PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING

PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
𝑥1 −2 𝑥2 −2
PASSING PACKAGE (2025EXAM) One-one: Consider 𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⇒ = ⇒ (𝑥1 − 2)(𝑥2 − 3) = (𝑥2 − 2)(𝑥1 − 3) 3 3 2  2 −1 2 
2. 1) If If A =   and B =   , verify that ( A + B ) = A + B
𝑥1 −3 𝑥3 −3

FOUR &FIVE MARKS MARK QUESTIONS ⇒ 𝑥1 𝑥2 − 3𝑥1 − 2𝑥2 + 6 = 𝑥1 𝑥2 − 3𝑥2 − 2𝑥1 + 6 ⇒ −3𝑥1 − 2𝑥2 = −3𝑥2 − 2𝑥1 4 2 0 1 2 4 
I.1)Let 𝒇: 𝑹 → 𝑹 be given by (𝒙) = 𝟒𝒙 + 𝟑 . Show that f is invertible and find ⇒ −𝑥1 = −𝑥2 ⇒ 𝑥1 = 𝑥2 ⇒ 𝑓is one-one. Solution:Let
the inverse of 𝒇. Onto: Let 𝑦 ∈ 𝐵. Suppose 𝑓(𝑥) = 𝑦 ⇒
𝑥−2
= 𝑦 ⇒ 𝑥 − 2 = 𝑥𝑦 − 3𝑦 ⇒ 𝑥 − 𝑥𝑦 = 2 − 3𝑦 3 3 2   2 −1 2  3 + 2 3 − 1 2 + 2  5 3 −1 4
𝑦−3
𝑥−3 A+ B =  + = = 
Solution: Let 𝑓: ℝ → ℝ Givn by 𝑓(𝑥) = 4𝑥 + 3 = y⇒ 4x = y – 3⇒ 𝑥 =
4 2 − 3𝑦 4 2 0  1 2 4   4 + 1 2 + 2 0 + 4  5 4 4
𝑦−3
⇒ 𝑥(1 − 𝑦) = 2 − 3𝑦 ⇒ 𝑥 = ∈𝐴
Let 𝑔: ℝ → ℝ as𝑔(𝑦) = 𝑥 = 1−𝑦
4  5 5
2−3𝑦
 
LHS= ( A + B )
4𝑥+3−3 4𝑥 𝑥−2 −2 2−3𝑦−2−2𝑦
Consider 𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(4𝑥 + 3) = =  3 − 1 4
1−𝑦
4
=
4
=𝑥 ⇒ ∀𝑦 ∈ 𝐵∃𝑥 ∈ 𝐴such that 𝑓(𝑥) = = 2−3𝑦 = =𝑦
𝑥−3 −3 2−3𝑦−3+3𝑦
 4 
1−𝑦

Consider (𝑓𝑜𝑔)(𝑦) = 𝑓(𝑔(𝑦)) = 𝑓 (

𝑦−3
) = 4(
𝑦−3
)+3 =𝑦−3+3= 𝑦  4 
4 4 ⇒ 𝑓is onto f is both one-one and onto. ∴ 𝑓𝑖𝑠𝑏𝑖𝑗𝑒𝑐𝑡𝑖𝑣𝑒.
⇒ 𝑔𝑜𝑓(𝑥) = 𝕀𝑅 and 𝑓𝑜𝑔(𝑦) = 𝕀𝑅 ⇒ 𝑓is invertible 5) Prove that the function f : R→R, given by f (x) = 2x, is one-one and onto.  3 4  2 1   3 + 2 4 + 1   5 5
     
Solution: Given 𝑓: ℝ → ℝ as𝑓(𝑥) = 2𝑥 A + B =  3 2  +  −1 2  =  3 − 1 2 + 2  =  3 − 1 4 
𝑥−3
⇒𝑔=𝑓 −1
𝑜𝑟𝑔(𝑥) = 𝑓 −1
(𝑥) ∴ 𝑓 −1 (𝑥)
= . RHS =
4
Consider,f (𝑥1 ) = f (𝑥2 ) ⇒2𝑥1 = 2𝑥2 ⇒𝑥1 = 𝑥2 .  2 0  2 4  2 + 2 0 + 4  4 
𝟐) State whether the function f is bijective. Justify your answer.
       4 
f : R →R defined by f (x) = 3 – 4x. ⇒ 𝑓is one-one.
Clearly LHS= RHS ⇒ ( A + B )
Solution: Given 𝑓 ∶ ℝ → ℝas𝑓(𝑥) = 3 – 4𝑥
Let 𝑦 ∈ ℝ, Suppose 𝑓(𝑥) = 𝑦 ⇒ 2𝑥 = 𝑦 ⇒ 𝑥 =
𝑦
∈ℝ = A + B
2
Consider 𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⇒ 3 − 4𝑥1 = 3 − 4𝑥2 ⇒ −4𝑥1 = −4𝑥2 ⇒ 𝑥1 = 𝑥2
𝑦 𝟎 𝟔 𝟕 𝟎 𝟏 𝟏 𝟐
∴ 𝑓is one-one ⇒ ∀𝑦 ∈ ℝ∗ ∃𝑥 ∈ ℝ∗ such that 𝑓(𝑥) = 2𝑥 = 2 ( ) = 𝑦 2) If𝑨 = [−𝟔 𝟎 𝟖] , 𝑩 = [𝟏 𝟎 𝟐] , 𝒂𝒏𝒅𝑪 = [−𝟐]. Calculat𝒆𝑨𝑪, 𝑩𝑪𝒂𝒏𝒅
2
𝟕 −𝟖 𝟎 𝟏 𝟐 𝟎 𝟑
Let 𝑦 ∈ ℝ Suppose 𝑓(𝑥) = 𝑦 ⇒ 𝑓is onto.
(𝑨 + 𝑩)𝑪. Also, verify that (𝑨 + 𝑩)𝑪 = 𝑨𝑪 + 𝑩𝑪.
⇒ 3 − 4𝑥 = 𝑦 ⇒ −4𝑥 = 𝑦 − 3 ⇒ 𝑥 =
3−𝑦
∈ℝ f is both one-one and onto.
4 0 7 8
3−𝑦 6) Show that the function 𝒇 ∶ 𝑹∗ → 𝑹∗ defined by 𝒇(𝒙) = is one-one and onto,
𝟏 Solution: Now A + B =  −5 0 10 
⇒ ∀𝑦 ∈ ℝ∗ ∃𝑥 ∈ ℝ∗ such that 𝑓(𝑥) = 3 – 4𝑥 = 3 − 4( )=𝑦 𝒙 
 8 −6 0 
4
where 𝑹∗ is the set of all non-zero real numbers. Is the result true, if the
⇒ 𝑓is onto. domain 𝑹∗ is replaced by N with co-domain being same as𝑹∗ ?
1 0 7 8   2   0 − 14 + 24  10 
∴f is both one-one and onto.∴ f is bijective Solution:Given 𝑓 ∶ 𝑅∗ → 𝑅∗ as 𝑓(𝑥) = LHS =
𝑥 ( A + B)C = 
 −5 0 10       
  −2  =  −10 + 0 + 30  =  20 
1 1
3) State whether the function f is bijective. Justify your answer. f : R →R One-one: Let 𝑥1 , 𝑥2 ∈ 𝑅∗ (Domain) ,Suppose 𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⇒ = ⇒ 𝑥1 = 𝑥2 
8 −6 0 3    16 + 12 + 0    28
𝑥1 𝑥2
defined by f (x) = 1 + 𝒙𝟐 . 0 6 7   2   0 − 12 + 21   9 
⇒ 𝑓is one-one
Also
Solution: Given 𝑓 ∶ ℝ → ℝas𝑓(𝑥) = 1 + 𝑥 2
1 1
AC =  −6 0 8   −2 =  −12 + 0 + 24  = 12 
Consider 1, −1 ∈ ℝ ⇒ 𝑓(1) = 1 + 12 = 2 and 𝑓(−1) = 1 + (−1)2 = 2 Onto: Let 𝑦 ∈ 𝑹∗ (Co-domain). Suppose, 𝑓(𝑥) = 𝑦 ⇒ = 𝑦 ⇒ 𝑥 = ∈ 𝑹∗  7 −8 0   3   14 + 16 + 0  30 
𝑥 𝑦

⇒ 𝑓(1) = 𝑓(−1), but 1 ≠ −1 ⇒ 𝑓is not one-one. ⇒ ∀𝑦 ∈ 𝑹∗ ∃𝑥 ∈ 𝑹∗ such that 𝑓(𝑥) = =

1 1
1 =𝑦 ⇒ 𝑓is onto. 0 1 1   2   0 − 2 + 3  1 
𝑥
𝑦 and BC = 1 2   −2  =  2 + 0 + 6  =  8 
Onto: For −2 ∈ ℝ , Suppose 𝑓(𝑥) = −2  0
1
For the function 𝑓 ∶ 𝑁 → 𝑅∗ as 𝑓(𝑥) = 1 2 0   3   2 − 4 + 0   −2
⇒ 1 + 𝑥 2 = −2 ⇒ 𝑥 2 = −3 ⇒ 𝑥 = ±√−3 ∉ ℝ 𝑥

⇒ 𝑓(𝑥) ≠ −2 ∀𝑥 ∈ ℝ ⇒ 𝑓is not onto One-one: Let 𝑥1 , 𝑥2 ∈ 𝑁(Domain), Suppose 𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⇒
1
=
1
⇒ 𝑥1 = 𝑥2  9   1  10 
AC + BC = 12  +  8  =  20 =RHS
𝑥1 𝑥2
⇒f is neither one-one nor onto.∴ 𝑓𝑖𝑠𝑛𝑜𝑡𝑏𝑖𝑗𝑒𝑐𝑡𝑖𝑣𝑒. ⇒ 𝑓is one-one
30  −2  28
4) Let A = R – {3} and B = R – {1}.Consider the function f : A →B defined by 2 2
Onto: Let ∈ 𝑹∗ (Co-domain), Suppose, 𝑓(𝑥) =
𝒙−𝟐
3 3
Clearly LHS = RHS ⇒ ( A + B ) C = AC + BC
f (x) =( ).Is f one-one and onto? Justify your answer. 1 2 3
𝒙−𝟑
𝑥−2
⇒ = ⇒ 𝑥 = ∉ 𝑵 ⇒ 𝑓is not onto⇒ 𝑓is one-one but not onto.
𝑥 3 2
Solution: Given 𝑓: 𝐴 → 𝐵 as𝑓(𝑥) = ( ), where A = R – {3} and B = R – {1}
𝑥−3

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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
3. 1. Solve the following system of linear equations using matrix ethod
3) If  1 
 and B = −1 2 1 , verify that ( AB ) = BA 1 2 −3 3 −1 2  4 1 2
 
A= 5) If A = 5 0 2  , B =  4
1) 𝟑𝒙 − 𝟐𝒚 + 𝟑𝒛 = 𝟖 , 𝟐𝒙 + 𝒚 − 𝒛 = 𝟏𝒂𝒏𝒅𝟒𝒙 − 𝟑𝒚 + 𝟐𝒛 = 𝟒
 −4 
  2 5  and C =  0 3 2  then x
 3 −2 3 8
 3 
1 −1 1   2 0 3  1 −2 3  Solution: Let A =[2 1 −1], X =[y] and B =[1]
 1 
Solution:We have 4 −3 2 z 4
A= A + B and B − C . Also, verify that A + ( B − C) = ( A + B) − C .

 −4  and B =  −1 2 1 compute 3 −2 3

 3  ∴ |A|=|2 1 −1| = 3(2-3)+2 (4+4)+3(-6-4) = -17.
3 −1 2   4 1 2   −1 −2 0 4 −3 2
 1   −1 2 1 
Solution: Now B − C =  4 2 5  − 0 2  =  4 −1 3
AB =   1 =  −4 
adj A
 −4   −1 2  4 −8   3 Given equation is AX = B ∴ X =A−1 B = ( |A| ) B

 3  
 −3 6 3 
  2 0 3  1 −2 3   1 2 0  −1 −5
−1
 −1 −3 𝑎𝑛𝑑 adj A = [ −8 −69]
4
= LHS 1 2 −3  −1 −2 0  0 0 −3
(AB)' =  −8 6 −10 17
2
 −4 3
 A+(B-C) = 5 0 2  +  4 −1 3 = 9 −1 5  =LHS adj A 1
−1 −5 −1 8
1
−8 − 5 − 4
1  ∴ X =A−1 B = ( |A| ) B= [ −8 −6 9 ] [1] = [−64 − 6 + 36]
1 −1 1   1 2 0  2 1 1  −17
−10 1 7 4
−17
−80 + 1 + 28
 −1 x x
−17 1
Now B ' =  2  and A' = 1 −4 3 4 1 −1 1
  ∴ [y ] = [−34]⇒[y] = [2]⇒ x = 1, y = 2, z = 3
A+ B =  7 
−17
 1  9 2  z −51 z 3

3 −1 4 2) 𝒙 − 𝒚 + 𝒛 = 𝟒 , 𝟐𝒙 + 𝒚 − 𝟑𝒛 = 𝟎𝒂𝒏𝒅𝒙 + 𝒚 + 𝒛 = 𝟐
 −1  −1 4 −3 1 −1 1 x 4
= RHS
BA =  
 2  1 −4 3 = 
2 −8 6 4 1 −1  4 1 2 0 0 −3 Solution:Let A =[2 1 −3], X =[y] and B =[0]
= RHS

1  
1 −4 3 (A+B)-C = 9 2 7  −  0 3 2  = 9 −1 5  1 1 1 z 2
 3 −1 4  1 −2 3  2 1 1  1 −1 1
Clearly LHS = RHS ⇒ ( AB ) = BA ∴ |A|=|2 1 −3| = 1(1+3)-(-1)(2+3)+1(2-1) = 4+5+1=10.

−𝟐
Clearly LHS = RHS ⇒ A + ( B − C ) = ( A + B) − C 1 1 1
adj A
Given equation is AX = B ∴ X =A−1 B = ( |A| ) B
4) If 𝑨 = [ 𝟒 ] , 𝑩 = [𝟏 𝟑 −𝟔],Verify that (𝑨𝑩)′ = 𝑩′ 𝑨′ . 0
𝟓 6) If A= [1] , 𝐵 = [1 5 7], thenverify that (𝑨𝑩)′ = 𝑩′ 𝑨′ . 4 2
2
2 𝑎𝑛𝑑 adj A = [−5 5]
0
 −2 
Solution:We have A =  4  and B = 1 3 0 0 0 0 1 −2
3
   −6
Solution:Let A= [1] 𝑎𝑛𝑑𝐵 = [1 5 7] ∴AB = [1 5 7] 4 2 2 4 16 + 0 + 4
adj A 1 1
 5  2 2 10 14 ∴ X =A−1 B =( |A| ) B= [−5 0 5] [0] = [−20 + 0 + 10]
10 10
0 1 2 1 −2 3 2 4+0+6
 −2   −2 −6 12  x 20 x 2
AB =   −6 =  −24  LHS = (𝐴𝐵)′ = [0 5 10]
 4  1
1
3  4 12  ∴ [y] = [−10]⇒[y] = [−1]⇒ x = 2, y = - 1, z = 1

 5  
 5 15 −30 

0 7 14 z
10
z
10 1
1
LHS =  −2 5  3) 𝟐𝒙 + 𝟑𝒚 + 𝟑𝒛 = 𝟓 , 𝒙 − 𝟐𝒚 + 𝒛 = −𝟒𝒂𝒏𝒅𝟑𝒙 − 𝒚 − 𝟐𝒛 = 𝟑.
4
Now 𝐵′ = [5] 𝑎𝑛𝑑𝐴′ = [0 1 2]
( AB ) =  −6 12 15  7
2 3 3 x 5

12 −24 −30 
 Solution:Let A = [1 −2 1 ], X =[y] and B =[−4]
0 1 2
3 −1 −2 z 3
 1  RHS = 𝐵′ 𝐴′ = [0 5 10]
Now 2 3 3
B =  
 3  and A =  −2
 4 5 0 7 14 ∴ |A|=|1 −2 1 | = 2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40.

 −6 
Clearly LHS = RHS ⇒ ( AB ) = BA
 3 −1 −2
adj A
Given equation is AX = B ∴ X =A−1 B = ( |A| ) B.
 1   −2 4 5  3 4
−1 2 1
7) If A= [−1 2] , 𝐵 = [ ], then verify that (𝑨 + 𝑩)′ = 𝑨′ + 𝑩′ 5 3 9
BA =  
 3   −2 4 5 = 
 −6 12 15 
=RHS
1 2 3 and adj A =[5 −13 1 ]
0 1

 −6 
 
12 −24 −30 
 3 4 5 11 −7
−1 2 1 5 3 9 5 25 − 12 + 27 40
8) If A= [−1 2] , 𝐵 = [ ], then verify that (𝑨 − 𝑩)′ = 𝑨′ − 𝑩′ 1 1 1
Clearly LHS = RHS ⇒ ( AB ) = BA
1 2 3 ∴ X =A−1 B = [5 −13 1 ] [−4]= [ 25 + 52 + 3 ] = [ 80 ]
0 1 40 40 40
5 11 −7 3 25 − 44 − 21 −40

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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
x 1 0 −1 2  I = 4 A − A2 = A(4 − A)
⇒[y] = [ 2 ]⇒ x = 1, y = 2, z = -1 𝑎𝑛𝑑 adj A = [2 −9 23]
z −1 1 −5 13  A−1I = A−1 A(4 − A) = I (4 − A)
4) 𝒙 − 𝒚 + 𝟐𝒛 = 𝟕, 𝟑𝒙 + 𝟒𝒚 − 𝟓𝒛 = −𝟓𝒂𝒏𝒅𝟐𝒙 − 𝒚 + 𝟑𝒛 = 𝟏𝟐. 1
0 −1 2 0 1 −2  A−1 = 4I − A
1 −1 2 x 7 ∴ A−1 = [2 −9 23]= [−2 9 −23]
Solution: Let A = [3 4 −5], X =[y] and B =[−5]
−1
1 −5 13 −1 5 −13  4 0   2 3   2 −3
 A−1 =  − = 
2 −1 3 z 12 The given system of equations can be written in form AX=B,  0 4  1 2   −1 2 
1 −1 2 2 −3 5 x 11 3.1. Find the area of the triangle whose vertices are (3, 8), (-4, 2) and (5, 1)
∴ |A|=|3 4 −5| = 1(12-5)-(-1)(9+10)+2(-3-8) = 7+19-22=4 where A =[3 2 −4] , X =[y] and B = [−5] 3 8 1
2 −1 3 1 1 −2 z −3 1
Solution:The area of the triangle is given by ∆ = |−4 2 1|
adj A
Given equation is AX = B ∴ X =A−1 B = ( |A| ) B. 0 1 −2 11 0−5+6 1 2
5 1 1
X = A−1 B = [−2 9 −23] [−5] = [−22 − 45 + 69] = [2] 1 1 61
7 1 −3 −1 5 −13 −3 −11 − 25 + 39 3 = [3(2-1) – 8 (-4-5) +1 (-4-10)]= (3+72-14)= sq.units.
2 2 2
And adj A = [−19 −1 11 ] x 1 2.Find the area of the triangle whose vertices are (1, 0), (6, 0) and (4,3)
−11 −1 7 ⇒ [y] = [2]⇒ x = 1, y = 2, z = 3. 1 0 1
7 1 −3 7 49 − 5 − 36 8 z 3 1
1 1 1 Solution: The area of the triangle is given by ∆ = |6 0 1|
∴X =A−1 B = [−19 −1 11 ] [−5] = [−133 + 5 + 132] = [ 4 ] 2
4 4 4  3 1 4 3 1
2.1) If A =   , show that A − 5 A + 7 I = 0 . Hence find A .
−11 −1 7 12 −77 + 5 + 84 12 2 −1
1 1 15
x 2  −1 2  = [1(0-3 -0 +1(18-0)] = (-3+18) = sq. units.
2 2 2
∴ [y] = [1]⇒ x = 2, y =1, z=3 3. Find values of k if area of triangle is 4 sq units and vertices are (-2, 0),
 3 1
z 3 Solution: Let A =  , (0, 4) and (0, k).
𝟑
5) 𝟐𝒙 + 𝒚 + 𝒛 = 𝟏 , 𝒙 − 𝟐𝒚 − 𝒛 = 𝒂𝒏𝒅𝟑𝒚 − 𝟓𝒛 = 𝟗.  −1 2  −2 0 1
𝟐 1
2 1 1 x 1  3 1   3 1   8 5 15 5 7 0 Solution: | 0 4 1| = ±4⇒ -2(4-k) -0 + 1(0-0) = ±8 ⇒ -8+2k = ±8
3 A = A A = 
2
 =  ,5A =[−5 10] and 7I = [0 7]
2
0 𝑘 1
Solution: Let A =[1 −2 −1], X =[y] and B =[ 2 ]  −1 2   −1 2   −5 3 ⇒ 2k = ±8 +8⇒ 2k = 16, 2k = 0 or K =8, k = 0
0 3 −5 z 9 LHS = A2 − 5 A + 7 I 4. If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4), then
2 1 1
∴ |A|=|1 −2 −1| = 2(10+3)-1(-5-0)+1(3-0)= 26+5+3 = 34 .  8 5 15 5  7 0  0 0  find values of k.
= − + =  = 0 = RHS
0 3 −5  −5 3  −5 10   0 7  0 0  1
2 −6 1
adj A Solution: | 5 4 1| = ±35 ⇒ 2(4-4)-(-6)(5-k)+1(20-4k) =±70
Given equation is AX = B ∴ X =A−1 B = ( |A| ) B. A − 5 A + 7I = 0  7 I = 5 A − A2
2 2
𝑘 4 1
13 8 1 1 ⇒ 0+30-6k+20-4k =±70 ⇒ -10k+50 =±70
and adj A = [ 5 −10 3 ]  A−1 = (5I − A)
7𝐴−1. = 5𝐼 − 𝐴
⇒ -10k =±70-50 ⇒ -10k = 20, -10k = -120 𝑂𝑟 k =-2, k = 12
3 −6 −5 7
5. Find equation of line joining (1, 2) and (3, 6) using determinants.
13 8 1 1 13 + 12 + 9 34 1  5 0  3 1   −1 1  2 −1
=   −   A =  1 2 1
7  1 3 
1 3 1 1
∴ X =A−1 B = [ 5 −10 3 ] [ 2 ] = [5 − 15 + 27] = [ 17 ] 7  0 5  −1 2  1
34 34 34 Solution: |3 6 1| =0 ⇒ 1(6-y) - 2(3-x) + 1(3y-6x) = 0
3 −6 −5 9 3 − 9 − 45 −51 2
𝑥 𝑦 1
x 1  2 3 ⇒ 6-y -6+2x+3y-6 =0 ⇒ -4x+2y = 0 ⇒ 2x-y=0.
2) Show that the matrix A =   satisfies the equation A − 4 A + I = 0 . Hence
1 3 2
⇒ [y] = [ 1/2 ]⇒ x = 1, y = , z = -
2 2 1 2 6. . Find equation of line joining (3, 1) and (9, 3) using determinants.
z −3/2
find A−1. 3 1 1
𝟐 −𝟑 𝟓 1
Solution: |9 3 1| =0 ⇒ 3(3-y) -1(9-x) + 1(9y-3x) = 0
6) If𝑨 = [𝟑 𝟐 −𝟒] , find𝑨−𝟏 . Using 𝑨−𝟏 solve the system of equations  2 3 2
𝟏 𝟏 −𝟐 Solution: Let A =   𝑥 𝑦 1
𝟐𝒙 − 𝟑𝒚 + 𝟓𝒛 = 𝟏𝟏, 𝟑𝒙 + 𝟐𝒚 − 𝟒𝒛 = −𝟓𝒂𝒏𝒅𝒙 + 𝒚 − 𝟐𝒛 = −𝟑. 1 2 ⇒9-3y-9+x+9y-3x = 0 ⇒ -2x+6y = 0 ⇒ x - 3y = 0.
2 −3 5  2 3   2 3   7 12  8 12 1 0 4.1.1)If 𝒚 = (𝐭𝐚𝐧−𝟏 𝒙)𝟐 then prove that (𝒙𝟐 + 𝟏)𝟐 𝒚𝟐 + 𝟐𝒙(𝒙𝟐 + 𝟏)𝒚𝟏 = 𝟐
Solution:LetA =[3 2 −4] We have, A2 = AA =   =  ,4A =[4 8 ] and I =[0 1]
1 2  1 2  4 7  Solution: Given 𝑦 = (tan−1 𝑥)2
1 1 −2 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑤. 𝑟. 𝑡 on both sides
2 −3 5
∴ |A| = |3 2 −4| = 2(-4+4)-(-3)(-6+4)+5(3-2) =0 - 6 + 5 = -1 2 tan−1 𝑥
7 12  8 12  1 0   0 0  𝑦1 = ⇒(1 + 𝑥 2 )𝑦2 = 2 tan−1 𝑥
1 1 −2  A2 − 4 A + I =  − +  = =0 1+𝑥 2
adj A  4 7   4 8  0 1 0 0  ⇒ (1 + 𝑥 2 )𝑦2 + 2𝑥𝑦1 =
2
⇒ (1 + 𝑥 2 )2 𝑦2 + 2𝑥(1 + 𝑥 2 )𝑦1 = 2
∴ A−1 = |A| 1+𝑥 2
Now A − 4 A + I = 0
2

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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
2) If 𝐲 = 𝐬𝐢𝐧−𝟏 𝐱 then prove that (𝟏 − 𝐱 𝟐 )𝐲𝟐 − 𝐱𝐲𝟏 = 𝟎. 6)If 𝒚 = 𝟑𝒆𝟐𝒙 + 𝟐𝒆𝟑𝒙 then prove that
𝒅𝟐 𝒚
−𝟓
𝒅𝒚
+ 𝟔𝒚 = 𝟎.
𝒅𝒙𝟐 𝒅𝒙
Solution:Given 𝑦 = sin−1 𝑥 2𝑥 3𝑥
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given f(x) is continuous at x = 3.
𝑑𝑦 1 𝑑𝑦
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given 𝑦 = 3𝑒 + 2𝑒
Differentiate w.r.t x, we get⇒ = ⇒ √1 − 𝑥 2 =1 ⇒ lim 𝑓(𝑥) = 𝑓(3)
𝑑𝑥 √1−𝑥 2 𝑑𝑥 𝑑𝑦 𝑥→3
= 6𝑒 2𝑥 + 6𝑒 3𝑥 = 6(𝑒 2𝑥 + 𝑒 3𝑥 )
Again Differentiate w.r.t x, we get 𝑑𝑥 ⇒ LHL = RHL ⇒ lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(3)
𝑥→3 𝑥→3
𝑑2𝑦 2𝑥 3𝑥 2𝑥 3𝑥 )
𝑑2𝑦 2𝑥 𝑑𝑦 ∴ = 12𝑒 + 18𝑒 = 6(2𝑒 + 3𝑒 ⇒lim(𝑎𝑥 + 1) = lim (𝑏𝑥 + 3) = 3a+1
√1 − 𝑥 2 𝑑𝑥 2 − =0 𝑑𝑥 2 𝑥→3 𝑥→3
2√1−𝑥 2 𝑑𝑥
𝑑2𝑦 𝑑𝑦
𝑑 𝑦2 𝑑𝑦 LHS= −5 + 6𝑦 = 6(2𝑒 2𝑥 + 3𝑒 3𝑥 ) − 30(𝑒 2𝑥 + 𝑒 3𝑥 ) + 6(3𝑒 2𝑥 + 2𝑒 3𝑥 ) ⇒ 3𝑎 + 1 = 3𝑏 + 3 = 3𝑎 + 1 ⇒ 3𝑎 − 3𝑏 = 3 − 1 = 2
⇒ (1 − 𝑥 2) 2 −𝑥 =0 𝑑𝑥 2 𝑑𝑥
2
𝑑𝑥 𝑑𝑥
= 12𝑒 2𝑥 + 18𝑒 3𝑥 − 30𝑒 2𝑥 − 30𝑒 3𝑥 + 18𝑒 3𝑥 + 12𝑒 2𝑥 = 0 = 𝑅𝐻𝑆. ⇒𝑎 − 𝑏 = .
3) If 𝒚 = 𝟑 𝐜𝐨𝐬(𝐥𝐨𝐠 𝒙) + 𝟒 𝐬𝐢𝐧(𝐥𝐨𝐠 𝒙) then prove that 𝒙𝟐 𝒚𝟐 + 𝒙𝒚𝟏 + 𝒚 = 𝟎. 3
2.1) Find the value of K, so that the function 𝟐 𝒊𝒇𝒙 ≤ 𝟐
Solution:Given𝑦 = 3 cos(𝑙𝑜𝑔𝑥) + 4sin (𝑙𝑜𝑔𝑥) 5) 𝒇(𝒙) = {𝑲𝒙 is continuous at 𝒙 = 𝟐.
𝒌𝒙 + 𝟏 𝒊𝒇 𝒙 ≤ 𝟓 𝟑 𝒊𝒇𝒙 > 2
Differentiating w.r.t x on both sides 𝑓(𝒙) = { is continuous at 𝒙 = 𝟓.
𝟑𝒙 − 𝟓 𝒊𝒇 𝒙 > 5 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:Given f(x) is continuous at x = 2.
3sin (𝑙𝑜𝑔𝑥) 4cos (𝑙𝑜𝑔𝑥)
⇒ 𝑦1 = − + 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given f(x) is continuous at x = 5. ⇒ lim 𝑓(𝑥) = 𝑓(2)
𝑥 𝑥 𝑥→2
⇒ lim 𝑓(𝑥) = 𝑓(5)
⇒ 𝑥𝑦1 = −3 sin(𝑙𝑜𝑔𝑥) + 4cos (𝑙𝑜𝑔𝑥) 𝑥→5 ⇒ LHL = RHL ⇒ lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(2)
𝑥→2 𝑥→2
Again differentiating on both sides we get ⇒ LHL = RHL= f(5) ⇒ lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)= f(5) ⇒lim𝑘𝑥 2 = lim 3 = 4k
𝑥→5 𝑥→5
𝑥→2 𝑥→2
3 cos(𝑙𝑜𝑔𝑥) 4𝑠𝑖𝑛 (𝑙𝑜𝑔𝑥) ⇒lim(𝑘𝑥 + 1) = lim 3𝑥 − 5=5k + 1 2
𝑥𝑦2 + (1)𝑦1 = − − 𝑥→5 𝑥→5 ⇒ 𝑘(2) = 3 = 4 ⇒ 4𝑘 = 3
𝑥 𝑥 9 3
⇒ 𝑥 2 𝑦2 + 𝑥𝑦1 = −[3 cos(𝑙𝑜𝑔𝑥) + 4sin (𝑙𝑜𝑔𝑥)] ⇒5𝑘 + 1 = 3(5) − 5 ⇒ 5𝑘 + 1 = 10⇒5𝑘 = 9 ⇒ 𝑘 = . ⇒𝑘 =
5 4
𝑥 2 𝑦2 + 𝑥𝑦1 = −𝑦 ⇒ 𝑥 2 𝑦2 + 𝑥𝑦1 + 𝑦 = 0 𝒌𝒙 + 𝟏 𝒊𝒇 𝒙≤ 6)Find the value of 𝒂𝒂𝒏𝒅𝒃 such that the function
2) 𝒇(𝒙) = { is continuous at x = 𝝅
𝒅𝒚 𝒅𝟐 𝒚 𝒅𝒚 𝟐 𝐜𝐨𝐬 𝒙 𝒊𝒇 𝒙> 𝟓 𝒊𝒇𝒙 ≤ 𝟐
4)If 𝒆𝒚 (𝒙 + 𝟏) = 𝟏, prove that = −𝒆𝒚 and hence prove that =( ) .
𝒅𝒙 𝒅𝒙𝟐 𝒅𝒙 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given f(x) is continuous at x = 𝜋 𝒇(𝒙) = { 𝒂𝒙 + 𝒃𝒊𝒇 𝟐 < 𝒙 < 10 is continuous function.
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: 𝑒 𝑦 (𝑥 + 1) = 1 ⇒ lim 𝑓(𝑥) = 𝑓(𝜋) 𝟐𝟏 𝒊𝒇𝒙 ≥ 𝟏𝟎
𝑥→𝜋
Differentiate w.r.t x on both sides,then we get 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given f(x) is continuous function.(continuous ∀ x ∈ 𝑅)
𝑑𝑦 𝑑𝑦 1
⇒ LHL = RHL ⇒ lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(𝜋)
𝑥→𝜋
𝑒 𝑦 + (𝑥 + 1)𝑒 𝑦 =0⇒ =− = −𝑒 𝑦 𝑥→𝜋
⇒ f(x) is continuous at x = 2 and at x = 10.
𝑑𝑥 𝑑𝑥 𝑥+1
⇒lim (𝑘𝑥 + 1) = lim 𝑐𝑜𝑠𝑥 = 4𝜋 + 1
Again differentiate w.r.t x, we get 𝑥→𝜋 𝑥→𝜋 ⇒ lim 𝑓(𝑥) = 𝑓(2)𝑎𝑛𝑑 lim 𝑓(𝑥) = 𝑓(10)
𝑥→2 𝑥→10
2 2 2 ⇒ k𝜋 + 1 = −1 = k𝜋 + 1 ⇒ k𝜋 = −1 − 1 = −2
𝑑 𝑦 𝑑𝑦 𝑑 𝑦 𝑑𝑦 ⇒ lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)= f(2) and lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(10)
= −𝑒 𝑦 ⇒ = ( ) ⇒𝑘 = − .
2 𝑥→2 𝑥→2 𝑥→10 𝑥→10
𝑑𝑥 2 𝑑𝑥 𝑑𝑥 2 𝑑𝑥 𝜋
𝒅𝟐 𝒚 𝒅𝒚 𝒌𝒄𝒐𝒔𝒙 
⇒lim 5 = lim(𝑎𝑥 + 𝑏) = 5 and lim (𝑎𝑥 + 𝑏) = lim 21 = 21
𝑥→2 𝑥→2 𝑥→10 𝑥→10
5)If 𝒚 = 𝑨𝒆𝒎𝒙 + 𝑩𝒆𝒏𝒙 then prove that − (𝒎 + 𝒏) + 𝒎𝒏𝒚 = 𝟎. 𝒊𝒇 𝒙≠ 
𝒅𝒙𝟐 𝒅𝒙
3) 𝒇(𝒙) = { −𝟐𝒙  is
𝟐
continuous at 𝒙 = . ⇒5 = 2𝑎 + 𝑏 ⇒ 2𝑎 + 𝑏 = 5 ……….(1) and 10𝑎 + 𝑏 = 21 ………(2)
𝟐
Solution: 𝑦 = 𝐴𝑒 𝑚𝑥 + 𝐵𝑒 𝑛𝑥 𝟑 𝒊𝒇 𝒙=
𝟐 Solving (1) and (2) ,we get a = 2 and b = 1.
Differentiating w.r.t x on both sides 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Given f(x) is continuous at 𝑥 =
𝜋

𝑑𝑦
2 5.1) Find the area enclosed by the circle x 2 + y 2 = a 2 
= 𝐴𝑚𝑒 𝑚𝑥 + 𝐵𝑛𝑒 𝑛𝑥 𝜋 𝑘𝑐𝑜𝑠 𝑥
Solution : Given circle x + y = a 
2 2 2
𝑑𝑥 ⇒ lim𝜋 𝑓(𝑥) = 𝑓 ( ) ⇒ lim𝜋 =3
𝑥→ 2 𝑥→ 𝜋 − 2𝑥
Again differentiate w.r.t x on both sides 2 2
⇒ 𝑥 = √𝑎2 − 𝑥 2 and center origin and radius a
𝜋
2 𝑠𝑖𝑛 ( 2 −𝑥)
𝑑 𝑦 ⇒𝑘 lim𝜋 =3
= 𝐴𝑚2 𝑒 𝑚𝑥 + 𝐵𝑛2 𝑒 𝑛𝑥 𝑥→
𝜋
2( −𝑥)
𝑑𝑥 2 2 2

𝑑2 𝑦 𝑑𝑦 𝑠𝑖𝑛 𝜃 𝑘
LHS = − (𝑚 + 𝑛) + 𝑚𝑛𝑦 ⇒ 𝑘 lim = 3 ⇒ (1) = 3
𝑑𝑥 2 𝑑𝑥
𝜃→0 2𝜃 2
= 𝑚2 𝐴𝑒 𝑚𝑥 + 𝑛2 𝐵𝑒 𝑛𝑥 − (𝑚 + 𝑛)(𝐴𝑚2 𝑒 𝑚𝑥 + 𝐵𝑛2 𝑒 𝑛𝑥 ) + 𝑚𝑛𝑦 ⇒𝑘 = 6
= 𝑚2 𝐴𝑒 𝑚𝑥 + 𝑛2 𝐵𝑒 𝑛𝑥 − 𝐴𝑚2 𝑒 𝑚𝑥 − 𝐵𝑚𝑛𝑒 𝑛𝑥 − 𝐴𝑚𝑛𝑒 𝑚𝑥 − 𝑛2 𝐵𝑒 𝑛𝑥 + 𝑚𝑛𝑦 4) Find the relationship between a and b so that the function f defined
= −𝐵𝑚𝑛𝑒 𝑛𝑥 − 𝐴𝑚𝑛𝑒 𝑚𝑥 + 𝑚𝑛𝑦 = −𝑚𝑛(𝐴𝑒 𝑚𝑥 + 𝐵𝑒 𝑛𝑥 ) + 𝑚𝑛𝑦 by
= −𝑚𝑛𝑦 + 𝑚𝑛𝑦 = 0 = RHS 𝒂𝒙 + 𝟏 𝒊𝒇𝒙 ≤ 𝟑
𝒇(𝒙) = { is continuous at 𝒙 = 𝟑.
𝒃𝒙 + 𝟑 𝒊𝒇𝒙 > 3

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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
a

4 6) Find the area bounded by the curve y = cos x between x = 0 and x = 2 
x x  
a
16
Required area = 4 (area AOB) = 4  y dx = 4 a 2 − x 2 dx = 3  16 − x 2 + sin −1  = 3 0 + 8 − 0 − 0 = 12 square unit. Solution : Given curve y = cos x
0 0
2 2 4 0  2 
 a 2 
a
x a2 x  x2 y2
= 4  a 2 − x 2 + sin −1  = 4  0 + sin −1 1 − (0 + 0)  4)Find the area of the region bounded by the ellipse + = 1
2 a 0 4 9
2  2   2 2
x y
a2  Solution : Given ellipse + = 1
=4  = a2 square unit. 16 9 From the figure, we get
2 2 3
⇒ 𝑥 = √4 − 𝑥 2 and center origin and vertices (0, 2) / 2 3 / 2 2

 cos x dx + 
2
2) Find the area enclosed by the ellipse
x2 y2
+ =1
Area = cos x dx +  cos x dx
a 2 b2 0 / 2 3 / 2

2 2
x y
Solution : Given ellipse + =1
= sin x 0 + [sin x]3/ 2/ 2 + sin x 3 / 2
/ 2 2
a 2 b2
𝑏
⇒ 𝑥 = √𝑎2 − 𝑥 2 and center origin and vertices (0, a) = 1 + 2 + 1 = 4 square unit.
𝑎

7)Find the area bounded by the curve y = sin x between x = 0 and x = 2 

Required area = 4 (area AOB)
2 2 Solution : Given curve y = sin x
3
= 4  y dx = 4  4 − x 2 dx
0 0
2

2
x 4 x  
= 6  4 − x 2 + sin −1  = 6 0 + 2 − 0 − 0
2 2 2 0  2 
= 6 square unit.
a a
From the figure, we get
b 5)Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the
 a
 2
Required area = 4 (area AOB) = 4 y dx = 4 a 2 − x 2 dx
0 0 ordinates x = −1, x = 1  Required area A = sin x dx =   sin x dx

Solution : Given line is y = 3x + 2
a 0
4b  x 2 a2 x 4b  a2  
a − x 2 + sin −1  = 0 +  − 0 − 0  =  − cos x 0 |  − cos x 
 2
= 
a 2 a 0 a  −2
2 2 2  line cuts the x-axis at x = in (−1, 1)
= ab square unit. 3 = 2 + 2 = 4 square unit.

x2 y2 SIX MARKS QUESTIONS

3)Find the area of the region bounded by the ellipse + = 1
16 9 1.Solve the following linear programming problem graphically:
x2 y2 Maximise Z = 4x + y ,subject to the constraints: x + y ≤50, 3x + y ≤90 ,
Solution : Given ellipse + = 1 and x ≥0, y ≥0.
16 9
3 Solution :Maximise Z = 4x + y
⇒ 𝑥 = √16 − 𝑥 2 and center origin and vertices (0, 4) x 0 50
4 Region represented by x + y  50  Consider x + y = 50
y 50 0
 Required area
−2/3 1 Clearly O (0, 0) satisfies x + y  50   Solution region is on origin side.
A= 
−1
(3x + 2) dx + 
−2/3
(3x + 2) dx Region represented by 3x + y  90 ,Consider 3x + y = 90 x 0 30
−2/3 1 y 90 0
 3x 2   3x 2 
= + 2x  + + 2x
Required area = 4 (area AOB)  2 −1  2  −2/3 Clearly O (0, 0) satisfies 3x + y  90  Solution region is on origin side.
4 4
3 And x ≥ 0, y ≥ 0 are called non negative constraints which represents the first
= 4  y dx 4  16 − x 2 dx 2 4 3 3 2 4  1 25 13
4 = − − +2 +  +2− +  = + = square unit. quadrant.
2 3 3 6 6
0 0
3 3 2 3
Now plotting the graph as follows

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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
Corner points A (0, 5), C (0, 6) and E (4, 3) Corresponding value of
Sl. No. Corner points
Corresponding value of z = 3x + 9 y
Sl. No. Corner points C (0, 10) 90
z = 200 x + 500 y 1.
2. G (5, 5) 60  Minimum
2500
1. A (0, 5) 3. H (15, 15) 180 Maximum
3000 4. A (0, 20) 180 Maximum
2. C (0, 6)
 The minimum value of Z is 60 at the point (5, 5). and maximum value of Z is 180 at
2300  Minimum all points on line segment joining the points A (0, 20) and H (15, 15).
3. E (4, 3)
 Solution lies in the region OAED. 4. Minimise Z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Corner points O (0, 0),A (0, 50), E (20, 30) and D (30, 0). Hence, minimum value of z is 2300 at the point (4, 3). Solution :To Minimise Z = - 3x + 4y
Corresponding value of 3. Solve the following problem graphically:Minimise and Maximise Z = 3x + 9y. Region represented byx + 2y ≤ 8, Consider, x + 2y = 8 x 0 8
Sl. No. Corner points y 4 0
z = 4x + y subject to the constraints: x + 3y ≤60, x + y ≥10 ,x ≤y .and x ≥0, y ≥0. Clearly O (0, 0) satisfies x + 2y ≤ 8 Solution region contain
Solution :Minimise and Maximise Z = 3x + 9y
1. O (0, 0) 0 the origin.
Region represented by x + 3 y  60 , x 0 60
A (0, 50) 50 Region represented by 3x + 2y ≤ 12 ,Consider,3x + 2y = 12 x 0 4
2. Consider the equation x + 3 y = 60 y 20 0
E (20, 30) 110 Clearly O (0, 0) not satisfies x + y  10  y 6 0
3. Clearly O (0, 0) satisfies x + 3 y  60  Solution region contain the origin.
D (30, 0) 120  Maximum  Solution region does not contain the origin.
4. Region represented by x + y  10  , x 0 10 And x ≥ 0, y ≥ 0 are called non negative constraints which represents the first
Hence, maximum value of Z is 120 at the point (30, 0). Consider the equation x + y = 10 . quadrant.
y 10 0
2.. Solve the following linear programming problem graphically:
Clearly O (0, 0) not satisfies x + y  10  Now plotting the graph as follows
Minimise Z = 200x + 500y .subject to the constraints:x+2y ≥10,3x + 4y ≤24 and
x ≥0, y ≥0.  Solution region does not contain the origin. x 0 10
Solution :Minimise Z = 200x + 500y Region represented by x  y i.e., x − y  0 ,Consider x = y i.e., y 0 10
: Region represented by x + 2 y  10  Consider x + 2 y = 10 x− y=0
x 0 10
y 5 0 Plot the points O(0, 0) and A(10, 10). Join OA and produce .
and (0, 10) satisfies x − y  0   Solution region contain the point (0, 10).
Clearly O (0, 0) not satisfies x + 2 y  10   Solution region does not contain the origin. And x ≥ 0, y ≥ 0 are called non negative constraints which represents the first
Region represented by 3x + 4 y  24 ,Consider 3x + 4 y = 24 quadrant.
x 0 8
Now plotting the graph as follows
y 6 0 Thus, the feasible region is OBCD,
Corner points D (0, 4), B (4, 0), C (2, 3) and O (0, 0)
Clearly O (0, 0) satisfies 3x + 4 y  24  Solution region contain the origin.
And x ≥ 0, y ≥ 0 are called non negative constraints which represents the first
quadrant.
Now plotting the graph as follows
Minimum value = -12 at (4, 0).
5. Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution : To Maximise Z =3x + 2y
Region represented by x + 2y ≤ 10, Consider, x + 2y = 10 x 0 10
Clearly O (0, 0) satisfies x + 2y ≤ 10 y 5 0
 Solution region contain the origin.
Region represented by 3x + y ≤ 15, Consider, 3x + y = 15 x 0 5
Thus, the feasible region is ACGH, Clearly O (0, 0) satisfies 3x + y ≤ 15 y 15 0
Corner points A (0, 20), C (0, 10), G (5, 5) and H (15, 15)  Solution region contain the origin.
Solution lies in the region ACE. And x ≥ 0, y ≥ 0 are called non negative constraints which represents the first
quadrant.
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PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E) PASSING PACKAGE II PU MATHEMATICS 2024 EXAM (PRT –D AND E)
Now plotting the graph as follows

Thus, the feasible region is ABCO,

Corner points A (5, 0), B (4, 3), C (0, 5) and O (0, 0) Thus, the feasible region is ABCD,
Thus, the feasible region is ABCD, Corner points A (0, 50), B (20, 40), C (50, 100) and O (0, 200)
Corner points A (60, 0), B (120, 0), C (60, 30) and O (40, 20)

Maximum value = 400 at (0, 200).

Maximum value = 18 at (4, 3). Minimum value = 100 at all points on the line segment joining the points (0,
6. Minimise and Maximise Z = 5x + 10y subject to 50) and (20, 40).
x + 2y ≤ 120, x + y ≥60, x – 2y ≥ 0, x, y ≥ 0.
Solution : To Maximise and minimise Z =5x + 10y. Maximum value = 600 at all points on the line segment joining the points 7.1 Find the angle between the vectors𝒊̂ − 𝟐𝒋̂ + 𝟑𝒌 ̂ and 𝟑𝒊̂ − 𝟐𝒋̂ + 𝒌
̂.
Region represented by x + 2y ≤ 120, (120, 0) and (60, 30). Solution:Let the given vectors be 𝑎⃗ = 𝑖̂ − 2𝑗̂ + 3𝑘̂and 𝑏⃗⃗ = 3𝑖̂ − 2𝑗̂ + 𝑘̂.
Consider, x + 2y = 120, x 0 120
y 60 0 Minimum value = 300 at (60, 0). ⃗⃗
𝑎⃗⃗∙𝑏 1×3+(−2)×(−2)+3×1 3+4+3 10
Clearly O (0, 0) satisfies x + 2y ≤ 120 Then, 𝑐𝑜𝑠𝜃 = |𝑎⃗⃗||𝑏⃗⃗ = = =
| √12 +(−2)2 +32 √32 +(−2)2 +12 14
7. Minimise and Maximise Z = x + 2y subject √14√14

 Solution region contain the origin. to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

5
⇒ 𝜃 = 𝑐𝑜𝑠 −1 ( ).
7
Region represented by x + y ≥ 60, Solution : To Maximise and minimise Z =x + 2y.
x 0 60 2. Find angle between the vectors𝒂 ̂and 𝒃
⃗⃗ = 𝒊̂ + 𝒋̂ − 𝒌 ⃗⃗ = 𝒊̂ + 𝒋̂ + 𝒌
̂.
Consider, x + y = 60 Region represented by x + 2y ≥ 100,
y 60 0 x 0 100 Solution:We have, 𝑐𝑜𝑠𝜃 = |𝑎⃗⃗||𝑏⃗⃗ =
⃗⃗
𝑎⃗⃗∙𝑏 1×1+1×1+(−1)×1
=
1
Clearly O (0, 0) not satisfies x + y ≥ 60 Consider, x + 2y = 100, y 50 0 | √12 +12 +(−1)2 √12 +12 +12 3
 Solution region not contain the origin. Clearly O (0, 0) not satisfies x + 2y ≥ 1
x 0 40 ⇒ 𝜃 = 𝑐𝑜𝑠 −1 ( ).
Region represented by x – 2y ≥ 0, 100 Solution region not contain the origin.
3

Consider, x – 2y = 0 y 0 20 3.Find the projection of the vector𝒊̂ + 𝟑𝒋̂ + 𝟕𝒌 ̂ on the vector 𝟕𝒊̂ − 𝒋̂ + 𝟖𝒌
̂.
Region represented by 2x + y ≤ 200,
Plot the points O(0, 0) and A(40, 20). x 0 100 Solution:Let the given vectors be 𝑎⃗ = 𝑖̂ + 3𝑗̂ + 7𝑘̂and 𝑏⃗⃗ = 7𝑖̂ − 𝑗̂ + 8𝑘̂.
Consider, 2x + y = 200,
Join OA and produce . Clearly O (0, 0) satisfies 2x + y ≤ 200 y 200 0 𝑎⃗⃗∙𝑏
Then projection of 𝑎⃗ on 𝑏⃗⃗ is given by 𝑝 = ⃗⃗ =
⃗⃗
(𝑖̂ +3𝑗̂ +7𝑘 )∙(7𝑖̂ −𝑗̂ +8𝑘 )
=
̂
7−3+56 ̂
2 2
|𝑏 | 2 √7 +(−1) +8 √114
and (20, 0) satisfies x – 2y ≥ 0 Solution region contain the point (20, 0).  Solution region contain the origin. 60
And x ≥ 0, y ≥ 0 are called non negative constraints which represents the x 0 20 Therefore, required projection = .
Region represented by 2x – y ≤ 0, √114
first quadrant. Consider, 2x – y = 0 y 0 40 4.Find the projection of the vector𝑎⃗ = 2𝑖̂ + 3𝑗̂ + 2𝑘̂ on the vector 𝑏⃗⃗ = 𝑖̂ + 2𝑗̂ + 𝑘̂.
Now plotting the graph as follows ⃗⃗ ̂ ̂
Plot the points O(0, 0) and A(20, 40). 𝑎⃗⃗∙𝑏
Solution:The projection of 𝑎⃗ on 𝑏⃗⃗is 𝑝 = ⃗⃗ =
(2𝑖̂+3𝑗̂ +2𝑘 )∙(𝑖̂+2𝑗̂ +𝑘 )
=
2+6+2
=
10
.
2 2 2
|𝑏 | √1 +2 +1 √6 √6
Join OA and produce .
5.Find the area of the parallelogram whose adjacent sides are determined by the
and (0, 20) satisfies x – 2y ≥ 0 Solution region contain the point (0, 20).
vectors𝒂 ̂and ⃗𝒃⃗ = 𝟐𝒊̂ − 𝟕𝒋̂ + 𝒌
⃗⃗ = 𝒊̂ − 𝒋̂ + 𝟑𝒌 ̂.
And x ≥ 0, y ≥ 0 are called non negative constraints which represents the
first quadrant. Solution:The area of a parallelogram with 𝑎⃗ and 𝑏⃗⃗ as its adjacent sides is given
Now plotting the graph as follows by |𝑎⃗ × 𝑏⃗⃗|.

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𝑖̂ 𝑗̂ 𝑘̂ 2
⇒ |𝑎⃗|2 + 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑐⃗ ∙ 𝑎⃗ + 𝑎⃗ ∙ 𝑏⃗⃗ + |𝑏⃗⃗| + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + |𝑐⃗|2 = 0
Now, 𝑎⃗ × 𝑏⃗⃗ = |1 −1 3| = (−1 − (−21))𝑖̂ − (1 − 6)𝑗̂ + (−7 − (−2))𝑘̂ = 20𝑖̂ + 5𝑗̂ − 5𝑘̂ . 2
2 −7 1 ⇒ |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2 + 2(𝑎⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗) = 0
21
Therefore, area of parallelogram |𝑎⃗ × 𝑏⃗⃗| = √202 + 52 + (−5)2 = √450 = 15√2sq units. ⇒ 1 + 16 + 4 + 2𝜇 = 0 ⇒ 2𝜇 = −21 ⇒𝜇=− .
2
6.Find the area of a parallelogram whose adjacent sides are given by the ⃗⃗&𝒄 ⃗⃗| = 𝟒, |𝒄
10.If 𝒂
⃗⃗,𝒃 ⃗⃗ are three vectors such that |𝒂
⃗⃗| = 𝟑, |𝒃 ⃗⃗| = 𝟓and each vector is
vectors𝒂 ̂ and 𝒃
⃗⃗ = 𝟑𝒊̂ + 𝒋̂ + 𝟒𝒌 ⃗⃗ = 𝒊̂ − 𝒋̂ + 𝒌
̂.
orthogonal to sum of the other two vectors then find |𝒂 ⃗⃗ + 𝒄
⃗⃗ + 𝒃 ⃗⃗|.
Solution:The area of a parallelogram with 𝑎⃗ and 𝑏⃗⃗ as its adjacent sides is given by
Solution:Given|𝑎⃗| = 3, |𝑏⃗⃗| = 4, |𝑐⃗| = 5 and 𝑎⃗ ∙ (𝑏⃗⃗ + 𝑐⃗) = 0, 𝑏⃗⃗ ∙ (𝑐⃗ + 𝑎⃗) = 0, 𝑐⃗ ∙ (𝑎⃗ + 𝑏⃗⃗) = 0.
|𝑎⃗ × 𝑏⃗⃗|. 2 2
Now, |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) = (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) ∙ (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗)
𝑖̂ 𝑗̂ 𝑘̂
Now, 𝑎⃗ × 𝑏⃗⃗ = |3 1 4| = (1 − (−4))𝑖̂ − (3 − 4)𝑗̂ + (−3 − 1)𝑘̂ = 5𝑖̂ + 𝑗̂ − 4𝑘̂. = 𝑎⃗ ∙ 𝑎⃗ + 𝑎⃗ ∙ (𝑏⃗⃗ + 𝑐⃗) + 𝑏⃗⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ (𝑐⃗ + 𝑎⃗) + 𝑐⃗ ∙ (𝑎⃗ + 𝑏⃗⃗) + 𝑐⃗ ∙ 𝑐⃗
1 −1 1 2
= |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2 = 9 + 16 + 25 = 50.
Therefore, area of parallelogram |𝑎⃗ × 𝑏⃗⃗| = √52 + 12 + (−4)2 = √25 + 1 + 16 = √42sq units.
Therefore, |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = √50 = 5√2.
7.Show that the position vector of the point R, which divides the line joining the
points P and Q having the position vectors𝒂 ⃗⃗internally in the ratio𝒎: 𝒏is
⃗⃗ and 𝒃
⃗⃗+𝒏𝒂
𝒎𝒃 ⃗⃗
.
𝒎+𝒏
Solution:Given ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ = 𝑏⃗⃗.
𝑂𝑃 = 𝑎⃗ and𝑂𝑄

𝑃𝑅 𝑚
R divides ⃗⃗⃗⃗⃗⃗
𝑃𝑄 internally in the ratio of 𝑚 ∶ 𝑛. =
𝑅𝑄 𝑛

⇒ 𝑛𝑃𝑅 = 𝑚𝑅𝑄 ⇒ 𝑚𝑅𝑄 ⃗⃗⃗⃗⃗⃗ = 𝑛𝑃𝑅

⃗⃗⃗⃗⃗⃗.
we have ⃗⃗⃗⃗⃗⃗ 𝑅𝑄 = 𝑂𝑄 𝑂𝑅 = 𝑏⃗⃗ − ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗⃗ 𝑃𝑅 = ⃗⃗⃗⃗⃗⃗
𝑂𝑅 and ⃗⃗⃗⃗⃗⃗ 𝑂𝑅 − ⃗⃗⃗⃗⃗⃗
𝑂𝑃 = ⃗⃗⃗⃗⃗⃗
𝑂𝑅 − 𝑎⃗
∴(𝑏⃗⃗ − ⃗⃗⃗⃗⃗⃗
𝑂𝑅 ) = 𝑛(𝑂𝑅 − 𝑎⃗) ⇒ 𝑚𝑏⃗⃗ − 𝑚𝑂𝑅 ⃗⃗⃗⃗⃗⃗ = 𝑛𝑂𝑅
⃗⃗⃗⃗⃗⃗ − 𝑛𝑎⃗
𝑚𝑏 +𝑛𝑎⃗⃗ ⃗⃗
⇒ 𝑚𝑏⃗⃗ + 𝑛𝑎⃗ = 𝑚𝑂𝑅
⃗⃗⃗⃗⃗⃗ + 𝑛𝑂𝑅
⃗⃗⃗⃗⃗⃗ = (𝑚 + 𝑛)𝑂𝑅
⃗⃗⃗⃗⃗⃗ or ⃗⃗⃗⃗⃗⃗
𝑂𝑅 = .
𝑚+𝑛
8. Find a unit vector perpendicular to each of the vectors𝒂 ⃗⃗and 𝒂
⃗⃗ + 𝒃 ⃗⃗, where𝒂
⃗⃗ − 𝒃 ⃗⃗ =
̂, 𝒃
𝒊̂ + 𝒋̂ + 𝒌 ⃗⃗ = 𝒊̂ + 𝟐𝒋̂ + 𝟑𝒌
̂.
Solution:Let 𝑎⃗ + 𝑏⃗⃗ = 2𝑖̂ + 3𝑗̂ + 4𝑘̂ and 𝑎⃗ − 𝑏⃗⃗ = −𝑗̂ − 2𝑘̂.
𝑖̂ 𝑗̂ 𝑘̂
Now, (𝑎⃗ + 𝑏⃗⃗) × (𝑎⃗ − 𝑏⃗⃗) = |2 3 4 | = −2𝑖̂ + 4𝑗̂ − 2𝑘̂ .
0 −1 −2
Now, |(𝑎⃗ + 𝑏⃗⃗) × (𝑎⃗ − 𝑏⃗⃗)| = √(−2)2 + 42 + (−2)2 = √4 + 16 + 4 = √24 = 2√6.
Therefore, the required unit vector is
̂ ⃗⃗)×(𝑎⃗⃗−𝑏
(𝑎⃗⃗+𝑏 ⃗⃗) 1
(𝑎⃗ + 𝑏⃗⃗) × (𝑎⃗ − 𝑏⃗⃗) = ⃗⃗+𝑏⃗⃗ ⃗⃗−𝑏⃗⃗ = (−𝑖̂ + 2𝑗̂ − 𝑘̂ ).
|(𝑎 )×(𝑎 )| √6

9.Three vectors𝒂 ⃗⃗, 𝒃⃗⃗ and 𝒄 ⃗⃗ satisfy the condition𝒂 ⃗⃗ + 𝒄

⃗⃗ + 𝒃 ⃗⃗. Evaluate the quantity
⃗⃗ = 𝟎
𝝁=𝒂⃗⃗. ⃗𝒃⃗ + ⃗𝒃⃗. 𝒄
⃗⃗ + 𝒄 ⃗⃗, if |𝒂
⃗⃗. 𝒂 ⃗⃗| = 𝟒 and |𝒄
⃗⃗| = 𝟏, |𝒃 ⃗⃗| = 𝟐.
Solution:Given |𝑎⃗| = 1, |𝑏⃗⃗| = 4, |𝑐⃗| = 2 and 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = 0
⃗⃗.
2
⇒ (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) = 0
⃗⃗2 ⇒ (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) ∙ (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) = 0
⇒ 𝑎⃗ ∙ 𝑎⃗ + 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑎⃗ ∙ 𝑐⃗ + 𝑏⃗⃗ ∙ 𝑎⃗ + 𝑏⃗⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗ + 𝑐⃗ ∙ 𝑏⃗⃗ + 𝑐⃗ ∙ 𝑐⃗ = 0

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