Class 12

Odisha State Board BSE Odisha 9th Class Maths Solutions Geometry Chapter 1 ରେଖା ଓ କୋଣ Ex 1(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 9 Maths Solutions Geometry Chapter 1 ରେଖା ଓ କୋଣ Ex 1(a)

Question 1.
ନିମ୍ନଲିଖ ପଦଗୁଡ଼ିକରୁ ସଂଜ୍ଞାବିହୀନ ଓ ସଂଜ୍ଞାବିଶିଷ୍ଟ (ଯାହାର ସଂଜ୍ଞା ଅଛି) ପଦଗୁଡ଼ିକୁ ଚିହ୍ନାଅ ।
ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ, ସ୍ଥାନାଙ୍କ, ଦୂରତା, ସରଳରେଖା, ରଶ୍ମି, ରେଖାଖଣ୍ଡ, ସମତଳ, ବିନ୍ଦୁ ।
ସମାଧାନ:
ସଂଜ୍ଞାବିହୀନ ପଦ – ସରଳରେଖା, ସମତଳ, ବିନ୍ଦୁ, ।
ସଂଜ୍ଞାବିଶିଷ୍ଟ ପଦ – ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ, ସ୍ଥାନାଙ୍କ, ଦୂରତା, ରଶ୍ମି, ରେଖାଖଣ୍ଡ ।

Question 2.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ସଂକ୍ଷିପ୍ତ ଉତ୍ତର ପ୍ରଦାନ କର ।
(କ) ଗୋଟିଏ ସରଳରେଖାରେ କେତୋଟି ବିନ୍ଦୁ ଥାଏ ?
ସମାଧାନ:
ଅସଂଖ୍ୟ

(ଖ) ଗୋଟିଏ ରେଖାଖଣ୍ଡରେ କେତୋଟି ବିନ୍ଦୁ ଥାଏ ?
ସମାଧାନ:
ଅସଂଖ୍ୟ

(ଗ) ଗୋଟିଏ ରେଖାଖଣ୍ଡରେ କେତୋଟି ପ୍ରାନ୍ତ ବିନ୍ଦୁ ଓ କେତୋଟି ମଧ୍ୟବିନ୍ଦୁ ଥାଏ ?
ସମାଧାନ:
ଦୁଇଟି ଓ ଗୋଟିଏ

(ଘ) ଗୋଟିଏ ରଶ୍ମି ଓ ତାହାର ବିପରୀତ ରଶ୍ମିର ସଂଯୋଗରେ କ’ଣ ଗଠିତ ହୁଏ ?
ସମାଧାନ:
ସରଳରେଖା

(ଙ) ଗୋଟିଏ ରଶ୍ମି ଓ ତାହାର ବିପରୀତ ରଶ୍ମିର ଛେଦରେ କେତୋଟି ବିନ୍ଦୁ ଥାଏ ?
ସମାଧାନ:
ଗୋଟିଏ

(ଚ) ତିନୋଟି ପୃଥକ୍ ସରଳରେଖା ପରସ୍ପରକୁ ଅତିବେଶିରେ କେତୋଟି ବିନ୍ଦୁରେ ଛେଦ କରିବେ ?
ସମାଧାନ:
ତିନୋଟି

(ଛ) ଚାରୋଟି ପୃଥକ୍ ସରଳରେଖା ପରସ୍ପରକୁ ଅତିବେଶିରେ କେତୋଟି ବିନ୍ଦୁରେ ଛେଦ କରିବେ ?
ସମାଧାନ:
ଛଅଗୋଟି

(ଜ) ଚାରୋଟି ପୃଥକ୍ ମଧ୍ୟବିନ୍ଦୁ ମଧ୍ୟରୁ କୌଣସି ତିନୋଟି ଏକରେଖୀ ହୋଇ ନଥିଲେ, ସେମାନଙ୍କ ଦ୍ବାରା କେତୋଟି ସରଳରେଖା ନିର୍ଦ୍ଧାରିତ ହୋଇ ପାରିବ ?
ସମାଧାନ:
ଛଅଗୋଟି

Question 3.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର । ଦତ୍ତ ଅଛି A – B – C
(i) A͞B ∪ \(\overrightarrow{AC}\) = ________
ସମାଧାନ:
\(\overrightarrow{AC}\)

(ii) \(\overrightarrow{BA}\) ∪ \(\overrightarrow{BC}\) = ________
ସମାଧାନ:
\(\overleftrightarrow{AC}\)

(iii) A͞B ∪ B͞C = ________
ସମାଧାନ:
AC

(iv) \(\overrightarrow{AB}\) ∪ \(\overrightarrow{AC}\) = ________
ସମାଧାନ:
\(\overrightarrow{AB}\) ବା \(\overrightarrow{AC}\)

(v) \(\overrightarrow{AB}\) ∩ \(\overrightarrow{BA}\) = ________
ସମାଧାନ:
AB

(vi) A͞C ∩ B͞C = ________
ସମାଧାନ:
BC

(vii) \(\overrightarrow{BA}\) ∩ \(\overrightarrow{BC}\) = ________
ସମାଧାନ:
{B}

(viii) AC – BC = ________
ସମାଧାନ:
AB

(ix) AC – AB = ________
ସମାଧାନ:
BC

Question 4.
L ସରଳରେଖା ଉପରିସ୍ଥ À ଓ B ବିନ୍ଦୁଦ୍ଵୟର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 3 ଓ 5 ହେଲେ AB କେତେ ?
ସମାଧାନ:
L ସରଳରେଖା ଉପରିସ୍ଥ A ଓ B ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 3 ଓ 5 ହେଲେ,
A͞B ର ଦୈର୍ଘ୍ୟ = AB = |-3 – 5| = |-8| = 8
ଅଥବା AB = |5 – (-3)| = |8| = 8

Question 5.
\(\overleftrightarrow{AB}\) ଉପରିସ୍ଥ A ଓ B ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 16 ଓ 20 ହେଲେ AB ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ କେତେ ? 
ସମାଧାନ:
 A ଓ B ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 16 ଓ 20 ।
∴ AB ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ = \(\frac{-16+20}{2}=\frac{4}{2}\) = 2

Question 6.
ନିମ୍ନସ୍ଥ ପ୍ରଶ୍ନଗୁଡ଼ିକରେ ସଂପୃକ୍ତ ସମସ୍ତ ବିନ୍ଦୁ ଏକରେଖୀ ଅଟନ୍ତି ।
(କ) A, B ଓ C ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 11, 4 ଓ 2 ହେଲେ, କେଉଁ ବିନ୍ଦୁଟି ଅନ୍ୟ ଦୁଇଟିର ମଧ୍ୟବର୍ତ୍ତୀ ?
ସମାଧାନ:
A, B, C ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 11, 4 ଓ 2 । ଏଠାରେ – 11 < 2 < 4 ଯୋଗୁଁ A – C – B
ଅର୍ଥାତ୍ C ବିନ୍ଦୁ A ଓ B ର ମଧ୍ୟବର୍ତ୍ତୀ ବିନ୍ଦୁ ହେବ ।

(ଖ) PQ = 8, QR = 5 ଓ RP = 3 ହେଲେ P, Q, R ମଧ୍ଯରେ କେଉଁ ବିନ୍ଦୁଟି ଅନ୍ୟ ଦ୍ଵୟର ମଧ୍ୟବର୍ତ୍ତୀ ?
ସମାଧାନ:
PQ = 8, QR = 5, RP = 3 10166 PQ = QR + RP
ଅର୍ଥାତ୍ R ବିନ୍ଦୁ Q ଓ P ର ମଧ୍ୟବର୍ତ୍ତୀ ବିନ୍ଦୁ ହେବ ।

(ଗ) A ର ସ୍ଥାନାଙ୍କ – 3, A – C – B, BC = 2 ଓ C ର ସ୍ଥାନାଙ୍କ – 4 ହେଲେ, B ର ସ୍ଥାନାଙ୍କ ଓ AB କେତେ ? 
ସମାଧାନ:
A ର ସ୍ଥାନାଙ୍କ – 3, BC = 2, C ର ସ୍ଥାନାଙ୍କ – 4 ଏବଂ A – C – B ।
∴ B ର ସ୍ଥାନାଙ୍କ = – 4 – 2 = -6

BC = 2 ଯୋଗୁଁ, AB = |-6 -3| = |-9|  = 9

(ଘ) ଓ B ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 11 ଓ 21 ହେଲେ, A͞B ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ କେତେ ଓ A ଠାରୁ ଏହାର ଦୂରତା କେତେ ?
ସମାଧାନ:
A ଓ B ର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ – 11 ଓ 21 । A ଓ B ର ମଧ୍ୟବିନ୍ଦୁ C ହେଲେ
∴ A͞B ର ମଧ୍ୟବିନ୍ଦୁ ‘C’ ର ସ୍ଥାନାଙ୍କ = \(\frac{-11+21}{2}=\frac{10}{2}\) = 5
∴ A ଠାରୁ ମଧ୍ୟବିନ୍ଦୁ ‘C’ ର ଦୂରତା = |-11 – 5| = |-6| = 6
∴ AC = 16

(ଙ) A ର ସ୍ଥାନାଙ୍କ – 5 ଓ ĀB ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ 0 ହେଲେ, B ର ସ୍ଥାନାଙ୍କ କେତେ ?
ସମାଧାନ:
A ର ସ୍ଥାନାଙ୍କ – 5 | AB ର ମଧ୍ୟବିନ୍ଦୁ (M) ର ସ୍ଥାନାଙ୍କ 0 |

ମନେକର B ର ସ୍ଥାନାଙ୍କ ‘x’
∴ 0 = \(\frac{-5+x}{2}\) ⇒ x = 5
∴ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ = 5

Question 7.
A, L ସରଳରେଖା ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ ଏବଂ A ର ସ୍ଥାନାଙ୍କ 5 ଅଟେ । A ଠାରୁ 2 ଏକକ ଦୂରତା ବିଶିଷ୍ଟ କେତୋଟି ବିନ୍ଦୁ L ଉପରେ ଅବସ୍ଥିତ ହେବ ଓ ସେମାନଙ୍କର ସ୍ଥାନାଙ୍କ କେତେ ହେବ ?
ସମାଧାନ:

L ସରଳରେଖା ଉପରିସ୍ଥ A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ 5 । A ଠାରୁ 2 ଏକକ ଦୂରତା ବିଶିଷ୍ଟ ଦୁଇଟି ବିନ୍ଦୁ L ଉପରେ ଅବସ୍ଥାନ କରିବ ।
ଆମେ ଜାଣୁ L ସରଳରେଖା ଉପରେ P ଏକ ବିନ୍ଦୁ ଏବଂ à ଏକ ଧନାତ୍ମକ ବାସ୍ତବ ସଂଖ୍ୟା ହେଲେ L 
ଉପରେ କେବଳ ଦୁଇଟି ବିନ୍ଦୁ ରହିବ; ଯାହାର ସ୍ଥାନାଙ୍କ P + a ଓ P – a ।
∴ L ଉପରେ ଦୁଇଟି ବିନ୍ଦୁ ରହିବ ଯଥା B ଓ C ଏବଂ ସେମାନଙ୍କର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ 5 + 2 = 7 ଓ 5 – 2 = 3 

Question 8.
ନିମ୍ନଲିଖ ପଦଗୁଡ଼ିକୁ ଉଦାହରଣ ମାଧ୍ୟମରେ ବୁଝାଅ ।
ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ, ବିପରୀତ ରଶ୍ମି, ବିନ୍ଦୁମାନଙ୍କ ମଧ୍ୟସ୍ଥ ଦୂରତା, ମଧ୍ୟବର୍ତ୍ତିତା ।
ସମାଧାନ:
ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ : ଏକ ରେଖାଖଣ୍ଡର ଗୋଟିଏ ମାତ୍ର ମଧ୍ୟବିନ୍ଦୁ ଥାଏ । ମନେକର A ଓ B ର ସ୍ଥାନାଙ୍କ
ଯଥାକ୍ରମେ x ଓ y ଏବଂ AB ର ମଧ୍ୟବିନ୍ଦୁ ‘C’ ହେଲେ C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ = \(\frac{x+y}{2}\) ହେବ ।
AC = CB = \(\frac{1}{2}\)AB 

‘C’ ବିନ୍ଦୁଟି AB ର ମଧ୍ୟବିନ୍ଦୁ ହେବ ।

ବିପରୀତ ରଶ୍ମି : L ରେଖା ଉପରିସ୍ଥ A, O ଓ B ବିନ୍ଦୁ ଯେପରିକି A – O – B ।
ଏଠାରେ \(\overrightarrow{\mathrm{OA}}\) ଓ \(\overrightarrow{\mathrm{OB}}\) କୁ ବିପରୀତ ରଶ୍ମି କୁହାଯାଏ ।

\(\overrightarrow{\mathrm{OA}}\) ଏବଂ \(\overrightarrow{\mathrm{OB}}\) ପରସ୍ପରର ବିପରୀତ ରଶ୍ମି ହେଲେ \(\overrightarrow{\mathrm{OA}}\) ∪ \(\overrightarrow{\mathrm{OB}}\) = \(\overleftrightarrow{\mathrm{AB}}\) = L
ଅର୍ଥାତ୍ ଦୁଇଟି ବିପରୀତ ରଶ୍ମି \(\overrightarrow{\mathrm{OA}}\) ଏବଂ \(\overrightarrow{\mathrm{OB}}\) ର ସଂଯୋଗ \(\overleftrightarrow{\mathrm{AB}}\) (AB ସରଳରେଖା) ଅଟେ ।

ବିନ୍ଦୁମାନଙ୍କ ମଧ୍ୟସ୍ଥ ଦୂରତା : A ଓ B ବିନ୍ଦୁଦ୍ଵୟର ସ୍ଥାନାଙ୍କ x ଓ y ହେଲେ,
A ଓ B ମଧ୍ୟସ୍ଥ ଦୂରତା = AB = |x – y| (∴ A ଓ B ର ସ୍ଥାନାଙ୍କ ଦ୍ଵୟର ଅଣଋଣାତ୍ମକ ଅନ୍ତର = AB) 
ଏ କ୍ଷେତ୍ରରେ AB ଗୋଟିଏ ଧନାତ୍ମକ ବାସ୍ତବ ସଂଖ୍ୟା ହେବ । ଯାହା ହେଉଛି A ଓ B ବିନ୍ଦୁ ମଧ୍ୟସ୍ଥ ଦୂରତା ।

ମଧ୍ୟବର୍ତ୍ତିତା : ତିନୋଟି ପୃଥକ୍ ବିନ୍ଦୁ A, B ଓ C ଯଦି ଏକ ସରଳରେଖା ଉପରେ ଅବସ୍ଥାନ

କରନ୍ତି ଓ AB + BC = AC ହୁଏ; ତେବେ B କୁ A ଓ C
(କିମ୍ବା C ଓ A )ର ମଧ୍ୟବର୍ତ୍ତୀ ବିନ୍ଦୁ କୁହାଯାଏ ।
ବିନ୍ଦୁ ତ୍ରୟର ଏ ପ୍ରକାର ଅବସ୍ଥାକୁ ସାଙ୍କେତିକ ଭାଷାରେ A – B – C କିମ୍ବା C – B – A ଭାବରେ ଲେଖାଯାଏ ।

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 13 Three Dimensional Geometry will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 13 Three Dimensional Geometry

Important Formulae:
Distance Formula:
The distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂)
= \(\sqrt{\left(x_1-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

Division Formula:
(i) Internal division:
If P (x, y, z) divides the line segment joining, A (x₁, y₁, z₁) and B (x₂, y₂, z₂) into the ratio m : n internally then

Remark:
(i) If P (x, y, z) divides the line segment joining the points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) into the ratio λ : 1 then

(ii) Co-ordinates of the mid-point of the line segment joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\)

Direction Cosines:
Suppose that a straight line makes angles α, β, γ with the positive directions of x-axis, y-axis and z-axis respectively.
Then direction cosines of the line are < cos α, cos β, cos γ >
We denote l = cos α, m = cos β and n = cos γ
Then l² + m² + n² = 1

Direction Ratios:
The direction ratios of a straight line are proportional to direction cosines.
If < a, b, c > are d. rs. and < l, m, n > are d.cs then

Direction ratios of a line segment joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are
< x₂ – x₁, y₂ – y₁, z₂ – z₁ >

The projection of a line segment joining the points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) onto the line ‘L’ with d.cs. < l, m, n >
= l (x₂ – x₁) + m (y₂ – y₁) + n (z₂ – z₁)

Angle between two lines:
Angle between two lines with d.cs.
< l₁, m₁, n₁ > and < l₂, m₂, n₂ > is given by cos θ = l₁l₂ + m₁m₂ + n₁n₂
(i) Two lines are parallel if their d.cs. are equal or d.r.s. are proportional.
(ii) Two lines are perpendicular if l₁l₂ + m₁m₂ + n₁n₂ = 0

Plane
Important formulae:
1. The general equation of the plane is ax + by + cz + d = 0
2. Equation of the plane passing through a poing (x₁, y₁, z₁) and having l, m, n direction cosines of the normal to the plane is l (x – x₁) + m (y – y₁) + n (z – z₁) = 0
3. Equation of the plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1
where a, b, c are the intercepts from the axes.
4. Equation of the plane in normal form is lx + my + nz = p
where < l, m, n > are d.cs of the normal and p is the length of the normal.
5. Equation of the plane passing through three points.
(x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is
\(\left|\begin{array}{rrr}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|\) = 0

6. (i) Angle between two planes is the angle between their normals.
(ii) If two planes are
a₁x + b₁y + c₁z + d₁ = 0 and
a₂x + b₂y + c₂z + d₂ = 0
then the direction ratios of their normal are < a₁, b₁, c₁ > and < a₂, b₂, c₂ >
(iii) If θ is the angle between two planes then
cos θ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}\)
(iv) Two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are parallel if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).
(v) The above two planes are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0.

7. The distance of a point (x₁, y₁, z₁) from a plane ax + by + cz + d = 0 is
\(\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|\)

8. Equations of the planes bisecting the angle between two planes
a₁x + b₁y + c₁z + d₁ = 0 and
a₂x + b₂y + c₂z + d₂ = 0 are
\(\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm \frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}\)

The Straight Line
Important formulae:
1. Unsymmetrical From:
The joint equation of two planes represent a stright line. Thus the equation of a straight line in unsymmetrical form a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

2. Symmetrical Form:
Equation of a straight line through a point (x₀, y₀, z₀) and having d.c. < l, m, n > is
\(\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}\)

3. Two-point Form:
The equation of a straight line passing through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is
\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

4. Condition that a line will lie on a Plane:
The straight line
\(\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}\) lie in a plane ax + by + cz + d = 0
if (i) al + bm + cn = 0
and (ii) ax₀ + by₀ + cz₀ + d = 0

5. Condition for Two Lines to be Coplanar:

6. Angle between a line and a plane:
The angle between the line

7. Distance of a Point from a Line:
The distance of a point (x₁, y₁, z₁) from a line

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 5 Determinants will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 5 Determinants

Evaluation of determinants:

Note: We can evaluate a determinant along any row or column.

Minors and cofactors:
Minor of an element a_ij is the determinant obtained by deleting ith row and jth column denoted by M_ij.
Example.

Properties of determinant:
(a) The value of a determinant remains unchanged by changing rows to columns and columns to rows.
(b) The interchange of two adjacent rows or columns of determinant changes the sign of a determinant without changing its numerical value

(c) If two rows or columns of a determinant are identical then the value of determinant is zero.
(d) If every element of any row or column is multiplied by a factor, then the determinant is multiplied by that factor.
(e) If every element of any row (or column) of a determinant is expressed as sum of two or more numbers, then the determinant can be expressed as the sum of two or more determinants.

(f) A determinant remains unchanged by adding k times the elements of any row (or column) to corresponding elements of any other row (or column) where k is any number or function.

Cramer’s rule to solve a system of linear equations:
Let
a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
is system of linear equations

Note:
(i) If D₁ = D₂ = D₃ = D = 0 then the system has infinitely many solutions.
(ii) If D = 0 and atleast one of D₁, D₂ or D₃ is not zero then the system has no solution.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 11 Differential Equations will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 11 Differential Equations

(a) Differential equation:
It is an equation involving independent variables, dependent variables and their derivatives.

(b) Ordinary D.E:
Differential equation that contains derivatives with respect to a single independent variable.

(c) Partial D.E:
Differential equation that contains derivatives with respect to more than one independent variable.

(d) Order of a D.E:
Order of a differential equation is the order of the highest order derivative occurs in the equation.
eg. \(\frac{d^3 y}{d x^3}+\left(\frac{d y}{d x}\right)^5\)+ y = 0 is of order – 3.

(e) Degree of D.E:
Degree of a differential equation is the power of the highest order derivative occurs in the equation after changing to integral powers.
In the above example the degree is ‘1’.

(f) Formation of a D.E:
(i) Differential equation for f(x, y, a) = 0 … (1)
can be obtained by eliminating a from (1) and
\(\frac{d}{d x}\)f(x, y, a) = 0 … (2)
(ii) Differential equation for
f(x, y, a, b) = 0 … (1)
can be obtained by eliminating a, b from (1),
\(\frac{d}{d x}\)f(x, y, a, b) = 0 … (2)
and \(\frac{{d}^2}{{dx}^2}\)f(x, y, a, b) = 0 … (3)

(g) Solution of a D.E:
A solution of a differential equation is a function which satisfies the given equation.
General Solution: A solution is a general solution if it contains as many arbitrary constants as the order of the differential equation.
Particular Solution: It is a solution that can be obtained by giving particular values to the arbitrary constants.
Singular Solution: The solutions which can not be obtained from the general solution are singular solution.

(h) First order and first degree differential equation:
Definition:
A first order first degree differential equation takes the form
\(\frac{d y}{d x}\) = f(x, y)

Standard types and methods of solution:
(i) Variable separable:
If we can express \(\frac{d y}{d x}\) = f(x, y) in the form N(y) dy = M(x) dx
then we can get a solution by direct integration. The reduced equation is called equation with separable variables.

(ii) Equations reducible to variables separable form:
If the equation is in the form
\(\frac{d y}{d x}\) = f(ax + by + c) then pur z = ax + by + c to reduce the equation to variable separable form.
The equations of the form \(\frac{d y}{d x}=\frac{a_1 x+b_1 y+c_1}{a_2 x+b_2 y+c_2}\) where \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) can be reduced to variable separable form by putting z = a₁x + b₂y.

(iii) Homogeneous differential equation:
A differential equation of the form \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\) where f(x, y) and g(x, y) are homogeneous function of x, y and of same degree is a homogeneous differential equation.
Solution process:
Put y = vx, \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) to reduce the equation to variable separable form. Then solve

(iv) Equations reduciable to homogeneous form:
Form of the equation:

Homogeneous equations:
Let f(x,y) and g(x, y) be homogeneous functions of x and y of same degree.
Then \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\) is called a homogeneous differential equation.
For homogeneous differential equation put y = vx and the proceed.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 8 Application of Derivatives will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 8 Application of Derivatives

Tangents and normals:
(a) If y = f(x) is the equation of any curve then = The slope of the tangent at P(x₁, y₁).
(b) Slope of the normal at (x₁, y₁)
(c) Equation of tangent at P(x₁, y₁) is y – y₁ =
(d) Equation of normal at P(x₁, y₁) is y – y₁ =
(e) Angle between two curves is the angle between two tangents at the point of contact.

Increasing and decreasing functions:
If y = f(x) is defined in [a, b] then
(i) f'(x) > 0, x ∈ (a, b)
⇒ f is strictly increasing on (a, b).
(ii) f'(x) > 0, x ∈ (a, b)
⇒ f is monotonic increasing on (a, b).
(iii) f'(x) < 0, x ∈ (a, b)
⇒ f is strictly decreasing on (a, b).
(iv) f'(x) < 0, x ∈ (a, b)
⇒ f is monotonic decreasing on (a, b).
(v) f'(x) = 0, x ∈ (a, b)
⇒ f is a constant function on (a, b).

Approximation:
(a) If y = f(x) is a function and δx is a very small change in x then the respective change in y is δy given by
δy = f'(x) δx => dy = f'(x) δx.
∴ The approximate value of y = f(x) at
x = a + δx is f(a + δx)
= f(a) + f(a) δx

Maxima and minima:
(a) First derivative criteria to find max/min of y = f(x)
Algorithm:
Step-1 : Put and solve for x.
Let x = a, b, c ……
Step-2 : If changes sign from (+ve) to (-ve) in then at x = a, ‘f’ has a local maximum. If changes sign from (-ve) to (+ve) in then at x = a, f has a local minimum. If in thenat x = a ‘f’ has neither maxima nor a minima (it may be a point of inflexion).

(b) Second derivative criteria
Algorithm:
Step-1 : Find the roots of f'(x) = 0.
Let they are a, b, c ……
Step-2 : Find f”(x) and put x = a, b, c ……
(i) If f”(a) > 0 then at x = a, f has a local minimum.
(ii) If f”(a) < 0 then at x = a, f has a local maximum.
(iii) If f”(a) = 0 and f”(x) changes sign in (a – δ, a + δ) then x = a is a point of inflexion.
(iv) If f”(a) = 0 and f”(x) does not change sign in then use first derivative criteria to check for maxima/minima.

Mean Value Theorems:
(a) Rolle’s theorem:
If a function f is
(i) continuous on the closed interval [a, b]
(ii) differentiable on the open interval (a, b) and
(iii) f(a) = f(b) then there exists a point c ∈ (a, b) such that f(c) = 0.
Geometrical interpretation:
If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b) then there exists atleast one point c ∈ (a, b) such that at x = c the tangent is parallel to x-axis.
Algebraic interpretation:
Between two roots ‘a’ and ‘b’ of f(x) there exists atleast one root of f'(x).

(b) Cauchy’s Mean Value theorem:
If ‘f’ and ‘g’ are two functions such that
(i) both are continuous on [a, b]
(ii) both are differentiable on (a, b) and
(iii) g'(x) ≠ 0 for any x ∈ (a, b) then there exists atleast one point c ∈ (a, b) such that
Geometrical interpretation:
The conclusion of Cauchy’s theorem can be written as i.e. the ratio of the mean rate of increase of two functions in an interval equals to the ratio of actual rate of increase at some point of the interval.

(c) Lagrange’s Mean Value theorem:
If a function f is
(i) continuous on the closed interval [a, b]
(ii) differentiable on the open interval(a, b) then there exists atleast one c ∈ (a, b), such that.
Geometrical interpretation:
Between two points A and B of the graph of y = f(x) there exists atleast one point c such that the tangent is parallel to the chord AB.

Indeterminate forms & L’Hospitals rule:
A limit is said to be in indeterminate form if it takes any of the forms.
Note:
If a limit is in indeterminate form then it can be evaluated using the following methods.
(i) Change the function to determinate form (by rationalisation, expansion or any other means) then find the limit.
Or, (ii) Bring to form then use L’Hospitals rule.
L’Hospitals rule:
Let f and g are two functions differentiable on some open interval containing ‘a’ such that g'(x) ≠ 0 for x ≠ a and g(a) = f(a) = 0, then provided the latter limit exists.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 9 Integration will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 9 Integration

Indefinite integral:
If \(\frac{d}{d x}\)F(x) = f(x) then the indefinite integral of f(x) w.r.t x is
∫f(x)dx = F(x) + C
which represents the entire class of anti-derivatives.

(a) Algebra of integrals:
(i) ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx
(ii) ∫af(x) dx = α ∫f(x) dx
(iii) ∫f(x) g(x) dx = f(x) ∫g(x) dx – ∫\(\left[(\frac{d}{d x} f(x)\right) \cdot \int g(x) d x]\) dx (Integration by parts)

(b) Some standard indefinite integrations:
(1) ∫x_n dx = \(\frac{x^{n+1}}{n+1}\) + C, n ≠ (-1)
(2) ∫\(\frac{d x}{x}\) = log_e|x| + C.
(3) ∫sin x dx = -cos x + C
(4) ∫cos x dx = sin x + C
(5) ∫tan x dx = -log |cos x| + C or log |(sec x)| + C
(6) ∫cot x dx = log |(sin x)| + C or -log |(cosec x)| + C
(7) ∫sec x dx = log |sec x + tan x| + C
(8) ∫cosec x dx = log |cosec x – cot x| + C
(9) ∫sec² x dx = tan x +C
(10) ∫cosec² x dx = -cot x + C
(11) ∫sec x tan x dx = sec x + C
(12) ∫cosec x cot x dx = -cosec x + C

[Note: To integrate by parts choose 1st function according to I LATE]
Where I → Inverse trigonometric functions.
L → Logarithmic function
A → Algebraic function
T → Trigonometric function
E → Exponential function

(c) Techniques of integration:

Definite integration:

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 10 Area Under Plane Curves will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 10 Area Under Plane Curves

Area of the portion bounded by x-axis the curve
y = f(x) and two ordinates at x = a and x = b.

= Area of the portion bounded by y-axis,
the curve x = f(y) and two abscissa at y = c and y = d.

Area between two curves y = f(x), y g(x) with g(x) < f(x) in [a, b] and between two ordinates
x = a and x = b is given by \(\int_a^b\){f(x) – g(x)}dx

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 7 Continuity and Differentiability will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 7 Continuity and Differentiability

Definition:
A function f is said to be continuous at a point a D_f if
(i) f(x) has definite value f(a) at x = a,
(ii) lim_x->a f(x) exists,
(iii) lim_x->a f(x) = f(a).
If one or more of the above conditions fail, the function f is said to be discontinuous at x = a. The  above definition of continuity of a function at a point can also be formulated as follows:
A function f is said to be continuous at x = a if
(i) holds and for a given ∈ > 0, there exists a δ > 0 depending on ∈ such that
|x – a| < 8 ⇒ |f(x) – f(a)| < ∈.
A function f is continuous on an interval if it is continuous at every point of the interval.
If the interval is a closed interval [a, b] the function f is continuous on [a, b] if it is continuous on (a, b),
lim_x->a+ f(x) = f(a) and lim_x->b- f(x) = f(b).

Differentiation of a function:
(a) Differential coefficient (or derivative) of a function y = f(x) with respect to x is

Fundamental theorems:

Then to get \(\frac{d y}{d x}\) it is convenient to take log of both sides before differentiation.

Derivative of some functions:

Higher order derivative:

Leibnitz Theorem:
If ‘u’ and ‘v’ are differentiable functions having ‘n’th derivative then
\(\frac{d^n}{d x^n}\)(u.v) = C₀u_nv + C₁u_n-1v₁ + C₂u_n-2v₂ + ….. + C_nuv_n

Partial derivatives and Homogeneous functions:
(a) If z = f(x, y) is any function of two variables then the partial derivative of z w.r.t. x and y are given below.

(b) Homogeneous function:
z = f(x > y) is a homogeneous function of degree ‘n’ if f(tx, ty) = tⁿ f(x, y).

Euler’s Theorem:
If z = f(x, y) is a homogeneous function of degree ‘n’ then \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}\) = nf(x, y).

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 6 Probability will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 6 Probability

Important terms:
(a) Random experiment: It is an experiment whose results are unpredictable.
(b) Sample space: It is the set of all possible outcomes of an experiment. We denote the sample space by ‘S’.
(c) Sample point: Each element of a sample space is a sample point.
(d) Event: Any subset of a sample space is an event.
(e) Simple event: It is an event with a single sample point.
(f) Compound event: Compound events are the events containing more than one sample point.
(g) Mutually exclusive events: Two events A and B are mutually exclusive if A ∩ B = φ (i.e. occurrence of one excludes the occurrence of other)
(h) Mutually exhaustive events: The events A₁, A₂, A₃ ……. A_n are mutually exhaustive if A₁ ∪ A₂ ∪ A₃ ∪ A_n = S.
(i) Mutually exclusive and exhaustive events.
The events A₁, A₂, A₃ ……. A_n are mutually exclusive and exhaustive if
(i) A₁ ∩ A₂ = φ for i ≠ j
(ii) A₁ ∪ A₂ ∪ …… A_n = S.
(j) Equally likely events: Two events are equally likely if they have equal chance of occurrence.
(k) Impossible and certain events: φ is the impossible and S is the sure or certain event.
(i) Independent events: The events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence of non-occurrence of other.

Probability of an event:
Let S be the sample space and A is an event then the probability of A is
\(P(A)=\frac{|A|}{|S|}=\frac{\text { No.of out comes favourable to } A}{\text { Total number of possible outcomes. }}\)
Note:
1. P(φ) = 0
2. P(S) = 1
3. P(A’) = P (not A) = 1 – P(A)
4. P(A) + P(A’) = 1

Odds in favour and odds against an event:
Let in an experiment is the number of cases favourable to A and ‘n’ is the number of cases not in favour of A then

Addition theorem:
If A and B are any two events then
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Note:
If A and B are mutually exclusive then
P(A ∪ B) = P(A) + P(B)
(∴ P(A ∩ B) = P(φ) – 0)

Conditional probability:
Let A and B are any two events and P(B) ≠ 0 then the conditional probability of A when B has already happened
P(A/B) = \( \frac{P(A \cap B)}{P(B)} \)
Note:
1. P(A ∩ B) = P(B) . P(A/B)
2. If A and B are mutually independent events then P(A/B) = P(A).
∴ P(A ∩ B) = P(A) . P(B)
3. If A and B are independent events then (i) A’ and B’ (ii) A’ and B (iii) A and B’ are also independent.
4. P(A₁ ∩ A₂ ∩ A₃ ….. ∩ A_n) = P(A₁) . P(A₂/A₁) . P(A₃/A₂ ∩ A₁) ….. P(A_n/A₁ ∩ A₂ ∩ …. ∩ A_n-1)
5. Let A₁, A₂ ….. A_n are mutually exhaustive and exclusive events and A is any event which occurs with A₁ or A₂ or A₃ … or A_n then
P(A) = P(A₁) . P(A/A₁) + P(A₂) . P(A/ A₂) + …….. + P(A_n) . P(A/A_n).
This is called the total conditional probability theorem.

Baye’s theorem:
If A₁, A₂ …… A_n are mutually exclusive and exhaustive events and A is any event which occurs with A₁ or A₂ or A₃ or …. A_n then
\( P\left(A_i / B\right)=\frac{P\left(A_i\right) \cdot P\left(A / A_i\right)}{\sum_{i=1}^n P\left(A_i\right) P\left(A / A_i\right)} \)

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 4 Matrices will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 4 Matrices

It is a system of mn numbers arranged in a rectangular system of m rows and n columns.
Example : \(\left(\begin{array}{lll}
a_{11} & a_{12} & a_{1 n} \\
a_{21} & a_{22} & a_{2 n} \\
a_{m_1} & a_{m_2} & a_{m_n}
\end{array}\right)\) is a m × n matrix.

Note : We write the above matrix in short as [a_ij]_m×n or [aij]_m×n or ||a_ij||_m×n.

Important types matrices:
(a) Square matrix:
It is a matrix where the number of rows are equal to the number of columns.

(b) Null (zero) matrix:
If all the entries of a matrix are zero, then the matrix is a zero matrix denoted by 0_m×n.

(c) Diagonal matrix:
It is a square matrix in which all the elements except those in main (or leading) diagonal are zero.

(d) Unit matrix:
It is a diagonal matrix where all the elements in the leading (main) diagonal are one.

(e) Scalar matrix:
It is a diagonal matrix with all the elements in the leading (main) diagonal are α (α ≠ 0 or 1).

(f) Singular matix:
A square matrix ‘A’ is singular iff |A| = 0, otherwise it is a non singular matrix.

(g) Symmetric matrix:
A square matrix is symmetric if A = A’ and skew-symmetric if A = -A’.

Matrix algebra:
(a) Addition and subtraction:
If A = [a_ij]_m×n and B = [b_ij]_m×n then A ± B = [a_ij ± b_ij]_m×n

(b) Scalar multiplication:
If A = [a_ij]_m×n then for any scalar ‘k’
kA = [ka_ij]_m×n

(c) Matrix multiplication:

Properties:
1. Matrix addition is commutative as well as associative.
2. 0_m×n is the additive identity.
3. A is the zero additive inverse of A.
4. We can add or subtract matrices if they are of same order.
5. We can multiply two matrices if the number of columns of 1st is equal to the number of rows of 2nd.
6. Matrix multiplication is non commutative but associative.
7. Matrix multiplication is distributive over addition.
8. AB = 0 ≠ A = 0 or B = 0 for two matrices A and B. Also AB = AC ≠ B = C.
9. If A is a square matrix of order n then
I_n =  is the multiplicative identity.

Transpose and adjoint of a matrix:
(a) The transpose of a matrix A = [a_ij]_m×n is A^T or A’ = [a_ji]_n×m.
Properties:
(i) (A’)’ = A
(ii) (A ± B)’ = A’ ± B’
(iii) (AB)’ = B’A’
(iv) (KA)’ = KA’
(v) Any matrix A can be expressed as sum of a symmetric and a skew-symmetric matrix as

(b) Adjoint of a matrix:
If A is a square matrix then Adj A = The transpose of the matrix of co-factors

Properties:
1. (Adj A) A = A (Adj A) = |A|I_n
∴ A^-1 = \(\frac{{Adj} A}{|~A|}\)
2. |Adj A| = A^n-1
3. Adj (AdjA) = |A|^n-2A
4. (AdjA)’ = Adj (A’)
5. (AB)^-1 = B^-1A^-1

Solution of system of linear equations (matrix method):
Let the system of linear equations is
a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
Let

Note: If the system is homogeneous i.e. d₁ = d₂ = d₃ = 0, then the system has a trivial solution x = 0, y = 0, z = 0 for |A| ≠ 0. In case |A| = 0 then the system has infinitely many solutions.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 3 Linear Programming will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 3 Linear Programming

Definition:
A general linear programming problem LPP is to obtain x₁, x₂, x₃ …… , x_n so as to

Optimize:
Z = c₁x₁ + c₂x₂ + c₃x₃ + ……. + c_nx_n … (A)
subject to
a₁₁x₁ + a₁₂x₂ + ……. + a_1nx_n  ≤ (or ≥) b₁
a₂₁x₁ + a₂₂x₂ + ……. + a_2nx_n  ≤ (or ≥) b₂ … (B)
where x₁, x₂, …… , x_n ≥ … 0 … (C)
and a_ij, b_i, c_j with i = 1, 2, … , m; j = 1, 2, … ,n are real constants.

In the LPP given above, the function Z in (A) is called the objective function. The variables x₁ x₃, ……, x_n are called decision variables. The constants c₁ c₂, ……, c_n are called cost coefficients. The inequalities in (B) are called constraints. The restrictions in (C) are called non-negative restrictions. The solutions which satisfy all the constraints in (B) and the non-negative restrictions in (C) are called feasible solutions.

The LPP involves three basic elements:

1. Decision variables whose values we seek to determine,
2. Objective (goal) that we aim to optimize,
3. Constraints and non-negative restrictions that the variables need to satisfy.

Types of Linear Programming Problems

As we have already discussed we come across different types of problems which we need solve depending on the objective functions and the constraints. Here we discuss a few important types of LPPs. before learning how to formulate them.

(i) Manufacturing Problem
A manufacturer produces different items so as to maximise his profit. He has to determine the number of units of products he must produce while satisfying a number of constraints because each unit of product requires availability of some amount of raw material, certain manpower, certain machine hours etc.

(ii) Diet Problem
Suppose a person is advised to take vitamins/nutrients of two or more types. The vitamins/nutrients are available in different proportions in different types of foods. If the person has to take a minimum amount of the vitamins/nutrients then the problem is to determine appropriate quantity of food of each type so that cost of food is kept at the minimum.

(iii) Allocation Problem
In this type of problem one has to allocate different resources/tasks to different units/persons depending on the nature of the gain or outcome.

(iv) Transportation Problem
These problems involve transporting materials from sources to destinations for sale or distribution of products or collection of raw materials etc. Here the aim is to use various options for transportation such as distance, time etc so as to keep the cost of transportation to a minimum.

Working procedure to solve LPP graphically
Step-1. Taking all inequations of the constraints as equations, draw lines represented by each equation and considering the inequalities of the constraint inequations complete the feasible region.
Step-2. Determine the vertices of the feasible region either by inspection or by and solving the two equations of the intersecting lines.
Step-3. Evaluate the objective function Z = ax + by at each vertex.
Case (i) F.R. is bounded: The vertex which gives the optional value (maximum or minimum) of Z gives the desired optional solution to the LPP.
Case (ii) F.R. is unbounded: When M is the maximum value of Z at a vertex Vmax, determine the open half plane corresponding to the inequation ax + by > M. If this open half plane has no points in common with the F.R. then M is the maximum value of Z and the point Vmax gives the desired solution. Otherwise, Z has no maximum value.

Similarly consider the open half plane ax+by < m when m is the minimum value of Z at the vertex Vmin. If this half-plane has no point common with the F.R. then m is the minimum value of Z and Vmin gives the desired solution. Otherwise Z has no minimum value.