Q. NO . 21.

( INVERSE TRIGONOMRTRIC FUNCTION) 2 mark

1.Write cot-1(
1
) in simplest form. 2. Prove that 2Sin-1 x = Sin-1 (2x √ 1−x 2 )
√ x −1
2
Put x = sin θ or θ = sin x -1

Put x = sec θ or θ = sec -1 x √ 1−x 2 = √ 1−sin2θ = cos θ

√ x2−1 = √ sec2θ −1 = tan θ RHS = Sin-1 (2x √ 1−x 2 )
1 1 = Sin-1 (2 sin θ cos θ)
cot-1( ) = cot-1( )
√ x −1
2 tan θ =Sin-1 ( sin 2 θ)
-1
= cot ( cot θ ) = 2 θ = 2 sin -1 x = LHS
= θ = sec -1 x 7π
4. Find the value of cos-1 (cos )
6
3. Prove that 2Cos x = Sin (2x √ 1−x )
-1 -1 2
7π 5π 5π
Cos = cos ( 2π - ) = cos
Put x = cos θ or θ = cos x -1 6 6 6

√ 1−x 2 = √ 1−cos 2θ = sin θ 7π 5π 5π

cos-1 (cos ) = cos-1 (cos ) =
RHS = Sin-1 (2x √ 1−x 2 ) 6 6 6
3 π
= Sin-1 (2 cos θ sin θ) 5. Find the value of tan-1 (tan )
4
3π
=Sin-1 ( sin 2 θ) Tan = tan ( π + (- π ) )
4 4
= 2 θ = 2 cos -1 x = LHS = tan (- π )
1 4
6. Find the value of tan-1 (2cos(2sin-1 ) 3π
2 tan-1 (tan ) = tan-1 [tan (- π ) ] = - π
4 4 4
tan-1 (2cos(2sin-1 1 ) = tan-1 (2cos(2 π )) 1
2 6 7. Write the simplest form of tan-1 ( )
-1
= tan (2cos π ) √ 2
x −1
3 Put x = sec θ or θ = sec -1 x
-1 1 -1 π √ x2−1 = √ sec 2θ −1 = tan θ
= tan (2 ) = tan (1)=
2 4 1 1
tan-1 ( ) = tan-1 ( )
√ tan θ
√
2
1−cos x x −1
8.Write the simplest form of tan-1 ( )
1+cos x = tan -1 (cot θ )= tan -1 [ tan ( π - θ) ]

√
2
2 x = π -θ = π - sec -1 x
2 sin
√ 1−cos x 2 2 2
Tan-1 ( )= tan-1 ( )
1+cos x
2 cos
2 x
9. Prove that 2 sin-1 3 = tan -1 24
2 5 7
x x 3 3 3
= tan -1 ( tan )= Let sin-1 = θ => sin θ = => tan θ =
2 2 5 5 4
10. Find the value of tan ( √ 3 ) - sec (-2)
-1 -1
2 tan θ 2.3 / 4 24
we know tan 2θ = = =
Let tan -1 ( √ 3 ) = π and sec-1 (-2) = π - π 1−tan2 θ 1−9/16 7
3 3
tan -1 ( √ 3 ) - sec-1 (-2)= π - (π - π ) 24 3 24
3 3 2θ = tan -1 or 2 sin-1 = tan -1
− π 7 5 7
=
3

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 22.( DETERMINANTS ) 2 mark

1. Find the area of the triangle whose vertices are 2. Find the area of the triangle whose vertices are
(3, 8) , (-4, 2) and (5, 1). (1, 0) , (6, 0) and (4, 3).

| | | | | | | |
x1 y1 1 3 8 1 x1 y1 1 1 0 1
1 1 1 1
Let ∆= x2 y2 1 = −4 2 1 Let ∆= x2 y2 1 = 6 0 1
2 2 2 2
x3 y3 1 5 1 1 x3 y3 1 4 3 1

1 1 15
= [ 3(2-1)- 8(-4-5) + 1(-4-10)] ∆ = [ 1(0-3)- 0 + 1(18-0)] = sq units
2 2 2
1 61 4. Find the area of the triangle whose vertices are
∆= [3+72-14 ] = sq units
2 2 (2, 7) , (1, 1) and (10, 8).

| | | |
3.Find the area of the triangle whose vertices are x1 y1 1 2 7 1
1 1
(-2, -3) , (3, 2) and (-1, -8). Let ∆= x2 y2 1 = 1 1 1
2 2
x3 y3 1 10 8 1

| | | |
x1 y1 1 −2 −3 1
1 1 1 47
Let ∆= x2 y2 1 = 3 2 1 ∆ = [ 2(1-8)- 7(1-10) + 1(8-10)] =
2 2 2 2
x3 y3 1 −1 −8 1
5. Find the equation of the line joining the (1, 2) and
1 (3, 6) using determinants.

| |
= [ -2(2+8)+3(3+1) + 1(-24+2)] 1 2 1
2
Let ∆ = 0 => 3 6 1 = 0
1 30 x y 1
∆= [-20+12-22 ] = =15 sq units
2 2 1( 6 – y ) -2 ( 3 – x ) + 1 ( 3y – 6x) = 0
6 -6y -6 + 2x + 3y – 6x =0
6. Find the equation of the line joining the (1, 3) -4x -3y = 0 or 4x + 3y = 0
and (0, 0) using determinants.
7. If the area of the triangle whose vertices are

| |
1 3 1 (-2 ,0) ,( 0,4) and (0, k) is 4 sq. units. Find value of k.
Let ∆ = 0 => 0 0 1 =0

| |
−2 0 1
x y 1 Given ∆ = ± 4 => 0 4 1 =±8
0 k 1
1( 0 – y ) -3 ( 0 – x ) + 1 ( 0 – 0) = 0
- y + 3x =0 OR y = 3x -2(4 – k ) - 0 + 1(0 – 0 ) = ± 8 => -8 + 2k = ± 8
8. Find the equation of the line joining the (3, 1) k=8 k=0
and (9, 3) using determinants. 9. Find the value of k if the area of the triangle is 4 sq.
units and vertices are ( k, 0) , (4, 0) and (0,2).

| |
3 1 1

| |
Let ∆ = 0 => 9 3 1 =0 k 0 1
1
x y 1 Given ∆ = ± 4 => 4 0 1 =±4
2
0 2 1
3( 3– y ) -1( 9 – x ) + 1 ( 9y– 3x) = 0 k(0 – 2) - 0 + 1(8 – 0 ) = ± 8
9 -3y – 9 + x + 9y - 3x =0 -2k + 8 = ± 8
-2x +6y = 0 or x – 3y = 0
=> -2k +8= 8 or - 2k+8 = -8
=> k = 0 OR k=8

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 32.( RELATIONS AND FUNCTIONS ) 3 mark

1. A relation R on the set A = {1, 2, 3......14} is defined 2. A relation R in the set N of natural number defined
as R = {(x, y) : 3x – y =0}. Determine whether R is as R = {(x, y) : y = x + 5 and x <4}. Determine whether
reflexive, symmetric and transitive. R is reflexive, symmetric and transitive.
R = {( 1 , 3 ) , ( 2 , 6 ) ,( 3 , 9 ),( 4 ,12 )} R = {( 1 , 6 ), ( 2, 7 ) , ( 3 , 8 )}
since (1 ,1)∉R R is not reflexive since (1 ,1)∉R , R is not reflexive
Let (1 ,3)∈R but (3 , 1)∉R Let (1 ,1)∈R but (6 , 1)∉ R
There fore R is not symmetric There fore R is not symmetric
Let (1 ,3)∈R and (3 , 9)∈R but (1 , 9)∉ R By definition of transitivity
There fore R is not transitive R is transitive
3. Determine whether R, in the set A of human beings 4. Relation R in the set Z of all integers is defined as
in a town at a particular time is given by R = {(x, y) : x – y is an integer}. Determine whether R
R = {(x, y) : x and y work at the same place} is reflexive, symmetric and transitive .
since x and x work at the same place Since x – x = 0 is an integer
..˙ ( x , x)∈ R Hence R is reflexive ..˙ ( x , x)∈ R Hence R is reflexive
Let ( x , y )∈ R => x and y work at same place Let ( x , x)∈ R => x – y is an integer
=> y and x also work at the same place ꞊> y – x is also an integer
=> ( y , x )∈R R is symmetric ꞊> ( y , x )∈ R , R is symmetric
Let ( x , y )∈ R and ( y , z )∈R Let Let (x , y )∈R and ( y , z )∈R
=>x and y work at the same place and
=> x – y is an integer and y – z is an integer
y and z work at the same place
=> x and z work at the same place => x- z is an integer
..˙ (x , z)∈R , R is transitive ..˙ ( x , z)∈R R is transitive
Hence R is an equivalence relation Hence R is an equivalence relation
5. A relation ‘R’ is defined on the set A = {1, 2, 3, 4, 5} 6. Show that the relation R in R, the set of reals
as R = {(x, y) : y is divisible by x}. Determine whether defined as R = {(a, b) : a ≤ b} is reflexive and
R is reflexive, symmetric, transitive . transitive but not symmetric.
Since x is divisible by x itself since a ≤ a is true
..˙ ( x , x)∈ R Hence R is reflexive ..˙ (a , a)∈R Hence R is reflexive
Let (1 ,3)∈R but (3 , 1)∉R Let (a , b)∈R => a≤b
Therefore R is not symmetric => but b ≤ a is not true
Let ( x , y )∈ R and ( y , z )∈R (b , a)∉R , R is not symmetric
꞊> y is divisible by x and z is divisible by y Let (a , b)∈R and (b , c)∈R
꞊>z is divisible by x ꞊> (x , z)∈R
=> a ≤ b and b ≤ c => a≤c
Therefore R is not transitive ..˙ (a , c)∈R R is transitive

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

7. Show the relation R in the set Z of integers given by 8. Show the relation R in the set A={ x ϵ Z: 0 ≤ x ≤ 12 }
R = {(a, b) : 2 divides (a – b)} is an equivalence given by R = {(a, b) : |a – b| is multiple of 4 } is an
relation. equivalence relation.
Since 2 divides a – a = 0 Since |a – a| = 0 is multiple of 4
..˙ (a , a)∈R Hence R is reflexive ..˙ (a , a)∈R Hence R is reflexive
Let (a , b)∈R => 2 divides a – b Let (a , b)∈R => | a – b | is multiple of 4
꞊> 2 also divides b – a ꞊> | b – a | is multiple of 4
꞊> (b , a)∈R => (b , a)∈R Hence R is symmetric
R is symmetric Let (a , b)∈R and (b , c)∈R
Let (a , b)∈R and (b , c)∈R => | a – b | is multiple of 4 and
=> 2 divides a – b and 2 divides b – c => | b – c | is multiple of 4
=> 2 divides a – c => | a – c | is multiple of 4
=> (a , c)∈R R is transitive ..˙ (a , c)∈ R , R is transitive
Hence R is an equivalence relation Hence R is an equivalence relation
9.Show that the relation R in the set A = {1, 2, 3, 4, 5} 10 .Let T be the set of triangles with R – a relation in T
given by R = {(a, b) : |a-b| is even} is an equivalence given by R ={(T1 , T2 ) : T1 is congruent to T2 } Show
relation. that R is an equivalence relation.
since | a – a | = 0 is even Since T1 is congruent to T1
..˙ (a , a)∈R Hence R is reflexive ..˙ (T 1 , T 1 )∈R Hence R is reflexive
Let (a , b)∈R => | a – b | is even Let (T 1 , T 2 )∈R => T1 is congruent to T2
꞊> | b – a | is also even ꞊> T2 is congruent to T1
꞊> (b , a)∈R R is symmetric ꞊> (T 2 , T 1 )∈R ,R is symmetric
Let (a , b)∈R and (b , c)∈R Let (T 1 , T 2 )∈R and (T 2 , T 3 )∈R
=> | a – b| even and => T1 is congruent to T2 and
| b – c | is even T2 is congruent to T3
=> | a – c | is also even => T1 is congruent to T3
=> (a , c)∈R R is transitive => (T 1 , T 3 )∈R R is transitive
Hence R is an equivalence relation Hence R is an equivalence relation
11. If R1 and R2 are 2 equivalence relations on a set, then prove that R1∩R2 is also an equivalence relation .
Since R1 and R2 are two equivalence relations (a, a)ϵR1 and (a, a)ϵ R2 => (a, a)ϵ R1∩R2
Hence R1∩R2 is reflexive
Let (a, b)ϵ R1 and (a, b)ϵ R2 => (b, a)ϵ R1 and (b, a)ϵ R2 => (b, a)ϵ R1∩R2
Hence R1∩R2 is symmetric
similarly, (a,b) ϵ R1∩R2 and (b,c) ϵ R1∩R2 => (a,c) ϵ R1 and (a,c) ϵ R2 S => (a, c) ϵ R1∩R2
Hence R1∩R2 is transitive Hence R1∩R2 is an equivalence relation

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 33.( INVERSE TRIGONOMRTRIC FUNCTION) 3 mark
1. Write the simplest form of 5. Write the simplest form of
3sin -1 x = sin-1(3x – 4x3 ) , x ∈[−1/ 2,1/ 2] cos x−sin x 3π
tan -1 ( ) , −π <x < .
put x = sin θ or θ = sin -1 x cos x+sin x 4 4
RHS = sin-1 (3 x – 4x3) cos x−sin x
Let G = tan -1 ( )
= sin-1 ( 3 sin θ – 4sin 3 θ) cos x+sin x
= sin -1 ( sin 3θ) = 3θ Dividing Nr and Dr by cos x ,
= 3 sin-1 x = LHS 1−tan x
we get G = tan -1 ( )
3sin -1 x = sin-1(3x – 4x3 ) 1+ tan x
2. Write the simplest form of = tan -1 ( tan ( π −x ) )
4
3 cos-1 x = cos-1 ( 4x3 – 3 x ) , x ∈[1/ 2, 1] π
G = ( −x )
put x = cos θ or θ = cos -1 x 4
RHS = cos-1 ( 4x3 – 3 x ) 6. Solve 2 tan -1 (cos x ) = tan -1 ( 2 cosec x )
= cos-1 ( 4cos 3 θ – 3 cos θ) Given 2tan -1 ( cos x ) = tan -1 ( 2 cosec x )
= cos-1 ( cos 3θ) = 3θ = 3 cos-1 x = LHS
3cos -1 x = cos-1( 4x3 –3x) 2 cos x
tan-1 ( 2
) = tan -1 ( 2 cosec x )
3. Write the simplest form of 1−cos x

tan -1 ( √ 1+ x 2−1 ) ,x≠0 2 cos x 2

x =
1−cos x 2
sin x
Let x = tan θ or θ = tan -1 x
2 cos x 2
√ 1+ x 2 = √ 1+tan 2θ = sec θ 2
=
sin x
sin x
tan -1 (
√ 1+ x 2−1 ) = tan -1 ( sec θ −1 ) => cos x = sin x => x = π
x tan θ 4
1−cos θ 4 12 33
= tan -1 ( ) 7. P. T . cos-1 +cos-1 = cos-1
sin θ 5 13 65
2 sin θ 4 12
2
2 Let cos -1 = A and cos -1 =B
= tan -1 ( ) 5 13
2 sin cos θ
θ 4 3
2 2 => cos A = sin A=
−1 5 5
tan x
= = tan -1 ( tan θ ) = θ = and cos B =
12
sin B =
5
2 2 2 13 13
cos x Now cos (A +B ) = cos A. cos B – sin A .sin B
4. Express tan -1 ( ),
1−sin x 4 12 3 5
−3 π = . - .
<x < π in the simplest form . 5 13 5 13
2 2 48 15
sin( π + x) (A + B ) = cos -1 (
65
-
65
)
cos x 2
Let = 4 12 33
1−sin x 1+cos( π + x ) ..˙ cos -1 + cos -1 = cos -1
2 5 13 65
x x 5 3 63
2 sin( π + )cos ( π + ) 8. P. T .sin -1 +cos-1 = tan -1
4 2 4 2 x 13 5 16
= = tan ( π + )
x 4 2 tan A+ tan B
2 cos ( π + )
2
using, tan (A + B ) =
4 2 1−tan A tan B
cos x 5/ 12+ 4 /3 63
tan -1( ) tan (A + B ) = =
1−sin x 1−5 /12 . 4 /3 16
x x 5 3 63
=tan-1(tan ( π + ) )= ( π + ) ..˙ sin -1 +cos-1 = tan -1
4 2 4 2 13 5 16
[ Do as in Q 7 ]

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 34.( MATRICES ) 3 mark
1. For any square matrix A with real number 2. If A and B are invertible matrices of same
entries ,A+A1 is a symmetric matrix and A- A1 is order , then (AB) -1 = B -1 A -1 .
a skew - symmetric matrix. We know ( AB) (AB) -1 = I
Let B = A+A1 then B 1 = ( A+A1 )1
= A1 + (A1)1 pre multiplying A -1 on both sides
= A1 + A A -1 ( AB) (AB) -1 = A -1 I
= A + A1
( A -1 A)B (AB) -1 = A -1
B1 = B
Therefore B = A+A1 is a symmetric matrix I B (AB) -1 = A -1

Now let C = A - A1 then C1 = ( A - A1 )1 B (AB) -1 = A -1

= A1 - (A1)1 pre multiplying B -1 on both sides
= A1 - A B -1 B (AB) -1 = B -1 A -1
1
= -(A - A )
C1 = - C I (AB) -1 = B -1 A -1
Therefore C = A - A1 is a skew-symmetric matrix Hence (AB) -1 = B -1 A -1
3. If A and B are symmetric matrices of same 4. If A and B are symmetric matrices , prove
order, then AB is symmetric if and only if A and B that AB - BA is a skew – symmetric matrix .
commute, that is AB = BA . A and B are symmetric matrices, A1 = A ,B1 = B
Since A and B are symmetric matrices, A1 = A , B1 Let C = AB – BA => C 1 = (AB – BA ) 1
=B
1
= (AB) 1 - (BA) 1
Let AB is symmetric , then (AB) = AB
1 1 1
= B1 A1 - A1 B1
But (AB) = B A = BA
= BA - AB
Therefore BA = AB
= - ( AB – BA )
Conversely , if AB = BA then
=- C
( AB ) 1 = ( BA) 1
..˙ C = A B- BA is a skew-symmetric matrix .
( AB ) 1 = A1 B 1 = AB
This shows that AB is symmetric .

5. Express A = [31 −15 ] as the sum of

6. Express A = [−11 52] as the sum of
symmetric and skew-symmetric matrix .
symmetric and skew-symmetric matrix .
Given A = [
3 5
1 −1 ]A1 =
3 1
5 −1 [ ] Given A = [
1 5
−1 2 ]
A1 =
1 −1
5 2 [ ]
A + A1 = [66 −26 ] and A - A1 = [−40 40] A + A1 = [42 44] and A - A1 = [−60 60]
1 1
A+ A A− A
we know A = + A+ A
1
A− A
1
2 2 we know A = +
2 2

[31 −15 ] = [33 −13 ] + [−20 20] [−11 52] = [12 22] + [−30 30 ]
↑ symmetric ↑ skew-symmetric
↑ symmetric ↑ skew-symmetric

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 23 , 35 &35.( CONTINUITY AND DIFFERENTIABILITY) 8 mark
dy dy
1. If 2x + 3y = sin x find 2. If 2x + 3y = sin y
dx dx
2x +3y = sin x 2x +3y = sin y
dy dy dy
2+3 = cos x 2+3 = cos y
dx dx dx
dy cos x−2 dy dy 2
= 2 = (cos y – 3) ═> =
dx 3 dx dx cos y−3
dy dy
3. If ax + by2 = cos y find 4. If xy + y2 = tan x + y find
dx dx
ax + by2 = cos y xy + y2 = tan x + y
dy dy dy dy dy
a + 2by = - sin y x + y.1 + 2y = sec2 x +
dx dx dx dx dx
dy dy
(2by + siny ) = -a ( x + 2y -1 ) = sec2 x - y
dx dx
dy −a dy sec 2 x− y
= =
dx 2 by +sin y dx x +2 y−1
dy dy
5. If x2 + xy + y2 = 100 find 6. If sin2 x + cos2 y = 1 find
dx dx
2 2
Given x2 + xy + y2 = 100 Given sin x + cos y = 1
dy dy dy
2x +x +y.1 + 2y =0 2 sinx cos x - 2 cos y sin y =0
dx dx dx
dy dy
(x + 2y ) = - 2x – y sin 2x – sin 2y =0
dx dx
dy −2 x− y dy sin 2 x
= =
dx x+2 y dx sin2 y
2x dy
-1 1−x2 dy 8 . If y = sin-1 ( ) find
7. If y= cos ( ) find 1+ x
2
dx
1+ x
2
dx
Put x = tan θ or θ = tan-1 x
Put x = tan θ or θ = tan-1 x
2 tan θ
1−x
-1
2
1−tan θ
2
y = sin -1( ) = sin-1 ( sin 2θ) = 2θ
y = cos ( 2
) =cos-1( ) 1+ tan 2θ
1+ x 1+ tan 2θ
dy 2
= cos-1 ( cos 2θ) = 2θ y = 2 tan-1 x ═> =
dx 1+ x 2
dy 2
y = 2 tan-1 x ═> = dy
dx 1+ x
2 10. If y = xsin x find
dx
1 dy
9. If y = sec-1 ( ) find y = xsin x taking log on both sides
2
2 x −1 dx
Put x = cos θ or θ = cos-1 x log y = sin x log x
1 1
y = sec-1( ) = sec-1 ( ) 1 dy 1
2 cos 2θ −1 cos 2θ = sin x + log x cos x
y dx x
y = sec-1( sec 2θ ) = 2θ
dy −2 dy 1
-1
y = 2 cos x ═> = = xsin x (sin x + log x cos x )
dx x
dx √1−x 2

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

 dy dy
11.If yx = xy find 12. If xy= e( x - y ) find
dx dx
yx = xy taking log on both sides
Given x y = e( x - y ) ---- (1)
x log y = y log x
dy dy
x dy 1 dy x + y .1 = e( x - y ) [ 1 - ] -----(2)
+ log y . 1 = y + log x dx dx
y dx x dx using (1) in (2) we get
dy x y dy dy
( - log x ) = - log y x + y .1 = xy [ 1 - ]
dx y x dx dx
dy
dy y ( y−x log y ) ( x + xy) = xy – y
= dx
dx x (x− ylog x) dy y ( x−1)
=> =
dy dx x( y+1)
13. Find if x = at2 , y = 2at dy
dx 14.Find if x = a cos θ , y = a sin θ
dx
x = at2 , y = 2at
dx dy x = a cos θ , y = a sin θ
= 2at = 2a dx dy
dt dt = -a sin θ = a cos θ
dθ dθ
dy 2a 1
= = dy a cosθ
dx 2 at t = = - cot θ
dx −a sin θ
dy dy
15.Find if x = a(θ+sin θ) ,y = a(1- cos θ) 16.Find if x =a(θ-sin θ) , y =a (1+cos θ)
dx dx
x = a(θ+sin θ) , y = a (1-cos θ ) x= a ( θ- sin θ ) , y = a (1+cos θ )
dx dy dx dy
= a (1 + cos θ ) = a sin θ = a (1 – cos θ) , = a ( - sin θ )
dθ dθ dθ dθ
dy a sinθ dy a (−sin θ )
= = =
dx a(1+cos θ ) dx a(1−cos θ )
−2 sin θ /2 cos θ /2
2 sin θ /2 cos θ /2 2
= - cot θ /2
= 2
= tan θ / 2 2 sin (θ /2)
2 cos (θ /2)
dy
18. Find if x = 2at2 and y = at4
17. Differentiate sin x 2
w.r.t. e cos x . dx
Let A = sin2 x and B = ecos x Given x = 2at2 , y = at4
dA dx dy
= 2 sin x cos x = 4at = 4at3
dx dt dt
dB dy 4at
3
= ecos x(-sin x) = = t2
dx dx 4 at
dA 2 sin x cos x 2 cos x
= = dy
√a √a
−1 −1
(sin x) (cos x)
dB −sin x e
(cos x)
−e
(cos x)
20. Find if x = ,y=
dx

19.Find
dy
dx
if x = 4t , y =
4
t xy = √a(sin
−1
x)
√a
−1
(cos x)
= √ a
(π )
2

4 differentiate w . r. t. x
x = 4t , y=
t
dy dy −y
dx dy −4 dy −1 x + y . 1= 0 OR =
=4 , = dx dx x
dt dt t
2
dx t
2

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 28 ( VECTOR ALGEBRA ) 2mark
1. The projection of the vector i +3 j +7 k on the vector 7 i - ^j +8 k^ .
^ ^ ^ ^
Let ⃗ a = ^i +3 ^j +7 k^ b = 7 ^i - ^j +8 k^
⃗
a . ⃗
⃗ b = 7-3+56 = 60 and |⃗b| = √ 49+1+64 = √ 114
⃗ = ⃗ a .⃗
b 60
Projection of ⃗
a on b =
⃗
|b| √ 114
2. Find the angle ‘θ’ between the vectors a = ^i + ^j - k^ and ⃗
⃗ b = ^i - ^j + k^ .
Given a = ^i + ^j - k^ and b
⃗ ⃗ = ^i - ^j + k^
|⃗a| = √3 , |⃗b| = √ 3 and a⃗ . ⃗b = 1 -1 -1 = -1
a⃗ . ⃗
b −1 −1
W.K.T cos θ = => cos θ = => θ = cos-1 ( )
⃗
|⃗a||b| √ 3 √3 3
3. Find the area of a parallelogram whose adjacent sides are determined by the vectors
a = ^i - ^j + 3 k^ and b
⃗ ⃗ = 2 ^i -7 ^j + k^ .

| |
^i ^i k^
Let ⃗ a X ⃗ b = 1 −1 3 = 20 ^i +5 ^j -5 k^ => |⃗a X ⃗b| = √ 400+25+25
2 −7 1
The area of a parallelogram = |⃗a X ⃗
b| = √ 450 or 15 √ 2 sq .units.
4. If ⃗a = 5 ^i - ^j -3 k^ and ⃗b = ^i +3 ^j -5 k^ , then show that the vectors a + ⃗
⃗ b
and a - ⃗
⃗ b are perpendicular .
Let ⃗ ⃗
a + b =6 i ^ ^j - 8 k^ and ⃗
+2 b = 4 ^i -4 ^j +2 k^
a - ⃗
Now ( ⃗ a + ⃗ b ).( ⃗a ⃗ -
b )= 24 -8 -16 = 0
Hence , the vectors ⃗a ⃗
b and a⃗ - ⃗
+ b are perpendicular .
5. Find the angle between two vectors ⃗a and b⃗ with magnitudes 1 and 2 respectively
and when ⃗a . ⃗b=1 .
Given |⃗a| = 1 , |⃗b| = 2 and ⃗a . ⃗b =1
a . b⃗
⃗ 1 1 π
W.K.T cos θ = => cos θ = => θ = cos-1 ( )=
|⃗a||⃗b| 1.2 2 3
6. If for a unit vector ⃗a , ( ⃗x - ⃗a ) . ( ⃗x + ⃗a ) = 12 then find |⃗x| .
Given ( ⃗x - ⃗ a ) = 12 and |⃗a| = 1
a ) . ( ⃗x + ⃗
2 2
|⃗x| - |⃗a| = 12 => |⃗x|2 = 13 => |⃗x| = √ 13
7.Find the area of a parallelogram whose adjacent sides are given by 3 ^i + ^j + 4 k^
and ^i - ^j + k^ .

| |
^i ^i k^
Let ⃗ a X b = 3 1 4 = 5 ^i + ^j - 4 k^
⃗ => |⃗a X ⃗b| = √ 25+1+16
1 −1 1
The area of a parallelogram = |⃗a X ⃗
b| = √ 42 sq .units.
8. Show that the vectors 2 ^i -3 ^j +4 k^ and -4 ^i +6 ^j -8 k^ are collinear .
Given a = 2 ^i -3 ^j +4 k^
⃗ and b = -4 ^i +6
⃗ ^j -8 k^
since ⃗ ^ ^
b = -2( 2 i -3 j +4 k )^ => ⃗b = -2 ⃗
a
Hence , the given vectors are collinear .

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

9. Find the angle between the vectors ⃗a and ⃗
b such that |⃗a|=3 and |⃗b|= √ 2 and
3
a X ⃗
⃗ b is a unit vector .
|⃗a|=3 , |⃗b|= √
2
Given and |⃗a X ⃗
b| = 1
3
|⃗a X ⃗b| 1 1 π
W.K.T sin θ = => sin θ = = => θ =
|⃗a||⃗
b| √2 √2 4
3.
3
10. Find |⃗a X ⃗b| , ifa = ^i -7 ^j + 7 k^ ,and
⃗ b = 3 ^i -2 ^j +2 k^ .
⃗

| |
i^ i^ k^
Let a X ⃗
⃗ b = 1 −7 7 = 0 ^i +19 ^j +19 k^ => |⃗a X ⃗
b| = √ 0+(19)2 +(19)2
3 −2 2
|⃗a X ⃗b| = 19 √ 2
b are such that |⃗a|=2 , |⃗b|=3 and ⃗
a and ⃗
11. If two vectors ⃗ b = 4 then find |⃗a−⃗b| .
a . ⃗
2 2
W.K.T |⃗a −⃗b| = |⃗a|2 + |⃗b| - 2 a⃗ . ⃗b
2
|⃗a −⃗b| = 4 + 9 - 2.4 = 5 => |⃗a−⃗b| = √ 5
12. Find the angle between two vectors ⃗ a and ⃗ b with magnitudes √ 3 and 2 respectively
and when ⃗a . ⃗b=√ 6 .
Given |⃗a| = √ 3 , |⃗b| = 2 and ⃗a . ⃗b = √ 6
a .⃗
=> cos θ = √ √3 √2
⃗ b 6 1
W.K.T cos θ = = => θ = cos-1 ( )= π
⃗
|⃗a||b| √ 3 .2 √3 . √2 . √2 √2 4
13. If ( 2 ^i +6 ^j +27 k^ ) X ( ^i +λ ^j +μ k^ ) = 0 , then find the value of λ and µ .
Given cross product of two vectors is 0. This is possible iff one vector is multiple of other.
Hence 2λ = 6 and 2µ = 27 => λ = 3 and µ = 27/ 2
14. The projection of the vector ⃗ a = 2 ^i +3 ^j +2 k^ on the vector ⃗ b = ^i +2 ^j + k^ .
Let ⃗ a =2 ^i +3 ^j +2 k^ b⃗ = ^i +2 ^j + k^
a . ⃗
⃗ b = 2+6+2 = 10 and |⃗b| = √ 1+4 +1 = √ 6
a .⃗
⃗ b 10
Projection of ⃗ a on ⃗ b = =
⃗
|b| √6
15. Find ⃗ a and ⃗ b , if a +⃗
(⃗ a −⃗
b) .(⃗ b)=8 and |⃗a|=8|⃗ b| .
2
a +⃗ a −⃗ b)=8 and |⃗a|=8|⃗ b| => |⃗a| =64|⃗ b|
2
Given (⃗ b).(⃗
2 2 2 2
|⃗a|2 - |⃗b| = 8 => 64|⃗
b| - |⃗b| = 8 => 63 |⃗b| = 8
=> |⃗b| =
√ 8
63
and
√
|⃗a| = 8
8
63
16. Show that the points A( -2 ^i +3 ^j +5 k^ ) , B ( ^i +2 ^j +3 k^ ) and
C( 7 ^i - k^ ) are collinear .
Let AB⃗ = OB ⃗ - OA ⃗ =3 ^i - ^j - 2 k^ , BC ⃗ = 6 ^i -2 ^j - 4 k^ , AC
⃗ = 9 ^i -3 ^j - 6 k^
⃗
|AB| = √ 14 , ⃗
|BC| = √ 56 = 2 √ 14 ⃗
, |AC| = √ 126 = 3 √ 14
⃗ ⃗
Clearly |AB| + |BC| = |AC| ⃗
Hence , the given points are collinear.

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 29. ( THREE DIMENTIONAL GEOMETRY ) 2 mark
1. Find the equation of a line passing through (1, 2 , 3) and parallel to the vector
3 ^i +2 ^j - 2 k^ in both vector and cartesian form .
Given ⃗ a = ^i +2 ^j +3 k^ or ( x1 , y1 , z1 ) = ( 1 , 2 , 3 ) and ⃗ b = 3 ^i +2 ^j - 2 k^
⃗r = ⃗ a +λ ⃗ b => ⃗r = ( ^i +2 ^j +3 k^ )+ λ ( 3 ^i +2 ^j - 2 k^ ) be the vector form .
( x−x 1 ) ( y − y1) ( z−z 1) ( x−1) ( y −2) ( z−3)
= = => = =
a b c 3 2 −2
2. Find the Cartesian equation of the line which passes through the point ( - 2 , 4 , - 5 ) and parallel to
x +3 y −4 z +8
the line given by = = .
3 5 6
For parallel lines , direction ratios are same and hence the required equation of line is

x +2 y −4 z−5
= = be the cartesian form of the line .
3 5 6
3. Find the angle between the pair of lines given by
⃗r = 3 ^i +2 ^j - 4 k^ +λ( ^i +2 ^j +2 k^ ) and ⃗r =5 ^i - 2 ^j + μ (3 ^i +2 ^j +6 k^ )
Given b⃗1 = ^i +2 ^j +2 k^ and b⃗2 = 3 ^i +2 ^j +6 k^
|b⃗1| = √ 9 =3 , |b⃗2| = √ 9+ 4+36 =7 , b⃗1 . b⃗2 = 3+4+12 =19
b⃗1 . b⃗2 19 19
W.K.T cos θ = => cos θ = => θ = cos-1 ( )
|b 1||b2|
⃗ ⃗ 3.7 21
4. Find the angle between the pair of lines given by
⃗r = 2 ^i -5 ^j + k^ + λ ( 3 ^i +2 ^j +6 k^ ) and ⃗r = 7 ^i - 6 k^ + μ ( ^i +2 ^j +2 k^ )
Given b⃗1 = 3 ^i +2 ^j +6 k^ and b⃗2 = ^i +2 ^j +2 k^
|b⃗1| = √ 9+ 4+36 =7 , |b⃗2| =3 , b⃗1 . b⃗2 = 3+4+12 =19
b⃗1 . b⃗2 19 19
W.K.T cos θ = => cos θ = => θ = cos-1 ( )
|b⃗1||b⃗2| 7.3 21
5. Find the angle between the pair of lines given by ⃗r = 3 ^i + ^j - 2 k^ + λ ( ^i - ^j - 2 k^ )
and ⃗r =2 ^i - ^j - 56 k^ + μ ( 3 ^i -5 ^j - 4 k^ )
Given b⃗1 = ^i - ^j -2 k^ and b⃗2 = 3 ^i -5 ^j -4 k^
|b⃗1| = √ 6 , |b⃗2| = √ 9+25+16 = 5 √ 2 , b⃗1 . b⃗2 = 3+5+8= 16
b⃗1 . b⃗2 16 8
W.K.T cos θ = => cos θ = => θ = cos-1 ( )
|b⃗1||b⃗2| √6 5 √2 5 √3
x +3 y −1 z +3 x +1 y −4 z−5
6. Find the angle between the pair of lines = = and = = .
3 5 4 1 1 2
⃗ ^ ^ ^ ⃗
Given b1 =3 i +5 j +4 k and b2 = i + j +2 k ^ ^ ^
|b⃗1| = 5 √ 2 , |b⃗2| = √ 6 , b⃗1 . b⃗2 = 3+5+8= 16
b⃗1 . b⃗2 16 8
W.K.T cos θ = => cos θ = => θ = cos-1 ( )
|b 1||b2|
⃗ ⃗ 5 √2 √6 5 √3
7. Find the angle between the pair of lines whose direction ratios are a , b , c and b –c , c –a , a –b .
Since the dot product of given vectors = a.(b – c) + b.(c –a) + c.(a – b) = a.b –a.c + b.c – c.b + c.a – c.b = 0
cos θ = 0 or lines are perpendicular and hence θ = π
2

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 42.( PROBABILITY) 3 mark
1. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two
bags is selected at random and a ball is drawn at random from the bag and it is found to be red. Find
the probability that the ball is drawn from first bag ?
1
Given P(B1) = P(B2)= and Let E be the event that that drawn ball is red .
2
E 4 E 2 B1
P( )= , P( )= , Now to find , P( ), By Bayes Theroem
B1 8 B2 8 E
E 1 4
P(B 1 ) P( ) .
B1 B1 2 8 4 2
P( )= = = =
E E E 1 4 1 2 6 3
P(B 1 ) P( )+ P(B 2 ) P( ) . + .
B1 B2 2 8 2 8
4
2. Probability that ‘A’ speaks truth is . A coin is tossed . ‘A’ reports that a head appears. What is
5
the probability that actually there was head?
Let E1: ‘ coin shows a head’ and E2 : ‘ coin shows a tail’ .
1 1
Then P(E1)= and P(E2)=
2 2
Let E be the event that ‘A’ reports that a head appears .
E 4 E 1 E1
P( )= , P( )= , Now to find , P( ), By Bayes Theroem
E1 5 E2 5 E
E 1 4
P( E 1 ) P( ) .
E1 E1 2 5 4
P( )= = =
E E E 1 4 1 1 5
P( E 1) P( )+ P( E 2) P( ) . + .
E1 E2 2 5 2 5
3. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six.
Find the probability that it is actually a six.
Let A: man speaks truth B : man lies
P(A) = 3/4 and P(B) = 1/4
Let E be the event that man throws a die and reports that it is a six.
E 1 E 5 A
P( )= , P( )= , Now to find , P( ), By Bayes Theroem
A 6 B 6 E
E 3 1
P( A) P( ) .
A A 4 6 3
P( )= = =
E E E 3 1 1 5 8
P( A) P( )+P( B) P( ) . + .
A B 4 6 4 6
4. Bag-I contains 3 red and 4 black balls. While another Bag -II contains 5 red and 6 black balls. One
ball is drawn at random from one of these bags and it is found to be red. Find the probability that it
was drawn from the Bag- II ?
1
Given P(B1) = P(B2)= and Let E be the event that that drawn ball is red .
2
E 3 E 5 B2
P( )= , P( )= , Now to find , P( ), By Bayes Theroem
B1 7 B2 11 E
E 1 5 5 5
P(B 2 ) P( ) .
B2 B2 2 11 11 11 35
P( )= = = = =
E E E 1 3 1 5 3 5 68 68
P(B 1 ) P( )+ P(B 2 ) P( ) . + . +
B1 B2 2 7 2 11 7 11 77

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 52.( Determinants ) 4 mark
1. Show that the matrix A = [21 32] satisfies the equation A2 – 4 A + I = O . where I is the 2 X 2
identity matrix and O is the 2 X 2 zero matrix . Using this equation , find A -1 .

Given A = [21 32] => A2 = [21 32] [21 32] = [42+2+3 6 +6

3+ 4 ] = [74 127 ]
Let A2 – 4 A + I = [74 127 ] - [48 128 ] + [10 01] = [00 00] = O , Hence , A2 – 4 A + I = O
Now I= 4A – A2
Xly A-1 , A I = 4 A-1 A - A-1 A.A
-1

A-1 = 4 I – A
A-1 =
4 0
0 4
– [ ] [ ]
2 3
1 2
=> A-1 =
2 −3
−1 2 [ ]
2. Show that the matrix A =
3 2
1 1 [ ]
satisfies the equation A2 – 4 A + I = O . where I is the 2 X 2
identity matrix and O is the 2 X 2 zero matrix . Using this equation , find A -1 .
Given A = [ ]
3 2
1 1
=> A2 =
3 2
1 1 [ ] [ ] [
3 2
1 1
=
9+2 6 +2
3+1 2+1
=
11 8
4 3 ] [ ]
Let A2 – 4 A + I = [114 83] – [124 84 ] + [10 01] = [00 00] = O
Hence , A2 – 4 A + I = O
Now I= 4A – A2
ly -1
X A , A I = 4 A-1 A - A-1 A.A
-1

A-1 = 4 I – A
A-1 =
4 0
0 4
– [ ] [ ]
3 2
1 1
=> A-1 =
1 −2
−1 3 [ ]
3. Show that the matrix A =
3 1
−1 2 [ ]
satisfies the equation A2 – 5 A + 7 I = O . where I is the
2 X 2 identity matrix and O is the 2 X 2 zero matrix . Using this equation , find A -1 .
Given A = [ 3 1
−1 2 ]
=> A2 =
3 1
−1 2 [ 3 1
−1 2 ] [ =
9−1
] [
−3−2 −1+4
3+2
=
8 5
−5 3 ] [ ]
Let A2 – 5 A + 7I = [−58 53] – [−515 105 ] + [70 70] = [00 00] = O
Hence , A2 – 5 A + 7I = O
Now 7I = 5A – A2
ly -1
X A , 7 A-1 I = 5 A-1 A - A-1 A.A
7 A-1 = 5 I – A
7 A-1 =
5 0
0 5
–[ ] 3 1
−1 2 [ ] => 7 A-1 = [21 −13 ]
[ ]
2 −1
A-1 =
1
7 [
2 −1
1 3
OR ] A-1 = 7
1
7
3
7 7

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 40 and 41.( vectors ) 2X 3 mark = 6 mark
1. Derive section formula .
Ans : Let P divides A and B in the ratio m: n as in fig .
Let OA ⃗ = ⃗ a , OB⃗ = ⃗ b and OP ⃗ = ⃗r then
⃗
AP m ⃗
OP−OA ⃗ m
= => =
⃗
PB n ⃗
OB− OP⃗ n

⃗r −⃗a m
= => n( ⃗r −⃗a ) = m( ⃗
b−⃗r )
⃗
b−⃗r n
m⃗ b +n ⃗a
=> ( m + n ) ⃗r = m ⃗
b +n ⃗
a => ⃗r =
m+n
2. Show that the vector ^i + ^j + k^ is equally inclined to the axes OX , OY and OZ.
Ans : Given ^i + ^j + k^ then direction ratios are 1 , 1 , 1 and
1 1 1
direction cosines are , ,
√3 √3 √3
1 1 1
i.e, cos ɑ = , cos β = , cos γ =
√3 √3 √3
1
=> ɑ = β = γ = cos -1
√3
^ ^ ^
Hence , the vector i + j + k is equally inclined to the axes OX , OY and OZ.
3. If ⃗ a , ⃗b , ⃗c are three unit vectors such that ⃗ a +⃗ b +⃗c =0 . Find the value of ⃗ a .⃗
b +⃗
b . ⃗c +⃗
c .⃗
a .
Ans : Given ⃗ a +⃗ b +⃗c =0 and |⃗a|=|⃗ b|=|⃗ c|=1
2
(⃗a + ⃗b+⃗c )2=0 => |⃗a| +|⃗ b| +|⃗c| +2 ( ⃗ a .⃗
b +⃗
2 2
b . ⃗c +⃗c .⃗
a )=0
−3
a .⃗
⃗ b +⃗ b . ⃗c +⃗c .⃗
a =
2
4. If ⃗ ⃗
a , b ,⃗c satisfy the condition ⃗ a +⃗b +⃗c =0 and |⃗a|=1 , |⃗b|=4 and |⃗c|=2 ,
Evaluate µ = ⃗ a .⃗
b +⃗ b .⃗c +⃗ c .⃗a .
Ans : Given ⃗ a +⃗ b +⃗c =0 and |⃗a|=1 , |⃗b|=4 and |⃗c|=2
2
(⃗a + ⃗b+⃗c ) =0 => |⃗a| +|⃗ b| +|⃗c| +2 ( ⃗ a .⃗b +⃗
2 2 2
b . ⃗c +⃗
c .⃗a )=0
1 + 16 + 4 + 2 ( ⃗ ⃗ ⃗
a . b + b . ⃗c +⃗c .⃗
a )=0
−21
Therefore , µ= ⃗ a .⃗b +⃗
b .⃗c +⃗ c .⃗a =
2
5. Let ⃗ ⃗ |⃗ |
a , b , ⃗c are three vectors such that |⃗a|=3 , , b =4 and |⃗c|=5 and each of them being
perpendicular to the sum of others two , find |⃗a + ⃗ b +⃗c|
Ans : Given ⃗ a ┴ ( ⃗ b +⃗ c ), ⃗ b ┴ ( ⃗c +⃗ a ) , ⃗c ┴ ( ⃗ a +⃗ b )
=> a . ( b +⃗
⃗ ⃗ ⃗
c ) = 0 , b . ( ⃗c +⃗ a ) = 0 , ⃗c . ( ⃗ ⃗
a +b ) = 0
adding these 2 ( ⃗ ⃗ ⃗
a . b + b . ⃗c +⃗ c .⃗a )=0
2 2 ⃗2
⃗
Now , |⃗a + b +⃗c| = |⃗a| +|b| +|⃗c| +2 ( ⃗ a .⃗
b +⃗
2
b . ⃗c +⃗c .⃗ a )
2
|⃗a + ⃗b +⃗c| =9 + 16 + 25 + 0 = 50 => |⃗a + ⃗ b+⃗c| = √ 50 or 5 √2
6. If ⃗ a = 2 ^i +2 ^j +3 k^ , ⃗ b = - ^i +2 ^j + k^ and ⃗c = 3 ^i + ^j are such that
a +λ ⃗
⃗ b is perpendicular to ⃗c , then find the value of λ .
Ans : Given ⃗ a +λ ⃗ b is perpendicular to ⃗c => ( ⃗ a +λ ⃗ b ) . ⃗c = 0
=> ^ ^
[ 2 i +2 j +3 k + λ (- i +2 j + k )] . ( 3 i + ^j ) = 0
^ ^ ^ ^ ^
=> ( 2 – λ ) 3 + (2 + 2λ ) 1+ ( 3 + λ ) 0 = 0
=> 6–3λ+ 2+2λ+0=0
=> – λ+8=0 => λ = 8

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

7 . Find a unit vector perpendicular to each of the vectors a⃗ + ⃗ b and ⃗ a −⃗b , where
a = ^i + ^j+ k^ and ⃗
⃗ b= ^i +2 ^j +3 k^
Ans : Let ⃗c = ⃗ a +⃗b = 2 ^i +3 ^j+4 k^ and ⃗ = ⃗
d a −⃗b = 0 ^i − ^j−2 k^

| |
^i ^i ^k
⃗c X ⃗ d = 2 3 4 = -2 ^i +4 ^j -2 k^ = 2( - ^i +2 ^j + k^ )
0 −1 −2
⃗
|⃗c X d| = 2 √ 1+4+1 = 2 √ 6
⃗c X ⃗
d 2
unit vector perpendicular to each of the vectors ⃗c and ⃗ d is = ( - ^i +2 ^j + k^ )
⃗
|⃗c X d| 2 √6
1
Therefore , required vector is ( - ^i +2 ^j + k^ )
√6
8. Find a unit vector perpendicular to each of the vectors a⃗ + ⃗ b and ⃗ a −⃗b , where
a =5 ^i− ^j−3 k^ and ⃗
⃗ b= ^i +3 ^j−5 k^ .
Ans : Let ⃗c = ⃗ a +⃗ b = 6 ^i +2 ^j−8 k^ and ⃗ = ⃗
d a −⃗b = 4 ^i −4 ^j+2 k^

| |
^i ^i ^k
⃗c X ⃗ d = 6 2 −8 = -28 ^i - 44 ^j -32 k^ = 4( -7 ^i -11 ^j -8 k^ )
4 −4 2
⃗
|⃗c X d| = 4 √ 49+121+64 = 4 √ 234
⃗c X ⃗
d 4
unit vector perpendicular to each of the vectors ⃗c and ⃗ d is = (-7 ^i -11 ^j -8 k^ )
⃗
|⃗c X d| 4 √ 234
1
Therefore , required vector is ( -7 ^i -11 ^j -8 k^ )
√ 234
9. Find the area of the triangle whose vertices A( 1 , 1 , 2 ) , B( 2 , 3 , 5 ) and C( 1 , 5 , 5 )
by using vector method.
Ans : Let AB ⃗ = OB ⃗ - OA ⃗ = (2 – 1 ) ^i +( 3 – 1 ) ^j +( 5 – 2 ) k^ = ^i + 2 ^j +3 k^

⃗ = OC
AC ⃗ = (1 – 1 ) ^i +( 5 – 1 ) ^j +( 5 – 2 ) k^ = 0 ^i + 4 ^j + 3 k^
⃗ - OA

| |
^i ^i k^
Now , ⃗ X
AB ⃗ =
AC 1 2 3 = - 6 ^i - 3 ^j + 4 k^
0 4 3
⃗ X AC
|AB ⃗ | = √ 36+9 +16 = √ 61

Area of triangle =
1 ⃗ | = √ 61
⃗ X AC
|AB
2 2
10. Find a unit vector perpendicular to each of the vectors ⃗ a +⃗
b and ⃗ a −⃗b , where
a =3 ^i +2 ^j+2 k^ and ⃗
⃗ b= ^i +2 ^j−2 k^
Ans : Let ⃗c = ⃗ a +⃗b = 4 ^i +4 ^j+0 k^ and ⃗ a −⃗b = 2 ^i +0 ^j+ 4 k^
d = ⃗

| |
^i ^i k^
⃗
⃗c X d = 4 4 0 = 16 ^i - 16 ^j -8 k^ = 8( 2 ^i -2 ^j - k^ )
2 0 4
⃗
|⃗c X d| = 8 √ 4+ 4+1 = 24
⃗c X ⃗d 8
unit vector perpendicular to each of the vectors ⃗c and ⃗ d is = ( 2 ^i -2 ^j - k^ )
|⃗c X ⃗d| 24
1
Therefore , required vector is ( 2 ^i -2 ^j - k^ )
3

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 43.(RELATIONS AND FUNCTIONS ) 5 mark
1. Verify the function f: R→R defined by f(x) = 3 – 4 x is one – one or onto or bijective .
Ans : Let x1 , x2 ϵ R f( x1 ) = f( x2 )
=> 3 – 4 x1 = 3 – 4 x2
=> – 4 x1 = – 4 x2
=> x1 = x2
Therefore f is one – one
Let y ϵ R ( co-domain) such that y = f(x)
=> y = 3 – 4 x
= > 4x = 3 – y
3− y
=> x = ϵ R ( Domain)
4
Therefore f is onto.
Since f is one- one and onto and hence bijective .
2. Verify whether the function f : N →Y defined by f(x) = 4x + 3 , Where Y = { y : y = 4x + 3 ,x ϵ N }
is bijective or not .
Ans : Let x1 , x2 ϵ R f( x1 ) = f( x2 )
=> 4 x1 +3 = 4 x2 + 3
=> 4 x1 = 4 x2
=> x1 = x2
Therefore f is one – one
Let y ϵ Y( co-domain) such that f(x) = y
=> 4 x + 3 = y
= > 4x = y – 3
y −3
=> x = ϵ N ( Domain) . Therefore f is onto.
4
Since f is one- one and onto and hence bijective .
3. Verify whether the function f : R→R defined by f(x) = 1 + x2 , is bijective or not .
Ans : Let x1 , x2 ϵ R f( x1 ) = f( x2 )
=> 1+ x12 = 1+ x2 2
=> x1 = ± x2
=> Therefore f is not one – one
Let y ϵ R( co-domain) such that f(x) = y
=> 1+ x2 = y => x = √ y−1 ϵ R ( Domain)
Therefore f is not onto.
Hence , f is not bijective function .
x−2
4. Let A = R – {3} and B = R – {1} . Consider f : A→ B defined by f(x) = . Is f one-one and
x−3
onto ? Justify your answer.
x 1−2 x 2−2
Ans : Let x1 , x2 ϵ A f( x1 ) = f( x2 ) => =
x 1−3 x 2−3
=> x1 x2 - 3 x1 -2x2 + 6 = x1 x2 - 2 x1 -3x2 + 6
=> - x1 = - x2
=> x1 = x2 Therefore f is one – one
Let y ϵ B( co-domain) such that f(x) = y
x−2
=> = y => x – 2 = xy – 3 y = > x( 1 – y ) = 2 – 3 y
x−3
2−3 y
=> x = € A ( Domain) Therefore f is onto.
1− y
Since f is one- one and onto and hence bijective .

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 46 (CONTINUTY AND DIFFERENTIABILITY ) 5 mark
2
1. If y = 3 cos ( log x ) + 4 sin ( log x ) Show that x y2 + x y1 + y = 0 .
Ans : Given y = 3 cos ( log x ) + 4 sin ( log x )
Diffrentiate , y1 = - 3 sin ( log x ) 1 + 4 cos ( log x ) 1
x x
xy1 = - 3 sin ( log x ) + 4 cos ( log x )
1 1
Diffrentiate , x y2 + y1 . 1 = -3 cos ( log x ) - 3 sin ( log x )
x x
Multiply by x , x2 y2 + x y1 = - 3 cos (log x ) - 4 sin ( log x )
=-y
Therefore , x2 y2 + x y1 + y = 0
2. If y = ( tan-1 x )2 then show that ( x2 + 1 )2 y2 + 2 x( x2 + 1) y1 = 2 .
Ans : Given y = ( tan-1 x )2
1
Diffrentiate , y1 = 2 ( tan-1 x ) . 2
1+ x
( 1+ x2 ) y1 = 2 ( tan-1 x )
1
Diffrentiate , ( 1+ x2 ) y2 + y1 . 2 x = 2 . 2
1+ x
Multiply by (1+x2) , ( x2 + 1 )2 y2 + 2 x( x2 + 1) y1 = 2
2
d y dy
3. If y = 3 e 2x + 2 e3x prove that 2
−5 +6 y=0 .
dx dx
dy
Ans : Given y = 3 e2x + 2 e3x => = 6 e2x + 6 e3x
dx
2
d y
=> 2
= 12 e2x + 18 e3x
dx
2
d y dy
Consider LHS = −5 +6 y = 12 e2x + 18 e3x - 5 (6 e2x + 6 e3x ) + 6 ( 3 e2x + 2 e3x )
dx
2
dx
= 12 e2x + 18 e3x - 30 e2x - 30e3x + 18 e2x + 12 e3x
2
d y dy
Hence , 2
−5 +6 y=0
dx dx
mx nx d2 y dy
4. If y = A e + B e prove that 2
−(m+n) +m n y=0 .
dx dx
dy
Ans : y = A e m x + B en x => = A e mx .m + B e nx .n
dx
2
d y
=> 2
= A e mx .m2 + B e nx .n2
dx
2
d y dy
Consider LHS = −(m+n) +m n y
dx
2
dx
= A e mx .m2 + B e nx .n2 - ( m + n )[A e mx .m + B e nx .n] + mn [A e m x + B en x ]

= A e mx [ m2 - ( m + n )m + mn ] + B e nx [ n2 - ( m + n )n + mn ]

= A e mx [ m2 - m2 - mn + mn ] + B e nx [ n2 - mn - n2 + mn ]
d2 y dy
Hence , 2
−(m+n) +m n y=0
dx dx

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

 2
-1 2 d y dy
5. If y = sin x then show that (1−x ) − x =0
dx 2
dx
dy 1 dy
Ans : Given y = sin -1 x => = => √ 1− x 2 =1
dx √ 1− x 2 dx

2
d y dy 1
=> √ 1− x 2 + (-2x) =0
dx 2 √ 1− x
2 2
dx
2
2 d y dy
=> Multiply by √ 1− x 2 , (1−x ) − x =0
dx 2
dx
2 2
6. If ey ( x + 1 ) = 1 then show that d y2=( dy ) .
dx dx
y
Ans : Given e ( x + 1 ) = 1 -------(1)
dy dy
Differentiate , e y( 1 ) + ( x + 1 ) e y = 0 => e y + 1. =0 using (1)
dx dx
dy
=-ey -----------(2)
dx
2
d y dy dy dy
Differentiate , =-ey = . using (2)
dx
2
dx dx dx
d2 y dy 2
=> =( ) Hence, the result .
dx
2
dx
2
−1
d y dy
7. If y = e acos x , ( -1 ≤ x ≤ 1 ) then show that (1−x 2 ) − x −a2 y=0 .
dx 2
dx
acos x
−1
dy −1
acos x −a
Ans : Given y = e ----(1) => = e .
dx √ 1− x 2
=> √ 1− x 2 dy dx
= -ay using (1)
2
=> Differentiate , √ 1− x 2 d y2 + dy 1
( - 2 x ) = -a
dy
dx dx 2 √ 1−x 2 dx
2
d y dy dy
=> Multiply by √ 1− x 2 , (1−x 2 ) −x = - a √ 1− x 2
dx
2
dx dx
2
2 d y dy
=> (1−x ) −x = - a ( -a y )
dx 2
dx
2
2 d y dy 2
=> (1−x ) − x −a y=0
dx
2
dx
2
8. If y = ( sin-1 x )2 then show that (1−x 2 ) d y2− x dy =2
dx dx
−1
dy 2 sin x dy
Given y = ( sin -1 x )2 => = => √ 1− x
2
= 2 sin- 1 x
dx √ 1− x 2 dx

dy 2
=> squaring on both sides (1 – x2 ) ( ) = 4 ( sin -1 x )2 = 4y
dx
2
dy d y dy 2 dy
=> Differentiate , (1 – x2 ) 2 +( ) (-2 x ) = 4
dx dx 2
dx dx
2
dy 2 d y dy
=> Dividing by 2 , we get (1−x ) − x =2
dx dx
2
dx

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 48 ( APPLICATION OF INTEGRALS ) 5 mark
2 2 2
1. Find the area of the circle x + y = a by using integration .
Ans : Given x2+ y2 = a2 => y = √ a 2−x 2
Required Area = 4 ( Shaded Area )
x =b a

=4 ∫ y dx =4 ∫ √ a 2− x2 dx
x =a 0
x a2 x
=4{
2
√ a −x
2 2
+
2
sin -1
a
}0a
2 2
a a π
=4{0+ sin -1 1 – (0 +0) } = 4
2 2 2
A = π a2
x2 y2
2. Find the area enclosed by the ellipse 2
+ 2 =1 by using integration .
a b
2 2
x y b
Ans : Given 2
+ 2
=1 => y =
a
√ a 2−x 2
a b
Required Area = 4 ( Shaded Area )
x =b a
b
= 4 ∫ y dx = 4 ∫ √ a2 −x 2 dx
x =a 0 a
b x a2 x
=4
a
{
2
√ 2
a −x +
2
2
sin -1
a
}0a
2
b a
=4 {0+ sin -1 1 – (0 +0) }
a 2
b a2 π
=4
a 2 2
A = π ab
3. Find the area of the region bounded by the line y = 3x + 2 the x – axis and the
ordinates x = -1 and x = 1 .
Ans : Given line is y = 3x + 2 cuts the x-axis at x = -2/3 in (-1 , 1 )

|∫ |
−2/3 1

Required Area A = (3 x +2) dx + ∫ (3 x +2) dx

−1 −2/ 3

|( 32x +2 x )|
2 2
3x
= +2 x) -2/31
-1
-2/3 +
(
2
2 4 3 3 2 4
= | - - +2 | + [ +2 - + ]
3 3 2 2 3 3
1 25 13
A = + = sq units
6 6 3
4. Find the area of the region bounded by the curve y2 = 4x , y – axis and the line y = 3 .
Ans : y2 = 4x , y -axis and the line y = 3
Required Area = shaded Area
3 3 2
y
= ∫ x dy = ∫ dy
0 0 4
1 27 9
= [ -0]=
4 3 4
Note : If area of circle in first quadrant means only shaded area ( remove 4 in Q1) and in
ellipse particular values for a and b are given then simply substitute the values in Q 2 .

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 50 ( THREE DIMENSIONAL GEOMETRY ) 5 mark
1. Derive the equation of a line passing through a given point and parallel to a given
vector in both vector and Cartesian form . ( 90% chance to be expected this question )
Ans : Let P be any point on the line and A( ⃗
a ) be given point
⃗
and parallel to a vector b
Now , ⃗ || ⃗
AP b
=> ⃗ =λ ⃗
AP b
=> ⃗ OA
OP− ⃗ =λ ⃗ b
=> ⃗r −⃗a = λ ⃗
b
=> a +λ ⃗
⃗r = ⃗ b be the vector form of the line .
Let A( x1 , y1 , z 1) be given point and P( x , y , z ) be any point on the line .
Let a , b , c are the direction ratio ’ s of ⃗ b . Then
a = x1 ^i + y1 ^j + z1 k^
⃗ , ⃗r = x ^i + y ^j + z k^ and ⃗ b = a ^i + b ^j + c k^
substitute these in ⃗r = a⃗ + λ ⃗ b we get
( x – x1) ^i + ( y – y1) ^j + ( z– z1) k^ = aλ ^i + bλ ^j +c λ k^
Comparing and equating the co-efficients of ^i , ^j , k^ and we get

x− x1 y − y1 z−z 1
= = be the cartesian form of the line .
a b c

Q. NO . 45 ( DETERMINANTS ) 5 mark
1. Solve by matrix method x + y + z = 6 , y + 3 z = 11 , x – 2y + z = 0 .

[ ] [] []
1 1 1 x 6
Ans : Given A = 0 1 3 , X= y , B= 11
1 −2 1 z 0
| A | = 1(1+6) – 1 ( 0 – 3 ) +1 (0 –1) = 9 ≠ 0

[ ]
+(1+6 ) −(0−3) +(0−1)
co-factor matrix = −(1+2) +(1−1) −(−2−1)
+(3−1) −(3−0) +(1−0)

[ ]
7 −3 2
adj A = 3 0 −3
−1 3 1

[ ] []
7 −3 2 6
1 1
X = A-1 B = adj A . B = 3 0 −3 11
|A| 9
−1 3 1 0

[] [ ] [ ] []
x 42−33+0 9 1
1 1
y = 18+0−0 = 18 = 2 x = 1 , y =2 , z = 3
9 9
z −6+33+0 27 3

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

2. Solve by matrix method x - y + z = 4 , 2x+ y - 3 z = 0 ,x+y+z=2 .

[ ]
[] []
1 −1 1 x 4
Ans : Given A = 2 1 −3 , X= y , B= 0
1 1 1 z 2
| A | = 1(1+3) + 1 ( 2 + 3 ) +1 (2 –1) = 10 ≠ 0

[ ] [ ] [ ]
+(1+3) −(2+3) +(2−1) 4 −5 1 4 2 2
co-factor matrix = −(−1−1) +(1−1) −(1+1) = 2 0 −2 adj A = −5 0 5
+(3−1) −(−3−2) +(1+2) 2 5 3 1 −2 3

[ ] []
4 2 2 4
1 1
X = A-1 B = adj A . B = −5 0 5 0
|A| 10
1 −2 3 2

[] [ ] [ ] []
x 16+0+ 4 20 2
1 1
y = −20+0+10 = −10 = −1 x = 2 , y =-1 , z = 1
10 10
z 4−0 +6 10 1
3. Solve by matrix method 3x -2 y + 3z = 8 , 2x+y - z = 1 , 4x - 3y +2 z = 4 .

[ ] [] []
3 −2 3 x 8
Ans : Given A = 2 1 −1 , X= y , B= 1
4 −3 2 z 4
| A | = 3(2–3) +2 ( 4+4 ) +3 ( –6–4 ) = – 17 ≠ 0

[ ] [ ] [ ]
+(2−3) −(4 +4) +(−6−4) −1 −8 −10 −1 −5 −1
C.M = −(−4+9) +(6−12) −(−9+8) = −5 −6 1 adj A = −8 −6 9
+(2−3) −(−3−6) +(3+ 4) −1 9 7 −10 1 7

[ ] []
−1 −5 −1 8
1
-1 1
X=A B= adj A . B = −8 −6 9 1
|A| −17
−10 1 7 4

[] [ ] [ ] []
x −8−5−4 −17 1
1 1
y = −64−6 +36 = −34 = 2 x = 1 , y =2 , z = 3
−17 −17
z −80+1+28 −51 3

4. Solve by matrix method 2x +3 y + 3z = 5 , x- 2y + z = -4 , 3x - y -2 z = 3 .

[ ] [] []
2 3 3 x 5
Ans : Given A = 1 −2 1 , X= y , B = −4
3 −1 −2 z 3
| A | = 2(4+1) - 3( -2 -3 ) +3 ( –1+6 ) = 40 ≠ 0

[ ] [ ] [ ]
+( 4+1) −(−2−3) +(−1+6) 5 5 5 5 3 9
C.M = −(−6+3) +(−4−9) −(−2−9) = 3 −13 11 adj A = 5 −13 1
+(3+6) −(2−3) +(−4−3) 9 1 −7 5 11 −7

[ ][]
5 3 9 5
1 1
X = A-1 B = adj A . B = 5 −13 1 −4
|A| 40
5 11 −7 3

[ ] [ ] []
25−12+27 40 1
1 1
X = 25 +52+3 = 80 = 2 x = 1 , y =2 , z = -1
40 40
25−44−21 −40 −1

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 52 ( CONTINUITY AND DIFFERENTIABILITY ) 4 mark
1. Find the value of k so that the function f by 2.Find the value of k so that the function f defined
f(x) = { kx+ 1if x≤5 is continuous at x = 5. by f(x) = { kx +1if x≤π is continuous at x = π .
3 x−5if x>5 cos x if x > π
Since f is continuous at x = 5 Since f is continuous at x = π
LHL =RHL = f(5) LHL =RHL = f(π)
lim kx+1 = lim 3 x−5 lim kx+ 1 = lim cos x
−¿ +¿ −¿ +¿
x →5 ¿ x →5 ¿ x →π ¿ x →π ¿
5k + 1 = 3(5) – 5 = 10 π k + 1 = cos π = -1
9 −2
5k=9 ═> k = π k = -2 ═> k = π
5
3. Find the value of k so that the function f 4. Find the values of a and b so that the function
defined by defined by
kcos x 5 , if x≤2
: x≠ π
f(x) = { π −2 x 2 is continuous at x = π f(x) = { ax +b if 2< x<10 is continuous function .
3 if x= π 2 21 ,if x≥10
2 ¿
Since f is continuous at x = π Since f is continuous fuction hence continuous at x =
2 2 , 10
π kcos x
lim f ( x ) = f( ) ═> lim =3 at x =2 lim f ( x) = f(2)
π
x→ ¿ 2 x → ¿ π −2 x
π x→ 2
2 2

π + h ═> x → π ═> h → 0 lim ax+b = 5 ═> 2a + b = 5 ------ (1)

put x = x→ 2
2 2
at x = 10 lim f ( x) = f(10)
kcos( π + h) x →10
2 k (−sin h)
lim = lim =3 lim ax +b = 21 ═> 10 a + b = 21 ------(2)
h→ 0 ¿ −2 h 2 h →0 ¿ −h x →10
k (2) – (1) ═> 8a = 16 ═> a = 2
.1= 3 ═> k = 6
2 (1) ═> b = 1
5. Find the relationship between a and b so that 6. Find the value of λ , if the function
the function by f(x) = { ax +1if x ≤3 is
2
f(x) = { λ ( x −2 x)if x≤0 is continuous at x = 0 .
bx +3 if x> 3 4 x +1if x >0
continuous at x = 3 .
Since f is continuous at x = 3 Since f is continuous at x = 0
LHL =RHL = f(3) LHL =RHL = f(0)
lim ax+1 = lim bx+3 2
lim λ (x −2 x) = lim 4 x +1
−¿ +¿
x →3 ¿ x →3 ¿ x →0 ¿
−¿ +¿
x →0 ¿
3a + 1 = 3b + 3 λ(0) = 1
3a – 3b = 2 ═> a – b = 2/3 Therefore , for no values of λ , f is continuous at x= 1
7. Find the value of k , if the function f defined 8. Find the value of k so that the function f by
f(x) = { kx+ 1if x≤5 is continuous at x = 5.
2
by f(x) = { kx if x ≤2 is continuous at x = 2 .
3 if x> 2 3 x−5if x>5
Since f is continuous at x = 2 Since f is continuous at x = 5
LHL =RHL = f(2) LHL =RHL = f(5)
lim kx2 = lim 3 lim kx+1 = lim 3 x−5
−¿ −¿ +¿
x →2 ¿ +¿
x →2 ¿ x →5 ¿ x →5 ¿
3 9
4k=3 => k= 5k + 1 = 3(5) – 5 = 10 => 5 k = 9 ═> k =
4 5

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

Q. NO . 51 ( LINEAR PROGRAMMING PROBLEM ) 6 mark
1. Solve the LPP graphically : Maximize Z= 4x +y S.t.C. x +y ≤ 50 , 3x +y ≤ 90 , x≥0 , y≥0 .
x+y = 50 3x+y = 90
x 0 50 x 0 30
y 50 0 y 90 0
The corner points are O(0,0) , A(30,0), B(20,30) , C(0,50)
Corner point Z =4x +y
O( 0 , 0 ) 0+0=0
A( 30 , 0 ) 120+0=120← max
B( 20 , 30 ) 80+30=110
C( 0 , 50 ) 0+50=50
The maximum value of Z is 120 at (30, 0 )
2. Solve the LPP graphically : Maximize Z= 250x +75y S.t.C. 5x +y ≤ 100 , x +y ≤ 60 , x≥0 , y≥0 .
5x+y = 100 x+y = 60
x 0 20 x 0 30
y 100 0 y 90 0
The corner points are O(0,0) , A(20,0), B(10,50) , C(0,60)
Corner point Z =250x +75y
O( 0 , 0 ) 0+0=0
A( 20 , 0 ) 5000+0=5000
B( 10 , 50 ) 2500+3750=6250 ←max
C( 0 , 60 ) 0+4500=4500
The maximum value of Z is 6250 at (10, 50 ).
3. Solve the LPP graphically : Minimize Z= 200x +500y S.t.C. x +2y ≥ 10 , 3x +4y ≤ 24 , x≥0 , y≥0 .
x+2y = 10 3x+4y = 24
x 0 10 x 0 8
y 5 0 y 6 0
The corner points are A(0,5), B(4,3) , C(0,6)
Corner point Z= 200x +500y
A( 0 , 5 ) 0+2500=2500
B( 4 , 3 ) 800+1500=2300← min
C( 0 , 6 ) 0+3000=3000
The minimum value of Z is 2300 at (4, 3)
4. Solve the LPP graphically : Minimize Z= - 3x +4y S.t.C. x +2y ≤ 8 , 3x +2y ≤ 12 , x≥0 , y≥0 .
x+2y = 8 3x+2y = 12
x 0 8 x 0 4
y 4 0 y 6 0
The corner points are O(0,0) , A(4,0), B(2,3) , C(0,4)
Corner point Z = - 3x + 4y
O( 0 , 0 ) 0
A( 4 , 0 ) -12 ← min
B( 2 , 3) 6
C( 0 , 4) 16

SMART PACKAGE GGPUC(BB0136) HOSAKOTE

5. Solve the LPP graphically : Maximize Z= 3x +2y S.t.C. x +2y ≤ 10 , 3x +y ≤ 15 , x≥0 , y≥0 .
x+2y = 10 3x+y = 15
x 0 10 x 0 5
y 5 0 y 15 0
The corner points are O(0,0) , A(0,5), B(4,3) , C(5,0)
Corner point Z = 3x + 2y
O( 0 , 0 ) 0+0=0
A( 0 , 5 ) 0+10=10
B( 4 , 3 ) 12+6=18
C( 5 , 0 ) 15+0 =15
The maximum value of Z is 18 at (4, 3 ).
6. Solve the LPP graphically : Maximize and minimise Z= 3x +9y S.t.C.
x +3y ≤ 60 , x +y ≥ 10 , x ≤ y , x≥0 , y≥0 .
x+3y = 60 x+y = 10
x 0 60 x 0 10
y 20 0 y 10 0
Draw the line y=x by taking (0,0) , (5, 5) or (10,10)
The corner points are A(0,10) , B(5,5), C(15,15) , D(0,20)
Corner point Z = 3x + 9y
A(0,10) 0+90=90
B(5,5) 15+45=60 ← min
C(15,15) 45+135=180← max
D(0,20) 0+180 =180← max
The minimum value of Z is 60 at B(5,5) . The maximum value of Z is 180 at C(15,15) and D(0,20) .
7. Solve the LPP graphically : Maximize Z= 5x +3 y S.t.C. 3x +5y ≤ 15 , 5x +2y ≤ 10 , x≥0 , y≥0 .
[ very less chance for exam -2024 ]
3x+5y = 15 5x+2y = 10
x 0 5 x 0 2
y 3 0 y 5 0
20 45
The corner points are O(0,0) , A(2,0), B(0, 3) , C( , )
19 19
Corner point Z= 5x +3 y
O( 0 , 0 ) 0+0=0
A( 2 , 0 ) 10+0=10
B( 0 , 3 ) 0+9=9
20 45 235
C( , ) =12.36 ← max
19 19 19
235 20 45
The maximum value of Z is at C( , ) .
19 19 19

SMART PACKAGE GGPUC(BB0136) HOSAKOTE