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Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(b)

Question 1.
ପ୍ରତିକଳ୍ପନ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖ୍ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ କର ।
(i) x + y – 8 = 0, 2x – 3y – 1 = 0
(ii) 3x + 2y – 5 = 0, x – 3y – 9 = 0
(iii) 2x – 5y + 8 = 0, x – 4y + 7 = 0
(iv) 11x + 15y + 23 = 0, 7x – 2y – 20 = 0
(v) ax + by – a + b = 0, bx – ay – a – b = 0
(vi) x + y – a = 0, ax + by – b² = 0
ସମାଧାନ ପ୍ରଣାଳୀ :
(i) ସହସମୀକରଣଦ୍ଵୟ ମଧ୍ୟରୁ ଗୋଟିକରୁ ‘x’ କିମ୍ବା ‘y’ର ମାନ ନେଇ ଯଥାକ୍ରମେ y କିମ୍ବା x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଏ ।
(ii) x କିମ୍ବା yର ମାନକୁ ଅନ୍ୟ ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ଯଥାକ୍ରମେ y କିମ୍ବା x ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଉ ।
(iii) ଉକ୍ତ ନିର୍ଣ୍ଣୟ ମାନକୁ (y କିମ୍ବା x) ନେଇ ଯେକୌଣସି ଗୋଟିଏ ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ଅନ୍ୟଟିର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
ସମାଧାନ :
(i) x + y – 8 = 0 …….(i) ଏବଂ
2x – 3y – 1 = 0 …….(ii)
ସମୀକରଣ (i)କୁ ବିଚାର କରି yକୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
x + y – 8 = 0 ⇒ y = 8 – x ……..(iii)
y ର ମାନକୁ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ ପାଇବା 2x – 3(8 – x) – 1 = 0
⇒ 2x – 24 + 3x – 1 = 0 ⇒ 5x = 25 ⇒ x = \(\frac{25}{5}\) = 5
x ର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ, y = 8 – x = 8 – 5 = 3
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (5, 3) ଅଟେ ।

(ii) 3x + 2y – 5 = 0 …….(i) ଏବଂ
x – 3y – 9 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ 3x = 5 – 2y ⇒ x = \(\frac{5-2y}{3}\) …….(iii)
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, \(\frac{5-2y}{3}\) – 3y – 9 = 0
⇒ \(\frac{5-2y-9y-27}{3}\) ⇒ 0 = -22 – 11y = 0
⇒ 11y = -22 = y = \(\frac{-22}{11}\) = -2
yର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ,
x = \(\frac{5-2y}{3}\) = \(\frac{5-2(-2)}{3}\) = \(\frac{5+4}{3}\) = \(\frac{9}{3}=3\)
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (3, -2) ଅଟେ ।

(iii) 2x – 5y + 8 = 0 …….(i) ଏବଂ
x – 4y + 7 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି x କୁ y ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
2x – 5y + 8 = 0 ⇒ 2x = 5y – 8 ⇒ x = \(\frac{1}{2}\)(5y – 8) …….(iii)
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, \(\frac{1}{2}\)(5y – 8) – 4y + 7 = 0
⇒ \(\frac{5y-8-8y+14}{2}\) = 0
⇒ -3y + 6 = 0 ⇒ y = \(\frac{-6}{-3}\) = 2
yର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ, x = \(\frac{1}{2}\)(5 × 2 – 8) = \(\frac{1}{2}\) × 2 = 1
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (1, 2) ଅଟେ ।

(iv) 11x + 15y + 23 = 0 …….(i) ଏବଂ
7x – 2y – 20 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
11x + 15y + 23 = 0 ⇒ 15y = -11x – 23
⇒ y = \(\frac{1}{15}\)(-11x – 23) …….(iii)
‘y’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, 7x – \(\frac{2}{15}\)(-11x – 23) = 20
⇒ \(\frac{105x+22x+46}{15}\) = 20
⇒ 127x = 300 – 46 ⇒ x = \(\frac{254}{127}\) = 2
x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = latex]\frac{1}{15}[/latex](-22 – 23) = \(\frac{1}{15}\) × -45 = -3
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (2, -3) ।

(v) ax + by – a + b = 0 …….(i) ଏବଂ
bx – ay – a – b = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
ax + by – a + b = 0 ⇒ by = -ax + a – b
⇒ y = \(\frac{1}{b}\)(-ax + a – b) …….(iii)
y ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, bx – \(\frac{a}{b}\)(-ax + a – b) = 0
⇒ \(\frac{b^2x+a^2x-a^2 +ab-ab-b^2}{b}\) = 0
⇒ x(a² + b²) = a² + b² ⇒ x = \(\frac{a^2+b^2}{a^2+b^2}\) = 1
x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = latex]\frac{1}{b}[/latex](-a + a – b) ⇒ y = \(\frac{-b}{b}\) = -1
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (1, -1) ଅଟେ।

(vi) x + y – a = 0 …….(i) ଏବଂ
ax + by – b² = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
x + y – a = 0 ⇒ y = a – x …….(iii)
y ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, ax + b(a – x) – b² = 0
⇒ ax + by – bx – b² = 0 ⇒ ax – bx = b² – ab
⇒ x(a – b) = -b(a – b) [a = b ତେଣୁ a – b ≠ 0]
ତେଣୁ x = -b
‘x’ ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = a + b
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (-b, a+b)।

Question 2.
ଅପସାରଣ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖ ସହ ସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(i) x – y – 3 = 0, 3x – 2y – 1 = 0
(ii) 3x + 4y = 10, 2x – 2y = 2
(iii) 3x – 5y – 4 = 0, 9x = 2y – 1
(iv) 0.4x – 1.5y = 6.5, 0.3x + 0.2y = 0.9
(v) √2x + √3y = 0, √5x + √2y = 0
(vi) ax + by = 0, x + y – c = 0 (a+b ≠ 0)
ସମାଧାନ ପ୍ରଣାଳୀ :
(i) ସମୀକରଣଦ୍ବୟରୁ ‘x’ ଅପସାରଣ କରାଯାଇ y ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
(ii) y ର ମାନକୁ ଯେକୌଣସି ସମୀକରଣରେ ପ୍ରୟୋଗ କରି x ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
(iii) ସେହିପରି ସମୀକରଣଦ୍ଵୟରୁ yକୁ ଅପସାରଣ କରାଯାଇ ‘x’ର ମାନ ନିର୍ଣ୍ଣୟ କରି ଏହାକୁ ଯେକୌଣସି ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ‘y’ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଇ ପାରିବ ।
ସମାଧାନ :
(i) x – y – 3 = 0 ……… (i) ଏବଂ
3x – 2y – 1 = 0 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ,
x – y – 3 = 0 ⇒ 4 – y – 3 = 0 ⇒ -y + 1 = 0 ⇒ y = 1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (4, 1) ଅଟେ।

(ii) 3x + 4y = 10 ……… (i) ଏବଂ
2x – 2y = 2 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
3x + 4y = 10 ⇒ 2 + 4y = 10 ⇒ 4y = 10 – 6 ⇒ y = \(\frac{4}{4}\) = 1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (2, 1) ଅଟେ।

(iii) 3x – 5y – 4 = 0 ……… (i) ଓ
9x = 2y – 1 ⇒ 9x – 2y + 1 = 0 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
3x – 5y – 4 = 0
⇒ 3 × (\(– \frac{1}{3}\)) – 5y – 4 = 0
⇒ -5y – 5 = 0 ⇒ y = \(\frac{5}{-5}\) = -1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (\(– \frac{1}{3}\), -1) ଅଟେ।

(iv) 0.4x – 1.5y = 6.5 ……… (i) ଓ
0.3x + 0.2y = 0.9 ………. (ii)

y ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
0.4x – 1.5y = 6.5 ⇒ 0.4x + 4.5 = 6.5
⇒ 0.4x = 6.5 – 4.5 ⇒ 1.4x = 2 ⇒ x = \(\frac{2}{0.4}\) = 5
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (5, -3) ଅଟେ।

(v) √2x + √3y = 0 ……… (i) ଓ
√5x + √2y = 0 ………. (ii)

c₁ = c₂ = 0 ଓ a₁b₂ – a₂b₁ = 0 ହେଲେ ସମୀକରଣଦ୍ଵୟର ସମାଧାନଟି (0, 0) ଅଟେ ।
ଏଠାରେ c₁ = c₂ = 0 ଏବଂ \(\frac{\sqrt{2}}{\sqrt{5}} \neq \frac{\sqrt{3}}{\sqrt{2}}\)
ତେଣୁ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ (0, 0) ଅଟେ ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ :

‘y’ ର ମାନକୁ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ,
√2x + √3 × 0 = 0 ⇒ √2x = 0 ⇒ x = 0
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (0, 0) ଅଟେ।

(v) ax + by = 0 ……… (i) ଏବଂ
x + y – c = 0 ………. (ii)

‘x’ ର ମାନକୁ ସମୀକରଣ (ii)ରେ ପ୍ରୟୋଗ କଲେ, y = c – x = c – \(\frac{bc}{b-a}\) =
= \(\frac{bc-ca-bc}{b-a}\) = \(\frac{-ca}{b-a}\) = \(\frac{ca}{a-b}\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (\(\frac{bc}{b-a}\), \(\frac{ca}{a-b}\)) ଅଟେ।

Question 3.
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖୂତ ସହ ସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(1) x + 2y + 1 = 0, 2x – 3y – 12 = 0
(ii) 2x + 5y = 1, 2x + 3y = 3
(iii) x + 6y + 1 = 0, 2x + 3y + 8 = 0
(iv) \(\frac{x}{a}+\frac{y}{b}\) = a+b, \(\frac{x}{a^2}+\frac{y}{b^2}\) = 2
(v) x + 6y + 1 = 0, 2x + 3y + 8 = 0
(vi) 4x – 9y = 0, 3x + 2y – 35 = 0
ବଜ୍ରଗୁଣନ ପୃତ୍ର : \(\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{x}{a_1b_2-a_2b_1}\)
ଯେଉଁଠାରେ a₁b₂ = a₂b₁ ≠ 0
ସମାଧାନ :
(i) x + 2y + 1 = 0 ……… (i) ଏବଂ
2x – 3y – 12 = 0 ………. (ii)
ସମୀକରଣମାନଙ୍କର a₁b₂ – a₂b₁ = (1)(-3) – (2)(2) = -7 ≠ 0
ତେଣୁ ସମାଧାନ ସମ୍ଭବ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ।

⇒ \(\frac{x}{-21}=\frac{y}{14}=\frac{1}{-7}\)
⇒ x = \(\frac{-21}{-7}\) ଓ y = \(\frac{14}{-7}\)
⇒ x = 3 ଓ y = -2
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (3, -2) ଅଟେ।

(ii) 2x + 5y = 1 ……… (i) ଏବଂ
2x + 3y = 3 ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ଦ୍ବୟରୁ 2x + 5y = 1, 2x + 3y = 3
ସମୀକରଣମାନଙ୍କର a₁b₂ – a₂b₁ = (2)(3) – (2)(5) = -4 ≠ 0
ତେଣୁ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ ସମ୍ଭବ ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ।

⇒ \(\frac{x}{15-(-3)}=\frac{y}{-2-(-6)}=\frac{1}{6-10}\)
⇒ \(\frac{x}{15+3}=\frac{y}{-2+6}=\frac{1}{6-10}\)
⇒ \(\frac{x}{-12}=\frac{y}{4}=\frac{1}{-4}\)
⇒ x = \(\frac{-12}{-4}=3\) ଓ y = \(\frac{4}{-4}=-1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (3, -2) ଅଟେ।

(iii) x + 6y + 1 = 0 ……… (i) ଏବଂ
2x + 3y + 8 = 0 ………. (ii)
ଏଠାରେ a₁b₂ – a₂b₁ = (1)(3) – (2)(6) = 3- 12 = -9 ≠ 0
ତେଣୁ ଦତ୍ତ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ ସମ୍ଭବ ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ

⇒ \(\frac{x}{48-3}=\frac{y}{2-86}=\frac{1}{3-12}\)
⇒ \(\frac{x}{45}=\frac{y}{-6}=\frac{1}{-9}\)
⇒ x = \(\frac{-45}{-9}=-5\) ଓ y = \(\frac{-6}{-9}=\frac{2}{3}\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (-5, \(\frac{2}{3}\)) ଅଟେ।

(iv) \(\frac{x}{a}+\frac{y}{b}\) = a+b ……… (i)
\(\frac{x}{a^2}+\frac{y}{b^2}=2\) ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ରୁ \(\frac{x}{a}+\frac{y}{b}-(a+b)=0\), \(\frac{x}{a^2}+\frac{y}{b^2}-2=0\)

∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (a², b²) ଅଟେ।

(v) x + 6y + 1 = 0 ……… (i)
2x + 3y + 8 = 0 ………. (ii)
ଏଠାରେ a₁ = 1
b₁ = 6
c₁ = 1
a₂ = 5
b₂ = 3
c₂ = 8
ପୁନଶ୍ଚ \({a_1}{a_2}=\frac{1}{2}\), \({b_1}{b_2}=\frac{6}{3}=2\)
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), ତେଣୁ ସହ-ସମୀକରଣଦ୍ୱୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ।
b₁c₂ – b₂c₁ = 6 × 8 – 3 × 1 = 48 – 3 = 45
c₁a₂ – c₂a₁ = 1 × 2 – 8 × 1 = 2 – 8 = -6
a₁b₂ – a₂b₁ = 1 × 3 – 2 × 6 = 3 – 12 = -9

⇒ \(\frac{x}{45}=\frac{y}{-6}=\frac{1}{-9}\)
⇒ x = \(\frac{-45}{-9}=-5\) ଓ y = \(\frac{-6}{-9}=\frac{2}{3}\)
∴ ସମାଧାନ (x, y) = (-5, \(\frac{2}{3}\) ) ।

(vi) 4x – 9y = 0 ……… (i)
3x + 2y – 35 = 0 ………. (ii)
ଏଠାରେ a₁ = 4
b₁ = -9
c₁ = 0
a₂ = 3
b₂ = 2
c₂ = -35
ତେଣୁ ସହ-ସମୀକରଣଦ୍ୱୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ; \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (\(\frac{4}{3} \neq \frac{-9}{2}\))
b₁c₂ – b₂c₁ = (-9)(-35) – 2 × 0 = 315
c₁a₂ – c₂a₁ = 0 × 3 – (-35) × 4 = 140
a₁b₂ – a₂b₁ = 4 × 2 – 3 × (-9) = 35 ≠ 0

⇒ \(\frac{x}{315}=\frac{y}{140}=\frac{1}{35}\)
⇒ \(\frac{x}{9}=\frac{y}{4}=1\) ⇒ x = 9 ଓ y = 4
∴ ସମାଧାନ (x, y) = (9, 4) ।

Question 4.
ନିମ୍ନଲିଖ୍ ସହସମୀକରଣମାନଙ୍କ ସମାଧାନ କର ।
(i) \(\frac{2}{x}+\frac{3}{y}=17, \frac{1}{x}+\frac{1}{y}=7(x \neq 0, y \neq 0)\)
(ii) \(\frac{5}{x}+6 y=13, \frac{3}{x}+20 y=35(x \neq 0)\)
(iii) \(2 x-\frac{3}{y}=9,3 x+\frac{7}{y}=2(y \neq 0)\)
(iv) 4x + 6y = 3xy, 8x + 9y = 5xy (x ≠ 0, y ≠ 0)
(v) (a – b)x + (a + b)y = a² – 2ab – b², (a + b)x+(a + b)y = a² + b²
(vi) \(\frac{2}{x}+\frac{3}{y}=2\), ax – by = a² – b²
(vii) \(\frac{5}{x+y}-\frac{2}{x-y}+1=0, \frac{15}{x+y}+\frac{7}{x-y}-10=0\)
(viii) \(\frac{xy}{x+y}=\frac{6}{5}, \frac{xy}{x+y}=6(x+y \neq 0, x-y \neq 0)\)
(ix) 6x + 5y = 7, x + 3y + 1 = 2 (x + 6y – 1)
(x) \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\)
(xi) \(\frac{x+y}{2}-\frac{x-y}{3}=8, \frac{x+y}{3}+\frac{x-y}{4}=11\)
(xii) \(\frac{x}{a}=\frac{y}{b}, ax + by=a^2+b^2 \)
ସମାଧାନ :
(i) \(\frac{2}{x}+\frac{3}{y}=17\) ⇒ \(\frac{2}{x}+\frac{3}{y}-17=0\) …….(1)
\(\frac{1}{x}+\frac{1}{y}=7\) ⇒ \(\frac{1}{x}+\frac{1}{y}-7=0\) …….(2)
ଏଠାରେ \(\frac{1}{x}=u\) ଏବଂ \(\frac{1}{x}=v\) ନେଲେ ଦତ୍ତ ସମୀକରଣଦ୍ବୟ
2u + 3v – 17 = 0, ଏବଂ u + v – 7 = 0 ହେବ ।
ଏଠାରେ a₁ = 2
b₁ = 3
c₁ = -17
a₂ = 1
b₂ = 1
c₂ = -7
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) ହୋଇଥିବାରୁ ସମୀକରଣ ଦ୍ଵୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ।
b₁c₂ – b₂c₁ = (3)(-7) – 1(-17) = -21 – 17 = -4
c₁a₂ – c₂a₁ = (-17) × 1 – (-7) × 2 = -17 + 14 = -3
a₁b₂ – a₂b₁ = 2 × 1 – 1 × 3 = 2 – 3 = -1 (≠ 0)

⇒ \(\frac{u}{-4}=\frac{v}{-3}=\frac{1}{-1}\) ⇒ u = 4 ଏବଂ v = 3
⇒ \(\frac{1}{x}=4\) ଏବଂ \(\frac{1}{y}=3\) ⇒ x = \(\frac{1}{4}\) ଏବଂ y = \(\frac{1}{3}\)
∴ ସମାଧାନ (x, y) = (\(\frac{1}{4}\), \(\frac{1}{3}\)) ।

‘x’ ର ମାନ ସମୀକରଣରେ (ii) ପ୍ରୟୋଗ କଲେ y = \(\frac{1}{3}\) ହେବ ।

(ii) \(\frac{5}{x}+6 y=13\) ⇒ \(\frac{5}{x}+6 y-13=0\) ……..(1)
\(\frac{3}{x}+20 y=35\) ⇒ \(\frac{3}{x}+20 y-35=0\) ……….(2)

ସମୀକରଣ (1) ରେ x = \(\frac{41}{25}\)

⇒ \(\frac{125}{41}+6 y-13=0 \Rightarrow 6 y=13-\frac{125}{41}=\frac{533-125}{41}\)
⇒ \(6 y=\frac{408}{41} \Rightarrow y=\frac{408}{41} \times \frac{1}{6}=\frac{68}{41}\)
∴ ସମାଧାନ (x, y) = (\(\frac{41}{25}\), \(\frac{68}{41}\)) ।

(iii) \(2 x-\frac{3}{y}=9,\) …….(i)
\(3 x+\frac{7}{y}=2\) …….(ii)

‘x’ ର ମାନ ସମୀକରଣରେ (ii) ପ୍ରୟୋଗ କଲେ 2 × 3 – \(\frac{3}{y}\) = 9 ⇒ – \(\frac{3}{y}\) = 9 – 6
⇒ 3y = -3 ⇒ y = -1
∴ ସମାଧାନ (x, y) = (3, -1)।

(iv) 4x + 6y = 3xy ………..(i)
8x + 9y = 5xy ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ର ଉଭୟ ପାର୍ଶ୍ଵକୁ xy ଦ୍ବାରା ଭାଗକଲେ,
\(\frac{4}{y}+\frac{6}{x}=3\) …..(iii)
\(\frac{8}{y}+\frac{9}{x}=5\)
ମନେକର \(\frac{1}{x}=u\) ଓ \(\frac{1}{y}=v\)। ତେଣୁ ସମୀକରଣ (iii) ଓ (iv) ରୁ
6u + 4v = 3 ……(v), 9u + 8v = 5 ……..(vi)

ସମୀକରଣ (v) ରେ u = \(\frac{1}{3}\) ବସୀକଲେ, 6 × \(\frac{1}{3}\) + 4v = 3 ⇒ 4v = 1 ⇒ v = \(\frac{1}{4}\)
u = \(\frac{1}{3}\) ⇒ \(\frac{1}{x}\) = \(\frac{1}{3}\) ⇒ x = 3, v = \(\frac{1}{4}\) ⇒ \(\frac{1}{y}\) = \(\frac{1}{4}\) ⇒ y = 4
∴ ସମାଧାନ (x, y) = (3, 4)।

(v) (a – b)x + (a + b)y = a² – 2ab – b² ……..(i)
(a + b)x+(a + b)y = a² + b² ……….(ii)
ସମୀକରଣ (ii) କୁ ସମୀକରଣ (i) ରୁ ବିପ୍ରୟୋଗ କଲେ,
x(a + b) – x(a + b) = -2ab – 2b²
⇒ x(a + b – a – b) = -2ab – 2b²
⇒ -2bx = -2b(a + b) ⇒ x = a + b
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, (a+b)(a+b) + (a+b) y = a² + b²
⇒ a² + b² + 2ab + (a + b) y = a² + b²
⇒ (a + b) y = -2ab ⇒ y = \(\frac{-2ab}{a+b}\)
∴ ସମାଧାନ (x, y) = (a+b, \(\frac{-2ab}{a+b}\))।

(vi) \(\frac{2}{x}+\frac{3}{y}=2\) ⇒ \(\frac{bx+ay}{ab}=2\)
⇒ bx + ay = 2ab ……(1), ax – by = a² – b² ………(2)

⇒ x(a² + b²) = a³ + ab² ⇒ x(a² + b²) = a(a² + b²) ⇒ x = a
ସମୀକରଣ (1) ରେ x = a ସ୍ଥାପନ କଲେ, b.a + ay = 2ab ⇒ ay = ab
⇒ y = b
∴ ସମାଧାନ (x, y) = (a, b)।

(vii) \(\frac{5}{x+y}-\frac{2}{x-y}+1=0\) ……..(i) ଏବଂ
\(\frac{15}{x+y}+\frac{7}{x-y}-10=0\) ……..(ii)
\(\frac{1}{x+y}=a\) ଏବଂ \(\frac{1}{x-y}=b\) ହେଲେ
ସମୀକରଣଦ୍ଵୟ 5a – 2b + 1 = 0 ………(iii) ଏବଂ 15a + 7b – 10 = 0 ……….(iv)

⇒ x – y = 1 …….. (v)
b ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗକଲେ,
5a – 2 × 1 + 1 = 0
⇒ 5a – 1 = 0 ⇒ 5a = 1 ⇒ a = \(\frac{1}{5}\)
⇒ \(\frac{1}{x+y}=\frac{1}{5}\) ⇒ x + y = 5 ……..(iv)
ସମୀକରଣ (v) ଓ (vi) ରୁ ୟୋଗକଲେ x + y + x – y = 5 + 1
⇒ 2x = 6 ⇒ x = 3
∴ ନିଶ୍ଚେୟ ସମାଧାନ (x, y) = (3, 2)।

(viii) \(\frac{xy}{x+y}=\frac{6}{5}\) …….(i)
\(\frac{xy}{x+y}=6\) ……..(ii)

⇒ \(\frac{1}{y}=\frac{2}{6}\) ⇒ \(\frac{1}{y}=\frac{1}{3}\) ⇒ y = 3
∴ ସମାଧାନ (x, y) = (2, 3)।

(ix) 6x + 5y = 7x + 3y + 1 = 2 (x + 6y – 1)
⇒ 6x + 5y = 7x + 3y + 1 ⇒ x – 2y + 1 = 0 ……(i)
ପୁନଶୃ 7x + 3y + 1 = 2 (x + 6y – 1)
⇒ 7x + 3y + 1 = 2x + 12y – 2 ⇒ 5x – 9y + 3 = 0 …..(ii)

ସମୀକରଣ (i) ରେ x = 3 ସ୍ଥାପନ କଲେ, 3 – 2y + 1 = 0
⇒ 2y = 4 ⇒ y = 2
∴ ସମାଧାନ ପେଟ୍ (x, y) = (3, 2)।

(x) \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\)
⇒ \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}\) ⇒ 3(x + y – 8) = 2(x + 2y – 14)
⇒ 3x + 3y – 24 = 2x + 4y – 28 ⇒ x – y = -4 ……..(i)
ପୁନଶୃ \(\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\) ⇒ 11(x + 2y – 14) = 3(3x + y – 12)
⇒ 11x + 22y – 154 = 9x + 3y – 36 ⇒ 2x + 19y = 118 ……….(ii)

ସମୀକରଣ (i) ରେ x = 2 ସ୍ଥାପନ କଲେ, 2 – y = -4 ⇒ y = 6
∴ ସମାଧାନ (x, y) = (2, 6)।

(xi) \(\frac{x+y}{2}-\frac{x-y}{3}=8\) ⇒ \(\frac{3(x+y)-2(x-y)}{6}=8\)
⇒ 3x + 3y – 2x + 2y = 48 ⇒ x + 5y = 48 ……..(i)
ପୁନଶୃ \(\frac{x+y}{3}+\frac{x-y}{4}=11\) ⇒ \(\frac{4(x+y)+3(x-y)}{12}=11\)
⇒ 4x + 4y + 3x – 3y = 132 ⇒ 7x + y = 132 ……..(ii)

ସମୀକରଣ (i) ରେ y = 6 ସ୍ଥାପନ କଲେ, x + 5 × 6 = 48 ⇒ x = 18
∴ ସମାଧାନ (x, y) = (18, 6)।

(xii) \(\frac{x}{a}=\frac{y}{b}\) ⇒ bx = ay ⇒ bx – ay = 0 ……..(i)
ଏବଂ \(ax + by=a^2+b^2 \) ………(ii)

x ର ମାନ ସମୀକରଣ (i) ରେ କଲେ, ba – ay = 0 ⇒ ay = ab ⇒ y = b

ବିକଳ୍ପ ସମାଧାନ :
\(\frac{x}{a}=\frac{y}{b}=k\) (ମନେକର) x = ak, y = bk
ax + by = a² + b² = a.ak + b.bk = a² + b²
k (a² + b²) = a² + b² ⇒ k = 1
∴ x = ak = a . 1 = a; y = bk = b . 1 = b

Question 5.
ନିମ୍ନଲିଖତ ଡିଟରମିନାଣ୍ଟର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
(i) \(\left|\begin{array}{ll}
2 & 5 \\
6 & 0
\end{array}\right|\)
(ii) \(\left|\begin{array}{ll}
2 & -1 \\
3 & 2
\end{array}\right|\)
(iii) \(\left|\begin{array}{ll}
0 & 4 \\
5 & -1
\end{array}\right|\)
(iv) \(\left|\begin{array}{ll}
\frac{1}{2} & 1 \\
\frac{3}{4} & \frac{1}{5}
\end{array}\right|\)
ସମାଧାନ :
(i) \(\left|\begin{array}{ll}
2 & 5 \\
6 & 0
\end{array}\right|\) = 2(0) – 6(5) = 0 – 30 = -30

(ii) \(\left|\begin{array}{ll}
2 & -1 \\
3 & 2
\end{array}\right|\) = 2 × 2 – 3 (-1) = 4 + 3 = 7

(iii) \(\left|\begin{array}{ll}
0 & 4 \\
5 & -1
\end{array}\right|\) = 0(-1) – 5 × 4 = 0 – 20 = -20

(iv) \(\left|\begin{array}{ll}
\frac{1}{2} & 1 \\
\frac{3}{4} & \frac{1}{5}
\end{array}\right|\) = \((\frac{1}{2})(\frac{1}{5})-(\frac{3}{4})(1)=\frac{1}{10}-\frac{3}{4}=\frac{2-15}{20}=\frac{-13}{20}\)

Question 6.
Cramer ଙ୍କ ନିୟମ ପ୍ରୟୋଗ କରି ନିମ୍ନ ସହସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(i) 2x + 3y = 5, 3x + y = 4
(ii) x + y = 3, 2x + 3y = 8
(iii) x – y = 0, 2x + y = 3
(iv) 2x – y = 3, x – 3y = -1
ସମାଧାନ :
(i) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
3 & 1
\end{array}\right|=2 \times 1-3 \times 3=2-9=-7\)
ଏଠାରେ ∆ ≠ 0 ତେଣୁ ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
5 & 3 \\
4 & 1
\end{array}\right|=5 \times 1-4 \times 3=5-12=-7\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 5 \\
3 & 4
\end{array}\right|=2 \times 4-5 \times 3=8-15=-7\)
x = \(\frac{∆_x}{∆}=\frac{-7}{-7}=1\), y = \(\frac{∆_y}{∆}=\frac{-7}{-7}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 1)

(ii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|=1 \times 3-2 \times 1=3-2=1\)
ଏଠାରେ ∆ ≠ 0 ତେଣୁ ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
3 & 1 \\
8 & 3
\end{array}\right|=3 \times 3-8 \times 1=9-8=1\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 3 \\
2 & 8
\end{array}\right|=1 \times 8-2 \times 3=8-6=2\)
x = \(\frac{∆_x}{∆}=\frac{1}{1}=1\), y = \(\frac{∆_y}{∆}=\frac{2}{1}=2\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 2)

(iii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
1 & -1 \\
2 & 1
\end{array}\right|=1 \times 1-2 \times -1=1+2=3\)
∆ ≠ 0 ତେଣୁ ସହ-ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
0 & -1 \\
3 & 1
\end{array}\right|=0 \times 1-3 \times -1=0+3=3\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 0 \\
2 & 3
\end{array}\right|=1 \times 3-2 \times 0=3-0=3\)
x = \(\frac{∆_x}{∆}=\frac{3}{3}=1\), y = \(\frac{∆_y}{∆}=\frac{3}{3}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 1)

(iii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
2 & -1 \\
1 & -3
\end{array}\right|=2 \times (-3)-1 \times (-1)=-6+1=-5\)
∆ ≠ 0 ତେଣୁ ସହ-ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
3 & -1 \\
-1 & -3
\end{array}\right|=3 \times -3-(-1) \times -1=-9-1=-10\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
1 & -1
\end{array}\right|=2 \times -1-1 \times 3=-2-3=-5\)
x = \(\frac{∆_x}{∆}=\frac{-10}{-5}=1\), y = \(\frac{∆_y}{∆}=\frac{-5}{-5}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (2, 1)

Odisha State Board BSE Odisha 10th Class Hindi Solutions Poem 1(c) तुलसीदास के दोहे Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Hindi Solutions Poem 1(c) तुलसीदास के दोहे

प्रश्न और अभ्यास (ପ୍ରଶ୍ନ ଔର୍ ଅଭ୍ୟାସ)

1. निम्नलिखित प्रश्नों के उत्तर दो-तीन वाक्यों में दीजिए।
(ନିମ୍ନଲିଖୂ ପ୍ରଶ୍ନୋ କେ ଉତ୍ତର ଦୋ-ତୀନ୍ ୱାର୍କୋ ମେଁ ଦୀଜିଏ।)
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନର ଉତ୍ତର ଦୁଇ-ତିନୋଟି ବାକ୍ୟରେ ଦିଅ।)

(क) कठोर वचन का क्यों परिहार करना चाहिए?
(କଠୋର୍ ବଚନ୍ କା ର୍ଯ୍ୟା ପରିହାର୍ କର୍‌ନା ଚାହିଏ?)
उत्तर:
कठोर वचन सबको दुःख पहुँचाता है। परिवेश को अशान्त कर देता है। कठोर वचन से दूसरों को पीड़ा पहुचती है इसलिए कठोर वचन को परिहार करके मीठे वचन बोलना चाहिए।

(ख) मीठे वचन से क्या लाभ होता है?
(ମୀଠ ବଚନ୍ ସେ କ୍ୟା ଲାଭ୍ ହୋତା ହୈ ?)
उत्तर:
मीठे वचन सबको प्रिय होते हैं। मीठी वाणी से हम सबको अपने वश में कर सकते हैं। मीठी वाणी से चारों ओर शांति बनी रहती है। सबको सुख मिलता है।

(ग) सन्तोष धन के सामने कौन-कौन से धन धूरि के बराबर माने जाते हैं?
(ସନ୍ତୋଷ୍ ଧନ୍ କେ ସାମ୍‌ନେ କୌନ୍-କୌନ୍ ସେ ଧନ୍ ଧୂରି କେ ବରାବର୍ ମାନେ ଜାତେ ହୈ ?)
उत्तर:
सन्तोष धन के सामने गोधन, गजधन, बाजीधन, रतनधन आदि धन धूरि के बराबर माने जाते हैं। क्योंकि इस प्रकार के धन से सुख शांति नहीं मिलती। मन चिंतित रहता है।

(घ) रोष या गुस्से के समय क्या नहीं खोलना चाहिए और क्यों?
(ରୋଷ୍ ୟା ଗୁସ୍‌ କେ ସମୟ କ୍ୟା ନହୀ ଖୋଲ୍‌ନା ଚାହିଏ ଔର୍ କ୍ୟା?)
उत्तर:
रोष या गुस्से के समय जीभ नहीं खोलनी चाहिए। क्योंकि क्रोध में मनुष्य कड़वी बातें बोल जाता है। कड़वी बातें तलवार से भी अधिक घाव कर देती है।

(ङ) मीठे वचन की तुलना वशीकरण मन्त्र से क्यों की गई है?
(ମୀଠ ବଚନ୍ କୀ ତୁଲନା ବଶୀକରଣ୍ ମନ୍ତ୍ର ସେ କୈ କୀ ଗଈ ହୈ ?)
उत्तर:
मीठे वचन की तुलना वशीकरण मन्त्र से की गई है क्योंकि मीठे वचन से हम सबको अपने वश में कर सकते हैं। मीठे वचन सबको प्रिय होते हैं, इससे सबको शांति और सुख मिलता हैं।

(च) हमें सोच विचार कर क्यों बोलना चाहिए?
(ହର୍ମେ ସୋଚ୍ ବିଚାର୍ କର୍ ଜ୍ୟୋ ବୋଲନା ଚାହିଏ ?)
उत्तर:
हमें सोच विचार कर हमेशा बोलना चाहिए। क्योंकि क्रोध में मनुष्य कड़वी बातें बोल जाता है। ये कड़वी बातें तलवार से भी अधिक घाव करती है। इसका प्रहार सीधे हृदय और मन पर होता है। मधुर वचन का परिणाम मधुर होता है।

निम्नलिखित अवतरणों का आशय दो-तीन वाक्यों में स्पष्ट कीजिए।
(ନିମ୍ନଲିଖ୍ ଅବତରର୍ଡୋ କା ଆଶୟ ଦୋ-ତୀନ୍ ୱାର୍କୋ ମେଁ ସ୍ପଷ୍ଟ କୀଜିଏ ।)
(ତଳଲିଖ୍ ଅବତରଣଗୁଡ଼ିକର ଆଶୟ ଦୁଇ-ତିନି ବାକ୍ୟରେ ସ୍ପଷ୍ଟ କର ।)

(क) तुलसी मीठे वचन ते, सुख उपजत चहुँओर।
(ତୁସୀ ମୀଠେ ବଚନ୍ ତେ, ସୁଖ୍ ଉପଜତ୍ ଚହୁଁଓର ।)
उत्तर:
इस पंक्ति में तुलसीदास यह बतलाते हैं कि मीठ वचन से सबको सुख मिलता है। चारों ओर शांति बनी रहती है। मीठे वचन सबको प्रिय होते हैं।

(ख) जब आवे सन्तोष धन, सब धन धूरि समान।
(ଜବ୍ ଆୱେ ସନ୍ତୋଷ୍ ଧନ୍, ସବ୍ ଧନ୍ ଧୂରି ସମାନ୍ ।)
उत्तर:
इस पंक्ति में कवि ने यह कहा है कि सन्तोष धन के सामने सब धन धूल के समान है। क्योंकि इस प्रकार के धन से सुख शांति नहीं मिलती, मन चिंतित रहता है।

(ग) रोष न रसना खोलिए, बरु खोलिओ तलवारि।
(ରୋସ୍ ନ ରସ୍‌ନା ଖୋଲିଏ ବରୁ ଖୋଲିଓ ତଲବାରି ।)
उत्तर:
गुस्से में जीभ नहीं खोलनी चाहिए। क्योंकि गुस्से में मनुष्य आपे से बाहर हो जाता है और कड़वी बातें बोल जाता है। ये कड़वी बाते दिल और मन को घायल करके अधिक कष्ट देती है।

3. निम्नलिखित प्रश्नों के उत्तर एक-एक वाक्य में दीजिए।
(ନିମ୍ନଲିଖତ୍ ପ୍ରକ୍ଷ୍ନୌ କେ ଉତ୍ତର୍ ଏକ୍-ଏକ୍ ୱାର୍କୋ ମେଁ ଦୀଜିଏ ।)
(ତଳଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଗୋଟିଏ-ଗୋଟିଏ ବାକ୍ୟରେ ଦିଅ ।)

(क) किससे चारों ओर सुख उपजता है?
(କିସ୍‌ ଚାର୍ରେ ଓର ସୁଖ୍ ଉପଚ୍ଚତା ହୈ ?)
उत्तर:
मीठे वचन से चारों ओर सुख उपजाता है।

(ख) वशीकरण का मंत्र क्या है?
(ବଶୀକରଣ୍ କା ମନ୍ତ୍ର କ୍ୟା ହୈ ?)
उत्तर:
मीठे वचन वशीकरण का मंत्र है।

(ग) हमें क्या परिहार करना या छोड़ना चाहिए?
(ହର୍ମେ କ୍ୟା ପରିହାର୍ କର୍‌ନା ୟା ଛାଡୁନା ଚାହିଏ?)
उत्तर:
हमें कटु वचन को परिहार करना या छोड़ना चाहिए।

(घ) कवि ने सन्तोष की तुलना किस से की है?
(କବି ନେ ସନ୍ତୋଷ୍ କୀ ତୁଲନା କିସ୍ ସେ କୀ ହୈ ?)
उत्तर:
कवि ने सन्तोष की तुलना धन से की है।

(ङ) कब रसना नहीं खोलनी चाहिए?
(କବ୍ ରସ୍‌ନା ନହୀ ଖୋଲ୍‌ନୀ ଚାହିଏ?)
उत्तर:
अधिक गुस्से में रसना नहीं खोलनी चाहिए।

(च) किस धन के सामने सारे धन तुच्छ माने जाते हैं ?
(କିସ୍ ଧନ୍ କେ ସାମ୍‌ ସାରେ ଧନ୍ ତୁଚ୍ଛ ମାନେ ଜାତେ ହେଁ ?)
उत्तर:
संतोष धन के सामने सारे धन तुच्छ माने जाते हैं।

(छ) सन्तोष धन के सामने सब धन किसके समान होते हैं?
(ସନ୍ତୋଷ୍ ଧନ୍ କେ ସାମ୍‌ ସବ୍ ଧନ୍ କିସ୍କେ ସମାନ୍ ହୋତେ ହେଁ ?)
उत्तर:
सन्तोष धन के सामने सब धन धूल के समान होते हैं।

(ज) विचार करके वचन कहने से क्या होता है?
(ବିଚାର୍ କର୍‌କେ ବଚନ୍ କହନେ ସେ କ୍ୟା ହୋତା ହୈ ?)
उत्तर:
विचार करके वचन कहने से उसका परिणाम मधुर होता है।

भाषा-ज्ञान (ଭାଷା-ଜ୍ଞାନ)

1. निम्नलिखित शब्दों के विपरीत शब्द लिखिए । ( 160 61 ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ବିପରୀତ ଶବ୍ଦ ଲେଖ ।)
मीठा, सुख, कठोर, छोड़ना, समान, खोलना
उत्तर:
मीठा – कड़वा/खट्टा
कठोर – मृदु/कोमल
समान – असमान
सुख – दुःख
छोड़ना – पकड़ना
खोलना – बंद करना

2. निम्नलिखित शब्दों के समानार्थक शब्द लिखिए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ସମାନାର୍ଥକ ଶବ୍ଦ ଲିଖିଏ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ସମାନାର୍ଥୀ ଶବ୍ଦ ଲେଖ )
वचन, सुख, कठोर, उपजना, गो, गज, बाजि
उत्तर:
वचन – वाणी/बात
कठोर – निर्दयी
गो – गाय/गऊ
सुख – आनन्द
उपजना – पैदा होना/जन्म होना
गज – हाथी

3. निम्नलिखित शब्दों के प्रयोग से सार्थक वाक्य बनाइए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ପ୍ରୟୋଗ୍ ସେ ସାର୍ଥକ ବାକ୍ୟ ବନାଇଏ ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ପ୍ରୟୋଗ କରି ସାର୍ଥକ ବାକ୍ୟ ଗଠନ କର।)
वसीकरण, कठोर, गोधन, सन्तोष, तलवार
उत्तर:
वसीकरण – मीठे वचन तो वशीकरण मंत्र के समान हैं।
कठोर – राहुल अत्यन्त कठोर स्वभाव का है।
गोधन – यशोदा गोधन की कसम खाकर कहती हैं कि कृष्ण ही उनका पुत्र है।
संन्तोष – सन्तोष रूपी धन के सामने बाकी सारा धन तुच्छ है।
तलवार – तलवार शरीर पर घाव करती है मगर कड़वी बातें दिल पर घाव कर देती हैं।

4. निम्नलिखित शब्दों के शुद्ध रूप लिखिए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ଶୁଦ୍ଧ ରୂପ ଲିଖିଏ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ଶୁଦ୍ଧ ରୂପ ଲେଖ।)
चहुँओर, वसीकरण, धूरि, तरवारि, परिनाम
उत्तर:
चहुँओर – चारों ओर
तरवारि – तलवार
वसीकरण – वशीकरण
धूरि – धूल/धूलि
परिनाम – परिणाम

5. निम्नलिखित शब्दों के साथ करण कारक ‘से’ चिह्न का प्रयोग करके वाक्य बनाइए:
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦା କେ ସାଥ୍ କରଣ୍ କାରକ୍ ‘ସେ’ ଚିହ୍ନ କା ପ୍ରୟୋଗ କର୍‌କେ ବାକ୍ୟ ବନାଇଏ)
(ନିମ୍ନଲିଖତ ଶବ୍ଦଗୁଡ଼ିକ ସହିତ କରଣ କାରକ ‘ସେ’ ଚିହ୍ନର ପ୍ରୟୋଗ କରି ବାକ୍ୟ ଗଠନ କର )
वचन, मंत्र, धन, तलवार
उत्तर:
वचन – मीठे वचन से सबको सुख मिलता है।
मंत्र – वशीकरण मंत्र से सभी को वश में किया जा सकता है।
धन – धन से सुख नहीं मिलता।
तलवार – तलवार से मत खेलो।

Very Short & Objective type Questions with Answers

A. निम्नलिखित प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. तुलसीदास का जन्म कब और कहाँ हुआ?
उत्तर:
तुलसीदास का जन्म सन् 1532 में उत्तर प्रदेश के राजापुर में हुआ।

2. तुलसीदास के गुरु कौन थे?
उत्तर:
श्री नरहरि दास तुलसीदास के गुरु थे

3. तुलसी जी किस-किस भाषा में लिखते थे?
उत्तर:
तुलसी जी अवधी और ब्रजभाषा में लिखते थे।

B. निम्नलिखित प्रश्नों के उत्तर एक शब्द / एक पद में दीजिए।

1. संतोष-धन आ जाता है तो बाकी सब धन किसके समान हो जाते हैं?
उत्तर:
धूल के

2. किसका परिणाम हितकर होता है?
उत्तर:
वशीकरण का

3. तुलसी ने श्रेष्ठ धन किसे कहा है?
उत्तर:
संतोष

4. तुलसीदास क्या परिहार करने को कहते हैं?
उत्तर:
कठोर वचन

5. जो वाणी सुनने में मधुर लगती है, उसका परिणाम क्या होता है?
उत्तर:
हितकर

6. किस अवस्था में रसना नहीं खोलनी चाहिए?
उत्तर:
गुस्से के समय

7. कवि तुलसीदास के अनुसार वशीकरण मंत्र का अर्थ क्या है?
उत्तर:
मीठी वाणी से सबको वश में करना

8. रोष के समय क्या नहीं खोलनी चाहिए ?
उत्तर:
रसना

9. हमे किस प्रकार की वाणी बोलनी चाहिए?
उत्तर:
मीठी

10. ‘परिनाम’ का अर्थ क्या है?
उत्तर:
परिणाम

C. रिक्त स्थानों की पूर्ति कीजिए।

1. विचार करके वचन बोलने से परिणाम ……………… होता है।
उत्तर:
हितकर

2. ………………… सी बात तलवार से अधिक घाव करती है।
उत्तर:
कड़वी

3. कड़वी बातें. ……………. घायल करती है।
उत्तर:
मन को

4. वशीकरण मंत्र ……………….. है।
उत्तर:
मधुर वाणी

5. तुलसी जी के गुरु ………………… है।
उत्तर:
रामानन्द

6. ………………… में रसना नहीं खोलनी चाहिए।
उत्तर:
अधिक गुस्से

7. तुलसी ने सन्तोष की तुलना …………………. से की है।
उत्तर:
धन

8. ………………. से चारों ओर सुख उपजता है।
उत्तर:
मिठे वचन

9. ………………. के सामने सारे धन तुछ माने जाते है।
उत्तर:
संन्तोष धन

10 ‘बरु’ का अर्थ है…………….. ।
उत्तर:
बल्कि

D. ठिक् या भूल लिखिए।

1. संतोष धन के आने से सारे धन धूल के समान हो जाते हैं।
उत्तर:
ठिक्

2. तुलसी के अनुसार विचार करके घूमना चाहिए।
उत्तर:
भूल

3. तलवार से रसना घातक होती है।
उत्तर:
ठिक्

4. चुटकुले चारों ओर सुख उपजाने में सहायक हैं।
उत्तर:
भूल

5. मीठे वचनों से सुख मिलता है।
उत्तर:
ठिक्

6. विचार करके कहानी कहना जरुरी है।
उत्तर:
भूल

7. ‘रामचरित मानस’ तुलसीदास ने लिखा है।
उत्तर:
भूल

8. बाजिधन के सामने सारे धन तुच्छ हैं
उत्तर:
ठिक्

9. ‘बाजी’ का अर्थ हाथी है।
उत्तर:
भूल

10. विनय पत्रिका तुलसीदास की रचना है।
उत्तर:
ठिक्

11. कवि ने संन्तोष की तुलना धन से की है।
उत्तर:
ठिक्

Multiple Choice Questions (mcqs) with Answers

सही उत्तर चुनिए : (MCQS)

1. संतोष-धन आ जाता है तो बाकी सब धन किसके समान हो जाते हैं?
(A) धूल के
(B) बाजि के
(C) रतन के
(D) प्राण के
उत्तर:
(A) धूल के

2. किसका परिणाम हितकर होता है?
(A) कटु वचन का
(B) रत्न-धन का
(C) वशीकरण का
(D) मधुर वचन का
उत्तर:
(C) वशीकरण का

3. तुलसी ने श्रेष्ठ धन किसे कहा है?
(A) रत्न-धन को
(B) राम – रत्न को
(C) संतोष-धन को
(D) सोने-चाँदीको
उत्तर:
(C) संतोष-धन को

4. तुलसीदास क्या परिहार करने को कहते हैं?
(A) कुसंग
(B) चिंत
(C) कामना
(D) कठोर वचन
उत्तर:
(D) कठोर वचन

5. जो वाणी सुनने में मधुर लगती है, उसका परिणाम क्या होता है?
(A) हितकर
(B) अहितकर
(C) भयानक
(D) खुशामद
उत्तर:
(A) हितकर

6. चहुँओर सुख उपजाता है।
(A) मीठे वचन से
(B) कटुवचन से
(C) अल्प वचन से
(D) धीमे वचन से
उत्तर:
(A) मीठे वचन से

7. तुलसीदास के अनुसार सबसे बड़ा धन है।
(A) गोधन
(B) गज धन
(C) संतोष धन
(D) रतन धन
उत्तर:
(C) संतोष धन

8. किस अवस्था में रसना नहीं खोलनी चाहिए।
(A) गुस्से के समय
(B) शान्ति के समय
(C) खाते समय
(D) राते समय
उत्तर:
(A) गुस्से के समय

दोहे (ଦୋହେ)

(i) तुलसी मीठे बचन ते, सुख उपजत चहुँओर।
वसीकरण यह मंत्र है, परिहरु बचन कठोर॥
ତୁଲସୀ ମୀଠେ ବଚନ୍ ତେ, ସୁଖ୍ ଉପଜତ ଚହୁଁଓର।
ବସୀକରଣ ୟହ ମନ୍ତ୍ର ହୈ, ପରିହରୁ ବଚନ୍ କଠୋର୍॥

हिन्दी व्याख्या:
तुलसीदास कहते हैं कि मीठे वचन सबको प्रिय होते हैं। मधुर वचन से हम सबको अपने वश में कर सकते हैं। मीठे वचन से चारों ओर शान्ति बनी रहती है। सबको सुख मिलता है। लेकिन कड़वी वाणी से सबको दुःख पहुँचता है। मीठे वचन तो वशीकरण मंत्र के समान है। इसलिए कड़वा वचन न बोलकर मीठे वचन बोलना चाहिए।

ଓଡ଼ିଆ ଅନୁବାଦ:
ମିଠା କଥା ସମସ୍ତଙ୍କୁ ଭଲ ଲାଗେ। ମିଠା କଥା କହି ଆମେ ସମସ୍ତଙ୍କୁ ଆପଣେଇ ପାରିବା। ମିଠା କଥାଦ୍ଵାରା ସବୁଆଡ଼େ ଶାନ୍ତି ଲାଗି ରହିଥାଏ। ସମସ୍ତଙ୍କୁ ସୁଖ ମିଳେ। କିନ୍ତୁ କଟୁକଥା ସମସ୍ତଙ୍କ ମନକୁ ଆଘାତ କରେ ଓ ଦୁଃଖ ଦିଏ। ମଧୁର ବଚନ ବଶୀକରଣ ମନ୍ତ୍ର ପରି। ଏଣୁ ଆମକୁ କଟୁ କଥା ନ କହି ମିଠା କଥା କହିବା ଦରକାର।

(ii) गोधन, गजधन, बाजिधन और रतनधन खान।
जब आवे सन्तोष धन सब धन धूरि समान॥
ଗୋଧନ୍, ଗଜଧନ୍, ବାଜିଧନ୍ ଔର୍ ରତନଧନ୍ ଖାନ୍।
ଜବ୍ ଆୱେ ସନ୍ତୋଷ୍ ଧନ୍, ସବ୍‌ ଧନ୍ ଧୂରି ସମାନ୍॥

हिन्दी व्याख्या :
कवि कहते हैं कि साधारणतः हमारी धारणा है कि जिसके पास पर्याप्त गाय-भैंस, हाथी और घोड़े होते हैं या धन रत्न आदि होते हैं, वह इस संसार में सबसे अधिक धनी है। लेकिन कवि के अनुसार ये सारे धन होते हुए भी अगर मन में सन्तोष नहीं है तो ये सबकुछ मूल्यहीन होता है। सन्तोष धन के सामने ये सब धन धूल के समान तुच्छ हैं। क्योंकि इस प्रकार के धन से सुख शान्ति नहीं मिलती। मन चिंतित रहता है।

ଓଡ଼ିଆ ଅନୁବାଦ:
ସାଧାରଣତଃ ଆମେ ଭାବିଥାଉ ଯେ ଯାହା ପାଖରେ ଗାଈ, ମଇଁଷି, ହାତୀ-ଘୋଡ଼ା ବା ଧନରତ୍ନ, ହୀରା-ମୋତି ଅଛି, ସେ ଜଣେ ବଡ଼ ଧନୀ ବ୍ୟକ୍ତି। କିନ୍ତୁ ତୁଳସୀ ଦାସଙ୍କ ଅନୁସାରେ ଏତେ ସବୁ ଧନ ଥାଇ ମଧ୍ୟ ଯଦି ତା’ମନରେ ଶାନ୍ତି ନାହିଁ ତାହାହେଲେ ସେ ଧନ ସବୁ ତା’ପାଇଁ ମୂଲ୍ୟହୀନ। ସନ୍ତୋଷ ରୂପୀ ଧନ ପାଖରେ ଏସବୁ ଧନରତ୍ନ ଧୂଳି ସଙ୍ଗେ ସମାନ ଓ ତୁଚ୍ଛ। ଏପରି ଧନ ସମ୍ପଦରେ ସୁଖ ଶାନ୍ତି ମିଳେ ନାହିଁ। ମନରେ ଚିନ୍ତା ଲାଗି ରହିଥାଏ।

(iii) रोष न रसना खोलिए, बरु खोलिओ तरबार।
सुनत मधुर परिनाम हित, बोलिओ बचन बिचारि॥
ରୋସ୍ ନ ରସନା ଖୋଲିଏ, ବରୁ ଖୋଲିଓ ତରବାରି।
ସୁନତ ମଧୁର୍ ପରିନାମ ହିତ, ବୋଲିଓ ବଚନ୍ ବିଚାରି॥

हिन्दी व्याख्या:
कवि तुलसीदास कहते हैं कि जब क्रोध अधिक हो तो जीभ नहीं खोलनी चाहिए। क्रोध में मनुष्य कड़वी बातें बोल जाता है। ये कड़वी बातें तलवार से भी अधिक घाव करती है। कड़वी बात का प्रहार सीधे दिल और दिमाग पर होता है। तलवार तो शरीर पर घाव करती है, लेकिन कड़वी बातें दिल और दिमाग को घायल करके अधिक कष्ट देती हैं।

ଓଡ଼ିଆ ଅନୁବାଦ :
ଯେତେବେଳେ ରାଗ ଅଧ୍ବକ ବଢ଼ିଚାଲେ ସେତେବେଳେ ଜିଭକୁ ନିୟନ୍ତ୍ରଣ ରଖିବାକୁ ପଡ଼ିବ। ରାଗରେ ମଣିଷ କଟୁ କଥା କହି ଚାଲିଥାଏ। ଏହି କଟୁ କଥା ଖଣ୍ଡାଠାରୁ ମଧ୍ୟ ଅଧିକ ଆଘାତ କରିଥାଏ। କଟୁ କଥାର ପ୍ରହାର ସିଧା ମନ ଓ ହୃଦୟ ଉପରେ ପଡ଼ିଥାଏ। ଖଣ୍ଡାର ଚୋଟରେ ଶରୀରରେ କ୍ଷତ ହୋଇଥାଏ। କିନ୍ତୁ କଟୁ କଥା ମନ ଓ ହୃଦୟକୁ ଆଘାତ କରେ ଏବଂ ଅଧିକ କଷ୍ଟ ଦିଏ।

शबनार: (ଶରାର୍ଥି)

मीठे – मीठा/मधुर (ମିଠା/ମଧୁର )।

उपजत – उपजना/पैदा होना (ଜନ୍ମହେବା/ଉପୁଜିବା)।

चहुँओर – चारों ओर (ଚାରିଆଡ଼େ)।

परिहरु – त्यागना/परित्याग करना (ତ୍ଯାଗ କରିବା)।

गोधन – गाय रूपी धन (ଗୋଧନ)।

बाजि – घोड़ा (ଘୋଡ଼ା)

ते – से/द्वारा (ଦ୍ଵାରା)।

वसीकरण – वशीभूत (ହାତି)।

गज – हाथी (ରତ୍ନ)।

रतन – रत्न (ବଶୀଭୂତ)।

खान – भंड़ार (ଭଣ୍ଡାର)।

धूरि – धूल यूरिकि)। (ଧୂଳି)।

रसना – जीभ (ଜିଭ)।

खोलिओ – खोले (ଖୋଲିବା)।

परिनाम – परिणाम (ପରିଣାମ)।

सुनत – सुनकर (ଶୁଣି)।

विचारि – विचार करके (ବିଚାର କରି)।

आवे – आए (ଆସେ)।

रोष – गुस्सा (ରାଗ)।

बरु – बल्कि ( ବରଂ)।

तरवारि – तलवार (ଖଣ୍ଡା)।

हित – मंगल (ହିତମଙ୍ଗଳ)।

बोलिअ – बोलो (କୁହ)।

कवि परिचय

भक्त कवि तुलसी दास का जन्म सन् 1532 में उत्तर प्रदेश के राजापुर में हुआ था और देहांत सन् 1633 में। पितामाता के स्नेह से वंचित होकर बचपन में उनको बड़ा कष्ट उठाना पड़ा। सौभाग्य से गुरु नरहरिदास ने उनकी बड़ी मदद की। तुलसी रामभक्त थे और यौवन काल में ही साधु बन गये। रामानंद उनके गुरु थे। वे हिन्दी और संस्कृत के बड़े पंड़ित थे। उस समय मुगलों का शासन था। देश की सामाजिक और धार्मिक परिस्थितियाँ अस्तव्यस्त थीं।

तुलसी दास ने रामचरित मानस लिखकर लोगों के सामने निष्कपट जीवन और आचरण का उदाहरण रखा। आज भी यह देश का अत्यंत लोकप्रिय ग्रंथ है। जन साधारण उसे बड़े चाव से पढ़ते हैं। दुःखी, निराश तथा भक्त लोगों को रामचरितमानस पढ़कर सुख शान्ति मिलती है। विनयपत्रिका, कवितावली, दोहावली, गीतावली आदि उनके अनेक ग्रंथ हैं। वे अवधी और ब्रजभाषा दोनों में लिखते थे।

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(b)

Question 1.
Expand in ascending power of x.
(i) 2^x
Solution:

(ii) cos x
Solution:

(iii) sin x
Solution:

(iv) \(\frac{x e^{7 x}-e^{-x}}{e^{3 x}}\)
Solution:

(v) \(\boldsymbol{e}^{e^x}\) up to the term containing x⁴
Solution:

Question 2.
If x = y + \(\frac{y^2}{2 !}+\frac{y^3}{3 !}\) + ….. then show that y = x – \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\) +….
Solution:

Question 3.
Find the value of \(x^2-y^2+\frac{1}{2 !}\left(x^4-y^4\right)+\frac{1}{3 !}\left(x^6-y^6\right)\) + ….
Solution:

Question 4.
Show that
(i) 2\(\left(\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots\right)=\frac{1}{e}\)
Solution:

(ii) \(\frac{9}{1 !}+\frac{19}{2 !}+\frac{35}{3 !}+\frac{57}{4 !}+\frac{85}{5 !}\) + …. = 12e – 5
Solution:

(iii) \(1+\frac{1+3}{2 !}+\frac{1+3+3^2}{3 !}+\ldots=\frac{1}{2}\left(e^3-e\right)\)
Solution:

(iv) \(\frac{1.3}{1 !}+\frac{2.4}{2 !}+\frac{3.5}{3 !}+\frac{4.6}{4 !}\) + …. = 4e
Solution:
t_n for L.H.S.
= \(\frac{n(n+2)}{n !}=\frac{n^2+2 n}{n !}\)

(v) \(\frac{1}{1.2}+\frac{1.3}{1.2 .3 .4}+\frac{1.3 .5}{1.2 .3 .4 .5 .6}\) + …. = √e – 1
Solution:

Question 5.
Prove that
(i) loge(1 + 3x + 2x²) = 3x – \(\frac{5}{2}\)x² + \(\frac{9}{3}\)x³ – \(\frac{17}{4}\)x⁴ + …..,|x| < \(\frac{1}{2}\)
Solution:

(ii) log_e(n + 1) – log_e(n – 1) = 2 \(\left[\frac{1}{n}+\frac{1}{3 n^3}+\frac{1}{5 n^5}+\ldots\right]\)
Solution:
There is a printing mistake in the question. The correct question is log_e(n + 1) – log_e(n – 1)

(iii) log_e(n + 1) – log_en = 2 \(\left[\frac{1}{2 n+1}+\frac{1}{3(2 n+1)^3}+\frac{1}{5(2 n+1)^5}+\ldots\right]\)
Solution:

(iv) log_em – log_en = \(\frac{m-n}{m}+\frac{1}{2}\left(\frac{m-n}{m}\right)^2\)
Solution:

= – [log n – log m]
= log m – log n = L.H.S.

(v) log_ea – log_eb = \(2\left[\frac{a-b}{a+b}+\frac{1}{3}\left(\frac{a-b}{a+b}\right)^3\right.\) \(\left.+\frac{1}{5}\left(\frac{a-b}{a+b}\right)^5+\ldots\right], a>b\)
Solution:

(vi) log_en = \(\frac{n-1}{n+1}+\frac{1}{2} \cdot \frac{n^2-1}{(n+1)^2}\)\(+\frac{1}{3} \cdot \frac{n^2-1}{(n+1)^3}\) + …..
Solution:

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(b)

Question 1.
²ⁿC₀ + ²ⁿC₂ + …… + ²ⁿC_2n = 2^2n-1 and ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1 = 2^2n-1
Solution:
We know that
(1 + x)²ⁿ = ²ⁿC₀ + ²ⁿC₁x + ²ⁿC₂x² + …..+ ²ⁿC_2nxⁿ …(1)
Putting x = 1
We get putting x = – 1  we get
∴ (²ⁿC₀ + ²ⁿC₂ + ²ⁿC₄ + ….. + ²ⁿC_2n) – (²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1) = 0
∴ ²ⁿC₀ + ²ⁿC₂ + ….. + ²ⁿC_2n
= ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1
= \(\frac{2^{2 n}}{2}\) = 2^2n-1

Question 2.
Find the sum of
(i) C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n
Solution:
C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n

(ii) C₀ + ²C₁ + ³C₂ + ….. + ⁽ⁿ⁺¹⁾C_n Hint: write (C₀ + C₁ + ….. + C_n) + (C₁ + ²C₂ + ….. + ⁿC_n) use (5) and exercise 1.

Question 3.
Compute \(\frac{(1+k)\left(1+\frac{k}{2}\right) \ldots\left(1+\frac{k}{n}\right)}{(1+n)\left(1+\frac{n}{2}\right) \ldots\left(1+\frac{n}{k}\right)}\)
Solution:

Question 4.
Show that
(i) C₀C₁ + C₁C₂ + C₂C₃ + ….. + C_n-1C_n = \(\frac{(2 n) !}{(n-1) !(n+1) !}\)
Solution:

(ii) C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\) Hint : Proceed as in Example 13. Compare the coefficient of a^n-1 to get (i) and the coefficient of a^n-r to get (ii)
Solution:


∴ C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\)

(iii) 3C₀ – 8C₁ + 13C₂ – 18C₃ + ….. + (n+1)^th term = 0
Solution:
3C₀ – 8C₁ + 13C₂ – 18C₃ + ….. + (n+1)^th term
= 3C₀ – (5 + 3)C₁ + (10 + 3) C₂ – (15 + 3) C₃ + …
= 3(C₀ – C₁ + C₂ …) + 5 (- C₁ + 2C₂ – 3C₃ …..)
But (1 – x)ⁿ =C₀ – C₁x + C₂x² – C₃x³ + … + (-1)ⁿ xⁿC_n … (1)
Putting x = 1 we get
C₀ – C₁ + C₂ ….. + (-1)ⁿC_n
and differentiating (1) we get n(1 – x)^n-1 (- 1)
= – C₁ + 2C₂x – 3C₃x² +…..
Putting x = 1 we get
– C₁ + 2C₂ – 3C₃ + ….. = 0
∴ 3C0 – 8C1 +13C2 …… (n+1) terms = 0

(iv) C₀n² + C₁(2 – n)² +C₂(4 – n)² + ….. + C_n(2n – n)² = n.2ⁿ
Solution:
C₀n² + C₁(2 – n)² +C₂(4 – n)² + ….. + C_n(2n – n)² = n.2ⁿ
= n²(C₀ + C₁ + + C_n) + (2²C₁ + 4²C₂ + ….. + (2n)²C_n) – 4n (C₁ + 2C₂ + 3C₃ + ….+ nC_n)
= n² . 2ⁿ + 4n(n+l)2^n-2 – 4 n . n . 2^n-1
= 2^n-1 (2n² + 2n² + 2n – 4n²)
= 2n . 2^n-1 = n2ⁿ

(v) C₀ – 2C₁ +3C₂ + ….. + (-1)ⁿ(n+1)C_n = 0
Solution:

(vi) C₀ – 3C₁ +5C₂ + ….. + (2n+1)C_n = (n+1)²ⁿ
Solution:

Question 5.
Find the sum of the following :
(i) C₁ – 2C₂ + 3C₃ + ….. + n(-1)^n-1C_n 
Solution:

(ii) 1.2C₂ + 2.3C₃ + ….. + (n-1)nC_n
Solution:

(iii) C₁ + 2²C₂ + 3²C₃ + ….. + n²C_n
Solution:

(vi) C₀ – \(\frac{1}{2}\)C₁ + \(\frac{1}{3}\)C₂ + ….. + (-1)ⁿ \(\frac{1}{n+1}\)C_n
Solution:

Question 6.
Show that
(i) C₁² + 2C₂² +3C₃² + ….. + nC_n² = \(\frac{(2 n-1) !}{\{(n-1) !\}^2}\)
Solution:

(ii) C₂ + 2C₃ + 3C₄ + ….. + (n – 1)C_n = 1 + (n – 2)2^n-1
Solution:

Question 7.
C₁ – \(\frac{1}{2}\)C₂ + \(\frac{1}{3}\)C₃ + ….. +(-1)ⁿ⁺¹ \(\frac{1}{n}\)C_n = 1 + \(\frac{1}{2}\) + …. + \(\frac{1}{n}\)
Solution:

Question 8.
C₀C₁ + C₁C₂ + ….. + C_n-1C_n = \(\frac{2^n \cdot n \cdot 1 \cdot 3 \cdot 5 \ldots(2 n-1)}{(n+1)}\)
Solution:

Question 9.
The sum \(\frac{1}{1 ! 9 !}+\frac{1}{3 ! 7 !}+\ldots+\frac{1}{7 ! 3 !}+\frac{1}{9 ! 1 !}\) can be written in the form \(\frac{2^a}{b !}\) find a and b.
Solution:

Question 10.
(a) Using binominal theorem show that 1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹ is divisible by 5 (Regional Mathematical Olympiad, Orissa – 1987)
Solution:
1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹
= 1 + (5 – 3)⁹⁹ + 3⁹⁹ + (5 – 1)⁹⁹ + 5⁹⁹
= 1 + (5⁹⁹ – ⁹⁹C₁5⁹⁸.3¹ + ⁹⁹C₂5⁹⁷.3² – …3⁹⁹) + 3⁹⁹ – (1 – ⁹⁹C₁5¹ + ⁹⁹C₂5² – …  5⁹⁹) + 5⁹⁹
= (3 × 5⁹⁹ – ⁹⁹C₁5⁹⁸.3¹ + ⁹⁹C₂5⁹⁷.3² – ….. + ⁹⁹C⁹⁸5¹.3⁹⁸) + (⁹⁹C₁5¹ – ⁹⁹C₂5² + …. – ⁹⁹C₉₈5⁹⁸) ….(1) which is divisible by 5 as each term is a multiple of 5.

(b) Using the same procedure show that 1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹ is also divisible by 3 so that it is actually divisible by 15.
Solution:
From Eqn. (1) above, it is clear that each term within the 1st bracket is divisible by 3 and the terms in the 2nd bracket are divisible by 99 and hence divisible by 3.
Each term in Eqn. (1) is divisible by 3. As it is divisible by 3 and 5, it is divisible by 3 × 5 = 15

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 2 Bamboo Curry

BSE Odisha 6th Class English Follow-Up Lesson 2 Bamboo Curry Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ You have read the Odia folk-tale “The Foolish Son-in-Law”. In other languages, there are similar stories. Let us read a similar Santal folk tale “Bamboo Curry”.
(ତୁମେ ଓଡ଼ିଆ ଲୋକକଥା ‘ନିର୍ବୋଧ ଜ୍ଵାଇଁ’’ ପଢ଼ିସାରିଛ । ଅନ୍ୟ ଭାଷାମାନଙ୍କରେ, ସେହି ଏକାପରି ଗପସବୁ ଅଛି । ଆସ ଆମେ ଏକାପରି ଏକ ସାନ୍ତାଳ ଲୋକକଥା ‘ବାଉଁଶ ତରକାରି’’ ପଢ଼ିବା ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
Text (ବିଷୟବସ୍ତୁ):

SGP-1:
Read paragraphs 1-2 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୧ – ୨ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. Once a foolish Santal son-in-law went to his in-law’s place. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from ?” Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” “Yes son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”.

2. Next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-laws’ house.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୧). ଏକଦା (ଥରେ) ଜଣେ ନିର୍ବୋଧ (ମୂର୍ଖ) ସାନ୍ତାଳ ଜ୍ଵାଇଁ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା । ତା’ର ଶାଶୁ ତା’ର ଜ୍ଵାଇଁ ପାଇଁ ସୁସ୍ୱାଦୁ ଖାଦ୍ୟ ରୋଷେଇ କଲା । ଖାଦ୍ୟଗୁଡ଼ିକ ମଧ୍ୟରୁ ଗୋଟିଏ ଥିଲା ବାଉଁଶ ଗଜାରେ ତିଆରି ହୋଇଥାଏ ତରକାରି । ଜ୍ଵାଇଁ ଜଣଙ୍କ ଏହାକୁ ବହୁତ ପସନ୍ଦ କଲା ଏବଂ ତା’ର ଶାଶୁକୁ ପଚାରିଲା, ‘‘ମାଆ, ତରକାରି ବହୁତ ସୁଆଦିଆ ହୋଇଛି । ତରକାରି କେଉଁଥିରେ ତିଆରି ହୋଇଛି ?’’ ତା’ର ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବା ପରିବର୍ତ୍ତେ, ସେ ବାଉଁଶ ଦରଜା (ତାଟି) ଆଡ଼କୁ ଆଙ୍ଗୁଳି ନିର୍ଦ୍ଦେଶ କରି ଠାରିଲେ । ଜ୍ଵାଇଁ ପଚାରିଲା, ‘ଏହା ବାଉଁଶରେ ତିଆରି ହୋଇଛି ?’’ ‘ହଁ ପୁଅ, ତରକାରିଟି ବାଉଁଶରେ ତିଆରି ହୋଇଛି ଏବଂ ଏଥିପାଇଁ ‘ବାଉଁଶ ତରକାରି’’ କୁହାଯାଏ ।’’
(୨) ତା’ପରଦିନ, ଜ୍ଵାଇଁ ତାଙ୍କ ଘରକୁ ବାହାରୁଥିଲେ । ତାଙ୍କର ମନକୁ ବାଉଁଶ ତରକାରି କଥା ଆସିଲା । କଥା ଭାବିଲେ । କିନ୍ତୁ ସେମାନଙ୍କର ବାଉଁଶ ନଥିଲା । (ତାଟି)କୁ ତା’ ନିଜ ଘରକୁ ବୋହିନେଲା । ସେ ତାଙ୍କ ନିଜ ଘରେ ବାଉଁଶ ତରକାରି କରିବା ତେଣୁ ସେ ତାଙ୍କ ଶ୍ଵଶୁର ଘରର ବାଉଁଶ ଦରଜା

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who went to his father-in-law’s house?
(କିଏ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା ?)
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question 2.
What curry did his mother-in-law cook?
(ତା’ର ଶାଶୁ କି ତରକାରି ତିଆରି କଲା ?)
Answer:
His mother-in-law cooked “Bamboo Curry”.

Question 3.
Did he like it?
(ଏହାକୁ ସେ ପସନ୍ଦ କଲା କି ?)
Answer:
Surely (ନିଶ୍ଚୟ), he liked it very much.

Question 4.
Why did he carry home the bamboo door of his in-law’s house?
(ସେ କାହିଁକି ଘରକୁ ଶ୍ୱଶୁର ଘରର ବାଉଁଶ ଦରଜା ବୋହିନେଲା ?)
Answer:
He carried home the bamboo door of his in-laws’ house because he thought of cooking bamboo curry at home, but they did not have any bamboo.

Session – 2 (ସୋପାନ – ୨):
SGP-2:

• Read paragraphs 3-4 silently and answer the questions that follow.
  (ଅନୁଚ୍ଛେଦ ୩ – ୪ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

3. Reaching home, he asked his wife to prepare the bamboo curry. He helped his wife in chopping the dry bamboo sticks. But the dry bamboo pieces did not get boiled. The pieces remained as hard and stiff as before. He asked his wife to put more water and boil.
4. That evening his in-laws came to their son-in-law’s house. The son-in-law offered them the bamboo curry. The in-laws laughed at their foolish son-in-law. They told him, “The bamboo curry is made from soft bamboo shoots and not from dry bamboo pieces”.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୩) ଘରେ ପହଞ୍ଚିସାରି, ସେ ତା’ର ସ୍ତ୍ରୀକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ସେହି ଶୁଖୁଲା ବାଉଁଶ କାଠିଗୁଡ଼ିକୁ ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବାରେ ସାହାଯ୍ୟ କଲା । କିନ୍ତୁ ସେହି ଶୁଖୁ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ । ସେହି ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକ ପୂର୍ବପରି କଠିନ (ଟାଣ) ହୋଇ ରହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ଅଧିକ ପାଣି ଦେଇ ସିଝାଇବାକୁ କହିଲା ।
(୪) ସେହିଦିନ ସନ୍ଧ୍ୟାରେ ତା’ର ଶାଶୁ-ଶ୍ଵଶୁର ସେମାନଙ୍କର ଜ୍ୱାଇଁ ଘରକୁ ଆସିଲେ । ଜ୍ଵାଇଁ ଜଣଙ୍କ ତାଙ୍କୁ ବାଉଁଶ ତରକାରି ଖାଇବାକୁ ଦେଲେ । ଶାଶୁ-ଶ୍ଵଶୁର ତାଙ୍କର ନିର୍ବୋଧ (ବୋକା) ଜ୍ୱାଇଁ ଆଡ଼କୁ ଅନାଇ ହସିଲେ । ସେମାନେ ତାକୁ କହିଲେ, ‘ବାଉଁଶ ତରକାରି କୋମଳ (ନରମ) ବାଉଁଶ ଗଜାରେ ତିଆରି ହୁଏ କିନ୍ତୁ ଶୁଖିଲା ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକରେ ନୁହେଁ ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who did he ask to prepare bamboo curry?
(ସେ କାହାକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା ?)
Answer:
He asked his wife to prepare bamboo curry.

Question 2.
How did he help his wife ?
(କିପରି ସେ ତା’ ସ୍ତ୍ରୀକୁ ସାହାଯ୍ୟ କଲା ?)
Answer:
He helped his wife in chopping the dry bamboo sticks.

Question 3.
When the bamboo did not boil what did he ask his wife to do?
(ଯେତେବେଳେ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ, ସେ ତା’ସ୍ତ୍ରୀକୁ କ’ଣ କରିବାକୁ କହିଲା ?)
Answer:
When the bamboo pieces did not boil, he asked his wife to put more water and boil.

Question 4.
Who came to his house?
(କିଏ ଜ୍ଵାଇଁ ଘରକୁ ଆସିଲେ ?)
Answer:
His in-laws came to his house.

Question 5.
Where from is the bamboo curry made?
(କେଉଁଥୁରୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ ହୁଏ ?)
Answer:
The bamboo curry is made from soft bamboo shoots.

Session – 3 (ସୋପାନ – ୩):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖ୍) :
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) What is the story about ?
(ଗପଟି କାହା ଉପରେ ଆଧାରିତ ?)
Answer:
The story is about a foolish Santal son-in-law.

(ii) What curry did the mother-in-law prepare?
(କେଉଁ ତରକାରି ଶାଶୁ ତିଆରି କଲେ ?)
Answer:
The mother-in-law prepared “Bamboo Curry”.

(iii) Is the son-in-law foolish ? Why ?
(ଜ୍ଵାଇଁଟି ନିର୍ବୋଧ ଥିଲା କି ? କାହିଁକି ?)
Yes, the ____________ because he asked ____________ out of dry bamboo.
Answer:
Yes, the son-in-law was very foolish because he asked his wife to prepare bamboo curry out of dry bamboo.

Word Note (ଶବ୍ଦାର୍ଥ):
(The words / phrases have been defined mostly on contextual meanings.) (ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଵଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

ashamed – feeling shame, ଲଜ୍ଜା ଅନୁଭବ
bamboo curry – a kind of dish (food) made out of young bamboo plants ବାଉଁଶ ଗଜା ତରକାରି
bamboo shoot- new-young bamboo plants, ବାଉଁଶ ଗଜା
chopping- cutting into small pieces, ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବା ସ୍ବାଦିଷ୍ଟ,ସୁସ୍ବାଦୁ
delicious- tasty (food), ସ୍ଵାଦିଷ୍ଟ, ସୁସ୍ୱାଦୁ
dishes- food items, curry, ତରକାରି, ସ୍ଵାଦିଷ୍ଟ ବ୍ୟଞ୍ଜନ
folk-tale- popular story of a community, କଥୁତଳ୍ପ
gentlest- very kind (behaviour) ବହୁତ ଦୟାଳୁ (ଆଚରଣ)
heaved a great sign of relief- feel relieved,ଆରାମ ଅନୁଭବ କର |
high sounding words- difficult words, କଠିନ ବା ବଡ଼ ବଡ଼ ଶବ୍ଦ
impolite- not good behaviour, rude, ଭଲ ବ୍ୟବହାର ନୁହେଁ
lamb- young sheep, ଛୋଟ ମେଣ୍
offered- gave, served, ଦେଲେ, (ଖାଦ୍ୟ) ପରିବେଷଣ କଲେ
piled high – kept (things) in a heap, ଗଦା, ସ୍ତୂପ
plucking- collecting (from a tree) ସଂଗ୍ରହ (ଏକ ଗଛରୁ)
pointed – showed hand towards (bamboo door), ହାତ ଦେଖାଇ ନିର୍ଦ୍ଦେଶିତ କଲେ
preferred – chose, ପସନ୍ଦ କଲେ ବା ଆଗ୍ରହ ଦେଖାଇଲେ
quack – a self claimed ignorant practitioner, ଶଠ ବଇଦ | ଠକ ବଇଦ
smeared- spread something (substance) on body, ବୋଳି ହୋଇଗଲା, ଲାଗିଗଲା
stiff- hard, କଠିନ
thought of- got an idea, ଗୋଟିଏ ଉପାୟ ଚିନ୍ତା କଲେ
thrashed- beat, ମାଡ଼ିଦେଲେ, ପିଟିଲେ

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 5 The Crab and the Fox

BSE Odisha 6th Class English Follow-Up Lesson 5 The Crab and the Fox Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ The teacher finds an activity to introduce the topic. S/he may use the pictures in the text for the purpose.
(ବିଷୟକୁ ଉପସ୍ଥାପନ କରିବାପାଇଁ ଶିକ୍ଷକ ଏକ କାର୍ଯ୍ୟ ସ୍ଥିରକରିବେ । ସେ (ପୁ/ସ୍ତ୍ରୀ) ବିଷୟବସ୍ତୁରେ ଥିବା ଛବିଗୁଡ଼ିକୁ ଏହି ଉଦ୍ଦେଶ୍ୟରେ ବ୍ୟବହାର କରିପାରନ୍ତି ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
→ Follow the three steps-teacher’s reading aloud two times followed by silent reading by the students.
(ତିନୋଟି ପର୍ଯ୍ୟାୟକୁ ଅନୁସରଣ କର – ଶିକ୍ଷକଙ୍କର ଦୁଇଥର ବଡ଼ପାଟିରେ ପଢ଼ିବା ଓ ଶିକ୍ଷାର୍ଥୀମାନେ ନୀରବରେ ତାଙ୍କ ପରେ ପଢ଼ିବା ।)

TEXT (ବିଷୟବସ୍ତୁ)

• Read the poem silently and answer the questions that follow.
  (କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

It was a very cool night
And there was no crab in sight.
The fox looked for one

But there was none.
“Where did they go?”
Not even one in sight!
They must be in their holes
If I’m right.”

Thinking so he got ready
(The hunger too made him greedy)
To go in search of a crab hole
And he straightened his tail like a pole.
Inside the hole his bushy tail he pushed
The crab, he thought, it slightly missed.
He waited long for a sweet pull
But for long there was none at all.
Finally, he pulled out his tail
But the crab was inside he could smell.
So he changed his plan and called “Brother Crab,
Let’s some song and dance have.
The weather calls for such merrymaking
What is life without dancing and singing ?”
The crab well understood
The fox’s real mood.
Thought he to himself ‘Am I a fool ?’

And answered from his hole :
“Who is going to sing and dance
In such weather cool?
I’ll rather eat and sleep well
in my cozy little hole.”

କବିତାଟିର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଇଟ୍ ୱାଜ୍ ଏ ଭେରି କୁଲ୍ ନାଇଟ୍
ଆଣ୍ଡ୍ ଦେୟାର୍ ୱାଜ୍ ନୋ କ୍ରାବ୍‌ ଇନ୍ ସାଇଟ୍ ।
ଦି ଫକ୍ସ ଲୁକ୍‌ ଫର୍ ୱାନ୍
ବଟ୍ ଦେୟାର ୱାଜ୍ ନନ୍ ।
‘‘ହୋୟାର୍ ଡିଡ଼୍ ଦେ ଗୋ ?’’
ନଟ୍ ଇଭିନ୍ ୱାନ୍ ଇନ୍ ସାଇଟ୍ !
ଦେ ମଷ୍ଟ ବି ଇନ୍ ଦେୟାର୍ ହୋଲ୍‌ସ୍
ଇଫ୍ ‘ଆଇ’ମ୍ ରାଇଟ୍ ।’’
ଥଙ୍କିଙ୍ଗ୍ ସୋ ହି ଗଟ୍ ରେଡ଼ି
(ଦି ହଙ୍ଗର୍ ଠୁ ମେଡ଼୍ ହିମ୍ ଗ୍ରିଡ଼ି)
ଟୁ ଗୋ ଇନ୍ ସର୍ଚ ଅଫ୍ ଏ କ୍ରାବ୍ ହୋଲ୍
ଆଣ୍ଡ ହି ଷ୍ଟେଟେଡ୍ ହିଜ୍ ଟେଲ୍ ଲାଇକ୍ ଏ ପୋଲ୍ ।
ଇସାଇଡ୍ ଦି’ ହୋଲ୍ ହିଜ୍ ବୁସି ଟେଲ୍ ହି ପୁସ୍‌
ଦି କ୍ରାବ୍, ହି ଥଟ୍, ଇଟ୍ ସ୍ଲାଇଟ୍‌ଲି ମିସିଡ୍ ।
ହି ୱେଟେଡ଼ ଲଙ୍ଗ୍ ଫର୍ ଏ ସୁଇଟ୍ ପୁଲ୍
ବଟ୍ ଫର୍ ଲଙ୍ଗ୍ ଦେୟାର୍ ୱାଜ୍ ନନ୍ ଆଟ୍ ଅଲ୍ ।
ଫାଇନାଲି ହି ପୁଲ୍‌ ଆଉଟ୍ ହିଜ୍ ଟେଲ୍
ବଟ୍ ଦି’ କ୍ରାବ୍ ୱାଜ୍ ଇନ୍‌ସାଇଡ୍ ହି କୁଡ଼ ସ୍କେଲ୍ ।
ସୋ ହି ଚେଞ୍ଜେଡ଼୍ ହିଜ୍ ପ୍ଲାନ୍ ଆଣ୍ଡ କଲ୍‌ ‘ବ୍ରଦର୍‌ କ୍ରାନ୍’’,
ଲେଟ୍ସ ସମ୍ ସଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ ହାଭ୍ ।
ଦି ୱେଦର୍ କଲସ୍ ଫର୍ ସବ୍ ମେରିମେକିଙ୍ଗ୍
ଦ୍ଵାଟ୍ ଇଜ୍ ଲାଇଫ୍ ଉଇଦାଉଟ୍ ଡ୍ୟାନ୍‌ସିଙ୍ଗ୍ ଆଣ୍ଡ ସିଙ୍ଗିଙ୍ଗ୍ ?’’
ଦି କ୍ରାବ୍‌ ୱେଲ୍ ଅଣ୍ଡରଷ୍ଟୁଡ୍
ଦି ଫକ୍ସ’ସ୍ ରିଅଲ୍ ମୁଡ଼ ।
ଥଟ୍ ହି ଟୁ ହିମ୍‌ସେଲୁ ‘ଆମ୍ ଆଇ ଏ ଫୁଲ୍ ?’
ଆଣ୍ଡ ଆନ୍‌ସର୍‌ଡ୍‌ ଫ୍ରମ୍ ହିଜ୍ ହୋଲ୍ :
“ହୁ ଇଜ୍ ଗୋଇଙ୍ଗ୍ ଟୁ ସିଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ,
ଇନ୍ ସଚ୍ ଏ ୱେଦର୍ କୁଲ୍ ?
ଆଇ’ଲ୍ ନ୍ୟାଦର୍ ଇଟ୍‌ ଆଣ୍ଡ ସ୍କ୍ରିପ୍ ୱେଲ୍
ଇନ୍ ମାଇଁ କୋଜି ଲିଟିଲ୍ ହୋଲ୍ ।’’

କବିତାର ଓଡ଼ିଆ ଅନୁବାଦ:
ଏହା ଏକ ବହୁତ ଶୀତଳ ରାତ୍ରି ଥିଲା
ଏବଂ କୌଣସି କଙ୍କଡ଼ା ଦୃଷ୍ଟିରେ ପଡୁନଥିଲା ।
କୋକିଶିଆଳ ଗୋଟାଏକୁ ଖୋଜୁଥିଲା
କିନ୍ତୁ ସେଠାରେ ଗୋଟିଏ ବି ନଥିଲା ।
‘ସେମାନେ କୁଆଡ଼େ ଗଲେ?’
ଏପରିକି ଗୋଟିଏବି ହେଲେ
ଦୃଷ୍ଟିରେ ପଡୁନାହାନ୍ତି !
ନିଶ୍ଚୟ ସେମାନେ ଥିବେ ସେମାନଙ୍କର ଗାତଗୁଡ଼ିକ ଭିତରେ
ଯଦି ମୁଁ ଠିକ୍ କହୁଥାଏ ।’
ଭାବି ସେ ପ୍ରସ୍ତୁତ ହୋଇଗଲେ |
(କ୍ଷୁଧା ମଧ୍ଯ ତାକୁ ଲୋଭୀ କରିଦେଲା)
ଅନ୍ଵେଷଣରେ ଯିବା ପାଇଁ ଗୋଟିଏ କଙ୍କଡ଼ା ଗାତ
ଏବଂ ସେ ସିଧା ବା ସଳଖ କରିଦେଲା ତା’ର ଲାଞ୍ଜକୁ ଗୋଟିଏ ଖୁଣ୍ଟ ପରି !
ଗାତ ଭିତରକୁ ତା’ର ବୁଦାଳିଆ ଲୋମଶ ଲାଞ୍ଚକୁ ସେ ଠେଲିଦେଲା
ସେ ଭାଙ୍ଗିଲା, କଙ୍କଡ଼ାଟିରେ ବାଜିବାରେ ଟିକେ ଭୁଲ୍ ହେଲା ।
ସେ ବହୁତ ସମୟ ଅପେକ୍ଷା କଲା ଗୋଟିଏ ମଧୁର ଟଣା ପାଇଁ
କିନ୍ତୁ ଦୀର୍ଘ ସମୟ ଧରି ସେପରି କିଛି ହେଲା ନାହିଁ ।
ଶେଷରେ ସେ ତା’ର ଲାଞ୍ଜକୁ ବାହାରକୁ ଟାଣି ଆଣିଲା
କିନ୍ତୁ କଙ୍କଡ଼ାଟି ଭିତରେ ଥ‌ିବାର ବାସନା ସେ ବାରିପାରିଲା ।
ତେଣୁ ସେ ତା’ର ଯୋଜନାକୁ ବଦଳାଇ ଦେଲା ଏବଂ ଡାକିଲା ‘କଙ୍କଡ଼ା ଭାଇ’’,
ଆସ ଆମେ କିଛି ଗୀତ ଏବଂ ନାଚ କରିବା ।
ଏପରି ମଉଜ କରିବାକୁ ପାଗ ଡାକୁଛି
ନାଚ ଓ ଗୀତ ବିନା ଜୀବନର ଅର୍ଥ କ’ଣ ଅଛି ?
କଙ୍କଡ଼ା ଭଲ ଭାବରେ ବୁଝିଗଲା
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
କୋକିଶିଆଳର ପ୍ରକୃତ ମନୋବୃତ୍ତି !
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
ଏବଂ ତା’ର ଗାତ ମଧ୍ୟରେ ଥାଇ ଉତ୍ତର ଦେଲା :
‘‘କିଏ ଯାଉଛି ଗୀତ ଗାଇବାକୁ ଓ ନାଚିବାକୁ
ଏପରି ଏକ ଶୀତଳ ପାଗରେ ?
ମୁଁ ବରଂ ଖାଇବି ଓ ଶୋଇବି ଭଲ ଭାବରେ
ମୋ’ର ଆରାମଦାୟକ ଛୋଟ ଗାତ ମଧ୍ଯରେ ।’’

Comprehension Questions (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ):

Question 1.
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇଛି ?)
Answer:
The story is about a crab and a fox.

Question 2.
What did the fox look for?
(କୋକିଶିଆଳ କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question 3.
Did he find one?
(ସେ ଗୋଟିଏ ହେଲେ କଙ୍କଡ଼ା ପାଇଲା କି ?)
Answer:
He did not find one.

Question 4.
Where did he push his tail? Why?
(ସେ ତା’ର ଲାଞ୍ଜକୁ କେଉଁଠାକୁ ଠେଲିଲା ? କାହିଁକି ?)
Answer:
He pushed his tail inside the hole. Because he wanted to pull the crab out of its hole.

Question 5.
How could he know that the crab was inside?
(କଙ୍କଡ଼ା ଭିତରେ ଅଛି ବୋଲି ସେ କିପରି ଜାଣିପାରିଲା ? )
Answer:
He could smell that the crab was inside the hole.

Question 6.
What was his new plan?
(ତା’ର ନୂଆ ଯୋଜନାଟି କ’ଣ ଥିଲା ?)
Answer:
His new plan was to make friends with the crab and to call him to come out of its hole for enjoying the fine weather.

Question 7.
Did the new plan work? Why?
(ନୂଆ ଯୋଜନାଟି କାମ କଲା କି ? କାହିଁକି ?)
Answer:
No, the new plan did not work well. Because the crab was cunning enough to the fox’s real mood.

Question 8.
Did the crab understand the intention of the fox?
(କୋକିଶିଆଳର ଉଦ୍ଦେଶ୍ୟକୁ କଙ୍କଡ଼ାଟି ବୁଝିପାରିଥିଲା କି ?)
Answer:
Yes, the crab understood the intention of the fox.

Question 9.
Did the crab come out of her hole?
(କଙ୍କଡ଼ାଟି ତା’ର ଗାତ ବାହାରକୁ ଆସିଥିଲା କି ?)
Answer:
No, the crab did not come out of her hole.

Question 10.
Who is clever?
(କିଏ ଚତୁର ଥିଲା ?)
Answer:
The crab was clever.

Session – 2 (ସୋପାନ – ୨):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖୁବା):
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i).
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇ ଅଛି ?)
Answer:
The story is about a crab and a fox.

Question (ii).
What did the fox look for?
(କୋକିଶିଆଳଟି କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question (iii).
Where did the fox push his tail?
(କୋକିଶିଆଳଟି ତା’ର ଲାଞ୍ଜକୁ କେଉଁଆଡ଼କୁ ଠେଲିଲା ?)
Answer:
The fox pushed his tail inside the crab hole.

Question (iv).
Who is clever?
(କିଏ ଚତୁର ଅଟେ ?)
Answer:
The crab is clever.

(b) Write the story by filling in the gaps:
(ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କରି ଗପଟିକୁ ଲେଖ ।)
(Question with Answer)
Once there lived a ____________ and a ____________. The fox looked for ____________. He____________ his tail inside the ____________. But the crab did not catch his ____________. The fox _____________his plan. He sang a song and called the crab to come out. But the ____________. She said, “Am I _________. I’ll _____________in my ____________.”
Answer:
Once there lived a crab and a fox. The fox looked for a crab. He pushed his tail inside the crab hole. But the crab did not catch his tail. The fox changed his plan. He sang a song and called the crab to come out. But the crab did not come out. She said, “Am I a fool ?” I’ll rather eat and sleep well in my cozy little hole.”

WORD NOTE (ଶବ୍ଦାର୍ଥ):
(The words/phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଯକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

dog — କୁକୁର
cat — ବିଲେଇ
thin — ପତଳା |
fat — ମେଦ
neither — ଏହା ନୁହେଁ କିମ୍ବା ତାହା ନୁହେଁ
pet — ଗୃହପାଳିତ ପଶୁ
mat — ଆସନ (ମସିଣା)
claimed — ଦାବି କଲେ
chased — ଗୋଡ଼ାଇଲା
retire — ବିଶ୍ରାମ ନେବା ବା ଶୋଇବା
hither — here, ଏଠାରେ
owner’s — ମାଲିକଙ୍କର,
cursed — ଅଭିଶାପିତ,
fate — ଭାଗ୍ୟ,
left — ବାମ,
As — ଯେପରି,
someone — କେହି ଜଣେ
pack — ପ୍ୟାକ୍ କରନ୍ତୁ |
sack — ଅଖା ବସ୍ତା
Hey – ହେ
gunny bag – ଛୋଟ ଅଖାଥଳି
grey — ଧୂସର ରଙ୍ଗ
wish — ଇଚ୍ଛା
religious — ଧାର୍ମିକ
carry — ବହନ କର
obey — ମାନ
nanny — ନାନୀ
funny — ମଜାଳିଆ
thought — ଭାବିଲା
rush — ଭିଡ଼, ଜନଗହଳି
Miss — ମିସ୍
hate — ଘୃଣା କରିବା
always — ସର୍ବଦା
late — ବିଳମ୍ବ, ଡେରି
cosy — ଉଷୁମ ଓ ଆରାମଦାୟକ
merry making — ହସଖୁସିରେ ତିଆରି
straightened — ସିଧା
fox — ଠେକୁଆ
crab — କଙ୍କଡ଼ା
greedy — ଲୋଭୀ
pole — ଖୁଣ୍ଟ

busy — ବୁଦାଳିଆ, ଲୋମଶ
pushed — ଠେଲି ହୋଇଗଲା
slightly — ଅଳ୍ପ ଟିକିଏ
missed — ମିସ୍
pull — ଟାଣନ୍ତୁ
smell — ଗନ୍ଧ ବା ବାସନା ଠଉରାଇବା
weather — ପାଣିପାଗ
understood — ବୋଧଗମ୍ୟ
real — ବାସ୍ତବ
mood — ମନ
himself – ତା’ ନିଜକୁ ନିଜେ
fool — ମୂର୍ଖ
hole — ଗାତ
cool — ଶୀତଳ
rather — ବରଂ
well — କୂପ

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(a)

Question 1.
The rows n = 6 and n = 7 in the pascal triangle have been kept vacant. Fill in the gaps
Solution:

Question 2.
Write down the expansion of (a + b)⁸ using Pascal’s triangle.
Solution:
The row n = 8 in Pascal’s triangle is 1, 8, 28, 56, 70, 56, 28, 8, 1.
∴ (a + b)⁸ = a⁸ + 8a^8-1b¹ + 28a^8-2b² + 56a^8-3b³ + 70a^8-4b⁴ + 56a^8-5b⁵ + 28a^8-6b⁶ + 8a^8-7b⁷ + b⁸
= a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸

Question 3.
Find the 3rd term in the expansion of \(\left(2 x^3-\frac{1}{x^6}\right)^4\) using rules of Pascal triangle.
Solution:

Question 4.
Expand the following :
(a) (7a + 3b)⁶
Solution:
(7a + 3b)⁶ = (7a)⁶ + ⁶C₁(7a)^6-1(3b)¹ + ⁶C₂(7a)^6-2(3b)² + ….. + (3b)⁶7⁶a⁶ + 6(7a)⁵(3b) + 15(7a⁴) × 9b² + …. + 3⁶b⁶
= 7a⁶ + 18 × 7⁵a⁵b + 135 × 7⁴a⁴b² + ….. + 3⁶b⁶

(b) \(\left(\frac{-9}{2} a+b\right)^7\)
Solution:

(c) \(\left(a-\frac{7}{3} c\right)^4\)
Solution:

Question 5.
Apply Binominal Theorem to find the value of (1.01)⁵.
Solution:
= 1 + ⁵C₁(0.01)¹ + ⁵C₂(0.01)² + ⁵C₃(0.01)³ + ⁵C₄(0.01)⁴ + (0.01)⁵
= 1 + 5 × 0.01 + 10(0.0001) + 10(0.000001) + 5(0.00000001) + 0.0000000001
= 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.000000001
= 1.0510100501

Question 6.
State true or false.
(a) The number of terms in the expansion of \(\left(x^2-2+\frac{1}{x^2}\right)^6\) is equal to 7.
Solution:
False

(b) There is a term independent of both x and y in the expansion of \(\left(x^2+\frac{1}{y^2}\right)^9\)
Solution:
False

(c) The highest power in the expansion of \(x^{40}\left(x^2+\frac{1}{x^2}\right)^{20}\) is equal to 40.
Solution:
False

(d) The product of K consecutive natural numbers is divisible by K!
Solution:
True

Question 7.
Answer the following :
(a) If the 6th term in the expansion of (x + *)ⁿ is equal to ⁿC₅x^n-10 find *
Solution:
Let the 6th term (x + y)ⁿ is ⁿC₅x^n-10
∴ ⁿC₅x^n-5y⁵ = ⁿC₅x^n-10 = ⁿC₅x^n-5.x^-5
⇒ y⁵ = x^-5 = \(\frac{1}{x^5}\)
∴ y = \(\frac{1}{x}\) . Hence * = \(\frac{1}{x}\)

(b) Find the number of terms in the expansion of (1 + x)ⁿ (1 – x)ⁿ.
Solution:
(1 + x)ⁿ (1 – x)ⁿ = (1 – x²)ⁿ
∴ The number of terms in this expansion is (n + 1)

(c) Find the value of \(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}\)
Solution:

(d) How many terms in the expansion of \(\left(\frac{3}{a}+\frac{a}{3}\right)^{10}\) have positive powers of a? How many have negative powers of a?
Solution:

Question 8.
Find the middle term(s) in the expansion of the following.
(a) \(\left(\frac{a}{b}+\frac{b}{a}\right)^6\)
Solution:
Here there is only one middle term i.e. the 4th term.
∴ 4th term i.e. (3 + 1)th term in the expansion of

(b) \(\left(x+\frac{1}{x}\right)^9\)
Solution:
Here there are two middle terms i.e. 5th and 6th terms.
∴ 5th term in the above expansion is

(c) \(\left(x^{\frac{3}{2}}-y^{\frac{3}{2}}\right)^8\)
Solution:
Here there is only one middle term i.e. 5th term.
∴ 5th term i.e. (4 + 1)th term in the expansion of

Question 9.
Find the 6th term in the expansion of \(\left(x^2+\frac{a^4}{y^2}\right)^{10}\)
Solution:
6th term i.e. (5 + 1)th term in the expansion of
(x² + \(\frac{a^4}{y^2}\))¹⁰ is ¹⁰C₅(x²)^10-5 (\(\frac{a^4}{y^2}\))⁵

Question 10.
(a) Find the fifth term in the expansion of (6x – \(\frac{a^3}{x}\))¹⁰
Solution:

(b) Is there a term independent of x? If yes find it out.
Solution:
Let (r + 1)th term in the expansion of  (6x – \(\frac{a^3}{x}\))¹⁰ is independent of x.
∴ (r + 1)th term = ¹⁰C_r(6x)^10-r(\(\frac{-a^3}{x}\))^r
= ¹⁰C_r6^10-rx^10-r(-1)^ra^3rx^-r
= (-1)^{r 10}C_r6^10-ra^3rx^10-2r
∴ x^10-2r = 1 = x⁰
or, 10 – 2r = 0 or, r = 5
∴ 6th term is term independent of x in the above expansion and the term is (-1)^{5 10}C₅6^10-5a^3.5
= – ¹⁰C₅6⁵a¹⁵ = – 252 × 6⁵a¹⁵

Question 11.
(a) Find the coefficient of \(\frac{1}{y^{10}}\) in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\)
Solution:

(b) Does there exist a term independent of y in the above expansion?
Solution:
Let (r + 1)th term is independent of y.
∴ y^30-8r = 1 = y⁰ or, 30 – 8r = 0
or, r = \(\frac{30}{8}\) = \(\frac{15}{4}\) which is not possible as r∈ N or zero.
∴ There is no term in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\) which is independent of y.

Question 12.
(a) Find the coefficient of x⁴ in the expansion of (1 + 3x + 10x²)(x + \(\frac{1}{x}\))¹⁰
Solution:

(b) Find the term independent of x in the above expansion.
Solution:

Question 13.
Show that the coefficient of a^m and aⁿ in expansion of (1 + a)^m+n are equal.
Solution:
(m + 1)th and (n + 1)th terms in the expansion of (1 + a)^m+n are ^m+nC_ma^m and ^m+nC_naⁿ
∴ The coefficient of a^m and aⁿ are ^m+nC_m and ^m+nC_n which are equal.

Question 14.
An expression of the form (a + b + c + d + …. ) consisting of a sum of many distinct symbols called a multinomial. Show that (a + b + c)ⁿ is
the sum of all terms of the form \(\frac{\boldsymbol{n} !}{\boldsymbol{p} ! \boldsymbol{q} ! \boldsymbol{r} !}\) a^pb^qc^r where p, q and r range over all positive triples of non-negative integers such that p + q + r = n.
Solution:

Question 15.
State and prove a multinominal Theorem.
Solution:
Multinominal Theorem:
(P₁ + P₂ + ……. + P_m)ⁿ
\(=\sum \frac{n !}{n_{1} ! n_{2} ! \ldots n_{m} !} p_1^{n_1} p_2^{n_2} \ldots p_m^{n_m}\)
where n₁ + n₂ + ……. + n_m = n
The proof of this theorem is beyond the syllabus.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(b)

Question 1.
Find the number of ways in which 5 different books can be arranged on a shelf.
Solution:
The number of ways in which 5 different books can be arranged on a shelf is 5! = 5. 4. 3. 2. 1 = 120

Question 2.
Compute ⁿP_r for
(i) n = 8, r = 4
Solution:
∴ ⁿP_r = \(\frac{n !}{(n-r) !}=\frac{8 !}{(8-4) !}\)
\(=\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 . !}{4 !}\) = 8.7.6.5 = 1680

(ii) n = 10, r = 3
Solution:
n = 10, r = 3
∴ ⁿP_r = \(\frac{n !}{(n-r) !}=\frac{10 !}{7 !}\)

(iii) n = 11, r = 0
Solution:
n = 11, r = 0
∴ ⁿP_r = ¹¹P₀ = 1

Question 3.
Compute the following :
(i) \(\frac{10 !}{5 !}\)
Solution:
\(\frac{10 !}{5 !}\) = 10. 9. 8. 7. 6 = 30240

(ii) 5! + 6!
Solution:
5 ! + 6! = 5 ! + 6.5 !
= 5 ! (1 + 6) = 120. 7 = 840

(iii) 3! × 4!
Solution:
3 ! × 4 ! = 6 × 24 = 144

(iv) \(\frac{1}{8 !}+\frac{1}{9 !}+\frac{1}{10 !}\)
Solution:

(v) 2!^3! = 2⁶ = 64
(vi) 2³! = 8! =40320

Question 4.
Show that 2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)
Solution:
2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)

Question 5.
Find r if P(20, r) = 13. P (20, r – 1).
Solution:

Question 6.
Find n if P(n, 4) = 12. P(n, 2).
Solution:
ⁿP₄ = 12 × ⁿP₂
or, \(\frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}\)
or, (n – 2)! = 12 (n – 4)!
or, (n – 2) (n – 3) (n – 4)! = 12 (n – 4)!
or, (n – 2) (n – 3) = 12
or, n² – 5n – 6 = 60
or, (n – 6) (n + 1) = 0
or, n = 6 – 1
Hence n = 6 as n is a natural number.

Question 7.
If P (n – 1, 3) : P (n + 1, 3) = 5: 12, Find n.
Solution:

Question 8.
Find m and n  if P(m + n, 2) = 56, P(m – n, 2) = 12
Solution:
^m+nP₂ = 56, ^m-nP₂ = 12
or, \(\frac{(m+n) !}{(m+n-2) !}=56, \frac{(m-n) !}{(m-n-2) !}=12\)
or, (m + n) (m + n – 1) = 8 × 7
(m – n) (m – n – 1) = 4 × 3
∴ (m + n) = 8, m – n = 4
∴ m = 6, n = 2

Question 9.
Show that
(i) P(n, n) = P(n, n – 1) for all positive integers.
Solution:

(ii) P(m, 1) + P(n, 1) = P(m + n, 1) for all positive integers m, n.
Solution:
^mp₁ + ⁿp₁ = ^m+np₁ ∀ m, n ∈ Z
∴ L.H.S.
= ^mp₁ + ⁿp₁ = m + n = ^m+np₁ = R.H.S.
(∴ ⁿp₁ = n)

Question 10.
How many two-digit even numbers of distinct digits can be formed with the digits 1, 2, 3, 4, and 5?
Solution:
Two-digit even numbers of distinct digits are to be formed with the digits 1, 2, 3, 4, and 5. Here the even numbers must end with 2 or 4. When 2 is placed in the unit place, the tenth place can be filled up by the other 4 digits in 4 different ways. Similarly, when 4 is placed in the unit place, the tenth place can be filled up in 4 different ways.
∴ The total number of two-digit even numbers = 4 + 4 = 8.

Question 11.
How many 5-digit odd numbers with distinct digits can be formed with the digits 0, 1, 2, 3, and 4?
Solution:
5-digit odd numbers are to be formed with distinct digits from the digits 0, 1, 2, 3, and 4. The numbers are to end with 1 or 3 and must not begin with 0.
The 5th place can be filled up by any one of 1 or 3 by 2 ways.

1st

2nd

3rd

4th

5th

Places
The 1st place can be filled by the rest 3 digits except 0 and the digit in 5th place.
The 2nd, 3rd, and 4th places can be filled up by the rest 3 digits in 3! ways.
So total no. of ways = 2 × 3 × 3 ! = 2 × 3 × 2 = 36 ways.

Question 12.
How many numbers, each less than 400 can be formed with the digits 1, 2, 3, 4, 5, and 6 if repetition of digits is allowed?
Solution:
Different numbers each less than 400 are to be formed with the digits 1, 2, 3, 4, 5, and 6 with repetition. Here the numbers are 1-digit, 2-digit, and 3-digit.
∴ The number of 1-digit numbers = 6.
The number of 2-digit numbers = 6² = 36.
The 3-digit number each less than 400 must begin with 1, 2, or 3. So the hundred’s place can be filled by 3 digits and ten’s and unit place can be filled by 6 digits each. So the number of 3 digit numbers = 3 × 6 × 6 = 108.
∴ The total number of numbers = 6 + 36 + 108 = 150.

Question 13.
How many four-digit even numbers with distinct digits can be formed out of digits 0, 1, 2, 3, 4, and 5, 6?
Solution:
Four-digit even numbers mean, they must end with 0, 4, 2, 6. When 0 is placed in last, the 1st place is filled by 1, 2, 3, 4, 5, 6, and the remaining 2 places can be filled by ⁵P₂ ways.
The number of numbers ending with 0 = ⁵P₂ × 6 = 120
Similarly, the number of numbers ending. with 2, 4 and 6 = ⁵P₂  × 5 × 3 = 300
∴ The total number of numbers = 420.

Question 14.
How many integers between 100 and 1000 (both inclusive) consist of distinct odd digits?
Solution:
Integers are to be formed with distinct odd digits between 100 and 1000.
The numbers between 100 and 1000 are 3-digited.
The odd digits are 1, 3, 5, 1, and 9.
The number of distinct 3-digit odd numbers = ⁵P₃ = 5.4.3 = 60.

Question 15.
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face, thrown thrice in succession. What is the total number of outcomes?
Solution:
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face is thrown thrice in succession.
∴ The total number of outcomes = 6³ = 216.

Question 16.
What is the total number of integers with distinct digits that exceed 5500 and do not contain 0, 7, and 9?
Solution:
The integer exceeding 55000 must be 4-digited, 5-digited, 6-digited, and 7-digited, as there are seven digits i.e. 1, 2, 3, 4, 5, 6, and 8 to be used for the purpose.
In the 4-digit integers, when 1st place is filled by 5 and 2nd place by .6, the rest two places can be filled by the remaining 5 digits in ⁵P₂ ways. Similarly, when 5 is in 1st place and 8 in 2nd place, the remaining 5 digits be used in ⁵P₂ ways. So the number of 4-digit integers beginning with 5 is 2 × 5P = 40.
When 6 is placed in 1st place, the remaining 3 placed be filled by the remaining 6 digits in 6P3 ways. Similarly, when 8 is placed in 1 st place, the remaining 3 places be filled by the remaining 6 digits in ⁶P₃ ways.
So the number of 4-digit integers starting with 6 and 8 is 2 × ⁶P₃  = 240.
∴ The total number of 4-digit numbers is 40 + 240 = 280.
The number of 5-digit integers is ⁷P₅ = 2520.
The number of 6-digit numbers is ⁷P₆ = 5040 and the number of 7 -digit numbers is ⁷P₆  = 5040.
∴ The total number of integers exceeding 5500 and not containing 0, 7, and 9 is 280 + 2520 + 5040 + 5040 = 12880.

Question 17.
Find the total number of ways in which the letters of the word PRESENTATION can be arranged.
Solution:
The number of letters in the word “PRESENTATION” is 12, out of which there are 2N’s, 2E’s, and 2T’s. So the total number of arrangements.
\(=\frac{12 !}{2 ! 2 ! 2 !}=\frac{1}{8}(12) !\)

Question 18.
Find the numbers of all 4-lettered words (not necessarily having meaning) that can be formed using the letters of the word BOOKLET.
Solution:
We have to form 4-lettered words using the letters of the word BOOKLET. The word contains 7 letters out of which there are 20’s. So there are 6 letters.
∴ The number of 4-lettered words = ⁷P₄ – ⁶P₄ = 7.6.5.4 – 6.5.4.3 = 480

Question 19.
In how many ways can 2 boys and3 girls sit in a row so that no two girls sit side by side?
Solution:
Two boys and 3 girls sit in a row so that no two girls sit side by side.
So the only possibility is boys should be situated in between the two girls. In between 3 girls there are 2 gaps where 2 boys will be site.
The girls will be arranged in 3! and boys in 2! ways.
∴ The total number of ways = 2! × 3!=2 × 6= 12

Question 20.
Five red marbles, four white marbles, and three blue marbles the same shape and size are placed in a row. Find the total number of possible arrangements.
Solution:
5 red, 4 white, and 3 blue marbles of the same size and shape are placed in a row.
∴ The total number of marbles is 12 out of which 5 are of one kind, 4 are of 2nd kind and 3 are of 3rd kind
∴ The total number of possible arrangements
\(=\frac{12 !}{5 ! 4 ! 3 !}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 3 \cdot 2}\)
= 12.11.10.3.7 = 27720.

Question 21.
Solve example 2.
Solution:
We have |A| = n, |B| = m
∴ The number of one-one functions from A to B is ^mP_n = \(\frac{m !}{(m-n) !}\) when m> n.
If m = n, the number of one-one functions is \(\frac{m !}{(m-m) !}=\frac{m !}{0 !}\) = m! = n!
If m < n, then there is no possibility of one-one functions.

Question 22.
In how many ways can three men and three women sit at a round table so that no two men can occupy adjacent positions?
Solution:
Since no 2 men are to sit together, there are 4 places available for them corresponding to any one way of sitting of 2 men i.e., two places between the women and 2 places at two ends.

Let m₁ be fixed m₂ can sit in 2 places, w₁ can sit in 3 places, m₃ can sit in 1 place, w₂ can sit in 2 places w₃ can sit in 1 place.
∴ The total number of ways = 2 × 3 × 1 × 2 × 1 = 12

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(c)

Solve the following systems of linear inequalities graphically.
Question 1.
2x – y ≥ 0, x – 2y ≤ 0, x ≤ 2, y ≤ 2 [Hint: You may consider the point (2, 2) to determine the SR of the first two inequalities.]
Solution:
2x – y ≥ 0
x – 2y ≤ 0
x ≤ 2
y ≤ 2
Step – 1: Let us draw the lines.
2x – y = 0, x – 2y = 0, x = 2, y = 2
2x – y = 0

X	0	1
y	0	2

x – 2y = 0

X	0	2
y	0	1

Step – 2: Let us consider point (1, 0) which does not line on any of these lines.
Putting x = 1, y = 0 in the inequations we get
2 ≥ 0 (True)
1 ≤ 0 (False)
1 ≤ 2 (True)
0 ≤ 2 (True)
Point (1, 0) satisfies all inequality except x – 2y < 0.
∴ Thus the shaded region is the solution region.

Question 2.
x – y < 1, y – x < 1
Solution:
x – y < 1
y – x < 1
Step – 1: Let us draw the dotted lines.
x – y = 1 and y – x = 1
x – y = 1 ⇒ y = x – 1

X	1	0
y	0	-1

y – x = 1 ⇒ y = x + 1

X	0	-1
y	1	0

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
0 < 1 (True)
0 < 1 (True)
∴ (0, 0) satisfies both the inequations.
∴ Thus the shaded region is the feasible region.

Question 3.
x – 2y + 2 < 0, x > 0
Solution:
x – 2y + 2 < 0, x > 0
Step – 1: Let us draw the dotted line x – 2y + 2 = 0
⇒ y = \(\frac{x+2}{2}\)

X	-2	0
y	0	1

Step – 2: Let us consider the point (1, 0) that does not lie on the lines  putting x = 0, y = 0 in the inequation, we get
2 < 0 (false)
1 > 0 (True)
⇒ (1, 0) satisfies x > 0 and does not satisfy x – 2y + 2 < 0.
∴ Thus the shaded region is the solution region.

Question 4.
x – y + 1 ≥ 0, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Solution:
x – y + 1 ≥ 0
3x + 4y ≤ 12
x ≥ 0, y ≥ 0
Step – 1: Let us draw the lines.
x – y + 1 = 0
3x + 4y = 12
Now, x – y + 1 = 0 ⇒ y = x + 1

X	0	-1
y	1	0

3x + 4y = 12

X	4	0
y	0	3

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 ≥ 0 (True)
0 ≤ 12 (False)
∴ (0, 0) satisfies both the inequations and x > 0, y > 0 is the first quadrant.
∴ Thus the shaded region is the solution region.

Question 5.
x + y > 1, 3x – y < 3, x – 3y + 3 > 0
Solution:
x + y > 1
3x – y < 3
x – 3y + 3 > 0
Step – 1: Let us draw the lines.
x + y = 1
3x – y = 3
x – 3y + 3 = 0
Now x + y = 1
⇒ y = 1 –  x

X	1	0
y	0	1

3x – y = 3
⇒ y = 3x – 3

X	1	0
y	0	-3

x – 3y + 3 = 0
⇒ y = \(\frac{x+3}{3}\)

X	-3	0
y	0	1

Step – 2: Let us consider the point (0, 0) that does not lie on these lines. Putting x = 0, y = 0 in the inequations we get,
0 > 1 (False)
0 < 3 (True)
3 > 0 (True)
Thus (0, 0) satisfies 3x – y < 3 and x – 3y + 3 > 0 but does not satisfy x + y > 1
∴ The shaded region is the solution region.

Question 6.
x > y, x < 1, y > 0
Solution:
x > y, x < 1, y > 0
Step – 1: Let us draw the dotted lines.

Step – 2: Let us consider a point (2, 1) that does not lie on any of the lines.
Putting x = 2, y = 2 in the inequations
we get,
2 > 1 (True)
2 < 1 (False)
1 > 0 (True)
⇒ (2, 1) satisfies x > y and y > 0 but does not satisfy x < 1.
∴ Thus the shaded region is the solution region.

Question 7.
x < y, x > 0, y < 1
Solution:
x < y
x > 0
y < 1
Step – 1: Let us draw the dotted lines.
x = y
x = 0
and y = 1

Step – 2: Let us consider point (1, 0) that does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 < 0 (False)
1 > 0 (True)
0 < 1 (True)
Clearly (1, 0) satisfies x > 0, y < 1 but does not satisfy x < y.
∴ The shaded region is the solution region.

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(b)

Question 1.
If Z₁ and Z₂ are two complex numbers then show that
\(\begin{aligned}
& \left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2 \\
& =\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)
\end{aligned}\)
Solution:
Let z₁ and z₂ be two complex numbers.
Let z₁ = a + ib, z₂ = c + id

Question 2.
If a, b, c are complex numbers satisfying a + b + c = 0 and a² + b² + c² = 0 then show that |a| = |b| = |c|
Solution:
Let a + b + c = 0 and a² + b² + c² = 0
Then (a + b + c)² = 0
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0
⇒ 2(ab + bc + ca) = 0
⇒ ab + bc = – ca
⇒ b(a + c) = – ca
⇒ b(- b) = – ca [a + b + c = 0]
⇒ b² = ca
⇒ b³ = abc
Similarly it can be shown that a³ = abc and c³ = abc
Thus a³ = b³ = c³
⇒ |a³|= |b³| = |c³|
⇒ |a|³ = |b|³ = |c|³
⇒ |a| = |b| = |c|

Question 3.
What do the following represent?
(i) { z : |z – a| + |z + a| = 2c } where |a| < c
Solution:
{ z : |z – a| + |z + a| = 2c } where |a| < c    …….(1)
Here z is a complex number.
Let z = x + iy.
∴ (x, y) is the point corresponding to the complex number z?
Let ‘a’ and ‘ – a’ be two fixed points.
∴ Eqn. (1) implies that the sum of the distances of the point (x, y) from two points la and a’ is constant i.e. 2c
∴ The locus is an ellipse.

(ii) {z : |z – a| – |z + a| = c }
Solution:
Here {z : |z – a| – |z + a| = c } implies that, the difference of the distances of the point (x, y) from two fixed points ‘ – a’ and ‘a’ is a constant i.e. c.
So the locus is a hyperbola.

(iii) What happens in (i) |a| > c?
Solution:
In(i), if |a| > c. then there is no locus. But if |a| = c, then the locus reduces to a straight line.

Question 4.
Given cos α + cos β + cos γ = sin α + sin β + sin γ = 0 Show that cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
Solution:
Let a = cos α + i sin α,
b = cos β + i sin β
c = cos γ + i sin γ
∴ a + b + c = ( cos α + cos β + cos γ) + i ( sin α + sin β + sin γ)
= 0 + i0 = 0
∴ a³ + b³ + c³ – 3 abc
= (a + b + c )( a² + b² + c² – ab – bc – ca) = 0
or, a³ + b³ + c³ = 3 abc
or, ( cos α + i sin α)³ + (cos β + i sin β)³ + ( cos γ + i sin γ)³
= 3( cos α + i sin α) (cos β + i sin β) ( cos γ + i sin γ)
or, cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ
= 3[cos (α + β + γ) + i sin (α + β + γ)]
or, (cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ)
= 3 cos (α + β + γ) + i 3 sin (a + β + γ)
∴ cos 3α + cos 3β + cos 3γ
= 3 cos (α + β + γ) and sin 3α + sin 3β + sin 3γ
= 3 sin (α + β + γ)

Question 5.
Binomial theorem for complex numbers. Show that (a+b)ⁿ = a^{n n}C₁a^n-1b +  …..+ ⁿc_ra^n-rb^r + …..+ bⁿ where a,b ∈ C and n, rule of multiplication of complex numbers and the relation ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)
Solution:
Let a and b be two complex numbers
Let a = α₁ + iβ₁, b = α₂ + iβ₂
(a + b)¹ = (α₁ + iβ₁ + α₂ + iβ₂)¹
= α₁ + iβ₁ + α₂ + iβ₂
= (α₁ + iβ₁)¹ + ¹C₁(α₁ + iβ₁)^1-1(α₁ + iβ₁)¹
= a¹ + ¹C₁ a^1-1 b¹
∴ P₁ is true
Let P_k be true
i.e., (a + b)^k = a^k + ^kC₁ a^k-1 b¹ + … + b^k
where a.b ∈ C
Now ( a + b)^k+1 = (a + b)^k (a + b)¹
= (a^k + ^kC₁ a^k-1b¹ +…+ b^k) (a + b)
= a^k-1 + ba^k+ ^kC₁ a^kb + ^kC₁a^k-1b² + … + b^k+1
= a^k+1 + a^kb(^kC₁+ 1)+ …+ b^k+1
= a^k+1 + ^k+1C₁a^kb¹ + ^k+1 C₂a^k-1b² +… + b^k+1
∴ P_k+1 is true
∴ P_n is true for all values of n ∈ N

Question 6.
Use the Binomial theorem and De Moiver’s theorem to show
cos 3θ = 4 cos ³ θ – 3 cos θ,
sin 3θ = 3 sin θ – 4 sin ³ θ
Express cos nθ as a sum of the product of powers of sin θ and cos θ. Do the same thing for sin nθ.
Solution:
We have (cos θ + i sin θ)³
= cos 3θ + i sin 3θ       …..(1)
But by applying the Binomial theorem, we have
(cos θ + i sin θ)³
= cos ³ θ + ³C₁ (cos θ)^3-1 (i sin θ)¹ + ³C₂ (cos θ)^3-2  (i sin θ)² + (i sin θ)³
= cos3θ + 3i cos2θ sin θ + 3i² cos θ sin² θ + i³ sin³ θ
= (cos³ θ – 3 cos θ sin² θ) + i(3 cos² θ sin θ – sin³ θ)
∴ cos ³ θ =cos³ θ – 3 cos θ(1 – cos² θ)
= cos³ θ – 3 cos θ + 3 cos ³ θ
= 4 cos³ θ – 3 cos θ and
sin 3θ = 3 cos² θ sin³ θ – sin³ θ
= 3 (1 – sin² θ) sin θ – sin³ θ
= 3 sin θ – 3 sin³ θ – sin³ θ
= 3 sin θ – 4 sin³ θ (Proved)
Again, (cos θ + i sin θ)ⁿ
= cos nθ + i sin nθ         ….(3)
Also, (cos θ + i sin θ)ⁿ
= cosⁿ θ + ⁿC₁ cos^n-1 θ ( i sin θ) + ⁿc₂ cos ^n-2 θ (i sin θ)² + …+ (i sin θ)
= cosⁿ θ – ⁿC₂ cos^n-2 θ sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ – …) + i (ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin ³ θ) + ⁿC₅ cos^n-5 θ sin⁵ θ – …)    …..(4)
Equating real part and imaginary parts in (1) and (3), we have
cos nθ = cosⁿ θ – ⁿC₂ cos^n-2 θ × sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ …
and sin nθ = ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin³ θ + ⁿC₅ cos^n-5 θ sin⁵ θ…

Question 7.
Find the square root of
(i) – 5 + 12 √-1
Solution:

(ii) – 11 – 60 √-1
Solution:
Let \(\sqrt{-11-60 \sqrt{-1}}\) = x + iy
Squaring both sides we get
– 11 – 60i = (x + iy)²

(iii) – 47 + 8 √-1
Solution:

As 2ab = 8 > 0, a and b have the same sign.
∴ \(\sqrt{-47+8 i}\)
\(=\pm\left(\sqrt{\frac{\sqrt{2273}-47}{2}}+i \frac{\sqrt{2273}+47}{2}\right)\)

(iv) – 8 + √-1
Solution:

(v) a² – 1 +2a √-1
Solution:
a² – 1 + 2a √-1
= a² + i² + 2ai = (a + i)²
∴ \(\sqrt{a^2-1+2 a \sqrt{-1}}\) = ±(a + i)

(vi) 4ab – 2 (a² – b²) √-1
Solution:
4ab – 2 (a² – b²) √-1
= (a + b)² – (a – b)² – 2(a² – b²)i
= (a + b)² + (a – b)²i² – 2(a + b)(a – b)i
= (a + b) – i(a – b)²
∴ \(\sqrt{4 a b-2\left(a^2-b^2\right) \sqrt{-1}}\)
= ±[(a + b) – i (a – b)]

Question 8.
Find the values of cos72° ….
Solution:
Let 18° = θ then 5θ = 90°
⇒ 3θ = 9θ – 2θ
⇒ cos 3θ = cos (9θ – 2θ) = sin 2θ
⇒ 4 cos³ θ – 3 cos θ = 2 sin θ cos θ
⇒ 4 cos² 0 – 3 = 2 sin 0 (∴ cos θ = cos 18° ≠ 0)
⇒ 4(1 – sin² θ) – 3 = 2 sin θ
⇒ 4 sin² θ + 2 sin θ – 1 = 0
⇒ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{2 \times 4}=\frac{-1 \pm \sqrt{5}}{4}\)
⇒ sin 18° = \(\frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4}\)
(∴ 18° is a cut)
Now cos 72° = cos (90° – 18°)
= sin 18° = \(\frac{\sqrt{5}-1}{4}\)
For other methods refer 148 pages of the text book.

Question 9.
Find the value of cos 36°.
Solution:
We have 36° = \(\frac{\pi^0}{5}\)
∴ cos 36° = cos \(\frac{\pi}{5}\)
Let α = cos \(\frac{\pi}{5}\) + i sin \(\frac{\pi}{5}\)
be the root of the equation x⁵ + 1 = 0
Again, if x⁵ + 1 = 0
or, x⁵ = – 1 = cosπ + i sinπ
or, x = (cos π + i sin π )^1/5
= [cos (π + 2kπ)+ i sin (π + 2kπ)]^1/5
or, x = cos \(\frac{(2 k+1) \pi}{5}+i \sin \frac{(2 k+1) \pi}{5}\)
where 2kπ is the period of sine and cosine and k = 0, 1, 2, 3, 4
∴ The eqn x⁵ + 1 = 0 has 5 roots out of which -1 is one root which corresponds to k = 2
Again,
x⁵ + 1 = (x + 1)(x⁴ – x³ + x² – x + 1)
So their 4 roots will be obtained on solving the eqn.
x⁴ – x³ + x² – x + 1 = 0
we have, x⁴ – x³ + x² – x + 1 = 0
or, x² – x + 1 – \(\frac{1}{x}+\frac{1}{x^2}\) = 0
(Dividing both sides by x²)

Re α i.e. cos \(\frac{\pi}{5}=\frac{1+\sqrt{5}}{5}=\frac{\sqrt{5}+1}{4}\)
and cos.108° = \(\frac{1-\sqrt{5}}{4}\)

Question 10.
Evaluate cos \(\frac{2 \pi}{17}\) using the equation x¹⁷ – 1 = 0
Solution:
x¹⁷ – 1 = 0
or, x¹⁷ = 1 = cos 0° + i sin 0°
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
or x = (cos 2kπ + i sin 2kπ)^1/17
= cos \(\frac{2k \pi}{17}\) + i sin \(\frac{2k \pi}{17}\)
If k = 1, x = cos \(\frac{2 \pi}{17}\) + i sin \(\frac{2 \pi}{17}\)
If k = 0 , x = 1
As x¹⁷ – 1 = (x – 1) (x¹⁶ + x¹⁵ + …. +1)
So one root of the eqn. x¹⁷ – 1 = 0 is 1 and all other roots are the roots of the eqn.
x¹⁶ + x¹⁵ + ….+ 1 = 0
∴ The value of cos \(\frac{2 \pi}{17}\) can be found from the roots of the eqn.  (1)

Question 11.
Solve the equations.
(i) z⁷ = 1
Solution:
z⁷ = 1 = cos 0 + i sin 0
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
∴ z = (cos 2kπ + i sin 2kπ)^1/7
= cos \(\frac{2k \pi}{7}\) + i sin \(\frac{2k \pi}{7}\)
where k = 0, 1, 2, 3, 4, 5, 6

(ii) z³ = i
Solution:

(iii) z⁶ = – i
Solution:

(iv) z³ = 1 + i
Solution:

Question 12.
If sin α + sin β + cos γ = 0
= cos α + cos β + cos γ = 0
Show that
(i) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution:
Refer to Q. No. 4

(ii) sin² α + sin² β + sin² γ = cos² α + cos² β + cos² γ =3/2
Solution:
Let x = cos α + i sin α,
y = cos β + i sin β
z = cos β + sin β
∴ x + y + z = (cos α + cos β + cos γ) + i ( sin α + sin β + sin γ) = 0 +i0= 0
∴ xy + yz + zx = xyz (\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)) = 0
Since \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) – i (sin α + sin β + sin γ) = 0 – i0 =0
∴ (x + y + z)² = x ² + y² +z² + 2(xy + yz + zx)
= x² + y² + z² + 0 = x² + x² + z²
or 0 = x² + y² + z²
∴ x² +y² + z² =0
or, (cos α + i sin α)² + (cos β + i sin β)² + ( cos γ + i sin γ)² = 0
or, cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
or, (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0
∴ cos 2α + cos 2β + cos 2γ = 0
or, cos² α – sin² α + cos² β – sin² β + cos² γ – sin² γ =0
or, (cos² α + cos² β + cos² γ) = (sin² α + sin² β + sin² γ)
But cos² α + sin² α + cos² β + sin² β + cos² γ + sin² γ = 1 + 1 + 1 = 3
∴ cos² α + cos² β + cos² γ = sin² α + sin² β + sin² γ = 3/2   (Proved)

Question 13.
If x + \(\frac{1}{x}\) = 2 cos θ
Show that \(x^n+\frac{1}{x^n}\) = 2 cos nθ

Question 14.
x_r = cos a^r + i sin a^r
r =1, 2, 3 and x₁ + x₂ + x₃ = 0 Show that \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\) = 0
Solution:

Question 15.
Show that \(\left(\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta}\right)^n\) = \(\cos \left(\frac{n \pi}{2}-n \theta\right)+i \sin \left(\frac{n \pi}{2}-n \theta\right)\)
Solution:

Question 16.
If α and β are roots x² – 2x + 4 = 0 then show that \(\alpha^n+\beta^n=2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
we have x² – 2x + 4 = 0

= \(2^n \times 2 \cos \frac{n \pi}{3}=2^{n+1} \cos \frac{n \pi}{3}\)

Question 17.
For a positive integer n show that
(i) (1 + i)ⁿ + (1 – i)ⁿ = \(2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}\)
Solution:

(ii) (1 + i√3)ⁿ + (1 – i√3)ⁿ = \(2^{n+1} \cos \frac{n \pi}{3}\)
Solution:

Question 18.
Let x + \(\frac{1}{x}\) = 2 cos α, y + \(\frac{1}{y}\) = 2 cos β, z + \(\frac{1}{z}\) = 2 cos γ. Show that
(i) 2 cos (α + β + γ) = xyz + \(\frac{1}{xyz}\)
Solution:
We can take x = cos α + i sin α
y = cos β + i sin β, z = cos γ + i sin γ
∴ xyz = (cos α + i sin α ) (cos β + i sin β ) (cos γ + i sin γ)
= cos (α + β + γ) – i sin (α + β + γ)
∴ \(\frac{1}{xyz}\) = cos (α + β + γ) – i sin(α + β + γ)
∴ xyz + \(\frac{1}{xyz}\) = 2 cos(α + β + γ)

(ii) 2 cos (pα + qβ + rγ) = \(x^p y^q z^r+\frac{1}{x^p y^q z^r}\)
Solution:

Question 19.
Solve x⁹ + x⁵ – x⁴ = 1
Solution:
x⁹ + x⁵ – x⁴ = 1
or, x⁵(x⁴ + 1) – (x⁴ + 1) = 0
or, (x⁵ – 1) (x⁴ + 1) = 0
x⁴ + 1 = 0 and x⁵ – 1 = 0
x⁴ = – 1 = cos π + i sin π
= cos (π + 2nπ) + i sin (π + 2nπ)
∴ x = [cos (2n + 1) π + 1 sin (2n + 1) π]^1/4
= \(\cos \frac{2 n+1 \pi}{4}+i \sin \frac{2 n+1 \pi}{4}\)
for n = 0, 1, 2, 3
Again, x⁵ – 1 = 0 or, x⁵ = 1
or, x⁵ = cos 0 + i sin 0
= cos 2nπ + i sin 2nπ
or, x = (cos 2nπ + i sin 2nπ)^1/5
= \(\cos \frac{2 n \pi}{5}+i \sin \frac{2 n \pi}{5}\)
Where n = 0, 1, 2, 3, 4.

Question 20.
Find the general value of θ if (cos θ + i sin θ) (cos 2θ + i sin 2θ),…..(cos nθ + i sin nθ) =1
Solution:

Question 21.
If z = x + iy show that |x| + |y| ≤ √2 |z|
Solution:
z = x + iy
∴ |z| = \(\sqrt{x^2+y^2}\)
∴ |z| = x² + y²
We have (|x| – |y|)² ≥ 0
⇒ |x|² + |y|² -2|x||y|> 0
⇒ 2(|x|² + |y|²) – 2|x||y| ≥ |x|² + |y|²
⇒ 2(|x|² + |y|²) ≥ |x|² + |y|² + 2|x||y|
⇒ 2|z|² ≥ (|x| + |y|)²
⇒ √2|z| ≥ |x| + |y|
⇒ |x| + |y| ≤ √2|z|

Question 22.
Show that
Re (Z₁Z₂) = Re z₁, Re z₂ – Im z₁, Im z₂
Im (Z₁Z₂) = Re z₁, Im z₂ + Re z₂ Im z₁
Solution:
Let z₁ = a + ib, z₂ = c + id
∴ z₁, z₂ = (a + ib) (c + id)
= ac + iad + ibc + i²bc
= (ac – bd) + i (ad + be)
∴ Re (z₁, z₂) = ac – bd – Re z₁. Re z₂
– Im z₁,. Im z₂
Again, Im z₁, z₂ = ad + be
= Re z₁. Im z₂ + Im z₁,. Re z₂

Question 23.
What is the value of arg ω + arg ω²?
Solution:
arg ω = arg ω² = arg (ω • ω²)
= arg (ω³) = arg (1) = 2nπ
∴ The principal agrument = 0.

Question 24.
If |z₁| ≤ 1, |z₂| ≤ 1 show that \(\left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2=\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)\) Hence and otherwise show that. \(\left|\frac{z_1-z_2}{1-z_1 z_2}\right|<1 \text { if }\left|z_1\right|<1,\left|z_2\right|<1\)
Solution:

Question 25.
If z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0 Show that |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|
Solution:
Let z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0
⇒ 2z₁² + 2z₂² + 2z₃² – 2z₁z₂ – 2z₂z₃ – 2z₃z₄ = 0
⇒ (z₁ – z₂)² + (z₂ – z₃)² + (z₃ – z₁)² = 0
Put a = z₁ – z₂, b = z₂ – z₃, c = z₃ – z₁
Then a + b + c = 0 and a² + b² + c² = 0 As in Q2 we can show that |a| = |b| = |c|
⇒ |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|

Question 26.
If |a| < |c| show that there are complex numbers z satisfying |z – a| = |z + a| = 2|c|
Solution:
Let z = x + iy
∴ |z – a| + |z + a| = 2c
or, |x + iy – a| + |x + iy + a| = 2c

Question 27.
Solve \(\frac{(1-i) x+3 i}{2+i}+\frac{(3+2 i) y+i}{2-i}=-i\) where x, y, ∈ R.
Solution:

Question 28.
If (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ, then prove that p₀ + p₃ + p₆ + …..+ 3^n-1
Solution:
Given, (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ
putting x = ω we get

Question 29.
Find the region on the Argand plane on which z satisfying
[Hint Arg (x + iy) =\(\frac{\pi}{2}\) = 0, y>0]
(i) 1 < |z – 2i| < 3
Solution:
Let z = x + iy
The given inequality is
1 < |x + i(y – 2)| < 3
⇒ \(1<\sqrt{x^2+(y-2)^2}<3\)

(ii) arg \(\left(\frac{z}{z+i}\right)=\frac{\pi}{2}\)
Solution:

As all are +ve we have
1 < x² + (y – 2)² < 9
x² + (y – 2)² < 9 is the region inside the circle with center (0, 2) and radius 1.
x² + (y – 2)² > 1 is the region outside the circle with center (0, 2) and radius.
∴ 1 < |z – 2i| < 3 is the region between two concentric circles with center (0, 2) and radius 1 and 3 which is shows below.

Odisha State Board BSE Odisha 7th Class English Solutions Test-2 Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Test-2

BSE Odisha 7th Class English Test-2 Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Your teacher will dictate ten words. Listen to him/her and write. [10]

Answer:
drowsy lullaby leap
mob stones heavier
weaver color
already float

2. Given below are some words. Your teacher will read aloud ten of them. Tick those s/he reads aloud. [10]
the study, family, paddy, wood, quack, worship, bow, arrow, collector, respect, grow, float, change, ride, spring.
Answer:
[Students tick the words that their teacher read aloud]

3. Your teacher will read aloud a paragraph. Listen to him/her and fill in the gaps. [10]
One day the ___________ man took ___________to the poor people living in ___________down the hills. He had in his mind to show his son how ___________he was in contrast to the ___________. He thought this would work like ______________.
Answer:
One day the rich man took his, son to the poor people living in small huts down the hills. He had in his mind to show his son how rich he was in contrast to the poor farmers. He thought this would work like medicine for his son’s sadness.

4. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]

Answer:
ବିରାକିଶୋର ଦାସ |  – Birakishore Das
ନିଲମାନି ମାର୍ଗ        – Nilamani Routray
ମଧୁସୂଦନ ଦାସ       – Madhusudan Das
ଗୋପାଳକୃଷ୍ଣ ଗୋଖଲେ | – Gopalkrushna Gokhle
ତିଲ୍କା ମାଜି             – Tilka Majhi
ଲକ୍ଷ୍ମଣ ନାୟକ |       – Laxman Nayak
ସୁରେନ୍ଦ୍ର ସାଇ          – Surendra Sai
ପଦ୍ମଲୋଚନ ବେହେରା | – Padmalochan Behera
ଲିଙ୍ଗରାଜ ନନ୍ଦା |      – Lingaraj Nanda
ରାଧନାଥ ରୟ        – Radhanath Roy

5. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]

Answer:
ବରିପଡା |  – Baripada
କେନ୍ଦୁଝର   – Keonjhar
କାକଟପୁର  – Kakatpur
ମୁମ୍ବାଇ |     – Mumbai
ରାୟାଗଡା | – Raygada
ବୋରିଗୁମା  – Boriguma
ସମ୍ବଲପୁର   – Sambalpur
କୋଲକାତା – Kolkata
ବେଙ୍ଗାଲୁରୁ  – Bengaluru
ଆହ୍ଲାବାଦ  – Allahabad

6. Match the pair of words that sound alike at the end.

Answer:

7. Order the letters to make meaningful words.
melca, mio, tresof, bitrba, cajkla, reed, veirr
Answer:
melca — camel
Inio — lion
tresof — forest
bitrba — rabbit
cajkla — jackal
reed — deer
veirr — river

8. Read the paragraph and answer the questions in complete sentences.[20]
There was a deep forest. In that deep forest lived a rabbit. One moonlit night the rabbit was walking happily near that forest. On his way, he came across a well. He looked into the well, and to his surprise, saw a big white ball. The white ball was floating on the water. The ball was nothing, but the reflection of the moon. But he thought it was a big cake.

(i) Where did the rabbit live?
Answer:
The rabbit lived in a deep forest.

(ii) Where was the rabbit walking?
Answer:
The rabbit was walking near the forest.

(iii) When was he walking near the forest?
Answer:
He was walking near the forest on a moonlit night.

(iv) What did he come across on his way?
Answer:
He came across a well on his way.

(v) What did he see in the well?
Answer:
He saw a big white ball floating on the water of the well.

(vi) What did he think?
Answer:
He thought that the big white ball was a big cake.

(vii) What was it?
Answer:
It was the reflection of the moon.

(viii)Was the rabbit clever? How do you know?
Answer:
No, the rabbit was not clever. Because he thought the reflection of the moon on the water was a big cake.

(ix) In the last line ‘it’ is used for
Answer:
In the last line ‘it’ is used for the reflection of the moon.

(x) What looked like a cake?
Answer:
The reflection of the moon on the water of the well looked like a cake.

9. Read the following poem and answer the questions in complete sentences. [14]

White sheep, white sheep
On a blue hill,
When the winds stop
You all stand still.

You all run away
When the winds blow.
White sheep, white sheep,
Where do you go?

Question 1.
What is the poem about?
Answer:
The poem is about clouds floating in the sky.

Question 2.
How many stanzas are there in this poem?
Answer:
There are two stanzas in this poem.

Question 3.
Where are the white sheep?
Answer:
The white sheep are on the blue hill.

Question 4.
When do they stand still?
Answer:
When the winds stop, they stand still.

Question 5.
When do they run away?
Answer:
When the winds blow, they run away.

Question 6.
Who is asking “Where do you go ?”
Answer:
The poet is asking “Where do you go ?”

Question 7.
Who is compared to the white sheep?
Answer:
The white clouds floating in the blue sky are compared to white sheep.