MATHEMATICS SOLUTION

(CBCGS SEM – 4 DEC 2018)

BRANCH – EXTC ENGINEERING
𝟐 𝟒
1 a) If 𝑨= ,then find the eigen values of of 6 𝑨-1 + 𝑨3 +2 𝑰. (05)
𝟎 𝟑

2 4
Ans.𝐴= ,
0 3

Since A is an upper triangular matrix, eigen values (𝜆) =diagonal elements =2,3
We know, 𝑓 (𝜆) is eigen value of 𝑓 (𝐴)

Let 𝑓 (𝐴) = 6 𝐴-1 + 𝐴-3 + 2 𝐼

∴ 𝑓 (𝜆) = 6 𝜆-1 + 𝜆³ + 2
6
= 𝜆 + 𝜆³ + 2
6
When 𝜆 =2 ,𝑓 (2) = 2 + 23 + 2 = 13
6
When 𝜆 = 3, 𝑓 (3) = + 33 + 2 =31
3

∴ Eigen values of 6 𝑨-1 + 𝑨3 + 2 𝑰 are 13, 31.

1 b) Find a vector orthogonal to both u = (-6,4, 2) and v (3,1,5). (05)

Ans. u = [- 6,4,2],v = [3,1,5]

Let 𝑤 = ( 𝑤 1 𝑤 2𝑤 3 ) be a vector which is orthogonal to both u and v.Type equation here.
∴ 𝑢 𝑤 = 0 ⇒ - 6𝑤 1 + 4𝑤 2 + 2𝑤 3 = 0 and,
∴ 𝑣 𝑤 = 0 ⇒ - 3𝑤 1 + 1𝑤 2 + 5𝑤 3 = 0

𝑤₁ −𝑤₂ 𝑤₃
By Crammer’s Rule, 4 2 = −6 2 = −6 4
1 5 3 5 3 1
𝑤₁ −𝑤₂ 𝑤₃
∴ 18 = −36 = −18

𝑤₁ 𝑤₂ 𝑤₃
∴ = = −1
1 1

𝑤₁ 𝑤₂ 𝑤₃
Let, = 1 = −1 = 𝑡
1
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∴ 𝑤 1= 𝑡; 𝑤 2 = 2𝑡; 𝑤 3 = -𝑡

∴ 𝑤 =[𝑡 2𝑡 -𝑡]
∴ For t = 1, = w = [1 2 -1]

Hence, a vector orthogonal to both u and v is w = [1 2 -1]

1 c) Evaluate ∫c log 𝒛𝒅𝒛 =2𝝅𝒊, where C is the until circle in the z plane. (05)

Ans.The unit circle in the z plane is |z| = 1 with Centre (0,0) and radius =1
Let 𝐼 = ∫c log 𝑧𝑑𝑧

Put 𝑧 = 𝑟𝑒 𝑖𝜃 = 1𝑒 𝑖𝜃 = 𝑒 𝑖𝜃

∴ 𝑑𝑧 = 𝑒 𝑖𝜃 .𝑖𝑑𝜃

2𝜋
∴𝐼= 0
𝑙𝑜𝑔( 𝑒 𝑖𝜃 ).𝑒 𝑖𝜃 . 𝑖𝑑𝜃

2𝜋
= 0
𝑖 𝜃log 𝑒. 𝑒 𝑖𝜃 𝑖𝑑𝜃

2𝜋
= 𝑖2 0
𝜃 𝑒 𝑖𝜃 𝑑𝜃

2𝜋
𝑒 𝑖𝜃 𝑒 𝑖𝜃
= 𝑖2 𝜃 − 1.
𝑖 𝑖2 0

2𝜋
= 𝑖𝜃 𝑒 𝑖𝜃 − 𝑒 𝑖𝜃 0

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 2𝜋
= 𝑒 𝑖𝜃 (𝑖𝜃 − 1) 0

= 𝑒 𝑖2𝜋 𝑖. 2𝜋 − 1 − 𝑒 0 (𝑖. 0 − 1)

= (cos 2π + 𝑖 sin 2 π)(2π 𝑖 -1) +1

= (1+0) (2 π 𝑖 -1) +1

= 2 𝜋 𝑖 -1 + 1

Hence, ∫c log 𝒛 𝒅𝒛 = 2 𝝅𝒊

1 d) Let x be continuous random variable with distribution (05)

𝒙 +𝒌 𝟎≤ 𝑥 ≤ 0
𝒇 𝒙 = 𝟔 ,
𝟎 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Evaluate k and find P (1 ≤×≤ 𝟐) and P (1 ≤×≤ 𝟑)

∞
Ans: For any probability density function , −∞
𝑓 𝑥 𝑑𝑥 = 0

3
𝑥
∴ + 𝑘 𝑑𝑥 = 1
6
0
3
1 𝑥2
∴ . 𝑘𝑥 =1
6 2 0

3²
∴ + 3𝑘 - 0 = 1
12

3
∴ 4 + 3 k =1

1
∴ 𝑘 = 12

𝑥 1
∴ 𝑓 (𝑥) = 6 + 12

2 𝑥 1
∴ P (1 ≤ 𝑋 ≤ 2) = 1 6
+ 𝑑𝑥
12

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 2
1 𝑥² 1
= . + 12 . 𝑥
6 2 1

2² 2 1² 1
= + 12 - + 12
12 12

1
= 3
3 𝑥 1
And, P(1≤ 𝑋 ≤ 3) = 1 6
+ 𝑑𝑥
12

3
1 𝑥² 1
= . + .𝑥
6 2 12 1

3² 3 1² 1
= + 12 - + 12
12 12

5
=6

𝟏 𝟓
Hence, 𝒌 = 𝟏𝟐 ; P (1 ≤ 𝑿 ≤ 𝟐) ; P (1 ≤ 𝑿 ≤ 𝟑 ) = 𝟔

2 a) Show that A is diagonalizable .Also find the transforming matrix M and the diagonal
−𝟗 𝟒 𝟒
matrix D where 𝑨 = −𝟖 𝟑 𝟒 (08)
−𝟏𝟔 𝟖 𝟕

Ans: Let 𝜆 be eigen value and X be corresponding eigen vector of matrix A.

Characteristic equation is |A – 𝝀 I | = 0
−9 − 𝜆 4 4
∴ −8 3−𝜆 4
−16 8 7−𝜆

On solving we get

𝝀𝟑 (sum of diagonal elements ) 𝝀𝟐 + (sum of the minors of diagonal elements ) 𝝀 − 𝑨 = 𝟎

∴ 𝜆³ - (-9 + 3 + 7 ) 𝜆² + (-11+ 1 + 5) 𝜆 − 3 = 0

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∴ 𝜆³ - 𝜆2 − 5𝜆 − 3 = 0

∴ Eigen values (𝜆) are -1 ,-1,3

Case 1: 𝜆 = −1
∴ 𝐴−𝜆𝐼 X=0

−8 4 4 𝑥₁ 0
∴ −8 4 4 . 𝑥₂ = 0
−16 8 8 𝑥₃ 0

−2 1 1 𝑥₁ 0
1
𝑅 2 - 𝑅₁ ;𝑅₃ - 2𝑅 1 ; 4 𝑅1 ⇒ 0 0 0 . 𝑥₂ = 0 → (1)
0 0 0 𝑥₃ 0

Numbers of unknowns (n) = 3

Rank (r) = number of non – zero rows =1

Algebraic multiplicity (A.M)

= No. of times “𝝀 = - 1” is repeated = 2

Geometric multiplicity ( G.M) = n-r = 3 -1 =2

∴A . M = G.M for “ 𝜆= - 1 ‘’

Expanding (1), -2𝑥1 + 𝑥2 + 𝑥3 = 0

Put 𝑥1 = 𝑡and 𝑥2 = 𝑠
∴ -2𝑡 + 𝑠 + 𝑥₃ = 0
∴ 𝑥₃ = 2𝑡 – 𝑠

𝑥₁ 𝑡 1𝑡 + 0𝑠 1 0
∴ Eigen vector 𝑥₂ = 𝑠 = 0𝑡 + 1𝑠 = 0 𝑡 + 1 𝑠
𝑥₃ 2𝑡 − 𝑠 2𝑡 − 1𝑠 2 −1

∴ Eigen vector 𝑋₁ = *1 0 2 +’&𝑋₂ = * 0 1 -1+’

Case 2; 𝜆 = 3
∴[ A – 𝜆 I] X = 0

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 −12 4 4 𝑥₁ 0
∴ −8 0 4 . 𝑥₂ = 0
−16 8 4 𝑥₃ 0

2 4 1
𝑅2 − 𝑅1 ; 𝑅3 − 𝑅1 ; 𝑅
3 3 4 1

−3 1 1 𝑥₁ 0
⇒ 0 −8/3 4/3 . 𝑥₂ = 0
0 8/3 −4/3 𝑥₃ 0

−3 1 1 𝑥₁ 0
3
𝑅3 + 𝑅2 ; 4 𝑅2 ⇒ 0 −2 1. 𝑥₂ = 0 → (3)
0 0 0 𝑥₃ 0

Here n = 3 and r = 1
A.M = No. of times “𝜆 = 3" is repeated =1
G.M = n-r = 3 – 2 = 1
∴ A.M = G.M .for “𝝀 = 𝟑 "

Expanding (3),

-3𝑥1 + 𝑥2 + 𝑥3 = 0 → 4 &
-2𝑥2 + 𝑥3 = 0 → (5)

From (5), 𝑥3 = 2𝑥₂

From (4), - 3𝑥1 + 𝑥2 + 2𝑥2 = 0
∴ 3 𝑥2 = 3𝑥1
∴ 𝑥1 = 𝑥2

𝑥₁ 𝑥₂ 1
∴ Eigen vector 𝑥₂ = 𝑥₂ = 1 𝑥₂
𝑥₃ 2𝑥₂ 2

∴ Eigen vector 𝑋2 = 𝐼 1 2 ‘

Since A.M = G.M for all eigen values, matrix A is diagonalizable .

∴ 𝑀-1 A𝑀 = 𝐷

So the given matrix A is diagonalized to Diagonal Matrix. D by the Transforming Matrix

M. where
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 −𝟏 𝟎 𝟎 𝟎 𝟏 𝟏
𝑫 = 𝟎 −𝟏 𝟎 and 𝑴 = 𝟏 𝟎 𝟏
𝟎 𝟎 𝟑 −𝟏 𝟐 𝟐

𝝅/𝟐
2 b)Find the extremal of the function 𝟎
(𝒚ˈ² − 𝒚𝟐 + 𝟐𝒚𝒙)𝒅𝒙 with y (0) = 0 &(06)

𝝅
𝒚 = 0.
𝟐

𝑥₁ 𝜋/2
Ans: Let 𝑥₁
𝐹𝑑𝑥 = 0
𝑦ˈ2 − 𝑦 2 + 2𝑦𝑥 𝑑𝑥

∴ 𝐹 = 𝑦ˈ2 − 𝑦 2 + 2𝑦𝑥
𝜕𝐹 𝜕𝐹
∴ = −2𝑦 + 2𝑥 − 0 ; & 2𝑦ˈ ;
𝜕𝑦 𝜕𝑦ˈ

By Euler’s Lagrange equation ,the necessary condition for the given functional to be
𝝏𝑭 𝒅 𝝏𝑭
extremum is − 𝒅𝒙 =0
𝝏𝒚 𝝏𝒚ˈ
𝑑
∴ 2 𝑥 – 2 𝑦 - 𝑑𝑥 (2𝑦ˈ) = 0

𝑑 𝑑𝑦
∴𝑥– 𝑦- = 0 (Dividing by 2)
𝑑𝑥 𝑑𝑥

𝑑²𝑦
∴ +𝑦 =𝑥
𝑑𝑥²

𝑑
∴ 𝐷2 𝑦 + 𝑦 = 𝑥 𝑤𝑕𝑒𝑟𝑒, 𝐷 =
𝑑𝑥
∴ 𝐷2 + 1 𝑦 = 𝑥

This is Liner Differential Equation

Auxiliary equation is 𝐷2 + 1 = 0
∴ 𝐷 = ±𝑖 = 0 ± 1𝑖

∴ Complimentary Function is
𝑦˳ = 𝑒 0𝑥 (𝑐₁ cos 1 𝑥 + 𝑐₂ sin )

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 1
Particular Integral is 𝑦𝑝 = 𝑋Where , X = x
𝑓(𝐷)

1
∴ 𝑦𝑝 = (𝐷 2 +1) ( 𝑥 )

= (1 + 𝐷2 )−1 (𝑥)

= ( 1 - 𝐷2 + 𝐷4 − ⋯. ) 𝑥

= 𝑥 - 𝐷2 𝑥 + 𝐷4 𝑥 − ⋯.

=𝑥+0+0+⋯

=𝑥

∴ The Complete Solution is 𝑦 = 𝑦𝑐 + 𝑦𝑝

Hence the solution is 𝑦 = 𝑐₁ cos 𝑥 + 𝑐2 sin 𝑥 + 𝑥 → (1)

Given, 𝑦 (0) = 0

∴ Put x = 0 and y = 0 in (1)

∴ 0 = ₁. cos0 + 𝑐2 .sin0 + 0

∴ 0 = 𝑐 1 . 1 + 𝑐2 .0

∴ 0 = 𝑐1

𝜋
Given, 𝑦 =0
2

𝜋
∴ Put 𝑥= 2 , 𝑐1 = 0 and y = 0 in (1)

𝜋 𝜋
∴ 0 = 0 + 𝑐2 sin 2 + 2

𝜋
∴ 0 = 𝑐2 . 1 + 2

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 𝜋
∴ 𝑐2 = - 2 → (3)

𝜋
∴ From (1), (2) and (3), 𝑦 = 0 − 2 sin 𝑥 + 𝑥
∴ The extremal of the given function is

𝝅
𝒚 = 𝒙 − 𝟐 𝐬𝐢𝐧 𝒙

2 c) Let 𝑹⁴ have the Euclidean inner product .use Gram -Schmidt process to transform the
basis {𝒖₁, 𝒖₂, 𝒖₃} in to an orthonormal basis where 𝒖₁ = (1,0,1,1); 𝒖₂ = (-1,0,-1,1); 𝒖₃ = (0,-
1,1,1).
(08)

Ans:Gram Schmidt orthogonalization:

𝑢₁ = (1,0,1,1); 𝑢₂ = (-1,0,-1,1); 𝑢₃ = (0,-1,1,1):

S1:
Let 𝑣1 = 𝑢1 = (1,0,1,1)

∴ 𝑣1 ² = (1)2 + (0)2 + (1)2 + (1)2 = 3 and

𝑢2, 𝑣1 = (-1) (1) + (0) (0) + (1) (-1) + (1)(1) = -1

S2:
𝑢 2, 𝑣1
Let 𝑣2 = 𝑢2 - 𝑣1
𝑣1 ²

(−1)
= (-1,0,-1,1)- ×( 1,0,1,1,)
3

1 1 1
= (-1,0,-1,1) + ,0 3,3
3

−2 −2 4
= ,0 ,3
3 3

−2 2 −2 2 4 2 8
∴ 𝑣2 ² = + 0 2+ + =3;
3 3 3

Now ,
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 𝑢3, 𝑣1 = (0)(1) + (-1)(0) + (1)(1) + (1)(1) =2 and
−2 −2 4 2
𝑢3, 𝑣2 = (0) + −1 0 + 1 + 1 =3
3 3 3

S3:
𝑢 3, 𝑣1 𝑢 3, 𝑣2
Let 𝑣3 = 𝑢3, - 𝑣1 - 𝑣2
𝑣1 ² 𝑣2 ²

2 2/3 −2 −2 4
= (0,-1,1,1) -3(1,0,1,1) -8 , 0, ,3
4/ 3 3 3

∴ Orthonormal Bases are:

→ 1 3 3 3
𝑣₁
→= = (1,0,1,1) = , 0, ,
𝑞₁ → 3 3 3 3
𝑣₁

→ 1 −2 −2 4
𝑣₂
→= = , 0, ,3
𝑞₂ → 8/3 3 3
𝑣₂

3 −2 −2 4 − 6 − 6 6
= , 0, ,3 = , 0, ,
8 3 3 6 6 3

→ 1 −1 1
𝑣₃
And,→ = = , −1, 2 , 0
𝑞₃ → 3/2 2
𝑣₃

2 −1 1 − 6 − 6 6
= , −1, 2 , 0 = , , ,0
3 2 6 3 6

∴ Orthonormal basis of the subspace S are

𝟑 𝟑 𝟑 − 𝟔 − 𝟔 − 𝟔 − 𝟔 − 𝟔 𝟔
, 𝟎, , ; , 𝟎, , ; , , ,𝟎
𝟑 𝟑 𝟑 𝟔 𝟔 𝟑 𝟔 𝟑 𝟔

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3 a) The number of accidents in a year attributed to taxi drivers in a city follows Poisson
distribution with mean 3 out of 1000 taxi drivers, find approximately the number of drivers
with (06)

1) No accidents in a year
2) More than 3 accidents in a year.
(Given: e-1 = 0.3679,e-2 = 0.1353,e-3 = 0.0498)

Ans: Let X denote the number of accidents in a year due to taxi drivers .

Given X follows Poisson distribution .

Average number of accidents in a year due to taxi drivers =m=3

N = 1000

𝑒 −𝑚 𝑚 𝑥 𝑒 −3 3𝑥
∴ 𝑃 𝑋=𝑥 = =
𝑥! 𝑥!

i) Probability a driver has no accidents in a year = P (X = 0 )

𝑒 −3 30
= 0!

= 0.04979

No. of drivers with no accidents in a year = N × P (X = 0)

= 1000 × 0.0.4979
= 49.8
= 50 drivers

ii) Probability a driver has more than 3 accidents in a year = P (X > 3)

= 1 – P ( X ≤ 3)

= 1 – [ P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + ( X = 3 )]

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 𝑒 −3 ×30 𝑒 −3 ×31 𝑒 −3 ×32 𝑒 −3 ×33
=1- + + +
0! 1! 2! 3!

1 3 9 27
= 1 - 𝑒 −3 +1+2+
1 6

= 0.3528

Number of drivers with more than 3 accidents in a year = N × P (X > 3)

= N × P ( X> 3 )

= 352.6

≈ 353 drivers

Hence, in a year

Number of drivers with no accidents = 50

Number of drivers with more than 3 accidents = 353

3 b) Calculate Rank Correlation co-efficient for the following data: (07)

𝑿 10 12 18 18 15 40

𝒀 12 18 25 25 50 25

Ans: Spearman’s rank correlation Co-efficient :

X Y 𝑅₁ 𝑅₂ 𝑑 I = ( 𝑅1 − 𝑅₂)2

10 12 6 6 0

12 18 5 5 0

18 25 2.5 3 0.25

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 18 25 2.5 3 0.25

15 50 4 1 9

40 25 1 3 4

Total 13.5

Here, 𝑚1 = 2; 𝑚2 = 3; n = 6

Spearman’s rank correlation Co-efficient is given by

𝟔 𝟏
𝑹 = 1 – 𝒏(𝒏𝟐 −𝟏) ∑𝒅𝟐𝒊 + 𝒎𝟑𝟏 − 𝒎𝟏 + 𝒎𝟑𝟐 − 𝒎𝟐 + ⋯ .
𝟏𝟐

6 1
= 1 - 6(62 −1) 13.5 + 23 − 2 + (33 − 3)
12

1 1
= 1 - 35 13.5 + 6 + 24
12

1
=1- 13.5 + 2.5
35

= 0.5429

Spearman’s rank correlation Co-efficient (R) = 0.5429

𝟏
3 c) Expand 𝒇 𝒛 = about z = 0 for (08)
𝒛²(𝒛−𝟏)(𝒛+𝟐)
i) |z|< 1;
ii) 1 < 𝒛 < 2;
iii) |z|> 2.

1 𝐴 𝐵 𝐶 𝐷
Ans: Let 𝑓 𝑧 = = 𝑧 + 𝑧² + 𝑧−1 + 𝑧+2 → (1)
𝑧²(𝑧−1)(𝑧+2)

−1 1 −1
By cover-up method, 𝐵= ;𝐶 = 3 ; 𝐷 = ;
2 12

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Put z= -1 in (1)

1 𝐴 1/2 1/3 1/12

= −1 − 2
+ −1−1 − −1+2
(−1)²(−1−1)(−1+2) −1

1 1 1 1
∴ - 2 = −𝐴 − −6 −
2 12

1
∴𝐴 = −
4
−1 1 1 1
∴𝑓 𝑧 = − + − → (2)
4𝑧 2𝑧 2 3 𝑧 − 1 12 (𝑧 + 2)

Case.1: For |z|< 1,

Obviously, |z|< 2

𝑧
∴ |z|< 1 and <1
2

∴ From (2)

−1 1 1 1
𝑓 (z) = − +3 −
4𝑧 2𝑧 2 𝑧−1 12×2 (𝑧/2+1)

−1 1 1 −1 1 𝑧 −1
= 4𝑧 − − 1−𝑧 − 1+2
2𝑧 2 3 24

−1 1 1 1 𝑧 𝑧2 𝑧2
= 4𝑧 − − 1 + 𝑧 + 𝑧 2 + 𝑧 3 + ⋯ − 24 1 − 2 + 22 − 23 + ⋯
2𝑧 2 3

Region of Convergence:

Above series is convergent for

|z|< 1 and |z|< 2 i.e. |z|< 1,

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Which is the interior of the circle with centre (0,0) and radius 1,

Case 2: For 1 < |z| < 2

Obviously, |z|> 1

i.e. 1 < |z| and |z| < 2

1 𝑧
∴ < 1and <1
𝑧 2

∴ From (2)

−1 1 1 1
𝑓 (𝑧) = 4𝑧 − − −
2𝑧 2 3𝑧 1−1/𝑧 12×2(𝑧/2+1)

−1 1 1 1 −1 1 𝑧 −1
= − + 1−𝑧 − 1+2
4𝑧 2𝑧 2 3𝑧 24

−1 1 1 1 1 1 𝑧 𝑧2
= − 2
− 1+ + +⋯ − 1− + −⋯
4𝑧 2𝑧 3𝑧 𝑧 𝑧² 24 2 22

−1 1 1 1 1 1 1 𝑧 𝑧2
= − − + 𝑧2 + 𝑧3 + ⋯ − 1− + 22 − ⋯
4𝑧 2𝑧 2 3 𝑧 24 2

Region of Convergence:

Above series is convergent for

1 𝑧
< 1& <1
𝑧 2

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i.e. 1 <|z| < 2, which is the annular region between the concentric circles with Centre (0,0) and
radii 1 & 2.

Case 3: For |z|> 2.

Obviously,|z| > 1

i.e. 1 <|z| and 2 <|z|

1 2
∴ < 1 and <1
𝑧 𝑧

∴ From (2)

−1 1 1 1
𝑓 (𝑧) = − + −
4𝑧 2𝑧 2 3 𝑧 1−1/𝑧 12𝑧 1+2/𝑧
−1 1 1 1 −1 1 2 −1
= − + 1−𝑧 − 1+𝑧
4𝑧 2𝑧 2 3𝑧 12𝑧

−1 1 1 1 1 1 2 22
= − − 1+𝑧+ +⋯ − 1−𝑧+ −⋯
4𝑧 2𝑧 2 3𝑧 𝑧2 12𝑧 𝑧2

−1 1 1 1 1 1 1 1 2 22
= − − + 𝑧2 + 𝑧3 + ⋯ − − 𝑧2 + 𝑧3 − ⋯
4𝑧 2𝑧 2 3 𝑧 12 𝑧

Region of Convergence:

Above series is convergent

1 2
For < 1 and <1
𝑧 𝑧

i.e |z| > 2, which is the exterior region of the circle with centre (0,0) and radius 3.

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 𝒔𝒊𝒏𝜽 𝒛
4.a) Evaluate 𝒄 𝒛−𝝅/𝟔 𝒏
𝒅𝒛 where c is circle |z| = 1 for n= 1, n = 3 (06)
Ans: Circle |z| =1 has Centre (0,0) and radius 1.

𝜋
Here, 𝑧0 = 6 lies inside the circle .

Case.I. n = 1

𝜋
𝑧0 = 6 is a simple pole.
𝜋
𝑅 1 = Residue of 𝑓(z) at “ 𝑧0 = 6 ”
= lim𝑧→𝑧0 𝑧 − 𝑧0 𝑓(𝑧)

𝑠𝑖𝑛 6 𝑧
= lim𝑧→𝜋/6 𝑧 − 𝜋/6 × 𝜋
(𝑧− )
6

𝜋
= 𝑠𝑖𝑛6 6

1
= 64

By Cauchy’s Residue theorem,

𝑐
𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖 𝑅1 + 𝑅 2 + ⋯

𝑠𝑖𝑛6 𝑧 1
∴ 𝑐
𝑑𝑧 = 2𝜋𝑖.
(𝑧 − 𝜋/6)1 64

𝒔𝒊𝒏𝟔 𝒛 𝝅𝒊
∴ 𝒄
𝒅𝒛 =
(𝒛 − 𝝅/𝟔)¹ 𝟑𝟐

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Case II: n =3
𝜋
𝑧0 =6 is a pole of order 3

𝜋
𝑅1 = Residue of 𝑓(𝑧) at “𝑧0 =6 ”
1 𝑑 𝑛 −1
= (𝑛−1)ǃ lim (𝑧 − 𝑧0 )𝑛 × 𝑓 (𝑧)
𝑧→𝑧0 𝑑𝑧 𝑛 −1

1 𝑑 3−1 𝑠𝑖𝑛 6 𝑧
= lim (𝑧 − 𝜋/6)3 ×
(3−1)ǃ 𝑧→𝜋/6 𝑑𝑧 3−1 (𝑧−𝜋/6)³

1 𝑑2
= lim 𝑠𝑖𝑛6 𝑧
2ǃ 𝑧→𝜋/6 𝑑𝑧 2

1 𝑑
= 2 lim (6 𝑠𝑖𝑛5 𝑧. cos 𝑧)
𝑧→𝜋/6 𝑑𝑧

1
= ×6 lim (sin5𝑧. – sin 𝑧 + cos 𝑧 .5 sin4𝑧 .cos 𝑧)
2 𝑧→𝜋/6
6𝜋 𝜋 𝜋
= 3 −𝑠𝑖𝑛 + 𝑐𝑜𝑠 2 6 .5 𝑠𝑖𝑛2 6
6

1 3 1
= 3 − 64 + . 5. 16
4

21
= 32

By Cauchy’s Residue theorem .

𝑐
𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖 𝑅1 + 𝑅2 + ⋯ 1

𝑠𝑖𝑛6 𝑧 21
∴ 𝑐
𝑑𝑧 = 2𝜋𝑖 .
(𝑧 − 𝜋/6)³ 32

𝒔𝒊𝒏𝟔 𝒛 𝟐𝟏𝝅𝒊
∴ 𝒄
𝒅𝒛 =
(𝒛 − 𝝅/𝟔)³ 𝟏𝟔

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4 b) Find the m.g.f of a random variable whose probability density function is (06)
𝟏 𝒙
𝑷 (𝑿 = 𝒙) = 𝒙 = 𝟏, 𝟐, 𝟑, …
𝟐
𝟎 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Hence ,find the mean variance.

Ans: By definition M.G.F. about origin is given by

𝑀𝑜 𝑡 = 𝐸 𝑒 𝑡𝑥

= ∑∞
𝑥=1 𝑃𝑥 . 𝑒
𝑡𝑥

1
= ∑∞
𝑥=1 . 𝑒 𝑡𝑥
2𝑥

𝑥
𝑒𝑡
= ∑∞
𝑥=1 2

1 2 3
𝑒𝑡 𝑒𝑡 𝑒𝑡
= + + + ⋯,
2 2 2
𝑒𝑡 𝑒𝑡
which is a Geometric Progression with 𝑎= 2 and r = 2

𝑒 𝑡 /2 𝑎
= 1−𝑒 𝑡 /2 , 𝐼𝑛 𝐺, 𝑃. , 𝑠∞ = 1−𝑟

𝑒𝑡
= 2−𝑒 𝑡

𝑒𝑡
=
𝑒 𝑡 𝑒2𝑡 −1

1
∴ 𝑀𝑜 𝑡 =
2𝑒 −𝑡 − 1

𝑑𝑟
Now, 𝑟 𝑡𝑕 Moments 𝜇𝑟 ′ = 𝑀𝑜 𝑡 → (1)
𝑑𝑡 𝑟 𝑡=0

𝑑1
∴ First Moment 𝜇𝑟 ′ = (2𝑒 −𝑡 − 1)−1
𝑑𝑡 1 𝑡=0

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 −2 𝑑
= −1 2𝑒 −𝑡 − 1 . 𝑑𝑡 2𝑒 −𝑡 − 1
𝑡=0

= −1 2𝑒 𝑡 − 1 −2
. 2𝑒 𝑡 . −1 𝑡=0 → (2)

= 2𝑒 0 − 1 −2
. 2𝑒 0

= 1-2× 2 × 1
=2

𝑑²
From (1),Second Moment 𝜇2 ′ = 𝑀𝑜 𝑡
𝑑𝑡² 𝑡=0
𝑑 −𝑡 −𝑡 −2
= . [2𝑒 2𝑒 −1 ] (From 2)
𝑑𝑡 𝑡=0

𝑑 −2 𝑑
= 2 𝑒 −𝑡 𝑑𝑡 2𝑒 −𝑡 − 1 −2
+ 2𝑒 −𝑡 − 1 𝑒 −𝑡
𝑑𝑡 𝑡=0

= 2 𝑒 −𝑡 . −2 2𝑒 −𝑡 − 1 −3
. 2𝑒 −𝑡 . −1 + 2𝑒 −𝑡 − 1 −2
. 𝑒 −𝑡 . −1 𝑡=0

= 2 2𝑒 0 2𝑒 0 − 1 −3
. 2𝑒 0 − 𝑒 0 2𝑒 0 − 1 −2

= 2 2 × 1−3. . 2 × 1 − 1 × 1−2

= 2[ 4-1]

=6
∴ Mean = 𝜇₁ˈ = 2

∴ Variance = 𝜇₂ˈ − 𝜇₁ˈ²

= 6 - 2²
=2
Hence
Mean = 2 and Variance = 2

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4 c) Verify the Cayley-Hamilton Theorem for matrix A and hence find 𝑨−𝟏 for (08)
𝟏 𝟒
𝑨= .
𝟐 𝟑
Hence find 𝑨𝟓 − 𝟐𝑨𝟒 − 𝟕𝑨𝟑 + 𝟏𝟏𝑨𝟐 − 𝑨 − 𝟏𝟎𝑰 interms of A.

Ans: Part I:

Let 𝜆 be eigen value of matrix A.

Characteristic equation is |A – 𝜆 I |=0

1−𝜆 4
∴ =0
2 3−𝜆

On solving we get

𝜆² - ( sum of diagonal elements ) 𝜆 − | A|= 0

∴ 𝜆2 + 1 + 3 𝜆 + (−5) = 0

∴ 𝜆2 + 4𝜆 − 5 = 0 → (1)

Cayley Hamilton Theorem states that the characteristic equation is satisfied by matrix A.
∴ 𝐴2 + 4𝐴 − 5 𝐼 = 0 → (2)
1 4 1 4 9 16
Now, 𝐴2 = 𝐴 × 𝐴 = × =
2 3 2 3 8 17
2
∴ LHS = 𝐴 + 4 𝐴 − 5 𝐼

9 16 1 4 1 0
= −4 −5
8 17 2 3 0 1

𝟎 𝟎
=
𝟎 𝟎

= RHS
∴ Cayley Hamilton Theorem is verified.

Part II:

Pre-multiply (2) by 𝐴−1

∴ 𝐴 − 4𝐼 − 5 𝐴−1 = 0

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∴ 5𝐴−1 = 𝐴 − 4 𝐼

1 4 1 0
∴ 5𝐴−1 = −4
2 3 0 1

−3 4
∴ 5𝐴−1 =
2 −1

𝟏 −𝟑 𝟒
∴ 𝑨−𝟏 =
𝟓 𝟐 −𝟏

Part III.
Consider,

Divided = Divisor × Quotient + Remainder

𝜆5 − 4𝜆4 − 7𝜆3 + 11𝜆2 − 𝜆 − 10 = 𝜆2 − 4𝜆 − 5 𝜆3 − 2𝜆 + 3 (𝜆 + 5)

= 0 + (𝜆 + 5) (from 1)
=𝜆 + 5

∴ 𝑨𝟓 − 𝟒𝑨𝟒 − 𝟕𝑨𝟑 + 𝟏𝟏𝑨𝟐 − 𝑨 − 𝟏𝟎𝑰 = 𝑨 + 𝟓𝑰

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5 a) Express 𝒑(𝒙) = 7 + 8𝒙 + 9𝒙² as a Liner combination of 𝒑1 = 2 + 𝒙 + 4𝒙2 ; 𝒑𝟐 =1- 𝒙 +
3𝒙2 𝒑₃ = 2 + 𝒙 + 5𝒙2 .
(06)

Ans:𝑝(𝑥) = 7 + 8𝑥+ 9𝑥²= (7,8,9)

𝑝₁ = 2 + 𝑥 + 4 𝑥² = (2,1,4)
𝑝₂ = 1 + 𝑥 + 3 𝑥² = (1,-1,3)
𝑝₃ = 2 + 𝑥 + 5 𝑥² = (2,1,5)

Let 𝑘₁, 𝑘₂, 𝑘₃ be scalers (not all zero)such that

𝑝(𝑥) = 𝑘₁𝑝₁(𝑥) + 𝑘₂𝑝₂(𝑥) + 𝑘₃𝑝₃(𝑥) → (1)

∴ (7,8,9) = 𝑘₁(2,1,4) + 𝑘₂ (1,-1,3) + 𝑘₃ (2,1,5)

∴ (7,8,9) =

(2𝑘₁,1𝑘1 , 4𝑘₁)+ (1𝑘₂,-1𝑘₂, 3𝑘₂)+(2𝑘₃,1𝑘₃, 5𝑘₃)

∴ (7,8,9) =

( 2𝑘₁ + 1𝑘₂ + 2𝑘₃ , 1𝑘₁ − 1𝑘₂ + 1𝑘₃.4𝑘₁ +3𝑘₂ + 5𝑘₃ )

∴ Comparing both sides, we get

7 = 2 𝑘₁ + 1𝑘₂ + 2𝑘₃
8 = 1 𝑘₁ − 1𝑘₂ + 1𝑘₃
9 = 4 𝑘₁ + 3𝑘₂ + 5𝑘₃ → (2)

Writing equation in (2) in the matrix form ,

2 1 2 𝑘₁ 7
1 −1 1 𝑘₂ = 8
4 3 5 𝑘₃ 9

1 −1 1 𝑘₁ 8
𝑅₂ ↔ 𝑅₁ ⇒ 2 1 2 𝑘₂ = 7
4 3 5 𝑘₃ 9

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 1 −1 1 𝑘₁ 8
𝑅₂ − 2𝑅₁; 𝑅₃ − 4𝑅₁;⇒ 0 3 0 𝑘₂ = −9
0 7 1 𝑘₃ −23

1 −1 1 𝑘₁ 8
1
3
𝑅₂;⇒ 0 1 0 𝑘₂ = −3
0 7 1 𝑘₃ −23

1 0 1 𝑘₁ 5
𝑅₃ - 7𝑅₂ : 𝑅₁ + 𝑅₂;⇒ 0 1 0 𝑘₂ = −3
0 0 1 𝑘₃ −2

On expansion ,

𝑘₃= −2, 𝑘₂ = −3 and ,

1𝑘₁ + 0𝑘₂ + 1𝑘₃; = 5
∴ 𝑘₁ + 0 −2 =5
∴ 𝑘₁ = 7

∴ From (1) , 𝒑 (𝒙)= 7 𝒑₁(𝒙) − 𝟑𝒑₂(𝒙) − 𝟐𝒑₃(𝒙)

Which is the required Linear combination.

𝟐𝝅 𝒄𝒐𝒔𝟐𝜽
5 b) Using Cauchy theorem ,evaluate 𝟎 𝟓 + 𝟒𝒄𝒐𝒔𝜽
(06)

Ans:Consider a circle |z| = 1 with Centre (0,0) & radius 1.

Put 𝑧 = 𝑟 𝑒 𝑖𝜃 = 1𝑒 𝑖𝜃 = 𝑒 𝑖𝜃
∴ 𝑑𝑧 = 𝑒 𝑖𝜃 .𝑖𝑑𝜃 = 𝑖𝑧𝑑𝜃
𝑑𝑧
∴ 𝑑𝜃 = 𝑖𝑧

𝑒 𝑖𝜃 +𝑒 −𝑖𝜃 𝑧+𝑧 −1 𝑧 2 +1
In Denominator ,cos𝜃 = = =
2 2 2𝑧

In Numerator,
cos 2𝜃 = R.P (𝑒 𝑖2𝜃 ) = R.P (𝑒 𝑖𝜃 )2 = R.P (𝑧 2 )

On substituting we get,

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 𝟐𝝅 cos 2𝜃 𝑅.𝑃.𝑧 2 𝑑𝑧
𝟎 5+4𝑐𝑜𝑠𝜃
𝑑𝜃 = ∫c (𝑧 2 +1)
. 𝑖𝑧
5×4×
2𝑧

𝑧.𝑧 2 𝑑𝑧
= R.P ∫c5𝑧+2(𝑧 2 +1) . 𝑖𝑧

𝑧2 𝑑𝑧
= R.P ∫c2𝑧 2 +5𝑧+2 . 𝑖

Now for singularity, 2𝑧 2 + 5𝑧 + 2

∴ 𝑧 = −0.5 𝑜𝑟 𝑧 = −2

Here,𝑧0 = -2 Lies outside while 𝑧0 = -0.5 Lies inside the circle |z|=1
𝑧0 = −0.5 is a simple pole.

𝑅₁ = Residue of 𝑓(𝑧) at “𝑧0 = − 0.5”

= lim𝑧→𝑧0 (𝑧 − 𝑧0 ) × 𝑓 (𝑧)

𝑧2
= lim (𝑧 + 0.5) × 𝑖×2(𝑧+0.5)(𝑧+2)
𝑧→0.5

(−0.5)²
= 𝑖×2(−0.5+2)

−𝑖
= 12

By Cauchy ‘s Residue theorem ,

∫c𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖(𝑅₁ + 𝑅₂ + ⋯ )

𝑧2 𝑑𝑧 −𝑖
∴ R.P. ∫c2𝑧 2 +5𝑧+2 . = 𝑅. 𝑃 2𝜋𝑖.
𝑖 12

2𝜋
𝑐𝑜𝑠2𝜃 𝜋
∴ 𝑑𝜃 = 𝑅. 𝑃.
5 + 4𝑐𝑜𝑠𝜃 6
0

𝟐𝝅
𝒄𝒐𝒔𝟐𝜽 𝝅
∴ 𝒅𝜽 =
𝟓 + 𝟒𝒄𝒐𝒔𝜽 𝟔
𝟎

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5 c) In an examination marks obtained by students in Mathematics ,physics and Chemistry are
normally distributed with means 51, 53 and 46 respectively and standard deviation 15,12,16
respectively .Find the probability of securing total marks. (08)
i) 180 or above
ii) 90 or below

Ans:
Let M,P and C denote marks in Mathematics, Physics and chemistry respectively
Given 𝑚𝑀 = 51; 𝑚𝑃 = 53; 𝑚𝐶 = 46; 𝜎𝑀 = 15; 𝜎𝑃 = 12; 𝜎𝐶 = 16

Let X = Total Marks of Mathematics ,Physics and Chemistry

∴X=M+P+C

Since,M,P,C are independent normal variates ,X is also a normal variate with,

Mean = 𝑚𝑀 + 𝑚𝑃 + 𝑚𝐶 = 51 + 53 + 46 = 150

Variance (𝜎𝑋 2 ) = 𝜎𝑀2 + 𝜎𝑃2 + 𝜎𝐶 2 = 15² + 12² + 16²= 625

∴ 𝜎𝑋 = 25

i)P (180 or above )= P ( X> 180)

𝑥−𝑚 180−150
=𝑃 >
𝜎 25

=P(z> 1.2)
= 0.5 – Area between ‘z=0’ to ‘z= 1.2’
= 0.5 – 0.3849
= 0.1151

ii) P (90 or below )= P ( X< 90)

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 𝑥−𝑚 90−150
=𝑃 <
𝜎 25

=𝑃 (𝑧 < −2.4)
= 0.5 – Area between ‘z=0’ to ‘z= - 2.4’
= 0.5 – 0.4918
= 0.0082

Hence,
i) Probability of securing total marks of 180 and above = 0.1151
ii) Probability of securing total marks of 90 and below = 0.0082

𝟏 𝟎 𝟎
6 a) If 𝑨 = 𝟏 𝟎 𝟏 then find 𝑨𝟓𝟎 .
𝟎 𝟏 𝟎

Ans: Let 𝜆 be eigen value of matrix A .

Characteristic equation is |A - 𝜆 I| = 0
1−𝜆 0 0
∴ 1 0−𝜆 1 =0
0 1 0−𝜆

On solving we get
𝜆3 − (Sum of diagonal elements)𝜆2 + (sum of the minors of diagonal elements ) 𝜆 − 𝐴 = 0

∴ 𝜆3 − 1 + 0 + 0 𝜆2 + 1 + 0 + 0 𝜆 − −1 = 0

∴ 𝜆3 − 𝜆2 + 𝜆 + 1 = 0

∴ Eigen values (𝜆) are 1,1,-1

Since,A is order 3× 3, let

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𝑓(𝐴) = 𝐴50 = 𝑎 𝐴2 + 𝑏𝐴 + 𝑐𝐼 → 1 ( where a, b,c are constants)

We assume 𝑓 (A) is satisfied by 𝜆

∴ 𝜆50 = 𝑎𝜆2 + 𝑏𝜆 + 𝑐 → (2)

Put 𝜆 = 1,

(−1)50 = 𝑎 −1 2
+ 𝑏 −1 + 𝑐
∴ 1 = a – b + c → (3)

Put 𝜆 = 1 𝑖𝑛 2

150 = 𝑎 1 2
+𝑏 1 +𝑐
∴ 1 = a + b + c → (4)

Differentiating (2) w. r.t 𝜆, 50 𝜆49 = 2𝑎𝜆 + 𝑏

49
When 𝜆 = −1 , 50 −1 = 2𝑎 −1 + 𝑏

∴ -50 = - 2a + b→ (5)

Solving (3), (4) and (5) simultaneously , we get

a = 25,b = 0, c = -24

Substituting in ( I ) ,we get ,

𝐴50 = 25𝐴2 + 0 𝐴 − 24 𝐼
∴ 𝐴50 = 25𝐴 × 𝐴 − 24𝐼
1 0 0 1 0 0
= 25 1 0 1 1 0 1 − 24 𝐼
0 1 0 0 1 0

1 0 0 1 0 0
= 25 1 1 0 − 24 0 1 0
1 0 1 0 0 1

1 0 0
= 25 1 0
25 0 1
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 𝟏 𝟎 𝟎
Hence, 𝑨𝟓𝟎 = 𝟐𝟓 𝟏 𝟎
𝟐𝟓 𝟎 𝟏

𝒆𝒛
6 b) Using residue theorem evaluate ∫c 𝟐 𝒅𝒛, 𝒘𝒉𝒆𝒓𝒆 𝑪 𝒊𝒔 𝒛 = 𝟒(06)
𝒛𝟐 +𝝅𝟐
Ans. The circle |z| = 4 has centre (0,0) & radius 4
For singularity, z² +π² = 0
∴ 𝑧²= -𝜋 2 = 𝑖 2 𝜋 2
∴𝑧=±𝑖𝜋

Here, “ 𝑧0 = ± I π”lies inside the circle

𝑧0 = “± i π” are poles of order 2

𝑅1 = Residue of 𝑓 (z) at “𝑧0 = 𝑖 π”

1 𝑑 𝑛−1
= lim 𝑛−1 (𝑧 − 𝑧0 )𝑛 × 𝑓(𝑧)
(𝑛 − 1)! 𝑧→𝑧0 𝑑𝑧

1 𝑑 𝑒𝑧
= lim𝑧→𝑖𝜋 𝑑𝑧 (𝑧 − 𝑖𝜋)² × (𝑧−𝑖𝜋 )²(𝑧+𝑖𝜋 )²
2−1 !

1 𝑑
= 1! lim 𝑒 𝑧 (𝑧 + 𝑖𝜋)−2
𝑧→𝑖𝜋 𝑑𝑧

= lim 𝑒 𝑧 . −2 𝑧 + 𝑖𝜋 −3
+ 𝑧 + 𝑖𝜋 −2
. 𝑒𝑧
𝑧→𝑖𝜋

−2𝑒 𝑖𝜋 𝑒 𝑖𝜋
= (𝑖𝜋 +𝑖𝜋 )³ + (𝑖𝜋 +𝑖𝜋 )³

Now, 𝑒 ±𝑖𝜋 = cosπ±𝑖 sinπ = −1 ± 0 = −1

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 2 1
∴ 𝑅1 = − 2
8𝑖³𝜋³ 4𝑖 𝜋²

Similarly, 𝑅2 = Residue of f (z) at “ 𝑧0 = −𝑖𝜋"

−2𝑒 −𝑖𝜋 𝑒 −𝑖𝜋

= (−𝑖𝜋 −𝑖𝜋 )³ + (−𝑖𝜋 −𝑖𝜋 )²

2 1
∴ 𝑅2 = − 2
8𝑖³𝜋³ 4𝑖 𝜋²

By Cauchy’s Residue theorem, ∫c𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖(𝑅₁ + 𝑅₂ + … )

𝑒𝑧 2 1 2 1
∴ ∫c 𝑧 2 +𝜋 2 2
𝑑𝑧 = 2𝜋𝑖 − 4𝑖 2 𝜋 2 − 8𝑖 3 𝜋 3 − 4𝑖 2 𝜋 2
8𝑖 3 𝜋 3

−2
= 2𝜋𝑖 ×
4𝑖²𝜋²

𝒆𝒛 𝒊
∴ ∫c(𝒛𝟐 +𝝅𝟐 )² 𝒅𝒛 = 𝝅

6 c) Using Rayleigh- Ritz method solve the boundary value problem

𝟏 𝟏
𝑰= 𝟎
𝒙𝒚 + 𝟐 𝒚ˈ² 𝒅𝒙; 𝟎 ≤ ×≤ 𝟏,given y (0)= y (1)= 0 where 𝒚 (𝒙) = 𝒄𝟎 + 𝒄𝟏 𝒙 + 𝒄𝟐 𝒙𝟐 .

1 1
Ans:𝐼 = 0
𝑥𝑦 + 2 𝑦ˈ² 𝑑𝑥 → (1)

Let the approximate solution be

𝑦 (𝑥) = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥 2 → (2)

Putting x =0, 𝑦 (0)= 𝐶0 + 0 + 0

∴ 0 = 𝐶0 → 3 ∵ 𝑦 0 = 0

Putting x = 1 in (2), 𝑦 (1) = 𝐶0 + 𝐶1 + 𝐶2 ∵ 𝑦 1 = 0

∴ 0 = 0 + 𝐶1 + 𝐶2 [𝐹𝑟𝑜𝑚 3]

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 ∴ 𝐶2 = −𝐶1 → (4)

Substituting (3) and (4) in (2),

𝑦 = 0 + 𝐶1 𝑥 − 𝐶1 𝑥² → (5)

Differentiating w.r.t. ‘x’, 𝑦ˈ = 𝑐₁ − 2𝑐₁𝑥 → (6)

Substituting (5) and (6) in (1), we get

1 1 2
𝐼= 0
𝑥 𝑐₁𝑥 − 𝑐₁𝑥² + 2 𝑐₁ − 2𝑐₁𝑥 𝑑𝑥

1 1
= 0
𝑐₁𝑥² − 𝑐₁𝑥³ + 2 𝑐₁² − 4𝑐₁²𝑥 + 4𝑐₁²𝑥² 𝑑𝑥

1 1
= 0
𝑐₁𝑥² − 𝑐₁𝑥³ + 2 𝑐₁² − 2𝑐₁²𝑥 + 2𝑐₁²𝑥² 𝑑𝑥

1
𝑐₁𝑥³ 𝑐1𝑥4 1 2𝑐₁²𝑥² 2𝑐₁²𝑥³
= − + 𝑐₁²𝑥 − +
3 4 2 2 3 0

𝑐₁ 𝑐₁ 1 2
= − + 2 𝑐₁2 − 𝑐₁2 + 3 𝑐₁2 − 0 − 0 + 0 − 0 + 0
3 4

𝑐₁ 1
=𝐼= + 𝑐₁²
12 6

𝑑𝐼
For maximum or minimum, =0
𝑑𝑐 ₁
𝑑𝐼 1 1
∴ 𝑑𝑐 ₁ = 12 + 6 × 2c₁=0

1 −1
∴ 3 c₁= 12

−1
∴ c₁ = 4

−1 1
∴ From (5), 𝑦 = 𝑥 +4 𝑥²
4

𝒙
Hence, the approximate solution is 𝒚= (𝒙 − 𝟏)
𝟒

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