, cofactors of |A| are 1 1+1 4 = [8 + 3] = 11 A11 = ( −1) −3 2 1+2 1 1 = −[ 2 − 1] = − 1 A12 = ( −1) 1 2

1 1

2/5 3969 + sin −1 4225 26 / 25

A21 = ( −1)2+1

−2 3 = − [ − 4 + 9] = −5 −3 2

4225 − 3969 5 + sin −1 4225 13

A22 = ( −1)2+2

1 3 = [ 2 − 3] = −1 1 2

256 5 + sin −1 4225 13

A23 = ( −11)2+3

1 −2 = − [ −3 + 2] = 1 1 −3

16 5 + sin −1 65 13



A31 = ( −1)3+1

[Using sin −1 x + sin −1 y = sin −1 ( x{ 1 − y 2 }+ y{ 1 − x 2 })] 16 15 + sin −1 65 13 2 2  16 5   5  16   1−  + 1−   = sin −1   13   65   13  65  12 5 63  −1  16 = sin  × + ×  65 13 13 65 \ sin −1

 16 ×12 + 315    13 × 65 

A13 = ( −1)1+3

= sin −1 1 − = sin −1



.

cos −1

−1

3  507  = sin −1 = RHS = sin −1   65 ×13  5 4. The given system of equations can be written in matrix form as 1 −2 3   x   6       1 4 1   y  = 12  1 −3 2   z   1 



OR

= sin



4 = [ −3 − 4] = − 7 −3

−2 3 = [ −2 − 12] = −14 4 1

A32 = ( −1)3+2

1 3 = − [1 − 3] = 2 1 1

A33 = ( −1)3+3

1 −2 = [ 4 + 2] = 6 1 4

\ Now,  A11  adj ( A ) =  A12  A13

A21 A22 A23

A31   A32  A33 

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

4  11  =  −1  −7



A −1 =

\

−5 −14   −1 2 1 6 





\

1 adj ( A ) | A|

AX = B



= ( x + y + z )2 [ xy + yz + zx + ( x 2 + y 2 + z 2 )]2



Hence, it is divisible by (x + y + z) and the quotient is ( x + y + z )[ xy + yz + zx − (x 2 + y 2 + z 2 )]2 . OR

       LHL = lim f(x) LHL = lim− f ( x ) p  p at x= x→ 

−1

\

[( yz − y 2 − z 2 + yz − yz + xz + xy − x 2 ]



 11 −5 −14  1   =  −1 −1 2   [| A|= −8] −8  − 7 1 6 

= ( x + y + z )2 [ xy + yz + zx − ( x 2 + y 2 + z 2 )]

 2

X=A B  11 −5 −14   6  1    = −1 −1 2  12  −8   − 7 1 6   1   66 − 60 − 14  1  −6 − 12 + 2  = −8   − 42 + 12 + 6 





2

= lim x→

3 cos2 x

p 2

= lim x→

(1 − sin 3 x )

p 2

(13 − sin 3 x ) 3(12 − sin 2 x )

(1 − sin x )(1 + sin x + sin 2 x ) p 3(1 − sin x )(1 + sin x ) x→

= lim

2

 −8   1  −1         = 8  −16  =  2   −24   3 



1 + sin x + sin 2 x p 3(1 + sin x ) x→

= lim

2

yz − x 2

zx − y 2

xy − z 2

p p    Put x = − h, As x → , h → 0 2 2 p  p  1 + sin  − h + sin 2  − h 2  2  = hlim →0   p 3 1 + sin  − h    2 

2 5. LHS = zx − y

xy − z 2

yz − x 2

xy − z 2

yz − x 2

zx − y 2



 x  1      y  =  2   [by multiplication property]  z   3 



Þ



Hence,   x = 1, y = 2 and z = 3.

Apply R2 → R2 – R1 and R3 → R3 – R1, 2

yz − x = ( x − y )( x + y + z ) ( x − z )( x + y + z )

2

2

xy − z zx − y ( y − z )( x + y + z ) ( z − x )( x + y + z ) ( y − x )( x + y + z ) ( z − y )( x + y + z )



Taking (x + y + z) common from R2 and R3,



yz − x 2 = (x + y + z) x − y x−z



Apply C1 → C1 + C2 + C3,

2

zx − y 2 y−z y−x

xy − z 2 z−x z−y

( xy + yz + zx ) − ( x 2 + y 2 + z 2 ) zx − y 2 xy − z 2 = (x+ y+ z)



2

0 0

y−z z−x y−x z−y

RHL = lim f ( x )

p   at x=  2



[( y − z )( z − y ) − ( z − x )( y − x )]

p   x →  2

= lim

q(1 − sin x )

p x→ 2

( p − 2 x )2

p π    Put x = + h , As x → , h → 0 2 2 p   1 − sin  + h 2  = q hlim 2 →0  p   p − 2  + h  2  

= q lim



1 − cos h

= q lim h→0 4h2 h  sin q  2 × 1 = q = lim   4 8 h h → 0 2    2  h→0

Expanding along C1,

= ( x + y + z )2 [ xy + yz + zx − ( x 2 + y 2 + z 2 )]

1 + cos h + cos2 h 1 + 1 + 1 1 = = h→0 3(1 + 1) 2 3 [1 + cos h ]

= lim

2 sin 2 4h2

h 2

Solutions

5

 p Also, f   = p  2



Þ

 p Since, f(x) is continuous at x = f    2

dy y 1 = −  log a ⋅ dt 2  1 − t2



\

dy y 2 1 − t2 y =− ⋅ =− . dx x 2 x 1 − t2

 p \ LHL = RHL = f    2 1 q \     = = p 2 8 1 Þ  q = 4 and p = . 2 6. Differentiating both sides of the given relation with respect to x, we get

{ (

d         log x 2 + y 2 dx

(

)} = 2 dxd tan 

)

1 d 2 Þ 2 × x + y2 = 2 2 dx x +y

( )

−1

 x    y   

d  y 2 dx   x   y 1+    x 1

( )

1 d 2  d 2 x + y  Þ 2 2  dx dx x +y     dy  x 2   x dx − y ×1  =2 2   x2  x + y2      dy  1 2   dy  Þ  − y 2 x + 2 y  = 2 x dx  x + y 2  dx x2 + y2   dy    dy  Þ         2 x + y  = 2 x − y dx dx     dy dy x+ y = x −y Þ dx dx dy Þ ( y − x ) = − (x + y ) dx dy x + y . Hence proved. = Þ dx x − y −1

7. x = asin t Squaring on both sides −1 x 2 = asin t Taking log on both sides, 2 log x = sin t log a



Þ Þ

2 dx 1 = log a ⋅ x dt 1 − t2 dx x  1 =  log a dt 2  1 − t2 −1

y = acos t Similarly Þ   2 log y = log a cos–1t



  

y = 3x − 2

Diff. w.r.t ‘x’,



dy 3 = dx 2 3x − 2



Since the tangent is parallel to the line



4x – 2y + 5 = 0 dy a \ =− =2 dx b 3 \ =2 2 3x − 2



Þ



Þ





dy    As, dx = 2

3 = 4 3x − 2

3 4 On squaring both sides, we get 9 Þ 3x − 2 = 16 9 3x = + 2 Þ 16 41 3x = Þ 16 41 x= Þ 48

When

3x − 2 =

x=

41 3 , y= 48 4

 41 3  \ The point is  ,   48 4  The eqn. of tangent is 3 41   y − =2x −  Þ  48  4



4 y − 3 2( 48 x − 41) = 4 48 Þ 24y – 18 = 48x – 41



Þ 48x – 24y = 23



Þ 48x – 24y – 23 = 0.



−1



8. Given,

  

Þ

9. Given equation is dy cos2 x + y = tan x dx dy y tan x + = dx cos2 x cos2 x 

[dividing both sides by cos2x]

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

6

dy + sec 2 xy = tan x . sec 2 x dx On comparing above equation with linear dy differential equation + Py = Q , we get dx 2 P = sec x and Q = tan x. sec2x

Þ

P dx sec = e∫ I.F. = e ∫ \ The solution is

2

x dx

= e tan x

y × I .F. = ∫ (Q × I .F.) dx + C   tan x

= ∫ tan x . sec x e 2

tan x



Þ ye



Put   tan x = t ⇒ sec 2 x dx = dt



\



\







Þ



Þ



Þ 



Þ

dx + C

II

 d(t ) t  ye tan x = t ∫ e t dt − ∫  ∫ e dt  dt + C  dt [using integration by parts]



= t e t − ∫ 1 . e t dt + C



= t e t − e t + C = e t (t − 1) + C

dv 2

2

=

1 dx 2∫ x

1  1  v −  −   2 2 1 1 v− − 1 2 2 = 1 log x + C Þ   log 1 1 1 2 2× v− + 2 2 2

ye tan x = ∫ t . e t dt + C I

∫

 1 x−a  dx = log  ∫ 2 2 2a x + a  x −a  v −1 1 log = log x + C 2 v y −1 y 1  x log = log x + C put v =  y x 2   x y−x 1 log = log x + C. 2 y

10. Suppose ABC is a right angled triangle whose area is S.

ye tan x = e tan x (tan x − 1) + C.[put t = tan x ] OR dy y y 2 = + ...(i) dx x x 2 It is a homogeneous differential equation. Put y = vx

Given : 2

dy dv = v+x dx dx \ Equation (i) becomes, dv   2 v + x  = v + v2 dx   dv 2v + 2x = v + v2 Þ dx dv 2x = v + v2 − 2v Þ dx dv 2x = v2 − v Þ dx dv 1 dx Þ = × 2 v −v 2 x Then,



On integrating both sides, we get dv 1 dx ∫ v2 − v = 2 ∫ x dv 1 dx = ∫ Þ∫ 1 1 v  2  2 x  v − 2 . + −  2 4 4

1 × Base × Height 2 1 …(i) = × y×x  2 Circumscribed circle of DABC is one which passes through points A, B and C. Let O be the centre of circumscribing circle. \ OA = OB = OC Þ OA = OC Since, O is the mid-point of AC. 1 OA = OC = AC \ 2 Now, radius of circumscribing circle 1 = × (Diameter) 2 1 1 = × AC = × x 2 + y 2 2 2  [using Pythagoras theorem] Area of circumscribing circle (A) = p × (Radius of circle)2

S=

2



1 2  = p× x + y2  2  p 2 …(ii) = (x + y2 )  4

Solutions

7

2S  [from eq. (i)] x On putting the value of y in Eq. (ii), we get y=

Now,

p 4S2  A =  x2 + 2  4 x 



On differentiating both sides w.r.t. x, we get  − 8S 2   dA p  = 2x +  3   dx 4   x   p 8S 2  =  2x − 3  4 x  dA must be zero. dx p 8S 2  2 x −  =0 4  x3 

For A being least,

i.e.,



Þ



Þ



Þ

x−

4S

2

=0 x3 x4 = 4S2 1 S = x 2 [taking square root] 2

1 On putting S = x 2 in eq. (i), we get 2 1 2 1 x = xy 2 2 Þ x=y Hence, area of circumscribed circle is least, when x = y.

11. I = ∫ ( 2 sin 2 x − cos x ) 6 − cos 2 x − 4 sin x dx       = ∫ ( 4 sin x − 1)

( sin x − 4 sin x + 5 ) cos x dx 2

      = ∫ ( 4t − 1) t 2 − 4t + 5 dt , where sin x = t

Þ cos x dx = dt

      = 2 ∫ ( 2t − 4 ) t 2 − 4t + 5 dt + 7 ∫ (t − 2 )2 + 1 dt Let m = t2 – 4t + 5 Þ dm = (2t – 4) dt On solving, we get



=2

2

3/2

(t − 4t + 5) t − 2 2 +7  t − 4t + 5 3  2 2 1  + log (t − 2 )+ t 2 − 4t + 5  + C 2 

3/2 4  (sin x − 2 ) = sin 2 x − 4 sin x + 5  + 7  3 2 

1 sin 2 x − 4 sin x + 5 + log|(sin x − 2 ) 2  + sin 2 x − 4 sin x + 5 + C  . 

p

12. I = ∫ p2 cos 2 x log(sin x ) dx 4

p

p

1 1 2 =  (log sin x ) ⋅ sin 2 x  − ∫ p2 cot x ⋅ sin 2 xdx p 2 2  4 4

 1  1  = 0 − log  −  2   ∫ 2 

p 2 2 p cos 4

x dx

p

1 1 = log 2 − ∫ p2 (1 + cos 2 x ) dx 4 2 4

p

1 1 sin 2 x  2 = log 2 −  x + 4 2 2  p 4

1 1  p   p 1  = log 2 −  + 0 −  +     4 2  4 2  2 1 p 1 = log 2 − + . 4 8 4 Let

I =∫

p

OR 2 x(1 + sin x )

dx 1 + cos2 x p p 2 x sin x 2x = ∫ dx + ∫ dx − p 1 + cos 2 x − p 1 + cos 2 x = I1 + I2 I1 = 0 (Being an odd function) p 2 x sin x I 2 = 2 ∫ dx 0 1 + cos 2 x  (Being an even function) \ I = I2 p 2 x sin x = 2 ∫ dx 0 1 + cos 2 x p x sin x = 4 ∫ dx 0 1 + cos 2 x p x sin x Let I3 = ∫ dx 0 1 + cos 2 x

−p

Apply property



=∫



=∫

a

p (p − x)

0

a

∫0 f ( x )dx = ∫0 f ( a − x )dx , sin( p − x )

1 + cos2 ( p − x )

p ( p − x )sin x

dx

dx 1 + cos2 x p π = ∫ p sin x dx − ∫ x sin x dx 0 1 + cos 2 x 0 1 + cos 2 x p sin x = p∫ dx − I 3 0 1 + cos 2 x p sin x \ 2 I 3 = p∫ dx 0 1 + cos 2 x 0

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

8

Putting cos x = t Þ –sin xdx = dt When x = 0; t = 1 and x = p; t = –1 −1 dt 2 I 3 = − p∫ 1 1+ t 2 b a = p[tan −1 x]1−1   ∫ f ( x )dx = ∫ f ( x )dx  0 0  

p2 = 2

p 4 Hence, I = p2.

Þ I 3 =

13. Given, Mean = np = 4 and variance = npq = 2 npq 2 = \ np 4 Þ



A wins, if he gets a ‘six’ in 1st or 4th or 7th… throw. His probability of getting a ‘six’ in first 1 throw is P( E) = . 6 A will get fourth throw if he fails in first, B fails in second and C fails in third throw.

\ Probability of winning of A in fourth throw

1  5 =  ×  6 6

Similarly,

Probability of winning of A in 7th throw 6

1  5 =   × and so on.  3 6





37  1 =   [ 28 + 8 + 1] = .  2 256

8

8

OR

Hence, Probability of winning of A 3

p=

3

1

2

1  2  1 \ P( X = 0 ) = 3 C0     =  3  3 27

B wins if he gets a ‘six’ in 2nd throw or 5th throw or 8th throw … \ Probability of winning of B

4

2



3

=

1

0

2 1 8 P( X = 3) = 3 C3     =  3  3 27 Hence probability distribution of X is :

7

 5 1  5 1  5 1 =   ⋅ +  ⋅ +  ⋅   6 6  6 6  6 6

2 1 2  2  1 P( X = 1) = 3 C1     = 3 × × =    3  3 3 9 9 4 1 4  2  1 P( X = 2 ) = 3 C2     = 3 × × =  3  3 9 3 9

6

1  5 1  5 1 +  ⋅ +  ⋅ +  6  6 6  6 6 1 36 6 = = 3 91  5 1−   6 =



Given that, total eggs = 15

10 2 =  (10 good eggs) 15 3 5 1 q= = , n=3    (5 bad eggs) 15 3 Let X denotes the number of good eggs in 3 draws. Then X can take values 0, 1, 2 and 3.



) () () ()

 

8

0

1 1 5 and P( E) = 1 − = 6 6 6

3

 1  1  1 = 8 C 6   + 8 C7   + 8 C 8    2  2  2



3 8/27

(





2 4/9

         P( E ) =

q=

8

1 2/9

= P E Ç E Ç E Ç E = P E P E P E P (E)

1 2 and p=1–q 1 1 =1 − = 2 2 1 Since, Mean = n × = 4 Þ n = 8 2 \ Required Probability

0 1/27

14. Let E be the event of ‘getting a six’ in a single throw of on unbaised die. Then,



2

X P(X)

 5 1   ⋅ 6 6  5 1−   6

3

=

30 91

Hence,

 36 30  Probability of winning of C = 1 −  +   91 91  =

25 . 91

Solutions

9 OR

Section ‘B’



15. We have, →



We have, →

→

   [ a − b

→

3a=2 b

→ →  Þ 3 ( x + 4 y ) a + ( 2 x + y + 1) b    → →  = 2 ( −2 x + y + 2 ) a + ( 2 x + 3 y − 1) b    →

Þ ( 3x + 12 y + 4 x − 2 y − 4 ) a

→

→

+ (6x + 3 y + 3 − 4 x + 6 y + 2) ⋅ b = 0



→

→

→

→

→

b− c

→ →  → → → → = ( a − b )×( b − c ) ⋅ ( c − a )     →

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→ → →

→

→

…(i)

2

 →  →  → Þ l | r | + m| r | + n| r | = 6 2 + ( −3)2 + 2 2       Þ  | r |2 (l 2 + m 2 + n 2 ) = 36 + 9 + 4 →

Þ           | r |2 = 49 ( l2 + m2 + n2 = 1) →

Þ          | r |= 7 →

Putting | r |= 7 in eq. (i), we obtain that the → 6 2 −3 direction cosines of r are l = , m = , n = . 7 7 7



→

→

→

→

→

→

→

→ → →



Þ   | a + b |2 =| c |2



Þ   | a |2 +| b |2 + 2 a ⋅ b =| c |2



Þ  | a |2 +| b |2 +2 | a |⋅| b |cos q =| c |2



Þ



Þ



Þ

→

→

→

→

→

→

→ →

→

→

→

1 2 p q= . 3

→

→

→→ →

[ Scalar triple product when any two vectors are equal is zero] → → →

→ → →

=[ a b c ] − [ a b c ]   → → →   →→ → [ b c a ] =[ a b c ]  

   = 0. 17. Clearly, height h of D ABC is the length of perpendicular from A(1, –1, 2) to the line x+2 y − 1 z − 0 which passes through = = 2 1 4 → P(–2. 1, 0) and is parallel to b = 2i + j + 4 k →

h=

→

| PA × b | →

A(1,–1,2)

h C

P(–2,1,0) M

B →

→

Now, PA = −3 i + 2 j − 2 k and b = 2i + j + 4 k

9 + 25 + 2( 3)( 5)cos q = 49 cos q =

→

→→ →

→

Þ    ( a + b ) ⋅ ( a + b ) =( − c ) ⋅ ( − c ) →

→

→

|b|



→

→

→ →→

→

→

→

→→ →

a + b =− c

Þ

→

− [ c a a ]+[ b c c ] − [ b c a ]



a+b+c =0

16. We have,

→

→ → →

→

→

→

→ → →

=[ a b c ] − [ b c a ]  

→

→

=[ a b c ] − [ a b a ]+[ c a c ]

the coordinate axes are l | r |, m| r |, n| r |, respectively

→

→

− ( c × a ) ⋅ a +( b × c ) ⋅ c − ( b × c ) ⋅ a

given vector r (say). Then, its projections on

2

→

→

 

2

→

[ b × b = 0 ]

 

→

→

→

 →

Let l, m, n be the direction cosines of the

→

→

=( a × b ) ⋅ c − ( a × b ) ⋅ a +( c × a ) ⋅ c

Þ 7 x + 10 y − 4 = 0  and  2 x + 9 y + 5 = 0 Þ x = 2, y = –1. OR

→

→

=( a × b + c × a + b × c ) ⋅ ( c − a )  

→

\      l | r |= 6 , m| r |= −3, n| r |= 2 

→

=( a × b − a × c − b × b + b × c ) ⋅ ( c − a )  

Þ (7 x + 10 y − 4 ) a + ( 2 x + 9 y + 5) b = 0

→

→

c − a]



\

i j k PA × b = −3 2 −2 2 1 3 →

→

= 10i +8 j − 7 k

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

10

→

→

Þ     | PA × b |= 100 + 64 + 49 = 213 →



\

h=

→

| PA × b | →

C

y =4

B

y=1

A X'

–2 3

1 1 71 \ Area of DABC= ( BC × h ) = × 5 × 2 2 7 1775 = sq. units. 28 OR We know that the shortest distance between →

→

→

→

→

→

the lines r = a1 + lb1 and r = a2 + µb2 is given by →

d=

→

→

→

→

→

( a2 − a1 ) ⋅ ( b1 × b2 ) | b1 × b2 |

comparing the given equations with the →

→

→

equations r = a 1 + lb1 respectively.

D (0, 4)

(0, 1)

71 7 It is given that the length of BC is 5 units =



y=9x2

213 21

=

|b |



Y

→

and      | b |= 4 + 1 + 16 = 21

→

→

→

→

→

r = a2 + lb2 ,

and →

a1 = 4i − j , a2 = i − j + 2 k , b1 = i + 2 j − 3k and

–1 3

At x = 0, At y = 1, Þ



and at y = 4,



Þ

→

\      a2 − a1 = −3i + 0 j + 2 k and i j k b1 × b2 = 1 2 −3 = 2i + − j + 0 k 2 4 −5 →

             →

→

→

→



= –6 + 0 + 0 = –6 →

→

and b1 × b2 = 4 + 1 + 0 = 5 →

\

Shortest distance =

→

→

→

→

→

( a2 − a1 ) ⋅ ( b1 × b2 ) | b1 × b2 |

=

−6 5

=

6 5

.

18. Given, y = 9x2 y x2 = i.e., 9 It represents a parabola with vertex at (0, 0)

1 3

x2 =

4 9

4

= ∫ xdy = ∫



1

4

1

y1 / 2 dy 3

3 / 2 4

1 y =   3  3 / 2  1 2 2 3/2 = ( 4 − 1) = ( 8 − 1) 9 9 14 sq units. = 9





→

\ ( a2 − a1 ) ⋅ ( b1 × b2 ) = ( −3i + 0 j + 2 k ) ⋅ ( 2i − j + 0 k )

Y'

y=0 1 x2 = 9

2 3 \ Required area = Area (ABCDA)



→

X

x=±

→

b2 = 2i + 4 j − 5k

2 3

1 3

x=±





O

Section ‘C’ 19. Let VAB be a conical funnel of semi-vertical p angle . At any time t the water in the cone 4 also forms a cone. Let r be its radius, l be the slant height and S be the surface area. Then, p VA ′ = l , O ′A ′ = r and ∠A ′VD′ = . 4 In D VO’A’, we have p VO ′ VO ′ p O ′A ′ O ′A cos = and sin = = = 4 VA ′ l 4 VA ′ l p p and O ′A ′ = l sin 4 4 The surface area S of the conical funnel is given by S = p (O ′A ′ ) (VA ′ ) Þ

VO ′ = l cos

Solutions

11 A

B

O A’

O’ ∏4 ∏4

B’

V



p  Þ   S = p  l sin  l  4



[ using: S = prl ]

p pl 2 = pl 2 sin = 2 2

Þ

ds 2 pl dl = dt 2 dt



Þ

−2 =



Þ

dl − 2 = dt pl



dl − 2 Þ   = cm/sec.  dt  l=4 4p

2 pl dl  ⋅ 2 dt

 ds  2  dt = −2 cm /sec   



d ( 600 − 4 x ) = − 4 dx d   Slope of MR = ( 600 − 8 x ) = − 8 dx \  Slope of MR = 2 × Slope of AR Slope of MR . or  Slope of AR = 2 OR Here, C = a + bx + cx2 Quantity = x    Slope of AR =

\ We know that, AC =

C a + bx + cx = x x

1 a ( MC − AC ) = c − 2  x x d d a  ( AC ) =  + b + cx   dx dx  x d a ( AC ) = − 2 + c  dx x From (i) and (ii), we get \



Thus, the rate of decrease in the slant height is 2 cm/sec. 4p 20. Given demand function p = 600 – 4x We know that Revenue function = px = (600 – 4x)x = 600x – 4x2 (a) Now, Marginal revenue function, d dR = ( 600 x − 4 x 2 )    MR = dx dx \ MR = 600 – 8x (b) Average revenue function R AR = x 600 x − 4 x 2 \    AR = x = 600 – 4x

2

1 a =  cx −  x x







a = + b + cx x d d and MC = (C ) = ( a + bx + cx 2 ) dx dx = b + 2cx 1 1 a  \ ( MC − AC ) =  b + 2 cx − − b − cx   x x x

…(i)

…(ii)

1 d ( AC )= ( MC − AC ) dx x Hence proved.

21. Given, Regression line y on x : y = x + 5 …(i)       Regression line x on y : 16x = 9y + 95…(ii) from eq. (i), bxy = coefficient of x = 1, positive 9 from eq. (ii), byx = coefficient of y = , positive 16 \ rxy = byx × bxy

= 1×

s r x = bxy Again,   sy sx =  

9 3 = = 0.75  16 4

bxy × s y g

=

9 4 × 16 3 4

4 9 = ×4× =3 16 3   OR Regression equation of x on y is : 3y – 5x + 180 = 0 Þ 5x = 3y + 180 3 x = y + 36  Þ 5 In Standard form s x − x = r x (y − y) sy

Þ

x=r

…(iii)

sx rs y+x − x y  sy sy

On comparing eq. (i) and (ii), we get, s 3 r x=  sy 5

…(i)

…(ii)



…(iii)

OSWAAL ISC Sample Question Papers, MATHEMATICS, Class-XII

12 and x − r

sx y = 36  sy

…(iv)

3 3 from eq. (iii),  r = 4 5 9  Given :  s 2x = s 2y 16 s 2x s 9 3 \   = Þ x= 2 16 4 s sy y

3 4 4 …(v) r = × = = 0.8  5 3 5 Again from eq. (iv) 3 x = y + 36 5 3 Þ x = × 44 + 36 = 26.4 + 36 = 62.4. 5  22. Let there be x units of food X and y units of food Y. Min Z = 24x + 36y Subject to the constraints x + 2y ≥ 10



2x + 2y ≥ 12 3x + y ≥ 8 x ≥ 0, y ≥ 0 x

y

10 0 1 2

0 8 5 4

Z (cost) 240 288 204 192

[ISC Marking Scheme, 2015]

Examiner's Comment Many candidates took incorrect inequality sign, hence they got incorrect feasible region and their corner points were also incorrect. Some candidates did not show any graphical representation of the inequalities. In some cases, the representation of the problem was not up to the mark, and the work was not systematic resulting in candidates missing the point of minimum cost.

ANSWERING TIP...

Students shared practice sketching of lines. They should practice to express the line in intercept from i.e. x/a + y/b = 1, so that sketching will be easy. Correct feasible region and its plotting is important.



Detailed Answer : Let x kg of food X and y kg of food Y are mixed together to make the mixture. Since one kg of food X contains one unit of vitamin A and one kg of food Y contain 2 units of vitamin A. Therefore, x kg of food X and y kg of food Y will contain x + 2y units of vitamin A. But the mixture should contain at least 10 units of vitamin A. Therefore, x + 2y ³ 10 Similarly, x kg of food X and y kg of food Y will produce 2x + 2y units of vitamin B and 3x + y units of vitamin C. But the minimum requirement of vitamins B and C are respectively of 12 and 8 units. \ 2x + 2y ³ 12 and 3x + y ³ 8 Since the quantity of food X and food y cannot be negative \ x ³ 0, y ³ 0 It is given that one kg of food X costs `6 and one kg of food Y costs `10. So, x kg of food X and y kg of food Y will cost ` (6x + 10y). Thus, the given linear programming problem is Minimize, Z = 6x + 10y Subject to, x + 2y ³ 10 2x + 2y ³ 12 3x + y ³ 8 and x ³ 0, y ³ 0 To solve this LPP, we draw the lines x + 2y = 10, 2x + 2y = 12 and 3x + y = 8 The feasible region of the LPP is shaded in the figure.

The coordinates of the vertices (corner-points) of shaded feasible region A1 P1 P2 B3 are A1(10, 0), P1(2, 4), P2(1, 5) and B3(0, 8). These points have been obtained by solving the equation of the corresponding intersecting lines, simultaneously.

Solutions

13 Y

B3 (0,8) 52 ) B2(0,6) (0, 10

P2(1,5)

B1 (0,5)

P1(2,4)

x+ 2

3

12 = 2y

y= 3x+

A3 ( 8 ,0)

+ 2x

0

y= 1

0 ( 52, 6 0)

6x A2 (6 +1 , 0y 0) =5 2

X A( 1 10, 0)

8



The values of the objective function at these points are given in the following table :

Point (x, y)

Value of the objective function Z = 6x + 10y

A1(10, 0)

Z = 6 × 10 + 10 × 0 = 60

A2(2, 4)

Z = 6 × 2 + 10 × 4 = 52

P2(1, 5)

Z = 6 × 1 + 10 × 5 = 56

B3(0, 8)

Z = 6 × 0 + 10 × 8 = 80

Clearly, Z is minimum at x = 2 and y = 4. The minimum value of Z is 52. We observe that the open half plane represented by 6x + 10y < 52 does not have points in common with the feasible region. So, Z has minimum value equal to 52. Hence, the least cost of the mixture is `52. 

nnn