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Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ଠିକ୍ ଉତ୍ତର ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(i) ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ ଥିବା △ABC ରେ ∠ABC = 90°
ଏବଂ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\), m∠ABD = ………
[m∠BAD, m∠DBC, m∠DCB, 2m∠BAD]

Solution:
m∠DCB

(ii) ଉପ6ର।କ୍ତ ରିତ୍ର6ର ଥିବା △ABC ରେ ∠ABC ସମ6କାଶ ଏବଂ BD AC ହେଲେ,
(a) AB2 = AD × ……….. [BC, CD, AC, BD]
(b) BC2 = AC × ……….. [DC, AD, BD, AB]
(c) BD2 = DC × ………. [AC, BC, AB, AD]
Solution:
(a) AC
(b) DC
(c) AD

Question 2.
ବମ୍ନ ଟିତ୍ର6ର ଥିର୍। △PQR ର m∠PQR = 90° ଏର୍ବ \(\overline{\mathbf{QM}}\) ⊥ \(\overline{\mathbf{PR}}\)
(i) QM = 12 6ସ.ମି. ଏବଂ PM = 6 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(ii) PQ = 6 6ସ.ମି. ଏବଂ PM = 3 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(iii) QR = 12 6ସ.ମି. ଏବଂ MR = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(iv) PQ = 12 6ସ.ମି. ଓ RM = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(v) PQ = 8 6ସ.ମି. ଓ QR = 15 6ସ.ମି. 6ଦ୍ର6କ , QM ଓ MR ନିଣ୍ଟଯ କର |

Solution:
(i) QM² = PM × RM ⇒ 12² = 6 × RM
⇒ RM = \(\frac{12^2}{6}\) 6ସ.ମି. = 24 6ସ.ମି.
PR = PM + RM = 6 6ସ.ମି. + 24 6ସ.ମି. = 30 6ସ.ମି.

(ii) PQ² = PM × PR
⇒ PR = \(\frac{\mathrm{PQ}^2}{\mathrm{PM}}\) = \(\frac{6^2}{3}\) 6ସ.ମି. = 12 6ସ.ମି.

(iii) QR² = MR × PR
⇒ PR = \(\frac{\mathrm{QR}^2}{\mathrm{MR}}\) = \(\frac{12^2}{9}\) 6ସ.ମି. = 16 6ସ.ମି.
PM = PR – MR = 16 6ସ.ମି. – 9 6ସ.ମି. = 7 6ସ.ମି. |

(iv) ମ6ନଜର PM = x 6ସ.ମି.
∴ PR = PM + MR = (x + 7) 6ସ.ମି. |
PQ² = PM . PR ⇒ 12² = x(x + 7)
⇒ x² + 7x – 144 = 0 ⇒ x² + 16x – 9x – 144 = 0
⇒ x(x + 16) -9(x + 16) = 0 ⇒ (x – 9) (x + 16) = 0
⇒ x = 9 ବ।, x = -16 (ଅସମ୍ମତ)
∴ PM = x 6ସ.ମି. = 9 6ସ.ମି.

(v) PR = \(\sqrt{\mathrm{PQ}^2+\mathrm{QR}^2}\) = \(\sqrt{8^2+15^2}\) = \(\sqrt{64+225}\) = \(\sqrt{289}\) = 17 6ସ.ମି.
QR² = MR × PR
⇒ MR = \(\frac{\mathrm{QR}^2}{\mathrm{PR}}\) = \(\frac{15 \times 15}{17}\) = \(\frac { 225 }{ 17 }\) = 13\(\frac { 4 }{ 17 }\) 6ସ.ମି.
QM² = PM × MR = (PR – MR) × MR
= (17 – \(\frac { 225 }{ 17 }\)) × \(\frac { 225 }{ 17 }\) = \(\frac { 64 }{ 17 }\) × \(\frac { 225 }{ 17 }\) ଦଙ୍ଗ6ସ.ମି.
∴ QM = \(\sqrt{\frac{64 \times 225}{17 \times 17}}\) = \(\frac{8 \times 15}{17}\) = \(\frac { 120 }{ 17 }\) = 7\(\frac { 1 }{ 17 }\) 6ସ.ମି. |

Question 3.
ବମ୍ନ ଟିତ୍ର6ର m∠ABC = m∠DCB = 90°, \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର ଛେଦଦିନ୍ଦୁ O ଏବଂ \(\overline{\mathbf{AC}}\) ⊥ \(\overline{\mathbf{BD}}\) | OC = 6 6ସ.ମି. ଦଙ୍ଗ6ସ.ମି.
(i) BO ନିଶ୍ଚୟ କର;
(ii) OA ନିର୍ଣ୍ଣୟ କର;
(iii) BC ନିର୍ଣ୍ଣୟ କର;
(iv) AB ନିର୍ଣ୍ଣୟ କର; ଏବଂ
(v) CD ନିର୍ଣ୍ଣୟ କର ।

Solution:

Question 4.
△ABC ଭେ ∠ABC ସପ6କାଣ ଏବଂ \(\overline{\mathbf{B D}}\) ⊥ \(\overline{\mathbf{AC}}\) | AD = p ଏକକ ପ୍ରତେକ, ପ୍ତମାଣ କର : BC = \(\frac{q \sqrt{\mathbf{p}^2+q^2}}{p}\)
Solution:

Question 5.
△ABC ଭେ, m∠ABC = 90° ଏବଂ BD ⊥ AC ଦ୍ରେଲେ , ପ୍ରତମାଣ କର ଯେ , AB² : BC² = AD : DC |
Solution:

Question 6.
△ABC ଭେ, ∠ABC ସପ6କାଶ , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD କ୍ତେଲେ , ପ୍ରମାଣ ଦର ଯେ \(\overline{\mathrm{BD}}\) 6ହର୍ଛି ∠ABC ତ ସପଦ୍ଦିଖଣ୍ଡକ |
Solution:

ଦର : △ABC ରେ , ∠B = 90° | \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD
ପ୍ରମ।ଣu : \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡକ |
ପ୍ରମ।ଣ : △BDC ~ △ABC ⇒ BC² = AC . AD ….(i)
କକୁ ଦଣ BC² = AC . AD ….(ii)
(i) ଓ (ii) ରୁ AC . AD = AC . BD ⇒ CD = BD ⇒ m∠DBC = m∠DCB … (iii)
କନ୍ତି △ABD ~ △BCD ⇒ m∠ABD = m∠DCB …(iv)
(iii) ଓ (iv) ର m∠DBC = m∠ABD ⇒ \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ କ | (ପ୍ରମାଣିତ)

Question 7.
ପାଣ୍ଡଣ୍ ଚିତ୍ରରେ ଥ୍ରନା ରହୁକ୍ତିଙ ABCD ରେ m∠ABC = m∠ADC = 90° ଏବଂ AB = AD | କଣ୍ତଦ୍ରଯତ 6ଛଦଦିନ୍ଦୁ M ଦ୍ତେକେ , ପ୍ରମାଣ କର ଯେ AM × MC = DM | (ପ୍ରମେଯ 1.4 ର ସ୍ତ ଯୋଗ କରି ପ୍ରମାଣ କମା | )
Solution:

Question 8.
△ABC ରେ m∠ABC = 90°, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ ପ୍ରମାଣ କର ଯେ AE² : EC² = AD : DC |
Solution:
ଦର : △ABC ରେ m∠ABC = 90° , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ

ପ୍ତାମାଣଧ୍ୟ : \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac { AD }{ DC }\)
ପ୍ତମାଣ : △ABC ରେ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ
⇒ \(\frac { AB }{ BC }\) = \(\frac { AE }{ EC }\) ⇒ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\)
△ABC ରେ m∠ABC = 90° ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\)
⇒ AB² = AD . AC ଓ BC² = CD . AC
∴ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\) = \(\frac{\mathrm{AD} \times \mathrm{AC}}{\mathrm{CD} \times \mathrm{AC}}\) = \(\frac { AD }{ CD }\) (ପ୍ରମାଣିତ)

Question 9.
△ABC ରେ , m∠BAC = 90° ଏବଂ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\) | ପ୍ରମାଣ କର ଯେ △ADC ର ଷ୍ଟ୍ରେତୃଫକ = \(\frac{\mathbf{A B} \times \mathbf{A C} \mathbf{C}^3}{2 \mathbf{B C} C^2}\) |

Solution:
ଦଉ : △ABC 6ର m∠BCA = 90° ଏବଂ \(\overline{\mathrm{AD}}\) | \(\overline{\mathrm{BC}}\) |
ପ୍ର।ମାଣ୍ୟ : △ADC ର ଷ୍ଟେର୍ଫଳ = \(\frac{\mathrm{AB} \times \mathrm{AC}^3}{2 \mathrm{BC}^2}\) |
ପ୍ରମାଣ : ABC ରେ m∠BAC = 90° ଓ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\)
⇒ AC² = CD . BC
△ABC ର ଷ୍ଟେର୍ଫଳ = \(\frac { 1 }{ 2 }\) AC × AB = \(\frac { 1 }{ 2 }\) BC × AD 

Question 10.
△ABC ର m∠ABC ସମକୋଣ,, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ଛେଦକରେ \(\overline{\mathrm{BD}}\) କୁ E ଦିନ୍ଦୁରେ ଛେଦକରେ | ପ୍ତପାଣ କର ଯେ BE² : DE² = AC : AD |
Solution:
ଦତ୍ତ : △ABC ରେ ∠ABC ସମକୋଣ, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ସମଦ୍ବିଖଣ୍ଡକ \(\overline{\mathrm{BD}}\) କୁ E ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) : \(\frac{\mathrm{AC}}{\mathrm{AD}}\)
ପ୍ରମାଣ : △ABC ରେ ∠B ସମକୋଣ ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) |
⇒ AB² = AD × AC
△ABD ରେ ∠BAD ର ସମଦ୍ଦିଖଣ୍ଡକ ରଶ୍ମି \(\overline{\mathrm{BD}}\) କୁ E ରେ ଛେଦକରେ ।
⇒ \(\frac { AB }{ AD }\) = \(\frac { BE }{ DE }\)
∴ \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{AD}^2}\) = \(\frac { AD × AC }{ AD × AD }\) = \(\frac { AC }{ AD }\) (∵ AB² = AD × AC) (ପ୍ରମାଣିତ)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(b)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ସଂଖ୍ୟା ଓ ଏହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି 2 । ସଂଖ୍ୟାଟିକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(ii) ଦୁଇଗୋଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଗୁଣଫଳ 20 । ସଂଖ୍ୟାଦ୍ଵୟ ମଧ୍ୟରୁ ଗୋଟିକୁ y ନେଇ ଆବଶ୍ୟକୀୟ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iii) ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି 18 ଏବଂ ସେମାନଙ୍କର ଗୁଣଫଳ 72 । ଗୋଟିଏ ସଂଖ୍ୟାକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iv) କୌଣସି ସଂଖ୍ୟା, ତାହାର ବର୍ଗ ସମାନ ହେଲେ ସଂଖ୍ୟାଟି ନିର୍ଣ୍ଣୟ କର ।
(v) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି S = \(\frac{n(n+1) }{2}\)। ଯଦି S = 120 ହୁଏ, ତେବେ nର ମୂଲ୍ୟ ନିରୂପଣ କରିବା ପାଇଁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(vi) \(\sqrt{x}\) + x = 6 କୁ ଏକ ଦ୍ବିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(vii) \(\sqrt{x+9}\) + 3 = x କୁ ଏକ ସ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(viii) x – 2\(\sqrt{2}\) – 6 = 0 ସମୀକରଣକୁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
ସମାଧାନ :
(i) ମନେକର ସଂଖ୍ୟାଟି x । ସଂଖ୍ୟାଟିର ବ୍ୟକ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}\) = 2 ⇒ x² + 1 = 2x ⇒ x² – 2x + 1 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ x² – 2x + 1 = 0

(ii) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା y । ଅନ୍ୟଟି = (y + 1)
ପ୍ରଶ୍ନନୁସାରେ, y (y + 1) = 20 ⇒ y² + y – 20 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ y + y – 20 = 0

(iii) ମନେକର ପ୍ରଥମ ସଂଖ୍ୟାଟି x ଓ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 18 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (18 – x) = 72 ⇒ 18x – x² = 72 = x² – 18x + 72 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ, x² – 18x + 72 = 0

(iv) ମନେକର ସଂଖ୍ୟାଟି x।
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟି ତା’ର ବର୍ଗ ସହ ସମାନ ହେବ । ଅର୍ଥାତ୍ x = x²
⇒ x² – x = 0 ⇒ x ( x – 1 ) = 0
⇒ x = 0 ବା x – 1 = 0 ⇒ x= 0 ବା x = 1
∴ ସଂଖ୍ୟାଟି 0 କିମ୍ବା 1 ।

(v) ପ୍ରଶ୍ନନୁସାରେ \(\frac{n(n+1) }{2}\) = 120 ⇒ n² + n = 240 ⇒ n² + n – 240 = 0

(vi) √x + x = 6 ⇒ √x = 6 – x
⇒ (√x)² =(6 – x)² ⇒ x = 36 + x² – 12x
⇒ x² – 12x – x + 36 = 0 ⇒ x² – 13x + 36 = 0

(vii) √(x + 9) + 3 = x ⇒ √(x + 9) = x – 3
⇒ (\(\sqrt{x+9}\))² = (x – 3)² ⇒ x² – 6x + 9 = x + 9
⇒ x² – 6x – x + 9 – 9 = 0 = x² – 7x = 0
ବିକଳ୍ପ ସମାଧାନ :
ମନେକର \(\sqrt{x+9}\) = t
⇒ x + 9 = t² ⇒ x = t² – 9
xର ମାନ ସମୀକରଣରେ ବସାଇଲେ k + 3 = t² – 9
⇒ t – t – 9 – 3 = 0 ⇒ t – t – 12 = 0

(viii) x – 2√2 – 6 = 0 ⇒ x – 6 = 2√2
⇒ (x – 6)² = (2√2)² ⇒ x² – 12x + 36 = 8
⇒ x² – 12x + 36 – 8 = 0 ⇒ x² – 12x + 28 = 0

Question 2.
ନିମ୍ନଲିଖ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ଧନାତ୍ମକ ସଂଖ୍ୟା ତାହାର ବର୍ଗମୂଳ ଅପେକ୍ଷା 12 ଅଧିକ ହେଲେ, ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
(ii) ଗୋଟିଏ ସଂଖ୍ୟା ଏବଂ ତାହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{41}{20}\) ହେଲେ ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
(iii) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମ ଭଗ୍ନସଂଖ୍ୟା ଦ୍ବୟର ଯୋଗଫଳ \(\frac{11}{30}\) ଦ୍ଵୟକୁ ନିରୂପଣ କରିବା ପାଇଁ ଆବଶ୍ୟକ ଦ୍ଵିଘାତ ସମୀକରଣଟି ଗଠନ କରି ସଂଖ୍ୟାଦ୍ଵୟ ନିରୂପଣ କର ।
(iv) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{25}{123}\) ହେଲେ ସଂଖ୍ୟାଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
(v) ଯଦି 51 କୁ ଦୁଇଭାଗ କଲେ ସେମାନଙ୍କର ଗୁଣଫଳ 378 ହୁଏ, ତେବେ ସଂଖ୍ୟା ଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(i) ମନେକର ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² । ଏହାର ବର୍ଗମୂଳ = √x² = x
ପ୍ରଶ୍ନନୁସାରେ, x² = x + 12 = x² -x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0 ⇒ x (x – 4) + 3 (x – 4) = 0
⇒ (x – 4) (x + 3) = 0 ⇒ x – 4 = 0 ବା x + 3 = 0
⇒ x = 4 ବା x = -3 (ଅସମ୍ଭବ)
∴ ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² = 4² = 16

(ii) ମନେକର ସଂଖ୍ୟାଟି x । ତାହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}=\frac{41}{20}\) = \(\frac{x²+1}{x}=\frac{41}{20}\)
⇒ 20x² + 20 = 41x ⇒ 20x² – 41x + 20 = 0
⇒ 20x² – 25x – 16x + 20 = 0 ⇒ 5x (4x – 5) – 4 (4x – 5) = 0
⇒ (4x – 5) (5x – 4)=0 = 4x – 5 = 0 ବା 5x – 4 = 0
⇒ x = \(\frac{5}{4}\)ବା x = \(\frac{4}{5}\)
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{5}{4}\) ହେବ ।
ଅଥବା x = \(\frac{5}{4}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{4}{5}\) ହେବ ।
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ବା \(\frac{5}{4}\) ହେବ ।

(iii) ମନେକର କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1
x ର ବ୍ୟୁତ୍‌କ୍ରମ \(\frac{1}{x}\) ଓ x + 1 ର ବ୍ୟକ୍ରମ \(\frac{1}{x+1}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{11}{30}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{11}{30}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{11}{30}\) ⇒ 11x² + 11x = 60x + 30
⇒ 11x² + 11x – 60x – 30 = 0 ⇒ 11x² – 49x – 30 = 0
⇒ 11x² – 55x + 6x – 30 = 0 ⇒ 11x (x – 5) + 6 (x – 5) = 0 ⇒ (x – 5) (11x + 6) = 0 ⇒ x – 5 = 0 ବା 11x + 6 = 0
⇒ x = 5 ବା x = \(\frac{-6}{11}\) (ଅସମ୍ଭବ)
∴ x = 5 ହେଲେ ଏହାର ପରବର୍ତ୍ତୀ
∴ ନିଶ୍ଚେୟ କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ଵୟ 5 ଓ 6 । କ୍ରମିକ ସଂଖ୍ୟା x + 1 = 6

(iv) ମନେକର କ୍ରମିକ ପୂର୍ବସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1 ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{23}{132}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{23}{132}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{23}{132}\) ⇒ 23x² + 23x = 264x + 132
⇒ 23x² + 23x – 264x – 132 = 0 ⇒ 23x² – 241x – 132 = 0
⇒23x² – 253x + 12x – 132 = 0 ⇒ 23x (x – 11) + 12 (x – 11) = 0
⇒ (x – 11) (23x + 12) = 0 ⇒ x – 11 = 0 ବା 23x + 12 = 0
⇒ x = 11 ବା x = \(– \frac{12}{23}\) (ଅସମ୍ଭବ) ⇒ x + 1 = 11 + 1 = 12
∴ ପୂର୍ଣ୍ଣସଂଖ୍ୟାଦ୍ଵୟ 11 ଓ 12 ।

(v) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା x ଓ ଅନ୍ୟଟି 51 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (51 – x) = 378 ⇒ 51x – x² = 378
⇒ x – 51x + 378 = 0 ⇒ x² – 42x – 9x + 378 = 0
⇒x (x-42) – 9 (x – 42) = 0 ⇒ (x – 42) (x – 9) = 0
⇒ x – 42 = 0 ବା x – 9 = 0 ⇒ x = 42 ବା x = 9
∴ ଯଦି x = 9 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 42 ହେବ ।
∴ ସଂଖ୍ୟାଦ୍ୱୟ 9 ଓ 42 ସେହିପରି x = 42 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 9 ହେବ ।

Question 3.
ଏକ ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କ ଦ୍ଵୟର ଗୁଣଫଳର 3 ଗୁଣ । ଏକକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଟି ଦଶକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଠାରୁ 2 ବୃହତ୍ତର । ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ x ଓ ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ y ।
∴ ସଂଖ୍ୟାଟି = 10y + x
ପ୍ରଶ୍ନନୁସାରେ 10y + x = 3xy …… (i) ଏବଂ x = y + 2 ……. (ii)
ସମୀକରଣ (ii)ରୁ x ର ମାନ ସମୀକରଣ (i) ରେ ବସାଇଲେ, 10y + y + 2 = 3y (y + 2)
⇒ 11y + 2 = 3y² + 6y ⇒ 3y² + 6y – 11y – 2 = 0
⇒ 3y² – 5y – 2 = 0 ⇒ 3y² – 6y+ y – 2 = 0
⇒ 3y (y – 2) + 1 (y – 2) = 0 ⇒ (y – 2) (3y + 1) = 0
⇒ y – 2 = 0 ବା 3y + 1 = 0 = y = 2 ବା y = (ଅସମ୍ଭବ)
∴ x = y + 2 = 2 + 2 = 4
ସଂଖ୍ୟାଟି = 10y + x = 10 × 2 + 4 = 24

Question 4.
ଗୋଟିଏ ପରିବାରରେ ଆଲ୍‌ଫାର ବୟସ, ବିଟା ଓ ଗାମାର ବୟସ୍‌ର ଗୁଣଫଳ ସହ ସମାନ । ଯଦି ବିଟା, ଗାମାଠାରୁ 1 ବର୍ଷ ବଡ଼ ହୁଏ ଏବଂ ଆଲ୍‌ଫାର ବୟସ 42 ହୁଏ, ତେବେ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ କେତେ ହେବ ?
ସମାଧାନ :
ମନେକର ପରିବାରରେ ଗାମାର ବୟସ = x ବର୍ଷ
∴ ବିଟାର ବୟସ = (x + 1) ବର୍ଷ । ଆଲ୍‌ଫାର ବୟସ = 42।
ପ୍ରଶ୍ନନୁସାରେ, x(x + 1) = 42 ⇒ x² + x = 42
⇒ x² + x – 42 = 0 ⇒ x² + 7x – 6x – 42 = 0
⇒ x(x + 7) – 6(x + 7) = 0 ⇒ (x + 7) (x – 6) = 0
⇒ x + 7 = 0 ବା x – 6 = 0 = x = -7 ବା x = 6
x = – 7 ବୟସ ଋଣାତ୍ମକ ହେବନାହିଁ ତେଣୁ ଏହା ଅସମ୍ଭବ। ତେବେ ଏଠାରେ x = 6 ଗ୍ରହଣୀୟ ।
∴ ଗାମାର ବୟସ = x ବର୍ଷ = 6 ବର୍ଷ ଏବଂ ବିଟାର ବୟସ = 6 + 1 = 7 ବର୍ଷ ।
∴ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ ହେବ = 7 + 5 = 12 ।

Question 5.
କୌଣସି ଏକ ଅରଣ୍ୟରେ ବାସ କରୁଥିବା ମର୍କଟମାନଙ୍କ ମଧ୍ୟରୁ ସେମାନଙ୍କ ସଂଖ୍ୟାର ଏକ ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରୀଡ଼ାରତ ଏବଂ ଅବଶିଷ୍ଟ ବାରଟି ମର୍କଟ ଏକ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ । ଅରଣ୍ୟରେ ସମ୍ଭବତଃ କେତେ ମର୍କଟ ଥିଲେ ?
ସମାଧାନ :
ମନେକର ଅରଣ୍ୟରେ ସମ୍ଭବତଃ xଟି ମର୍କଟ ଥିଲେ ।
ସେମାନଙ୍କ ଏକ- ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରିଡ଼ାରତ ଥିଲେ । କ୍ରୀଡ଼ାରତ ଥିବା ମର୍କଟଙ୍କ ସଂଖ୍ୟା = (\(\frac{x}{8}\))² = \(\frac{x²}{64}\)
ଅବଶିଷ୍ଟ 12 ଟି ମର୍କଟ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x²}{64}+12\) = x = \(\frac{x²+768}{64}=x\)
⇒ x² + 768x = 64 x ⇒ x² – 64x + 768 = 0
⇒ x² -48x – 16x + 768 = 0 ⇒ x (x – 48) – 16 (x – 48) = 0
⇒ (x – 48) (x – 16)=0 = x – 48 = 0 ବା x – 16 = 0
⇒ x = 48 ବା x = 16
∴ ଅରଣ୍ୟରେ ସମ୍ଭବତଃ 48 କିମ୍ବା 16ଟି ମର୍କଟ ଥିଲେ ।

Question 6.
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ 30 ବ. ସେ.ମି. । ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ହେଲେ, ଭୂମିର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 30 ବର୍ଗ ସେ.ମି.
ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ।
ମନେକର ଭୂମିର ଦୈର୍ଘ୍ୟ X ସେ.ମି. । ଉଚ୍ଚତା = x + 7 ସେ.ମି.
∴ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) ଭୂମି × ଉଚ୍ଚତା = \(\frac{1}{2}\)x(x+7)
⇒ 30 = \(\frac{1}{2}\)(x² + 7x) ⇒ x² + 7x = 60
⇒ x²+7x-60 = 0 ⇒ x² + 12x – 5x – 60 = 0
⇒ x (x + 12) – 5 (x + 12) = 0 ⇒ (x + 12) (x – 5)= 0
⇒ x + 12 = 0 ବା x – 5 = 0 = x=- 12 (ଅସମ୍ଭବ) ବା x = 5
∴ ତ୍ରିଭୁଜର ଭୂମିର ଦୈର୍ଘ୍ୟ 5 ସେ.ମି. ।

Question 7.
ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜର ସମକୋଣର ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଓ (3x – 1) ସେ.ମି. ଓ କ୍ଷେତ୍ରଫଳ 60 ବର୍ଗ ସେ.ମି. । ତେବେ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ସମକୋଣ ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଏବଂ (3x – 1) ସେ.ମି. ।
ସମକୋଣୀ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 60 ବର୍ଗ ସେ.ମି. (ଦତ୍ତ)
⇒ \(\frac{1}{2}\) × ସମକୋଣ ସଂଲଗ୍ନ ଦୈର୍ଘ୍ୟର ବାହୁଦ୍ୱୟର ଗଣଫଳ = 60
⇒ \(\frac{1}{2}\) . 5x . (3x – 1) = 60 ⇒ x(3x – 1) = 60 × \(\frac{2}{5}\)
⇒ 3x² – x = 24 ⇒ 3x² – x – 24 = 0
ଏଠାରେ a = 3, b = -1, c = – 24

∴ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x = 5 × 3 = 15 ସେ.ମି. ।
ଓ 3x – 1 = 3 × 3 – 1 = 8 ସେ.ମି. ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ : 3x² – x – 24 = 0 ର ସମାଧାନ ଉତ୍ପାଦକୀକରଣ ଦ୍ଵାରା ସମ୍ଭବ ।

Question 8.
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଡ ସଂଖ୍ୟା \(\frac{n}{2}\)n(n-3)। ଯଦି ବହୁଭୁଜର 54ଟି କର୍ଣ୍ଣ ରହିବ, ତେବେ ବହୁଭୁଜର ବାହୁର ସଂଖ୍ୟା କେତେ ?
ସମାଧାନ :
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଣ୍ଣ ସଂଖ୍ୟା = \(\frac{n}{2}\)n(n-3)
ଏହାର କର୍ଣ୍ଣ ସଂଖ୍ୟା 54 1
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{n}{2}\)n(n-3) = 54 ⇒ n² – 3n = 108
→n² – 3n-108 = 0 ⇒ n² – 12n + 9n – 108 = 0
→ n (n – 12) + 9 (n – 12) = 0 → (n – 12) (n + 9) = 0
→ n – 12 = 0 ବା n + 9 = 0
⇒ n =12 ବା n = -9 (ଅସମ୍ଭବ)
∴ ବହୁଭୁଜର ବାହୁ ସଂଖ୍ୟା 12 ।

Question 9.
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବ.ମି. ଏବଂ ପରିସୀମାଦ୍ବୟର ଅନ୍ତର 24 ମି. ହେଲେ ପ୍ରତ୍ୟେକର ବାହୁର ଦୈର୍ଘ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବର୍ଗ ମି. । ପରିସୀମାଦ୍ବୟର ଅନ୍ତର = 24 ମିଟର
ମନେକର ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ a ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ b ମି. ।
ପ୍ରଶ୍ବାନୁସାରେ, a² + b² = 468 ଏବଂ 4a – 4b = 24
⇒ 4(a – b) = 24 ⇒ a – b = \(\frac{24}{4}\) = 6 ………(i)
2ab = (a² + b²) – (a – b)² = 468 – 36 = 432
(a + b)² = a² + b² + 2ab = 468 + 432 = 900
⇒ a + b = 30 ……..(ii)
ସମୀକରଣ (i) ଓ (ii)କୁ ଯୋଗକଲେ a – b + a + b = 6 + 30 = 36
⇒ 2a = 36 ⇒ a = = 18 ମି., b = 30 a 30 18 = 12 ମି.
∴ ବହୁଭୁଜର ବାହୁ ଦୈର୍ଘ୍ୟ 18 ମି. ଓ 12 ମି. ।

ବିକଳ୍ପ ପ୍ରଣାଳୀ : a² + b² = 468 ଏବଂ a – b = 6
a – b = 6 = b = a – 6
∴ a² + (a – b)² = 468⇒ 2a² – 12a + 36 = 468
2a² – 12a – 432 = 0 = a² – 6a – 216 = 0
⇒ (a – 18) (a + 12) = 0 ⇒ a = 18 ବା -12
ଏଠାରେ a = 18 ଗ୍ରହଣୀୟ ଏବଂ a = -12 ଅଗ୍ରହଣୀୟ
ଯଦି a = 18 ହୁଏ, ଦେବେ b = 18 – 6 = 12
∴ ବର୍ଗକ୍ଷେତ୍ର ବାହୁର ଦୈର୍ଘ୍ୟ 18 ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ 12 ମି. ।

Question 10.
ଜଣେ ବ୍ୟକ୍ତି ତାଙ୍କ ଚାଲିବାର ବେଗକୁ ଯଦି ଘଣ୍ଟା ପ୍ରତି 1 କି.ମି. ବୃଦ୍ଧି କରେ, ତେବେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବା ପାଇଁ 10 ମିନିଟ୍ କମ୍ ସମୟ ନେଇଥା’ନ୍ତା । ତେବେ ବ୍ୟକ୍ତିର ଚାଲିବାର ଘଣ୍ଟା ପ୍ରତି ବେଗ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ = x କି.ମି. ।
ଏହି ବେଗରେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାପାଇଁ ସମୟ ଲାଗିଲା = \(\frac{2}{x}\) ଘଣ୍ଟା ।
ଯଦି ତାଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 1 କି.ମି. ବୃଦ୍ଧି ହୁଏ, ତେବେ ତାଙ୍କର ବେଗ = (x+1) କି.ମି. ।
ବେଗ ବୃଦ୍ଧି ହେଲେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାକୁ ସମୟ ଲାଗିବ = \(\frac{2}{x+1}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{2}{x}-\frac{2}{x+1}=\frac{10}{60}\) (∵ 10 ମିନିଟ = \(\frac{10}{60}\) ଘଣ୍ଟା ।)

x = – 4 ଗ୍ରହଣୀୟ ନୁହେଁ (କାରଣ ବେଗ ଋଣାତ୍ମକ ହେବା ଅସମ୍ଭବ) ।
∴ ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 3 କି.ମି. ।

Question 11.
ଏକ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି. ପ୍ରତି ଘଣ୍ଟା । ଏହା ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଅତିକ୍ରମ କରି ପୁନଶ୍ଚ (ଅନୁକୂଳରେ) ଫେରି ଆସିବାକୁ 4 ଘଣ୍ଟା 30 ମି. ସମୟ ନେଲା । ତେବେ ସ୍ରୋତର ଘଣ୍ଟା ପ୍ରତି ବେଗ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଗୋଟିଏ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି.|ଘଣ୍ଟା ।
ମନେକର ସ୍ରୋତର ବେଗ x କି.ମି./ଘଣ୍ଟା ।
ସ୍ରୋତର ଅନୁକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 + x) କି.ମି. ।
ସ୍ରୋତର ପ୍ରତିକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 – x) କି.ମି. ।
ସ୍ରୋତର ଅନୁକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15+x}\) ଘଣ୍ଟା । ଏବଂ
ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15-x}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନ ନୁସାରେ, \(\frac{30}{15+x}+\frac{30}{15-x}=4\frac{1}{2}\)
⇒ 30(\(\frac{15-x+15+x}{(15+x)(15-x)}\)) = \(\frac{9}{2}\)
⇒ \(\frac{30×30}{225-x²}=\frac{9}{2}\) ⇒ 1800 = 9 (225 – x²)
⇒ 225 – x² = \(\frac{1800}{9}\) = 200 ⇒ 225 – 200 = x² ⇒ x² = 25 ⇒ x= √25 = 5
∴ ସ୍ରୋତର ବେଗ 5 କି.ମି./ ଘଣ୍ଟା ।

Question 12.
ଗୋଟିଏ ଶ୍ରେଣୀର ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟକ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବଣ୍ଟାଗଲା । ଯଦି 25 ଜଣ ଛାତ୍ର ଅଧ‌ିକ ହୋଇଥା’ନ୍ତେ, ତେବେ ପ୍ରତ୍ୟେକ 0.50 ଟଙ୍କା ଲେଖାଏଁ କମ୍ ପାଇଥା’ନ୍ତେ । ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା ସ୍ଥିର କର ।
ମନେକର ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା x ଜଣ ।
x ଜଣ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବାଣ୍ଟିଲେ ଜଣକା ପାଇବେ = \(\frac{250}{x}\) ଟଙ୍କା
25 ଜଣ ଛାତ୍ର ଅଧ୍ଵ ହେଲେ ଛାତ୍ର ସଂଖ୍ୟା = (x + 25) ଜଣ ।
25 ଜଣ ଛାତ୍ର ଅଧ୍ୟା ହେଲେ ଜଣକା ପାଇବେ = (\(\frac{250}{x}\) – 0.50) ଟଙ୍କା
ପ୍ରଶ୍ନନୁସାରେ, (x + 25)(\(\frac{250}{x}\) – 0.50) = 250
⇒ (x + 25)(500 – x) = 500x ⇒ 500x – x² + 12500 – 25x = 500x
⇒ x² + 25x – 12500 = 0

x = – 125 ଗ୍ରହଣୀୟ ନୁହେଁ, କାରଣ ଛାତ୍ର ସଂଖ୍ୟା ଋଣାତ୍ମକ ହେବନାହିଁ ।
⇒ x = 100 ∴ ଶ୍ରେଣୀର ଛାତ୍ରସଂଖ୍ୟା 100 ।

Question 13.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମିଟର ଅଧୀକ । କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 240 ବର୍ଗ ମିଟର ହେଲେ କ୍ଷେତ୍ରଟିର ପରିସୀମା କେତେ ?
ସମାଧାନ :
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମି. ଅଧ୍ବକ ।
ମନେକର ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = x ମି., ଦୈର୍ଘ୍ୟ = (x + 8) ମିଟର,
ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = (x + 8) x ବର୍ଗ ମିଟର ।
ପ୍ରଶ୍ନନୁସାରେ, (x + 8) x = 240
⇒x² + 8x = 240 ⇒ x² + 8x – 240 = 0
⇒ x² + 20x – 12x – 240 = 0
⇒ x(x + 20) – 12(x + 20) = 0
⇒ (x + 20) (x – 12) = 0
⇒ x + 20 = 0 ବା x – 12 = 0
⇒x = -20 (ଅସମ୍ଭବ) ବା x = 12
∴ ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = 12 ମି., ଦୈର୍ଘ୍ୟ = x + 8 = 12 + 8 = 20 ମି.
ଆୟତକ୍ଷେତ୍ରର ପରିସୀମା = 2 (ଦୈର୍ଘ୍ୟ + ପ୍ରସ୍ଥ) = 2 (20 + 12) = 2 × 32 = 64 ମିଟର ।

Question 14.
ଏକ ରେଳଗାଡ଼ି 300 କି.ମି. ଦୀର୍ଘ ଯାତ୍ରା ପଥରେ ସମାନ ବେଗରେ ଗତି କରୁଥିଲା । ଯଦି ଗାଡ଼ିର ବେଗ ଘଣ୍ଟାପ୍ରତି 5 କି.ମି. ଅଧ୍ଵ ହୋଇଥା’ନ୍ତା, ତେବେ ଗାଡ଼ିଟି ନିର୍ଦ୍ଦିଷ୍ଟ ସମୟର 2 ଘଣ୍ଟା ପୂର୍ବରୁ ଯଥା ସ୍ଥାନରେ ପହଞ୍ଚିନ୍ତା । ତେବେ ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ x କି.ମି. ।
ଏହି ବେଗରେ ଗାଡ଼ିକୁ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x}\) ଘଣ୍ଟା ।
ମାତ୍ର ଘଣ୍ଟାକୁ x + 5 କି.ମି. ବେଗରେ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x+5}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{300}{x}-\frac{300}{x+5}=2\) ⇒ 300(\(\frac{1}{x}-\frac{1}{x+5}\) = 2

Question 15.
ଏକ ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ 25 ମିଟର, ପ୍ରସ୍ଥ 16 ମିଟର ଓ ପଡ଼ିଆର ଚତୁଃପାର୍ଶ୍ବରେ ସମାନ ଚଉଡ଼ାର ଏକ ରାସ୍ତା ଅଛି । ଯଦି ଚତୁଃପାର୍ଶ୍ଵରେ ଥ‌ିବା ରାସ୍ତାର କ୍ଷେତ୍ରଫଳ 230 ବର୍ଗମିଟର ହୁଏ, ତେବେ ରାସ୍ତାର ଚଉଡ଼ା ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର ରାସ୍ତାର ଚଉଡ଼ା = x ମିଟର
ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = 25 ମିଟର, ପ୍ରସ୍ଥ = 16 ମିଟର ।
.. ବାହାର ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = (25 + 2x) ମି. ଓ ପ୍ରସ୍ଥ = (16 + 2x) ମି.
ପ୍ରଶ୍ନନୁସାରେ, (25 + 2x) (16 + 2x) – 25 × 16 = 230
⇒ 400 + 50x + 32x + 4x² – 400 = 230
⇒ 4x² + 82x – 230 = 0 ⇒ 2x² + 41x – 115=0
ଏଠାରେ a = 2, b = 41, c = -115

∴ ରାସ୍ତାର ଚଉଡ଼ା 2.5 ମିଟର ।
ବି.ଦ୍ର. : ଉତ୍ପାଦକୀକରଣଦ୍ୱାରା 2x² + 41x – 115 = 0 ର ସମାଧାନ ମଧ୍ୟ ସମ୍ଭବ ।

Question 16.
କେତେକ ଛାତ୍ରଛାତ୍ରୀ ଏକ ବଣଭୋଜିର ଆୟୋଜନ କଲେ । ଖାଦ୍ୟ ଅଟକଳ (Budget) 480 ଟଙ୍କା ଥିଲା । ସେମାନଙ୍କ ମଧ୍ୟରୁ 8 ଜଣ ବଣଭୋଜିକୁ ଗଲେ ନାହିଁ; ଯାହା ଫଳରେ ଖାଦ୍ୟ ବାବଦ ଖର୍ଚ୍ଚ ଜଣପିଛା 10 ଟଙ୍କା ବଢ଼ିଗଲା । ତେବେ କେତେଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିକୁ ଯାଇଥିଲେ ?
ସମାଧାନ :
ମନେକର x ଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିର ଆୟୋଜନ କରିଥିଲେ ।
ଖାଦ୍ୟ ଅଟକଳ = 480 ଟଙ୍କା ଥିଲା ।
ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଥାନ୍ତା = \(\frac{480}{x}\)
8 ଜଣ ବଣଭୋଜିକୁ ନଯିବାରୁ ପିଲା ସଂଖ୍ୟା = x – 8
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8
x – 8 ଜଣ ପିଲାଙ୍କ ପାଇଁ ଖର୍ଚ୍ଚ 480 ଟଙ୍କା ହେଲେ, ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଲା = \(\frac{480}{x-8}\) ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{480}{x-8}-\frac{480}{x}=10\) ⇒ 480(\(\frac{x-x+8}{(x-8)x}\)) = 10
⇒ \(\frac{8}{x²-8x}=\frac{1}{48}\) ⇒ x² – 8x = 384
⇒ x² -8x – 384 = 0 ⇒ x² – 24x + 16x – 384 = 0
⇒ x (x – 24) + 16 (x – 24) = 0 ⇒ (x – 24) (x + 16) = 0 ⇒x-24=0 Ql x + 16 = 0
⇒x= 24 Q1 x = – 16
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8 = 24 – 8 = 16 ଜଣ।

Question 17.
ସମାଧାନ ଜର :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
(v) (3x² – 8)² – 23 (3x² – 8) + 76 = 0
(vi) 5 (5^x + 5^-x) = 26
(vii) (x² – 2x)² – 4 (x² – 2x) + 3 = 0
(viii) x^-4 – 5x^-2 + 4 = 0
(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)
(x) \(\frac{3}{√2x}-\frac{√2x}{5}=5 \frac{9}{10}\)
(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\) (x ≠ 0, x ≠ – 1)
(xii) x (2x + 1) (x – 2) (2x – 3) = 63
(xiii) \(\frac{x-3}{x+3}+\frac{x+3}{x-3}=6 \frac{6}{7}\) (x ≠ – 3, 3)
(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)
(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
(xvi) \(\sqrt{2x+9}\) + x = 13
(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4
ସମାଧାନ :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
= (x+1) (x+4) (x + 2) (x + 3) = 120 ⇒ (x² + 5x + 4) (x² + 5x + 6) = 120
ମନେକର x² + 5x = p
ପ୍ରଭେ ସମୀକରଣରଟି (p + 4) (p + 6) = 120
⇒ p² + 10p+ 24 = 120 ⇒ p² + 10p + 24 – 120 = 0
⇒p² + 10p – 96 = 0 ⇒ p² + 16p – 6p – 96 = 0
⇒p(p + 16) – 6(p + 16) = 0 ⇒ (p + 16) (p – 6) = 0
⇒p + 16 = 0 କିମ୍ବା p – 6 = 0 ⇒ p = -16 କିମ୍ବା p=6
p ପରିବର୍ତ୍ତେ x + 5x ନେଲେ x² + 5x = – 16
⇒ x² + 5x + 16 = 0 ଏଠାରେ D = b² – 4ac = 25 – 64 = -39
⇒ D < 0 ବୀଜଦ୍ଵୟ ଅବାସ୍ତବ, ଗ୍ରହଣୀୟ ନୁହେଁ ।
ପୁନଶ୍ଚ x² + 5x = 6
⇒ x²+5x – 6 = 0, ଏଠାରେ a = 1, b = 5, c = – 6

∴ ନିର୍ଦେୟ ସମାଧାନ – 6 ଓ 1 ।

(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
ମନେକର \(\sqrt{\frac{3}{x}}\) = p = \(\sqrt{\frac{x}{3}}=\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣରଟି 5p + \(\frac{7}{p}\) = \(\frac{68}{3}\) ⇒ \(\frac{5p²+7}{p}=\frac{68}{3}\)
⇒ 15p² + 21 = 68p ⇒ 15p² – 68p + 21 = 0, ଏଠାରେ a = 15, b = -68, c = 21

ବି.ଦ୍ର. : x ର ମାନ \(\frac{25}{147}\) ଦେଇ ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି କି ନାହିଁ ପରୀକ୍ଷା କରି ଦେଖ । ଯଦି ନ ହେଉଥାଏ ତେବେ ସମାଧାନ କେବଳ 27 ହେବ । ଏଠାରେ \(\frac{25}{147}\) ଅଗ୍ରହଣୀୟ ।

(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
⇒ \(\frac{48x²+5-32x}{16x}=0\) ⇒ 48x² + 5 – 32x = 0

(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
ମନେକର (2x+1)= p ଦଉ ସମୀକରଣ p⁴ – 6p² + 8 = 0 => p⁴ – 4p² – 2p² + 8 = 0
⇒ p²(p² – 4) – 2(p² – 4) = 0 ⇒ (p² – 4)(p² – 2) = 0
⇒ p² – 4 = 0 ବା p² – 2 = 0 ⇒ p² = 4 ବା p² = 2 ⇒ p = ±2 ବା p = ±√2
\(\frac{2x+1}{x+1}=2\), x ର ମାନ ନିର୍ଦୟ ଅସମ୍ଭବ |

ବି.ଦ୍ର. :
(\(\frac{2x+1}{x+1}\))² = p ମନେକର
ଦଉ ସମୀକରଣ p² – 6p + 8 = 0 ⇒ p = 4 ବା p = 2
\(\frac{2x+1}{x+1}\) = 4 ⇒ \(\frac{2x+1}{x+1}\) = ±2 ନେଇ x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।
ପୁନଶ୍ଚ (\(\frac{2x+1}{x+1}\))² = 2 = \(\frac{2x+1}{x+1}\) = ±√2 ନେଇ
x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।

(v) ମନେକର 3x² – 8 = p
ପ୍ରଭେ ସମୀକରଣ p² – 23p + 76 = 0 । ଏଠାରେ a = 1, b = 23, c = 76

(vi) ମନେକର 5^x = p
ପ୍ରଭେ ସମୀକରଣ 5(p + \(\frac{1}{p}\)) = 26 ⇒ \(\frac{5(p²+1)}{p}\) = 26
⇒ 5p² + 5 = 26p ⇒ 5p² – 26p + 5 = 0
⇒ 5p² – 25p – p + 5 = 0 ⇒ 5p (p -5) – 1(p – 5) = 0
⇒ (p – 5) (5p – 1) = 0 ⇒ p – 5 = 0 କିମ୍ବା 5p – 1 = 0
⇒p = 5 କିମ୍ବା p = \(\frac{1}{5}\) ⇒5^x = 5¹ କିମ୍ବା 5^x = 5^-1
⇒ x = 1 କିମ୍ବା x = -1
∴ ନିର୍ଦେୟ ସମାଧାନ 1 ଓ -1 ।

(vii) ମନେକର x² – 2x = p
ପ୍ରଭେ ସମୀକରଣ, p² – 4p + 3 = 0
p² – 3p – p + 3 = 0 p(p – 3) – 1(p – 3) = 0
⇒(p – 3) (p – 1) = 0 ⇒ p – 3 = 0 ବା p – 1 = 0
⇒ p = 3 ବା p = 1
x² – 2x = 3 ⇒ x² – 2x – 3 = 0
⇒ x – 3x + x – 3 = 0 ⇒ x (x – 3) + 1 (x – 3) = 0
⇒ (x – 3) (x + 1) = 0 ⇒ x – 3 = 0 ବା x + 1 = 0
⇒ x = 3 ବା x = – 1
ପୁନଶ୍ଚ x² – 2x = 1 ⇒ x² – 2x – 1 = 0 । ଏଠାରେ a = 1, b = – 2, c = – 1 ବା

(viii) ମନେକର x^-2 = p। ଦଉ ସମୀକରଣ, p² – 5p + 4 = 0

(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)

ପୁନଶ୍ଚ x + \(\frac{1}{x}\) = -1 ⇒ \(\frac{x²+1}{x}\) = -1 ⇒ x² + 1 = -x ⇒ x² + x + 1 = 0
ସମୀକରଣ x² + x + 1 = 0 ରେ D < 0 ହେତୁ ବାସ୍ତବ ବୀଜ ସମ୍ଭବ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 2 ଓ \(\frac{1}{2}\) ।

(x) ମନେକର \(\frac{1}{√2x}\) = p; √2x = \(\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣ, 3p – \(\frac{1}{5p}=\frac{59}{10}\)
⇒ \(\frac{15p²-1)}{5p}=\frac{59}{10}\) ⇒ \(\frac{15p²-1)}{p}=\frac{59}{2}\)
⇒ 30p² – 2 = 59p ⇒ 30p² – 59p – 2 = 0
⇒30p² – 60p + p – 2 = 0 ⇒ 30p (p – 2) + 1 (p – 2) = 0
⇒ (p – 2) (30p+1) = 0 ⇒ p – 2 = 0 ବା 30p + 1 = 0
⇒p = 2 ବା p = \(– \frac{1}{30}\)
p = 2 ⇒ \(\frac{1}{√2x}\) = 2 ⇒ \(\frac{1}{2x}\) = 4 ⇒ x = \(\frac{1}{8}\)
p = \(– \frac{1}{30}\) ⇒ \(\frac{1}{√2x}=- \frac{1}{30}\) = \(\frac{1}{2x}=\frac{1}{900}\)
⇒ 2x = 900 ⇒ x = 450
x = 450 ପୁନଶ୍ଚ ସମୀକରଣରେ ବସାଇଲେ ଏହା ସମୀକରଣକୁ ସିଦ୍ଧ କରିବ ନାହିଁ ।
ତେଣୁ ଏହା ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ xର ଏକମାତ୍ର ମୂଲ୍ୟ \(\frac{1}{8}\) ଅଟେ ।

(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\)
⇒ \(\frac{x²+(x+1)²}{x(x+1)}=\frac{34}{15}\) ⇒ \(\frac{x²+x²+2x+1}{x²+x)}=\frac{34}{15}\)
⇒ \(\frac{2x²+2x+1}{x²+x}=\frac{34}{15}\) ⇒ 34 (x² + x) = 15 (2x² + 2x + 1)
⇒ 34x² + 34x = 30x² + 30x + 15
⇒ 34x² – 30×2 + 34x – 30x – 15 = 0 ⇒ 4x² + 4x – 15 = 0
⇒ 4x² + 10x – 6x – 15=0 ⇒ 2x (2x + 5) – 3 (2x + 5) = 0
⇒ (2x + 5) (2x-3)=0 ⇒ x = \(\frac{-5}{2}\) ବା x = \(\frac{3}{2}\)
∴ ନିର୍ଦେୟ ସମାଧାନ \(\frac{3}{2}\) ଓ \(\frac{-5}{2}\) ।

(xii) x (2x + 1) (x – 2) (2x – 3) = 63 ⇒ {x (2x – 3)} {(2x + 1) (x – 2)} = 63
⇒ (2x² – 3x) (2x² -3x – 2) = 63
ମନେକର 2x² – 3x = p
∴ p (p – 2) = 63 ⇒ p² – 2p – 63 = 0
⇒p² – 9p + 7p – 63 = 0 ⇒ p (p – 9) + 7 (p – 9) = 0
(p – 9) (p + 7) =0 p – 9 = 0 p + 7 = 0
⇒ p = 9 ବା p = -7
∴ p = 9 ⇒ 2x² – 3x = 9 ⇒ 2x² – 3x – 9 = 0
⇒ 2x² – 6x + 3x – 9 = 0 ⇒ 2x (x – 3) + 3 (x – 3) = 0
⇒ (x – 3) (2x + 3) = 0 ⇒ x = 3 ବା x = \(\frac{-3}{2}\)
p = -7 ⇒ 2x² – 3x + 7 = 0
ଏଠାରେ b² – 4ac = 9 – 56 = – 47
D < 0 xର ଅବାବ ମାନ ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 3 ଓ \(\frac{3}{2}\) ।

(xiii) ମନେକର \(\frac{x-3}{x+3}=p\), ପ୍ରଭଦ ସମୀକରଣଟି p – \(\frac{1}{p}=\frac{48}{7}\)
⇒ \(\frac{p²-1}{p}=\frac{48}{7}\) ⇒ 7(p²-1) = 48p ⇒ 7p² – 48p – 7 = 0
ଏଠାରେ a = 7, b = -48, c = -7

∴ ନିର୍ଦେୟ ସମାଧାନ -4 ଓ \(\frac{9}{4}\) ।

(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)

(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
ମନେକର \(\frac{x+1}{x-1}=p\)
ଦଉ ସମାଜରଣ p² – p – 2 = 0
ଏଠାରେ a = 1, b = -1, c = -2

ଯଦି p = 2 ⇒ \(\frac{x+1}{x-1}\) = 2 ⇒ x + 1 = 2(x – 1) ⇒ x + 1= 2x – 2
⇒ x – 2x = -2 -1 ⇒ -x = -3 ⇒ x = 3
ଯଦି p = 1 ⇒ \(\frac{x+1}{x-1}\) = -1 ⇒ x + 1 = -1(x – 1) ⇒ x+ 1 = -x + 1
⇒ x + x = 1 – 1 ⇒ 2x = 0 ⇒ x = 0

(xvi) \(\sqrt{2x+9}\) + x = 13

x = 8 ଦ୍ବାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = 20 ହେଲେ, \(\sqrt{2x+9}\) + x ≠ 13
∴ ଦତ୍ତ ସମୀକରଣର ସମାଧାନ x = 8 ।

(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4

x = 6 ଦ୍ଵାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = \(\frac{21}{2}\) ହେଲେ,
\(\sqrt{2 x+\sqrt{2 x+4}}\) ≠ 4 (ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉନାହିଁ)
∴ ନିଶ୍ଚେ ସମୀକରଣର ସମାଧାନ 6।

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect

SECTION – 1

Look at the sentences below.
(a) I reached the hostel in the morning and found that somebody had broken into my room during the night.
(b) She said that her friend had published a book.
(c) He had lived in this town for ten years; then he migrated to Japan.
Can you find the Past Perfect Tense in each sentence? Note that the sentence in which it occurs refers to two actions — the action expressed by the Past Perfect and another action expressed by the Past Simple. Of the two actions, which takes place earlier and which takes place later? List them below.

(a) (1) I reached the hotel in the morning and found. (later).
(2) that somebody had broken (earlier action) into my room during the night.
(b) (1) She said _________(later action).
(2) that her friend had published a book _________(earlier action).
(c) (1) _________then he migrated to Japan _________(later action).
(2) He had lived in this town for ten years, (earlier action).
Can you answer now? Which action does the Past Perfect refer to — the earlier one or the later one? Which action does the Past Simple refer to?
Answer:
The earlier action refers to Past Perfect and the later one refers to Past Simple Tense.

Activity – 23
Combine each pair of sentences below into a single sentence, using the Past Perfect to show which action took place earlier. (You may have to use words like after; when etc.
(a) (i) I finished my homework.
(ii) Then I went to buy a pen.
_____________________________________.
(b) (i) Then the doctor gave some medicine to the patient,
(ii) Then the patient regained his senses.
_____________________________________.
(c) (i) I read a few pages from the book.
(ii) After that I returned it to the librarian.
_____________________________________.
(d) (i) I worked in the garden for some time.
(ii) After that I had my breakfast.
____________________________
(e) (i) He left the place in a hurry.
(ii) After that his friend arrived.
____________________________
(f) (i) The young girl finished shopping.
(ii) Then she met with an accident.
____________________________
(g) (i) The thief ran away with the gold.
(ii) After that the police arrived.
____________________________
Answer:
(a) After I had finished my homework, I went to buy a pen.
(b) After the doctor had given some medicine to the patient, the patient regained his senses.
(c) After I had read a few pages from the book, I returned to the librarian.
(d) I had worked in the garden for some time before I had my breakfast.
(e) When he had left the place in a hurry. his friend arrived.
(f) After the young girl had finished shopping. she met with an accident.
(g) The thief had run away with the gold before the police arrived.

Activity — 24
Jatin arrived late at different places yesterday. What did he find when he arrived
at each place?
Example — When he arrived at the cricket stadium, the game had ended.
(a) the hank it / already I close.
____________________________
(b) his uncle’s house his uncle I go the sleep.
____________________________
(c) the bus stops the bus I already / leave.
____________________________
(d) book shop the book he wanted/sold out already.
_________________________________
(e) the club his friends/leave.
_________________________________
(f) the hostel everyone / go to bed.
_________________________________
Answer:
(a) When he arrived at the bank, it had already closed.
(b) When he got to his uncle’s house, his uncle had gone to sleep.
(c) When he reached the bus stop, the bus had already left.
(d) When he came to the bookshop, the book he wanted had been sold out already.
(e) When he arrived at the club, his friends had left.
(f) When he came to the hostel, everyone had gone to bed.

Activity – 25
Use the verb supplied in brackets in the appropriate form.
(a) We went to Anil’s house and _____________(knock) on the door but there _____________ (be) no answer. Either he _____________ (go) out or he _____________ (not want) to see anyone.
(b) Sadhan _____________(go) for a walk yesterday because the doctor _____________(tell) him last week that he _____________(need) exercise.
(c) A : _____________(Seema / arrive) at the party in time last night ?
B: No, she was late. By the time, we got there, everyone _____________(leave).
Answer:
(a) We went to Anil’s house and knocked on the door but there was no answer. Either he had gone out or he did not want to see anyone.
(b) Sadhan went for a walk yesterday because the doctor told him last week that he needed exercise.
(c) A: Did Seema arrive at the party in time last night?
B: No, she was late. By the time, we got there, everyone had left.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 5 Past Simple and Past Progressive Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 5 Past Simple and Past Progressive

SECTION – 1

Study the sentences below :
(a) It started to rain while we were walking home.
(b) My sister was tidying my room when I saw your letter.
(c) Anita was walking along the road when suddenly she heard footsteps behind her. Someone was following her. She was frightened and started to run.
What do you think the use of the Past Simple and the Past Progressive indicates in these sentences?
(Hint: Think of a point of time and a duration of time in the past and relate them to the action.)

Note:
The Past Simple tells us that the work/action started and finished in the past. The speaker has a definite time in mind. But in Past Progressive the time of beginning or completion of the activity is not mentioned. The activity was in progress for that hour.

Activity – 21
Put the verbs into the correct form, Past Progressive or Past Simple.
(a) My friend ______________(meet) Anima and Amiya at the bus stop four days ago. They ______________(go) to Paradeep and my friend ______________ (go) to Bolangir. They ______________(have) a chat while they ______________(wait) for their buses.
(b) My brother ______________ (cycle) to school last Monday when suddenly an old woman ______________(step) out into the road in front of him. He ______________(go) quite fast but luckily he ______________ (manage) to stop in time and ______________(not / hit) her.
Answer:
(a) My friend met Anima and Amiya at the bus stop four days ago. They were going to Paradeep and my friend was going to Bolangir. They had a chat while they were waiting for their buses.
(b) My brother was cycling to school last Monday when suddenly an old woman stepped out into the road in front of him. He was going quite fast but luckily he managed to stop in time and did not hit her.

Activity – 22
Here is a true story.
An old couple …………………………living in a flat in Bhubaneswar. ………………………… locked up in one room ………………………….. Some unknown people took away everything ………………………… police arrived ………………………… climbed ………………………… rescued ………………………… broke open a door ………………………… one dacoit was killed ………………………… detective was called …………………………interviewed a witness.
Imagine that you are being questioned by the police as if you were a witness to the crime. A police Inspector is recording your statements in a notebook. Think about the situation and write the appropriate answers.
Inspector: Where were you standing at that time?
Answer: _____________________________.
Inspector: Why did you come here?
Answer: _____________________________.
Inspector: What was the old man doing at the time?
Answer: _____________________________.
Inspector: How did you see that?
Answer: _____________________________.
Inspector: How long were you standing there?
Answer: _____________________________.

Answer:
An old couple _____ living in a flat in Bhubaneswar, _____ Orissa locked up in one room _____. Some unknown people took away everything. _____ police arrived _____, limbed _____ rescued _____ , broke open a door _____one dacoit killed, _____detective was called _____ interviewed a witness.
Inspector: Where were you standing at that time?
Answer: I was standing near the flat.
Inspector: Why did you come here?
Answer: I came here to play.
Inspector: What was the old man doing at that time?
Answer: The old man was shouting and trembling out of fear.
Inspector: How did you see that?
Answer: I heard him shouting and saw him trembling.
Inspector: How long were you standing there?
Answer: I stood there till your arrival.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 4 Present Perfect and Past Simple Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 4 Present Perfect and Past Simple

Study the dialogue given below.
Susant: Have you ever ridden a horse?
Subir: Yes, I have.
Susant: When was that?
Subir: I rode one last summer.
Susant : What was it like?
Subir: Oh, it was awful.
Susant : Why? What happened?
Subir: I fell off and hurt my back.
Identify the Present Perfect and Past Simple sentences and examine their use carefully. How are they different in meaning?
{Hint: One of them answers the question ‘When’? and the other does not)
Except for the first two i.e. “Have you ever ridden a horse ?” and “Yes, I have”, the rest of the sentences in the above dialogue belong to Past Simple constructions.
When an action/event took place in the past but its result is still operative at the present moment of speaking/time, we generally use a Present Perfect tense and the Past Simple means that the action/happening occurred before the present moment.

Activity – 18
1. Complete the dialogue using the hints given.
(i) A: ever / see /a lion _______________?
B: Yes, _______________.
A: Where _______________?
B: In the zoo _______________.
A: What/look _______________?
B: terrible _______________.
A: You / afraid _______________?
B: No, _______________?

(ii) A : ever / be to / Dhauligiri ………………………?
B: Yes, _______________.
A: What! see /there _______________?
B: A temple I top/hill _______________.
A See / the inscriptions _______________?
B: Yes, _______________?
A: Able to read the inscriptions _______________?
B: No, _______________.

Answer:
(i) A: Have you ever seen a lion?
B: Yes, I have.
A: Where did you see it?
B: I saw it in the zoo.
A: What did it look like?
B: Yes, it was very terrible to look at.
A: Were you afraid?
B: No, I wasn’t.

(ii) A: Have you ever been to Dhauligiri?
B: Yes, I have been two times.
A: What did you see there?
B: I saw a temple at the top of the hill.
A: Did you see the inscriptions there?
B: Yes, I saw the inscriptions there.
A: Were you able to read the inscriptions there?
B: No, I wasn’t.

Activity – 19
Choose the right verb for each blank space and put it into the correct tense.
(do, wear, carry, ask, say, think)
A : _______________your grandfather _______________ something really crazy ?
B: He _______________ something really silly last summer. One one of the hottest days he _______________ a raincoat and _______________ an umbrella. Everyone _______________ him why. He _______________he _______________ it was going to rain.
Answer:
A: Did your grandfather wear something really crazy?
B: He wore something really silly last summer. On one of the hottest days, he wore a raincoat and carried an umbrella. Everyone asked him why. He said he thought it was going to rain.

Activity – 20
Complete the sentences, using the verbs in brackets either in Past Simple or Present Perfect form.
(a) She _______________ up her mind (made). She’s going to look for another college.
(b) Amulya : _______________me his pen but I’m afraid I _______________ it. (give, lose)
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What _______________ you _______________? (say)
(d) Where is my bike? It _______________ outside the classroom. It _______________! (be, disappear)
(e) Did you know that Umesh _______________ a new scooter? (buy)
(f) I did Sanskrit at school but I _______________ most of it. (forget)
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We _______________ already ___________ (meet).

Answer:
(a) She has made up her mind. She’s going to look for another college.
(b) Amulya gave me his pen but I’m afraid I have lost it.
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What did you say?
(d) Where is my bike? It was outside the classroom. It has disappeared!
(e) Did you know that Umesh has bought a new scooter?
(f) I did Sanskrit at school but I have forgotten most of it.
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We have already met.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 3 Past Simple Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 3 Past Simple

SECTION – 1
Read the passage below carefully.
Ramesh was bon in Baleswar in 1960. He was brought up in Cuttack by his uncle because of his parents in Bangalore. Then he went to Hyderabad to complete his studies. He got a first-class M.Sc. degree in Physics and became a lecturer at a college in Orissa. In 1985 he joined a university but soon went to the U.S.A. on a scholarship. He came back in 1990 and got married. He bought a house in Bhubaneswar in 1993. The happenings above took place in the past time. The basic element of meaning is: “the happening takes place before the present moment.” This means that the present moment is excluded.

Activity – 16
Answer the following questions which relate to the things you do every day. Answer in full sentences.
1. When do you wake up?
2. What do you eat before you go to college?
3. When do you leave home?
4. How do you get to college?
5. What do you pass on the way?
6. How long does it take you?
7. When do your classes start?
Answer:
1. I wake up early at 6 a.m. in the morning.
2. I usually eat rice, dal, and curry before I go to college.
3. I leave home at half past 9 (o’clock).
4. I get to college by bicycle.
5. I pass people, motor cars, etc. on the way to college.
6. It takes me twenty minutes time.
7. Our classes start at 10 o’clock.

Imagine that a friend of yours wants to know from you what you did last Wednesday, which was a very typical day in your life. What kind of questions would he ask you and what answers would you give him? Here are a few questions and answers for you to write.

Question 1.
When did you wake up last Wednesday?
Answer:
I woke up at 6. a.m. last Wednesday.

Question 2.
How did you enjoy the day?
Answer:
I went on merry-making with my friends as it was my birthday.

Question 3.
What did you do in the morning that day?
Answer:
I took a clean bath first and put on a new pair of clothes.

Question 4.
What did you do after wearing the pair of new clothes?
Answer:
I went to the temple to worship God.

Question 5.
What did you give your friends to eat?
Answer:
I gave them some cake, sweets, and other sumptuous food to eat.

Question 6.
What did they give you on the day?
Answer:
They presented me with fabulous gifts.

Question 7.
Were you really happy on that day?
Answer:
Yes. I was very happy on that day.

Activity – 17
The following years were related to important events in Gandhiji’s life. Can you write a sentence on each of these years? One has been done for you.
1. (1869) Gandhiji was born.
2. (1888) ____________________________________________.
3. (1891) ____________________________________________.
4. (1893) ____________________________________________.
5. (1906) ____________________________________________.
6. (1915) ____________________________________________.
7. (1917) ____________________________________________.
8. (1931) ____________________________________________.
9. (1942) ____________________________________________.
10. (1948) ____________________________________________.
Answer:
1. (1869) Gandhiji was bom.
2. (1888) He went to London to study law.
3. (1891) He passed the law examination and was admitted to the bar.
4. (1893) He went to South Africa.
5. (1906) He wrote a significant letter to his brother Laxmidas Gandhi.
6. (1915) He left for Rajkot and Porbandar to meet his relations.
7. (1917) He received a summons to appear before the sub-divisional officer on April 18.
8. (1931) Gandhi-Irwin pact was signed in Delhi.
9. (1942) He attended the All-India Congress Committee meeting at Wardha.
10. (1948) He was shot dead by Nathuram Binayak Godse an R.S.S. member.

Activity – 18
Imagine that you went for a picnic last Sunday with some friends. Write a letter to your friend telling him/her about the picnic. You may follow the hints given below.
1. when and how it was planned
2. the place selected
3. how you went there
4. what you saw on the way
5. what you did there
6. what you enjoyed most
7. when you returned

Dear ___________,
It was really nice to hear from you again. Thanks for telling me about your plans for an excursion. I’m afraid I won’t be able to make it. I have exams to sit for. But I have already made up for the loss.
Yesterday we went on a picnic, and I feel I must share the excitement of it all with you.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________

Yours truly,
___________.

Answer:
Dear Suresh,
It was really nice to hear from you again. Thanks for telling me about your plans for an excursion. I’m afraid I won’t be able to make it. I have exams to sit for. But I have already made up for the loss.
Yesterday we went on a picnic, and I feel I must share the excitement of it all with you.
Our classmates sat together a week ahead and decided to go on a picnic to Nandankanan. We got the permission of the Principal and gave each one hundred rupees for the purpose. We were 50 students and three lecturers too consented to go with us. All of us were of the opinion to see Nandankanan, the zoo. We hired a bus and it was about two hours journey from our college. We saw a lot of buses, cars, and trees full of flowers on our way to Nandankanan. We arrived at Nandankanan by 9 o’clock in the morning. On getting there, we first had our breakfast. Mohan, Sanjay, and Pravakar voluntarily agreed to take charge of cooking. They discharged their duty pretty well. Others went to see the beautiful march of colorful sights and sounds. Our menu was very simple. We had rice, mutton curry, and fruit salad. The preparation of the food was superb. We all enjoyed the food to our heart’s content. After finishing our lunch, we took a little rest and then went round the zoo again. We first returned to our college at 6 p.m. Then the day scholars went to their houses and we went to our hostel. It was really a pleasant outing. I felt your absence there.

With love
Yours truly
Sandeep.

Note:
We can use sentences in the past simply when ‘the speaker/writer has a definite time in mind’.
Examples :
Once this town was a beauty spot.
Mahatma was bom in 1869.
“Past simple” can also be used for “habit in the past”.

Example :
He always carried an umbrella.
They never drank tea.
Note that in most cases we use used to if we wish to emphasize that the habit has been discontinued.

Example :
He used to smoke, (which means that he smoked at one time but he doesn’t smoke now.)
Past Simple can also be used for hypothetical meaning.

Example :
It’s time we had a holiday.
If you caught the 9 o’clock train, you would get there by lunchtime, etc.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(a)

Question 1.
\(\lim _{x \rightarrow 3}\)(x + 4)
Solution:
Clearly, if we take x very close to 3, x + 4 will go very close to 7.
Now let us use ε – δ technique to confirm the result.
Given ε > 0, we seek for δ > 0 depending on ε such that
|x – 3| < δ ⇒ |(x + 4) – 7|< ε
Now |(x + 4) – 7| < ε
if |x – 3| < ε
∴ We can choose ε = 8
Hence for given ε > 0, there exist 8 = ε > 0
such that |x – 3| < δ ⇒ |(x + 4) – 7| < ε
∴ \(\lim _{x \rightarrow 3}\)(x + 4) = 7

Question 2.
\(\lim _{x \rightarrow 1}\)(4x – 1)
Solution:
By taking very close to 1 we have 4x- 1 tends to 3.
Let us use ε – δ technique to confirm the result.
Given ε > 0. We shall find δ > 0 depending on ε such that
|x – 1| < 5 ⇒ |(4x – 1) – 3| < ε
Now |(4x – 1 ) – 3| < ε
if |4x – 1| < ε i.e.|x – 1| < \(\frac{\varepsilon}{4}\)
Let us choose δ = \(\frac{\varepsilon}{4}\)
∴ For given ε > 0 there exists δ = \(\frac{\varepsilon}{4}\) > 0
such that |x – 1| < δ
⇒ |(4x – 1) – 3| < ε
∴ \(\lim _{x \rightarrow 1}\)(4x – 1) = 3

Question 3.
\(\lim _{x \rightarrow 1}\)(√x + 3)
Solution:
As x → 1 we see √x + 3 → 4
We will confirm the result using ε – δ technique
Let ε > 0, we will choose δ > 4
such that |x – 1| < 8 ⇒ |√x + 3 – 4| < ε
Now |√x + 3 – 4| = |√x – 1|
\(=\frac{|x-1|}{|\sqrt{x}+1|}\)
But |√x + 1| > 1
⇒ \(\frac{1}{|\sqrt{x}+1|}\) < 1
⇒ \(\frac{|x-1|}{|\sqrt{x}+1|}<\frac{\delta}{1}\)
∴ (√x + 3) – 4 < \(\frac{\delta}{1}\)
We can take δ < ε i.e. δ = min {1, ε}
∴ |x – 1| < δ ⇒ |(√x + 3) – 4| < ε
for given ε > 0 and (δ = ε)
⇒ \(\lim _{x \rightarrow 1}\)(√x + 3) = 4

Question 4.
\(\lim _{x \rightarrow 0}\) (x² + 3)
Solution:
As x → 0 we observe that x³ + 3 → 3
Let us use ε – δ technique to confirm the result.
Let ε > 0, we seek for a δ > 0 such that
|x – 0| < ε ⇒ |x² + 3 – 3| < ε
Let |x| < 8
Now |x² + 3 – 3| < ε
We have |x|² < ε ⇒| x| < √ε
(∴ |x| and ε are positive.)
∴ we can choose δ = √ε
∴ We have for given δ > 0 there exists
δ = √ε > 0 such that |x| < δ ⇒ |x² + 3 – 3| < ε
∴ \(\lim _{x \rightarrow 0}\) (x² + 3) = 3

Question 5.
\(\lim _{x \rightarrow 0}\) 7
Solution:
If x → 0 we observe that 7 → 7.
Let us use e- 8 technique to confirm the limit.
Let f(x) = 7
Given ε > 0, we will choose a δ > 0
such that |x – 0| < δ ⇒ |f(x) – 7| < ε
Now |f(x) – 7| < ε
If f(x) ∈ (7 – ε . 7 + ε)
But for every x, f(x) = 7
⇒ for|x| < δ also f(x) = 7 ∈ (7 – ε . 7 + ε)
∴ Choosing ε = δ we have
|x| < δ ⇒ |f(x) – 7| < ε
∴ \(\lim _{x \rightarrow 0}\) (7) = 7

Question 6.
\(\lim _{x \rightarrow 1} \frac{(x-1)^3}{(x-1)^3}\)
Solution:
We guess the limit is 1.
Let us confirm using ε – δ technique.
Let ε > 0, f(x) = \(\frac{(x-1)^3}{(x-1)^3}\)
We will choose a δ > 0 such that
|x – 1| < δ ⇒ |f(x) – 1)| < ε
Now |f(x) – 1| < ε
if 1 – ε < f(x) < 1 + ε
∴ We will choose a δ > 0 such that
x ∈ (1 – δ, 1 + δ) – { 1 }
⇒ f(x) ∈ ( 1 – ε, 1 + ε)
As f(x) = for x ≠ 1
We have f(x) ∈ (1 – ε. 1 + ε) for all x ∈ (1 – δ, 1 + δ) – [1]
∴ We can choose δ = ε
for given ε > 0, there exists δ = ε
s.t. |x – 1| < δ ⇒ |f(x) – 1| < ε
∴ \(\lim _{x \rightarrow 1}\) f(x) = 1

Question 7.
\(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\)
Solution:
If we take x very close to 3 (≠ 3)
we have \(\frac{x^3-9}{x-3}\)
= \(\frac{(x-3)\left(x^2+3 x+3^2\right)}{2}\) → 27
Let ε > 0 and x ≠ 3
Now |\(\frac{x^3-9}{x-3}\) – 27| = |x² + 3x +9 – 27|
=|x² – 9 + 3(x – 3)| ≤ |x² – 9| + 3|x – 3|
= |x – 3| [|x + 3| + 3] ≤ |x – 3| [|x + 6| < |x – 3| [|x – 3 + 9|]]
If |x – 3| < δ and δ < 1 then |x – 3| [x – 3 + 9| < δ {1 + 9} = 10 δ
Let δ = min {1, \(\frac{\varepsilon}{10}\)}
∴ For given ε > 0 we have a δ = min {1, \(\frac{\varepsilon}{10}\)} >0 such that
|x – 3| < δ ⇒ |\(\frac{x^3-9}{x-3}\) – 27|
∴ \(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\) = 27

Question 8.
\(\lim _{x \rightarrow 1} \frac{3 x+2}{2 x+3}\)
Solution:
we observe that as x → 1, \(\frac{x+2}{2 x+3}\) → 1
To establish this let ε > 0,
we seek a δ > 0,

Question 9.
\(\lim _{x \rightarrow 0}|x|\)
Solution:
We see that when x → 0,|x| → 0
Let us establish this using ε – δ technique.
Let ε > 0 we seek a δ > 0 depending on
ε s.t.|x – 0| < ε ⇒ ||x| – 0| < ε
Now ||xl – 0| = ||x|| = |x| < δ
By choosing ε = δ we have |x| < ε ⇒ ||x| – 0| < ε
∴ \(\lim _{x \rightarrow 0}|x|\) = 0

Question 10.
\(\lim _{x \rightarrow 2}(|x|+3)\)
Solution:
We see that as x → 2, |x| + 3 → 5
Let ε > 0 we were searching for a, δ > 0
such that |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
Now ||x|| + 3 – 5| = ||x| – 2| < |x – 2| < δ
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ \(\lim _{x \rightarrow 2}(|x|+3)\) = 5

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Exercise 12(a)

Question 1.
Fill in the blanks by choosing the correct answer from the given alternatives :
(a) The center of the circle x² + y² + 2xy – 6y + 1 = 0 is _____________. [(2, -6), (-2, 6), (-1, 3), (1, -3)]
Solution:
(-1, 3)

(b) The equation 2x² – ky² – 6x + 4y – 1 = 0 represents a circle if k = ____________. [2, -2, 0, 1]
Solution:
-2

(c) The point (-3, 4) lies ______________ the circle x² + y² = 16 [outside, inside, on]
Solution:
Outside

(d) The line y = x + k touches the circle x² + y² = 16 if k = _______________. [±2√2, ±4√2, ±8√2, ±16√2]
Solution:
±4√2

(e) The radius of the circle x² + y² – 2x + 4y + 1 = 0 is _______________. [1, 2, 4, √19]
Solution:
2

Question 2.
State (with reasons), which of the following is true or false :
(a) Every second-degree equation in x and y represents a circle.
Solution:
Every 2nd-degree equation in x and y represents a circle if the coefficients of x and y are equal and the equation does not contain xy term (False)

(b) The circle (x – 1)² + (y – 1)² = 1 passes through origin.
Solution:
(0 – 1)² + (0 – 1)² = 1 + 1 = 2 ≠ 1.
So the circle does not pass through the origin. (False)

(c) The line y = 0 is a tangent to the circle (x + 1)² + (y – 2)²  = 1.
Solution:
The line y = 0 is a tangent to the circle centre at (-1, 2) and the radius is 1. (True)
∴ The distance of the centre from the line y = 0 is 1 which is equal to its radius.

(d) The radical axis of two circles always passes through the centre of one of the circles,
Solution:
As radical axis is the common chord of the circles, which should not pass through the centre of one of the circles. (False)

(e) The circle x² + (y – 3)² = 4 and (x – 4)² + y² = 9 touch each other.
Solution:
The distance between the centres is \(\sqrt{(0-4)^2+(3-0)^2}\) = 5 which is equal to the sum of the radii. (True)

Question 3.
Find the equation of circles determined by the following conditions.
(a) The centre at (1, 4) and passing through (-2, 1).
Solution:

(b) The centre at (-2, 3) and passing through origin.
Solution:
Centre at (-2, 3) and circle passes through origin.
∴ Radius of the circle = \(\sqrt{(-2)^2+3^2}=\sqrt{13}\)
∴ Equation of the circle is (x – h)² + (y – k)² = a²
or, (x + 2)² + (y – 3)² = 13

(c) The centre at (3, 2) and a circle is tangent to x – axis.
Solution:

(d) The centre at (-1, 4) and circle is tangent to y – axis.
Solution:

(e) The ends of diameter are (-5, 3) and (7, 5).
Solution:
The endpoints of the diameter of the circle are (-5, 3) and (7, 5).
∴ Equ. of the circle is
(x – h)² + (y – k)² = a²
(x- x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
or, (x + 5)(x – 7) + (y – 3)(y – 5) = 0
or, x² – 7x + 5x – 35 + y² – 5y – 3y + 15 = 0
or, x² + y² – 2x – 8y – 20 = 0

(f) The radius is 5 and circle is tangent to both axes.
Solution:
As the circle is tangent to both axes, we have its centre at (5, 5).
∴ Equation of the circle is
or, (x ± 5)² + (y ± 5)² = 25

(g) The centre is on the x-axis and the circle passes through the origin and the point (4, 2).
Solution:

∴ \(\sqrt{(4-a)^2+4}\) = a
or, (4 – a)² + 4 = a²
or, 16 + a² – 8a + 4 = a²
or, 8a = 20 or, a = \(\frac{20}{8}=\frac{5}{2}\)
∴ Equation of the circle is
(x – h)² + (y – k)² = a²
or, (x – \(\frac{5}{2}\))² + (y – 0)² = (\(\frac{5}{2}\))²
or, x² + \(\frac{25}{4}\) – 5x + y² = \(\frac{25}{4}\)
or, x² + y² – 5x = 0

(h) The centre is on the line 8x + 5y = 0 and the circle passes through the points (2, 1) and (3, 5).
Solution:
Let the equation of the circle be x² + 2gx + y² + 2fy + c = 0
∴ Its centre at (- g, -f). As the centre lies on the line 8x + 5y = 0
We have -8g – 5f = 0      …..(1)
Again, as the circle passes through points (2, 1) and (3, 5)
We have
4 + 4g + 1 + 2f + c = 0  …..(2)
and 9 + 6g + 25 + 10f + c = 0   …..(3)
Now from (1), we have g = \(\frac{-5 f}{8}\)
From equation (2), 4g + 2f + c + 5 = 0
or, 4 \(\frac{-5 f}{8}\) + 2f + c + 5 = 0
or, -5f + 4f + 2c + 10 = 0
or, f = 2c + 10    …..(4)
(2) 6g + 10f + c + 34 = 0
or, 6\(\frac{-5 f}{8}\) + 10f + c + 34 = 0
or, -15f + 40f + 4c + 136 = 0
or, 25f = -4c – 136
or, f = \(\frac{-4 c-136}{25}\)
∴ 2c + 10 = \(\frac{-4 c-136}{25}\)
or, 25 (c + 5) = -2c – 68
or, 25c + 2c = -68 – 125
or, 27c = -193 or, c = \(\frac{-193}{27}\)
∴ f = 2C + 10 = 2(\(\frac{-193}{27}\)) + 10
= \(\frac{-386+270}{27}=\frac{-116}{27}\)
∴ g = \(\frac{-5 f}{8}=\frac{-5}{8} \times\left(\frac{-116}{27}\right)=\frac{145}{54}\)
Eqn. of the circle is x² + y² + 2 × \(\frac{145}{54}\) x + 2 \(\frac{-116}{27}\) y + \(\frac{-193}{27}\) = 0
or, 27x² + 27y² + 145x – 232y – 193 = 0

(i) The centre is on the line 2x + y – 3 = 0 and the circle passes through the points (5, 1) and (2, -3).
Solution:
Let the eqn. of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through (5, 1) and (2, -3),
we have 25 + 1 + 10g + 2f + c = 0 …(1)
and 4 + 9 + 4g – 6f + c = 0    …..(2)
Again as the centre lies on the line 2x + y – 3 = 0,
we have- 2g – f – 3 = 0 or, f= -2g – 3
∴ From equation (1)
10g + 2 (-2g – 3) + c + 26 = 0
or, 10g – 4g – 6 = -c – 26
or, 6g = -c – 20
or, g = \(\frac{-c-20}{6}\)
∴ From equation (2)
4g – 6 (-2g – 3) + c + 13 = 0
or, 4g + 12g + 18 + c + 13 = 0
or, 16g = -c – 31
or, g = \(\frac{-c-31}{16}\)
∴ \(\frac{-c-20}{6}=\frac{-c-31}{16}\)
or, -8c – 160 = -3c – 93
or, 5c = -160 + 93 = -67
or, c = –\(\frac{67}{5}\)
∴ g = \(\frac{-c-20}{6}=\frac{\frac{67}{5}-20}{6}=\frac{67-100}{5 \times 6}\)
= \(\frac{-33}{5 \times 6}=\frac{-11}{10}\)
∴ f = -2g – 3 = (-2)\(\left(\frac{-11}{10}\right)\)
= \(\frac{11-15}{5}=\frac{-4}{5}\)
∴ Eqn. of the circle is x² + y² + 2 (\(\frac{-11}{10}\))x + 2(\(\frac{-4}{5}\))y – \(\frac{67}{5}\) = 0
or, 5x² + 5y² – 11x – 8y – 67 = 0

(j) The circle is tangent to the line x + 2y – 9 = 0 at (5, 2) and also tangent to the line 2x – 3y – 7 = 0 at (2, -1).
Solution:

(k) The circle touches the axis of x at (3, 0) and also touches the line 3y – 4x = 12.
Solution:
Let the centre be at (3, k)
Radius = k

or, 3k – 24 = ±5k or, 2k = -24
or, k = -12
Also k = 3
∴ Equation of the Circle is
(x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y2 – 6x – 2 (-12)y = 0
or, x² + y² – 6x + 24y + 9 = 0
and x² + y² – 6x – 6y + 9 = 0

(l) Circle is tangent to x – axis and passes through (1, -2) and (3, -4).
Solution:
Let the centre be at (h, k).
So the radius is k.

∴ Equation of the circle is (x – h)² + (y – k)² = k²
or, (x + 5)² + (y + 10)² = 100 and (x – 3)² + (y + 2)² = 4

(m) Circle passes through origin and cuts of intercepts a and b from the axes.
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(n) Circle touches the axis of x at a distance of 3 from the origin and intercepts a distance of 6 on the y-axis.
Solution:
Let the centre be at (3, k).
So the radius is k.
∴ Equation of the circle is (x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y² – 6x – 2xy + 9 = 0

∴ |y₂ – y₁| = 2\(\sqrt{k^2-9}\) = 6
or, \(\sqrt{k^2-9}\) = 3
or, k² = 18, or, k = ±3√2
∴ Equation of the circle is x² – y² – 6x ± 6y√2 + 9 = 0

Question 4.
Find the centre and radius of the following circles:
(a) x² + y² + 6xy – 4y – 12 = 0
Solution:
x² + y² + 6xy – 4y – 12 = 0
∴ 2g = 6, 2f = – 4, c = -12
∴ 8 = 3, f = -2
Centre of (-g, -f) = (-3, 2) and radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{9+4+12}\) = 5

(b) ax² + ay² + 2gx + 2fy + k = 0
Solution:
ax² + ay² + 2gx + 2fy + k = 0
or, x² + y² + \(\frac{2 g}{a}\)x + \(\frac{2 f}{a}\)y + \(\frac{k}{a}\) = 0
∴ Centre of \(\left(\frac{-g}{a}, \frac{-f}{a}\right)\)
and radius = \(\sqrt{\frac{g^2}{a^2}+\frac{f^2}{a^2}-\frac{k}{a}}=\sqrt{\frac{g^2+f^2-a k}{a}}\)

(c) 4x² + 4y² – 4x + 12y – 15 = 0
Solution:
4x² + 4y² – 4x + 12y – 15 = 0
or, x² + y² – 4 + 3y  – \(\frac{15}{4}\) = 0
∴ 2g = -1, 2f = 3, c = \(\frac{15}{4}\)
∴ g = – \(\frac{1}{2}\), f = \(\frac{3}{2}\)
∴ Centre at (-g, -f) = (\(\frac{1}{2}\), \(\frac{-3}{2}\)) and radius \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{\frac{1}{4}+\frac{9}{4}+\frac{15}{4}}=\frac{5}{2}\)

(d) a(x² + y²) – bx – cy = 0
Solution:
a(x² + y²) – bx – cy = 0
or, x² + y² – \(\frac{b x}{a}\) – \(\frac{c y}{a}\) = 0

Question 5.
Obtain the equation of circles passing through the following points and determine the coordinates of the centre and radius of the circle in each case:
(a) the points (3, 4) (4, -3) and (-3, 4).
Solution:
Let the centre be at (h, k)

(b) the points (2, 3), (6, 1) and (4, -6).
Solution:
Let the centre be at (h, k).

we have \(|\overline{\mathrm{PC}}|=|\overline{\mathrm{QC}}|=|\overline{\mathrm{RC}}|\)
∴ (h – 4)² + (k + 6)² = (h – 6)² + (k – 1)² and (h – 4)² + (k + 6)² = (h – 2)² + (k – 3)²
∴ h² + 16 – 8h + k² + 36 + 12k
= h² + 36 – 12h + k² + 1 – 2k
and h² + 16 – 8h + k² + 36 + 12k
= h² + 4 – 4h + k² + 9 – 6k
or, 14k = -4h – 15 and 18k = 4h – 39
or, k = \(\frac{-4 h-15}{14}\) and k = \(\frac{4 h-39}{18}\)

(c) the points (a, 0), (-a, 0) and (0, b).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0. As it passes through the points (a, 0), (-a, 0) and (0, b). We have
a² – 2ga + c = 0   …..(1)
a² + 2ga + c = 0   …..(2)

(d) the points (-3, 1), (5, -3) and (-3, 4).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points. we have (-3, 1), (5, -3) and (3, 4).
We have 9 + 1 – 6g + 2f + c = 0    …..(1)
25 + 9 + 10g – 6df + c = 0      …(2)
9 + 16 – 6g + 8f + c = 0     …..(3)

Question 6.
Find the equation of the circles circumscribing the triangles formed by the lines given below :
(a) the lines x = 0, y = x, 2x + 3y = 10
Solution:

∴ The coordinates. C are (0, 0) of
Lastly, solving \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)
we have y = x, 2x + 3y = 1 0
we have 5x = 10
or, x = 2 and y = 2.
∴ The coordinates of A are (2, 2).
∴ The circle passes through the points (2, 2), (0, \(\frac{10}{3}\)) and (0, 0)
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points A, B, C we have c = 0, 4 + 4 + 4g + 4f = 0,
\(\frac{100}{9}\) + 2. f. \(\frac{10}{3}\) + 0 = 0
∴ f = \(\frac{-100}{9} / \frac{20}{3}=\frac{-5}{3}\)
and g = \(\frac{-4 f-8}{4}=\frac{-4\left(\frac{-5}{3}\right)-8}{4}\)
= \(\frac{20-24}{3 \times 4}=\frac{-1}{3}\)
∴ Equation of the circle is  x² + y² + 2(\(\frac{-1}{3}\))x + 2 \(\frac{-5}{3}\)y + 0 = 0
or, 3(x² +  y²) – 2x – 10y = 0

(b) The lines x = 0, 4x + 5y = 35, 4y = 3x + 25
Solution:



or, 4x² + 4y² – 24x – 53y + 175 = 0

(c) The lines x = 0, y = 0, 3x + 4y – 12 = 0
Solution:
The coordinates of A, B and C are (4, 0), (0, 3) and (0, 0).
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(d) The lines y = x, y =2 and y = 3x + 2
Solution:

(e) the lines x + y = 6, 2x + y = 4 and x + 2y = 5
Solution:

Question 7.
Find the coordinates of the points where the circle x² + y² – 7x – 8y + 12 = 0 meets the coordinates axes and hence find the intercepts on the axes. [Hint: If a circle intersects a line at points A and B, then the length AB is its intercepts on line L]
Solution:
x² + y² – 7x – 8y + 12 = 0
Putting x = 0, we have y² – 8y + 12 = 0 or, (y – 6) (y – 2) = 0, or, y = 6, 2.
∴ The circle meets the Y-axis at (0, 6) and (0, 2) and its Y-intercept is 6 – 2 = 4.
Again putting y = 0,
we have x² – 7x + 12 = 0
or, (x – 4)(x – 3) = 0 or, x = 4, x = 3.
∴ The circle meets the X-axis at (4, 0) and (3, 0) and its x-intercept is 4 – 3 = 1.

Question 8.
Find the equation of the circle passing through the point (1, -2) and having its centre at the point of intersection of lines 2x – y + 3 = 0 and x + 2y – 1 =0
Solution:

Question 9.
Find the equation of the circle whose ends of a diameter are the points of intersections of the lines and x + y – 1 = 0, 4x + 3y + 1 = 0 and 4x +y + 3 = 0, x – 2y +3 = 0.
Solution:
Solving x + y – 1 = 0, 4x + y + 3 = 0

∴ The endpoints of the diameter are (-4, 5) and (-1, 1).
∴ Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
or, (x + 4) (x + 1) + (y – 5) (y – 1) = 0
or, x² + x + 4x + 4 + y² – y – 5y + 5 = 0
or, x² + y² + 5x – 6y + 9 = 0.

Question 10.
Find the equation of the circle inscribed inside the triangle formed by the line \(\frac{x}{4}+\frac{y}{3}\) = 1 and the coordinate axes.
Solution:
The circle is inscribed in the triangle formed by x = 0, y = 0 and \(\frac{x}{4}+\frac{y}{3}\) = 1
∴ If (h, k) is the centre and r is the radius of the circle then h = k = r.
The perpendicular distance of the centre (h, h) from the line 3x + 4y = 12 is the radius.
⇒ \(\left|\frac{3 h+4 h-12}{5}\right|\) = h
⇒ 7h – 12 = ±5h
⇒ 2h = 12 or 2h = 12
⇒ h = 6 or h = 1
But h can not be 6 thus the circle has equation (x – 1)² + (y – 1)² = 1
⇒ x² + y² – 2x – 2y + 1 =0

Question 11.
(a) Find the equation of the circle with its centre at (3, 2) and which touches to the line x + 2y – 4 = 0.
Solution:

(b) The line 3x + 4y + 30 = 0 is a tangent to the circle whose centre is at (\(-\frac{12}{5},-\frac{16}{5}\)). Find the equation of the circle.
Solution:

(c) Prove that the points (9, 7), and (11, 3) lie on a circle with centre at origin. Find the equation of the circle.
Solution:

(d) Find the equation of the circle which touches the line x = 0, x = a and 3x + 4y + 5a = 0.
Solution:

(e) If a circle touches the co-ordinate axes and also touches the straight line \(\frac{x}{a}+\frac{y}{b}\) = 1 and has its centre in the 1st quadrant, And its equation.
Solution:
Let the centre be at (k, k) and the radius is k.

Question 12.
ABCD is a square of side ‘a’ If AB and AD are taken as co-ordinate axes, prove that the equation of the circle circumscribing the square is
x² + y² = a(x + y)
Solution:

or, x² + \(\frac{a^2}{4}\) – ax + y² + \(\frac{a^2}{4}\) – ay = \(\frac{a^2}{2}\)
or, x² + y² – ax – ay = 0
or, x² + y² = a(x + y)

Question 13.
(a) Find the equation of the tangent and normal to the circle x² + y² = 25 at the point (3, -4).
Solution:
Equation of the tangent to the circle x² + y² = 25 at the point (3, -4) is
xx₁ + yy₁ = a²
3x – 4y = 25
Equation of the normal is x₁y = xy₁
or, 3y = -4x or, 4x + 3y = 0

(b) Find the equation of the tangent and normal, to the circle, x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3).
Solution:
Equation of the tangent of the circle x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, -2x + 3y – \(\frac{3}{2}\) (x – 2) + 2(y + 3) – 31 = 0
or, – 4x + 6y – 3x + 6 + 4y + 12 – 62 = 0
or, -7x + 10y – 44 = 0
or, 7x – 10y + 44 = 0
Equation of the normal is x(f + y₁) – y(g + x₁) fx₁ + gy₁ = 0
or, x(2 + 3) – y(\(-\frac{3}{2}\) – 2) – 2(-2) – \(\frac{3}{2}\) × 3 = 0
or, 5x + \(\frac{7y}{2}\) + 4 – \(\frac{9}{2}\) = 0
or, 10x + 7y – 1 = 0

(c) Find the equation of the tangents to the circle x² + y² + 4x – 6y – 16 = 0 at the point where it meets the y – axis.
Solution:
Putting x = 0 in the circle equation, we have
y² – 6y – 16 = 0
or, y² – 8y + 2y – 16 = 0
or, y(y – 8) + 2(y – 8) = 0
or, (y – 8)(y + 2) = 0
y = 8 or, -2
The circle meets y – axis at (0, 8) and (0, -2).
Eqn. of the tangents are
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, 0 + 8y + 2 (x + 0) – 3(y + 8) – 16 = 0
or, 8y + 2x – 3y – 24 – 16 = 0
or, 2x + 5y = 40 and
x × 0 – 2y + 2 (x + 0) – 3 (y – 2) – 16 = 0
or, -2y + 2x – 3y + 6 – 16 = 0
or, 2x – 5y – 10 = 0

(d) Find the condition under which the tangents at (x₁, y₁) and (x₂, y₂) to the circle x² + y² + 2gx + 2fy + c = 0 are perpendicular.
Solution:
Equation of tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at the point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, (g + x₁)x + y(f + y₁) + gx₁ + fy₁ + c = 0
Again equation of the tangent to the circle at (x₂, y₂) is
x(g + x₂) + y(f + y₂) + gx₂ + fy₂ + c = 0
As the tangent (1) and (2) are perpendicular, we have the product of their slopes is -1.
∴ \(\frac{g+x_1}{f+y_1} \times \frac{g+x_2}{f+y_1}\) = -1
or, (g + x₁)(g + x₂) = -(f + y₁)(f + y₂)
or, (g + x₁)(g + x₂) + (f + y₁)(f + y₂) = 0

(e) Calculate the radii and distance between the centres of the circles, whose equations are, x² + y² – 16x – 10y + 8 = 0; x² + y² + 6x – 4y – 36 = 0. Hence or otherwise prove that the tangents drawn to the circles at their points of intersection are perpendicular.
Solution:
x² + y² – 16x – 10y + 8 = 0;
x² + y² + 6x – 4y – 36 = 0.
g₁ = -8, f₁ = -5, c₁ = 8,
g₂ = 3, f₂ = -2, c₂ = -36
The centres are (-g₁, -f₁) and (-g₂, -f₂)

Question 14.
(a) Find the equation of the tangents to the circle x² + y² = 9 perpendiculars to the line x – y – 1 = 0
Solution:

(b) Find the equation of the tangent to the circle x² + y² – 2x – 4y = 40, parallel to the line 3x – 4y = 1.
Solution:


(c) Show that the line x – 7y + 5 = 0 a tangent to the circle x² + y² – 5x + 5y = 0. Find the point of contact. Find also the equation of tangent parallel to the given line.

Solution:
we have the line is x – 7y + 5 = 0
or, y = \(\frac{x+5}{7}\)
Now putting the value of y in the circle, we have x² + y² – 5x + 5y = 0
or, x² + (\(\frac{x+5}{7}\))² – 5x + 5 \(\frac{x+5}{7}\) = 0
or, 49x² + x² + 25 + 10x – 245x + 35x + 175 = 0
or, 50x² – 200x + 200 = 0
or, x² – 4x + 4 = 0
∴ a = 1, b = -4, c = 4
∴ b² – 4ac = (-4)² – 4 × 1 × 4
= 16 – 16 = 0
∴ The line x – 7y + 5 = 0

x – 7y – 45 and x – 7y + 5 = 0

(d) Prove that the line ax + by + c = 0 will be the tangent to the circle x² + y² = r² if r²(a² + b²) = c².
Solution:
We know that a line is a tangent to the circle if the distance of the line from the centre is equal to the radius.
Now the circle is x² + y² = r²
⇒ Centre is at (0, 0) and radius r. The distance of (0, 0) from ax + by + c = 0 is

(e) Prove that the line 2x + y = 1 tangent to the circle x² + y² + 6x – 4y + 8 = 0.
Solution:

(f) If the line 4y – 3x = k is a tangent to the circle x² + y² + 10x – 6y + 9 = 0 find ‘k’. Also, find the coordinates of the point of contact.
Solution:
Center of the circle is (-5, 3) and the radius is \(\sqrt{25+9-9}\) = 5
Distance of the centre from the line 4y – 3x – k = 0

Question 15.
(a) Find the length of the tangent, drawn to the circle x² + y² + 10x – 6y + 8 = 0 from the centre of the circle x² + y² + 4x = 0.
Solution:
Center of the circle x² + y² + 4x = 0 is (2, 0)
∴ Length of the tangent drawn from the point (2, 0) to the circle x² + y² + 10x – 6y + 8 = 0
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 g \mathrm{~g}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+0+10 \times 2+0+8}=\sqrt{32}=4 \sqrt{2}\)

(b) Find the length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0
Solution:
Length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0 is
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+1+(-6) \times 2+10(-1)+18}\)
= \(\sqrt{5-12-10+18}\) = 1

(c) Find the length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15.
Solution:
Length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15 is \(\sqrt{16+49-15}\) = √50 = 5√2

Question 16.
(a) Prove that the circle given by the equations x² + y² + 2x – 8y + 8 = 0 and x² + y² + 10x – 2y + 22 = 0 touches each other externally. Find also the point of contact
Solution:
x² + y² + 2x – 8y + 8 = 0
g₁ = 1, f₁ = -4, c₁ = 8
Hence centre = c₁(-g₁, -f₁) = c₁(-1, 4)
Radius = r₁ = \(\sqrt{1+16-8}\) = 3
Again x² + y² + 10x – 2y + 22 = 0
g₂ = 5, f₂ = -1, c₂ = 22
Centre c₂(-g₂, -f₂) = c₂(-5, 1)
Radius r₂ = \(\sqrt{25+1-22}\) = 2
Now

(b) Prove that the circle is given by the equations x² + y² = 4 and x² + y² + 6x + 8y – 24 = 0, touch each other and find the equation of the common tangent.
Solution:
x² + y² = 4,
x² + y² + 6x + 8y – 24 = 0
Their centres are (0, 0) and (-3, -4) and radii are 2 and \(\sqrt{9+16+24}\) = 7
∴ Distance between the centres is \(\sqrt{(-3)^2+(-4)^2}\) = 5, which is equal to the difference between the radii.
∴ The circles touch each other internally.
∴ Equation of the common tangent is S₁ – S₂ = 0
or, (x² + y² + 6x + 8y – 24) – (x² + y² – 4) = 0
or, 6x + 8y – 20 = 0
or, 3x + 4y = 10

(c) Prove that the two circle x² + y² + 2by + c² = 0 and x² + y² + 2ax + c² = 0,  will touch each other \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\).
Solution:
x² + y² + 2by + c² = 0,
x² + y² + 2ax + c² = 0
g₁ = 0, f₁ = b, c₁ = c².
g₂ = a, f₂ = 0, c₂ = c².
The centres of the circle are (0, -b) and (-a, 0) and radii are \(\sqrt{b^2-c^2}\) and \(\sqrt{a^2-c^2}\). As the circles touch each other, we have the distance between the centres is equal to the sum of the radii.

(d) Prove that the circles given by x² + y² + 2ax + 2by + c = 0, and x² + y² + 2bx + 2ay + 2c = 0, touch each other, if (a + b) = 2c.
Solution:
x² + y² + 2ax + 2by + c = 0
x² + y² + 2bx + 2ay + 2c = 0,
The centre of the circle is (-a, -b) and (-b, -a). the radii of the circle are \(\sqrt{a^2+b^2-c}\) and \(\sqrt{b^2+a^2-c}\). As the circles touch each other we have, the distance between the centres is equal to the sum of the radii.

Question 17.
Find the equation of the circle through the point of intersection of circles x² + y² – 6x = 0 and x² + y² + 4y – 1 = 0 and the point (-1, 1).
Solution:
Let the equation of the circle be (x² + y² – 6x) + λ(x² + y² + 4y – 1) = 0
As it passes through the point (-1, 1),
we have (1 + 1 + 6) + λ(1 + 1 + 4 – 1) = 0
or, 8 + 5λ = 0 or, λ = \(\frac{-8}{5}\)
∴ Equation of the circle is (x² + y² – 6x) – \(\frac{8}{5}\) (x² + y² + 4y – 1) = 0
or, 5x² + 5y² – 30x – 8x² – 8y² – 32y + 8 = 0
or, 3x² + 3y² + 30x + 32y – 8 = 0

Question 18.
Find the equation of the circle passing through the intersection of the circles, x² + y² – 2ax = 0 and x² + y² – 2by = 0 and having the centre of the line \(\frac{x}{a}-\frac{y}{b}\) = 2
Solution:

Question 19.
Find the radical axis of the circles x² + y² – 6x – 8y – 3 = 0 and 2x² + 2y² + 4x – 8y = 0
Solution:
x² + y² – 6x – 8y – 3 = 0
2x² + 2y² + 4x – 8y = 0
x² + y² – 6x – 8y – 3 = 0
x² + y² + 2x – 4y = 0
∴ The equation of the radical axis is S₁ – S₂ = 0
or, (x² – y²– 6x – 8y – 3) – (x² + y² + 2x – 4y) = 0
or, -6x – 8y- 3 – 2x + 4y = 0
or, -8x – 4y – 3 = 0
or, 8x + 4y + 3 = 0

Question 20.
Find the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0 Prove that the radical axis is perpendicular to the line joining the centres of the two circles.
Solution:
Equation of the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0
(x² + y² – 6x + 8y – 12) – (x² + y² + 6x – 8y + 12) = 0
or, -12x + 16y – 24 = 0
or, 3x – 4y + 6 = 0
Again, slope of the radical axis is \(\frac{3}{4}\) = m₁ (say)
Centres of the circles are (3, -4) and (-3, 4).
Slope of the line joining the centres is \(\frac{4+4}{-3-3}=\frac{8}{-6}=-\frac{4}{3}\) = m₂ (say)
m₁. m₂ = \(\frac{3}{4}\left(-\frac{4}{3}\right)\) = -1
∴ The radical axis is perpendicular to the line joining centres of the circles. (Proved)

Question 21.
If the centre of one circle lies on or inside another, prove that the circles cannot be orthogonal.
Solution:
The orthogonality condition for two circles.
x² + y² + 2g₁x + 2f₁y + C₁ = 0   …..(1)
and x² + y² + 2g₂x + 2f₂y + C₂ = 0    …..(2)
is 2(g₁g₂ + f₁f₂) – C₁ – C₂ = 0
Let us consider two circles
Case-1. Let the centre of (2) which is C (-g₂, -f₂) lies on the circle (1). Hence it satisfies the equation (i)
i.e., g₂² + f₂² – 2g₁g₂ – 2f₁f₂ + C₂ = 0
⇒ 2g₁g₂ + 2f₁f₂ – C₁ – C₂ = g₂² + f₂² – C₂
Its right-hand side is the square of the radius of 2nd circle which can not be equal
to zero i.e., 2(g₁g₂ + f₁f₂) – C₁ – C₂ ≠ 0
Hence circles are not orthogonal.
Case-2. Let the centre of (2) which is (-g2, -f2) lies inside the circle (1).
Distance between their centres < radius of the first circle.
i,e. \(\sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}<\sqrt{g_1^2+f_1^2-C_1}\)
⇒ g₁² – 2g₁g₂ + f₁² + f₂² + 2f₁f₂ < g₁² + f₁² – C₁
⇒ 2g₁g₂ – 2f₁f₂ – C₁ – C₂ > g₂² + f₂² – C₂
= square of the radius of 2^nd circle. Hence greater than 0.
⇒ 2(g₁g₂ + f₁f₂) – C₁ – C₂ > 0
So two circles are not orthogonal. By case -1 and case -2 we conclude that if the centres of one circle lie on or inside another, then circles cannot be orthogonal.

Question 22.
If a circle S intersects circles S₁ and S₂ orthogonally. Prove that the centre of S lies on the radical axis of S₁ and S₂. [Hints: Take the line of centres of S₁ and S₂ as x – axis and the radical axis as y – axis. Use conditions for the orthogonal intersection of S, S₁ and S, S₂ simultaneously and prove that S is centred on the y – axis.]
Solution:
Let the equation of the circle S, S₁ and S₂ are
x² + y² + 2gx + 2fy + C = 0      …(1)
x² + y² + 2g₁x + 2f₁y + C = 0      …(2)
and x² + y² + 2g₂x + 2f₂y + C = 0      …(3)
According to the question, the circle S intersects circles S₁ and S₂ orthogonally.
Hence 2 (g₁g + f₁f) – C₁ – C = 0 …(4)
and 2 (g₂g + f₂f) – C₂ – C = 0  ….(5)
Subtracting (4) from (3) we get
2g(g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0 …(6)
Now radical axis of circles S₁ and S₂ is S₁ – S₂ = 0
i, e. 2x (g₁ – g₂) + 2y (f1 – f2)+ C₁ – C₂ = 0 ….(7)
The centre of the circle S is (-g, -f).
If it lies in the radical axis then equation (7) will be satisfied by the centre.
i,.e, 2g (g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0
which is nothing but equation (5). Hence centres of S lie on the radical axis of S₁ and S₂.

Question 23.
R is the radical centre of circles S₁, S₂ and S₃. Prove that if R is on/inside/outside one of the circles then it is similarly situated with respect to the other two.
Solution:
Given R is the radical centre of S₁, S₂ and S₃
The radical centre is the intersection point of three radical axes whose equations are
S₁ – S₂ = 0
S₂ – S₃ = 0    …..(1)
S₃ – S₁ = 0
Let S₁ : x² +y² + 2g₁x + 2f₁y + C₁ =0
S₂ : x² + y² + 2g₂x + 2f₂y + C₂ =0
S₃ : x² + y² + 2g₃x + 2f₃y + C₃ =0
Now equations of radical axes by set of equation (1) are
2x(g₁ – g₂) + 2y(f₁ – f₂) + C₁ – C₂ =0 …(2)
2x(g₂ – g₃) + 2y(f₂ – f₃) + C₂ – C₃ =0 …(3)
and 2x(g₃ – g₁) + 2y(f₃ – f₁) + C₃ – C₁ = 0 …(4)
Let the co-ordinate of R be (x1, y1) the
point R must satisfy (2), (3) and (4).
i.e., 2x₁(g₁ – g₂) + 2y₁(f₁ – f₂) + C₁ – C₂ = 0 …(5)
2x₁(g₂ – g₃) + 2y₁(f₂ – f₃) + C₂ – C₃ =0 …(6)
2x₁(g₃ – g₁) + 2y₁(f₃ – f₁) + C₃ – C₁ =0 …(7)
Subtracting (6) for (5) we get
2x₁(g₁ – g₃) + 2y₁(f₁ – f₃) + C₁ – C₃ =0
⇒ 2g₁x₁ + 2f₁y₁ + C₁ =2g₃x₁ + 2f₃y₁ + C₃
Similarly subtracting (7) from (6) we get
2g₂y₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁
Combining the above two equations we get
2g₂x₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁ = 2g₃x₁ + 2f₃y₁ + C₃
If R x₁ y₁ lies on / inside / outside of S₁ …(8) then x₁² + y₁² + 2g₁x₁ + 2f1y1 + C₂ (= / < / >)0 respectively.
⇒ x₁² + y₁² + 2g₂x₂ + 2f₂y₂ + C₂(=/</>) 0
⇒ x₁² + y₁² + 2g₃x₃ + 2f₃y₃ + C₃(=/</>) 0
respectively by Eqn (8).
This concludes that if R is on /inside/outside. One of the circles then it is similarly situated with respect to the other two.

Question 24.
Determine a circle which cuts orthogonally to each of the circles.
S₁: x² + y² + 4x – 6y + 12 = 0
S₂: x² + y² + 4x + 6y + 12 = 0
S₃: x² + y² – 4x + 6y + 12 = 0
[Hints: The centre of the required circle S must be the radical centre R (why?), which lies outside all the circles. Then show that the radius of S must be the length of the tangent from R to any circle of the system.
Solution:
Let the equation of the required circle is x² + y² + 2gx + 2fy + C = 0    …..(1)
We know if two circles
x² + y² + 2g₁x + 2f₁y + C₂ = 0 and
x² + y² + 2g₂x + 2f₂y + C₂ = 0
are orthogonal then
2(g₁g₂ + f₁f₂) – C₁ – C₂ =0   ….(2)
According to the question circle (1) is orthogonal to the circles
S₁: x² + y² – 4x – 6y + 12 = 0     ….(3)
S₂: x² + y² + 4x + 6y + 12 = 0   ….(4)
S₃: x² + y² – 4x + 6y + 12 = 0     ….(5)
For these circles equation (2) will be
2(-2g – 3f) – C – 12 = 0     ….(6)
2(2g + 3f) – C – 12 = 0     …..(7)
2(-2g + 3f) – C – 12 = 0     …..(8) respectively.
Now subtract eqn. (7) from (6) and (8) from (7) we get
2(- 4g – 6f) = 0
⇒ 2(4g) = 0 ⇒ g = 0 , and f = 0
Using the value of g and f in eq. (6) we get
C = -12
Using g = 0, f= 0 , C = -12 in (1) we get
x² + y² – 12 = 0 is the required equation of the circle.

Question 25.
Prove that no pair of concentric circle can have radical axes.
Solution:
Let the centre of pair of concentric circles is C (h, k) and radii are r₁ and r₂.
So equation of the circles are
S₁: (x – h)² + (y – k)² = r₁²
S₂: (x – h)² + (y – k)² = r₂²
Equation of the radical axis is S₁ – S₂ = 0
⇒ r₁² – r₂² = 0
which is not a straight line as r₁ and r₂ are constants.
Hence it concludes that no pair of concentric circles have a radical axis.

Odisha State Board BSE Odisha 10th Class Maths Solutions Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(a)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ସମ୍ଭାବ୍ୟ ଉତ୍ତର ମଧ୍ୟରୁ ସଠିକ୍ ଉତ୍ତରଟି ବାଛ ।
(i) 1, 2, 3, 4, …….. ଅନୁକ୍ରମରେ t₈ = …………..
(a) 6
(b) 7
(c) 8
(d) 9
ଉ-
(c) 8

(ii) 2, 4, 6, 8, …….. ଅନୁକ୍ରମରେ t₇ = …………..
(a) 12
(b) 14
(c) 16
(d) 18
ଉ-
(b) 14

(iii) – 5, – 3, – 1, 1, ……. ଅନୁକ୍ରମରେ t₁₁ = …………..
(a) 13
(b) 15
(c) 17
(d) 19
ଉ-
(b) 15

(iv) 3, 6, 9, ……… ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 3
(b) 4
(c) 5
(d) 6
ଉ-
(a) 3

(v) -4, -2, 0, 2, A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) – 2
(b) -3
(c) 2
(d) 3
ଉ-
(c) 2

(vi) 10.2, 10.4, 10.6, 10.8, …. ରେ ସାଧାରଣ ଅନ୍ତର t₅ = …………..
(a) 11.0
(b) 11.2
(c) 11.4
(d) 11.6
ଉ-
(a) 11.0

(vii) 2.5, 2.9, 3.3, 3.7, ….. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 1.5
(b) 1.4
(c) 0.5
(d) 0.4
ଉ-
(d) 0.4

(viii) 3, x, 9, ଏକ A.P. ହେଲେ x = …………
(a) 4
(b) 5
(c) 6
(d) 7
ଉ-
(c) 6

(ix) 1.01, 1.51, 2.01, 2.51, ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 1
(b) 0.5
(c) 1.5
(d) 1.05
ଉ-
(b) 0.5

(x) 5, 0, -5, 10, ………. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) – 5
(b) 5
(c) – 10
(d) 10
ଉ-
(a) – 5

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) 1, 2, 3, 4, …….. ଅନୁକ୍ରମର t₈ = 8
(କାରଣ ଏଠାରେ t₁ = 1, t₂ = 2, t₃ = 3, t₈ = 8)

(ii) 2, 4, 6, 8 ……. ଅନୁକ୍ରମର t = 14
(କାରଣ t₁ = 2 × 1, t₂ = 2 × 2, t₃ = 2 × 3, t₇ = 2 × 7 = 14)

(iii) -5, -3, – 1, 1 ଅନୁକ୍ରମର t₁₁ = 15
(t₁₁ = 5 + (11 – 1) 2 = -5 + 20= 15).

(iv) 3,6,9, 60 ରେ ସାଧାରଣ ଅନ୍ତର d = 3
(କାରଣ 6 – 3 = 9 – 6 = 3)

(v) -4, -2, 0, 2 ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = 2
(କାରଣ -2 – (-4) = 0 -(-2) = 2)

(vi) 10.2, 10.4, 10.6, 10.8…… ରେ t₂ = 11
(କାରଣ 10.2 + (5 – 1) × 0.2 = 10.2 + 0.8 = 11)

(vii) 2.5, 2.9, 3.3, 3.7 …….. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = 0.4
(କାରଣ 2.9 – 2.5 = 3.3 – 2.9 = 0.4)

(viii) 3, x, 9 ….. A.P. ହେଲେ x = 6
(କାରଣ x = \(\frac{3+9}{2}\) = \(\frac{12}{2}\) =6)

(ix) 1.01, 1.51, 2.01, 2.51, ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(କାରଣ 1.51 – 1.01 = 2.01 – 1.51 = 0.50)

(x) 5, 0, -5, 10, ………. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(କାରଣ 0 – 5 = -5 – 0 = -5)

Question 2.
ନିମ୍ନଲିଖୂତ ଅନୁକ୍ରମ ମଧ୍ୟରୁ କେଉଁଗୁଡ଼ିକ A.P, ସେଗୁଡ଼ିକୁ ଚିହ୍ନଟ କର :
(i) 1, 4, 7, 10, 15, 16, 19, 22
(ii) 1, 8, 15, 22, 29, 36, 43, 50
(iii) 1, 6, 11, 15, 22, 28, 34, 40
(iv) 1, 4, 7, 9, 11, 14, 17, 20
(v) – 5, -3, -1, 0, 2, 4, 6, 8
(vi) a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d
(vii) 0.6, 0.8, 1.0, 1.5, 1.7, 1.8, 1.9, 2.0
(viii) -7,-4,- 1, 2, 5, 8, 11, 14
ଉ –
ନିମ୍ନଲିଖୂତ ଅନୁକ୍ରମ ମଧ୍ୟରୁ (ii), (vi) ଏବଂ (viii) ଗୁଡ଼ିକ A.P. ଅଟନ୍ତି ।

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) 1, 4, 7, 10, 15, 16, 19, 22 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 4 – 1 ≠ 16 – 15, ଏଠାରେ ଯେକୌଣସି ପଦରୁ ତା’ର ପୂର୍ବ ପଦକୁ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ ସମାନ ରହୁନାହିଁ ।)

(ii) 1, 8, 15, 22, 29, 36, 43, 50 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
(କାରଣ ଏଠାରେ ସାଧାରଣ ଅନ୍ତର (d) = 7 ଅଟେ ।)

(iii) 1, 6, 11, 15, 22, 28, 34, 40 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 6 – 1 ≠ 15 – 11, ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(iv) 1, 4, 7, 9, 11, 14, 17, 20 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 4 – 1 ≠ 11 – ୨ ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(v) -5, -3, -1, 0, 2, 4, 6, 8 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ – 3 – (- 5) = 2 କିନ୍ତୁ 0 – (- 1) = 1, ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(vi) a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଅନୁକ୍ରମର t₁ = a ଏବଂ ସାଧାରଣ ଅନ୍ତର = d)

(vii) 0.6, 0.8, 1.0, 1.5, 1.7, 1.8, 1.9, 2.0 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ d ସବୁ କ୍ଷେତ୍ରରେ ସମାନ ନୁହେଁ ।)

(viii) -7,-4,-1, 2, 5, 8, 11, 14 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ d = -4 – (-7) = -1 – (-4) = 2 – (-1) = 8 – 5 = 11 – 8 = 14 – 11 = 3)

Question 3.
ପ୍ରଶ୍ନ 2ରେ ଯେଉଁଗୁଡ଼ିକ A.P. ସେମାନଙ୍କ କ୍ଷେତ୍ରରେ ସାଧାରଣ ଅନ୍ତର ନିରୂପଣ କର ।
ସମାଧାନ :
(ii) 1, 8, 15, 22, 29, 36, 43, 50 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଏହି ଅନୁକ୍ରମର ସାଧାରଣ ଅନ୍ତର (d) = 8 – 1 = 15 – 8 = 7

(vi) , a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଏଠାରେ ସାଧାରଣ ଅନ୍ତର (d) = a + 4d – a – 3d = d

(viii) -7,-4,-1, 2, 5, 8, 11, 14 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଅନୁକ୍ରମଟିର ସାଧାରଣ ଅନ୍ତର (d) = – 4 – (- 7) = 3

Question 4.
ପ୍ରଥମ ପଦ a = 5 ନେଇ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ ଲେଖ ଯେପରିକି ସାଧାରଣ ଅନ୍ତର
(i) d = 5
(ii) d = 4
(iii) d = 2
(iv) d =-2
(v) d=-3 ହେବ |
ସମାଧାନ :
(i) ପ୍ରଥମ ପଦ a = 5, d = 5 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 5 = 10, t₃ = a + 2d = 5 + 2 × 5 = 15
t₄ = a + 3d = 5 + 3 × 5 = 20
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 10, 15 ଓ 20 1

(ii) a = 5, d = 4 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 4 = 9, t₃ = a + 2d = 5 + 2 × 4 = 13
t₄ = a + 3d = 5 + 3×4 = 17
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 9, 13 ଓ 17 !

(iii) a = 5, d = 2 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 2 = 7, t₃ = a + 2d = 5 + 2 × 2=9
t₄ = a + 3d = 5 + 3 × 2 = 11
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 7, 9 ଓ 11 1

(iv) a = 5, d= – 2 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + (- 2) = 3, t₃ = a + 2d = 5 + 2 (- 2) = 1
t₄ = a + 3d = 5 + 3 (- 2) = – 1
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 2, – 1 ଓ – 4 ।

(v) a = 5, d = -3 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + (- 3) = 2, t₃ = a + 2d = 5 + 2 (- 3) = – 1
t₄ = a + 3d = 5 + 3 (- 3) = – 4
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 2, – 1 ଓ – 4 ।

Question 5.
ଏକ A.P.ର n ତମ ପଦ । ନିମ୍ନରେ ପ୍ରଦତ୍ତ ହୋଇଛି । ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ t₅, t₈ ଓ t₁₀ କେତେ ନିରୂପଣ କର।
(i) t_n = \(\frac{n+1}{2}\)
(ii) t_n = -10 + 2n
(iii) t_n = 10n + 5
(iv) t_n = 4n – 6
ସମାଧାନ :
(i) ଏକ A.P. ର t_n = \(\frac{n+1}{2}\)
t₅ = \(\frac{6+1}{2}=\frac{6}{2}=3\)

t₈ = \(\frac{8+1}{2}=\frac{9}{2}=4.5\)

t₁₀ = \(\frac{10+1}{2}=\frac{11}{2}=5.5\)

(ii) t_n = -10 + 2n
t₅ = -10 +2 × 5
= – 10 + 10 = 0,

t₈ = -10 + 2 × 8
= -10 + 16 = 6

t₁₀ = -10 + 2 × 10
= 10 + 20= 10

(iii) t_n = 10n + 5
t₅ = 10 × 5 + 5 = 55,
t₈ = 10 × 8+ 5 = 85,
t₁₀ = 10 x 10 + 5 = 105

(iv) t_n = 4n – 6,
t₅ = 4 × 5 – 6
= 20 – 6 = 14

t₈ = 4 × 8 – 6
= 32 – 6 = 26

t₁₀ = 4 × 10 – 6
= 40 – 6 = 34

Question 6.
ନିମ୍ନଲିଖୂ A.P. ଗଠନ କର (କେବଳ ଦ୍ବିତୀୟ, ତୃତୀୟ ଓ ଚତୁର୍ଥ ପଦ ତ୍ରୟ ଆବଶ୍ୟକ) ଯେଉଁଠାରେ :
(i) ପ୍ରଥମ ପଦ a = 4, ସାଧାରଣ ଅନ୍ତର d =3
(ii) ପ୍ରଥମ ପଦ a = -8, ସାଧାରଣ ଅନ୍ତର d = – 2
(iii) ପ୍ରଥମ ପଦ a = 7, ସାଧାରଣ ଅନ୍ତର d = – 4
(iv) ପ୍ରଥମ ପଦ a = 10, ସାଧାରଣ ଅନ୍ତର d = 5
(v) ପ୍ରଥମ ପଦ a = \(\frac{1}{2}\), ସାଧାରଣ ଅନ୍ତର d = \(\frac{3}{2}\)
(vi) ପ୍ରଥମ ପଦ a = \(\frac{1}{2}\), ସାଧାରଣ ଅନ୍ତର d = -1
ସମାଧାନ :
(i) ଏକ A.P.ର ପ୍ରଥମ ପଦ a = 4, ସାଧାରଣ ଅନ୍ତର d = 3
t₂ = a + d = 4 + 3 = 7,
t₃ = a + 2d = 4 + 2 × 3 = 10
t₄= a + 3d = 4 + 3 × 3 = 13

(ii) a = -8, d = -2
t₂ = a + d = -8 + (- 2) = – 10
t₃ = a + 2d = – 8 + 2 × (- 2) = – 8 – 4 = – 12
t₄ = a + 3d = -8 + 3 (- 2) = – 8 – 6=-14

(iii) a = 7, d=-4
t₂ = a + d = 7+ (- 4) = 3,
t₃ = a + 2d = 7 + 2 (- 4) = 7 – 8 = -1
t₄ = a + 3d = 7+ 3 (- 4) = 7 – 12 = – 5

(iv) a = 10, d = 5
t₂ = a + d = 10 + 5 = 15,
t₃ = a + 2d = 10 + 2 × 5 = 20
t₄ = a + 3d = 10 + 3 × 5 = 10 + 15 = 25

(v) a = \(\frac{1}{2}\), d = \(\frac{3}{2}\)
t₂ = a + d = \(\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2\)
t₃ = a + 2d = \(\frac{1}{2}+2 × \frac{3}{2}=\frac{1}{2}+3=\frac{7}{2}\)
t₄ = a + 3d = \(\frac{1}{2}+3 × \frac{3}{2}=\frac{1}{2}+\frac{9}{2}=\frac{10}{2}=5\)

(vi) a = \(\frac{1}{2}\), d = -1
t₂ = a + d = \(\frac{1}{2}-1=-\frac{1}{2}\)
t₃ = a + 2d = \(\frac{1}{2}+2(-1)=\frac{1}{2}-2=\frac{1-4}{2}=\frac{-3}{2}\)
t₄ = a + 3d = \(\frac{1}{2}+3(-1)=\frac{1}{2}-3=\frac{1-6}{2}=\frac{-5}{2}\)

Question 7.
ନିମ୍ନରେ ପ୍ରଦତ୍ତ ଉକ୍ତିଗୁଡ଼ିକ ଭୁଲ୍ ବା ଠିକ୍ ଲେଖ ।
(a) 1, 2, 3, 4 ……. ସମାନ୍ତର ପ୍ରଗତି ସୃଷ୍ଟି କରନ୍ତି ।
(b) 1, – 1, 1, -1, -1, ……. ଅନୁକ୍ରମଟି ସମାନ୍ତର ପ୍ରଗତି ଅଟେ ।
(c) 2, 1, – 1, – 2 ସଂଖ୍ୟା ଚାରିଗୋଟି ସମାନ୍ତର ପ୍ରଗତିରେ ବିଦ୍ୟମାନ ।
(d) ଯେଉଁ ଅନୁକ୍ରମର t = n – 1, ତାହା ଏକ A.P. ଅଟେ ।
(e) ଯେଉଁ ଅନୁକ୍ରମର S_n = \(\frac{n(n-1)}{2}\) ତାହା A.P. ଅଟେ ।
(f) ଯଦି କୌଣସି ତ୍ରିଭୁଜର କୋଣ ତ୍ରୟର ପରିମାଣର ଅନୁପାତ 2 : 3 : 4 ହୁଏ, ତେବେ କୋଣତ୍ରୟର ପରିମାଣ ଗୋଟିଏ A.P. ଗଠନ କରିବେ ।
(g) ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ ଗୋଟିଏ A.P.ରେ ରହିପାରିବେ ।
(h) ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ A.P. ଗଠନ କରନ୍ତି ନାହିଁ ।
(i) 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ଏକ A.P. ଅଟନ୍ତି ।
(j)5, x, ୨ ସଂଖ୍ୟାତ୍ରୟ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିଲେ x = 6 I
ଉ –
(a) ଠିକ୍, (b) ଭୁଲ୍, (c) ଭୁଲ୍, (d) ଠିକ୍, (e) ଠିକ୍, (f) ଭୁଲ୍, (g) ଭୁଲ୍, (h) ଭୁଲ୍, (i) ଠିକ୍, (j) ଭୁଲ

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର :
(a) 1, 2, 3, 4 ………. ସମାନ୍ତର ପ୍ରଗତି ସୃଷ୍ଟି କରନ୍ତି । (ଠିକ୍)
(କାରଣ ଏହି ଅନୁକ୍ରମର ସାଧାରଣ ଅନ୍ତର (d) = 1)
(b) 1, – 1, 1, – 1, ……. ଅନୁକ୍ରମଟି ସମାନ୍ତର ପ୍ରଗତି ଅଟେ । (ଭୁଲ୍)
(କାରଣ ଏହି ଅନୁକ୍ରମର ଓଁ ସମାନ ନୁହେଁ ।)
(c) 2, 1, – 1, – 2 ସଂଖ୍ୟାଟି ଚାରିଗୋଟି ସମାନ୍ତର ପ୍ରଗତିରେ ବିଦ୍ୟମାନ । (ଭୁଲ୍)
(କାରଣ 1 – 2 = -1, -1 – 1 = -2 ଏଠାରେ d ଅସମୀନ)
(d) t = n – 1 ଏକ A.P. ଅଟେ । (ଠିକ୍)
(କାରଣ ଅନୁକ୍ରମର ସାଧାରଣ ପଦ t_n = n – 1)
(e) ଯେଉଁ ଅନୁକ୍ରମର S_n = \(\frac{n(n-1)}{2}\) ତାହା A.P. ଅଟେ । (ଠିକ୍)
(f) ଯଦି କୌଣସି ତ୍ରିଭୁଜର କୋଣ ତ୍ରୟର ପରିମାଣର ଅନୁପାତ 2 : 3 : 4 ହୁଏ, ତେବେ କୋଣତ୍ରୟର ପରିମାଣ ଗୋଟିଏ A.P. ଗଠନ କରିବେ । (ଠିକ୍)
(କାରଣ କୋଣତ୍ରୟ 40°, 60°, 80° ଏହା A.P. ଅଟେ ।)
(g) ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ ଗୋଟିଏ A.P.ରେ ରହିପାରିବେ । (ଭୁଲ୍)
(କାରଣ 3, 4, 5; 6, 8, 10 ଇତ୍ୟାଦି ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁ; କିନ୍ତୁ ଯେକୌଣସି ସମକୋଣୀ ତ୍ରିଭୁଜ ପାଇଁ ଠିକ୍ ନୁହେଁ ।)
(h) ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ A.P. ଗଠନ କରନ୍ତି ନାହିଁ । (ଭୁଲ୍)
(3, 5, 7, 9 A.P. ଗଠନ କରନ୍ତି ।)
(i) 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ଏକ A.P. ଅଟନ୍ତି । (ଠିକ୍)
(କାରଣ 5, 10, 15, 20 A.P. ଅଟନ୍ତି ।)
(j) 5, x, ୨ A.P. ରେ ରହିଲେ x = 6 । (ଭୁଲ୍) ( କାରଣ x = \(\frac{9+5}{2}=\frac{14}{2}=7\)

‘ଖ’ ବିଭାଗ

Question 8.
(a) 1 + 2 + 3 + ….. ରେ S₃₀ କେତେ ?
(b) 1 + 3 + 5 + …….. ରେ S₁₀ କେତେ ?
(c) 2 + 4 + 6 + …….ରେ S₁₅ କେତେ ?
(d) 1 – 2 + 3 – 4 + ….. ରେ S₃₀ କେତେ ?
(e) 1 – 2 + 3 – 4 + …….. ରେ S₄₁ କେତେ ?
(f) 1 + 1 + 2 + 2 + 3 + 3 ……… ରେ S₁₇ କେତେ ?
(g) 1 + 2 + 3 + 2 + 3 + 4 + 3 + 4 + 5 ….. ରେ S₃₉ କେତେ ?
(h) -7 – 10 – 13 – ……… ରେ S₂₁ କେତେ ?
(i) 10 + 6 + 2 + ……. ରେ S₁₅ କେତେ ?
(j) 20 + 9 – 2 + …….. ରେ S₂₅ କେତେ ?
(k) n + (n – 1) + (n – 2) + ……… ରେ S_n କେତେ ?
(l) \(5+4 \frac{1}{3}+3 \frac{2}{3}\) + ……… ରେ S₂₀ କେତେ ?
ସମାଧାନ :
(a) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 1, d= 3 – 2 = 2 – 1 = 18 ଓ n = 30
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
S₃₀ = \(\frac{30}{2}\) {2.1 + (30 – 1) .1} = 15(2 + 29) = 15 × 31 = 465

(b) ଏଠାରେ a = 1, d= 3 – 1 = 5 – 3 = 2, n = 10
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
⇒ S₁₀ = \(\frac{10}{2}\) {2.1 + (10 – 1) 2} = 15(2 + 29) = 5 × 20 = 100

(c) ଏଠାରେ a = 2, d= 4 – 2 = 6 – 4 = 2, n = 15
S₁₅ = \(\frac{15}{2}\) {2 × 2 + (15 – 1) 2} = \(\frac{15}{2}\)(4 + 28) = 15 × 16 = 240

(d) 1 – 2 + 3 – 4 + ……… S₃₀
ଦତ୍ତ ଅନୁକ୍ରମଟି A.P.ରେ ନାହିଁ ।
ଦୁଇଟି ପଦକୁ ଗୋଟିଏ ପଦରେ ପରିଣତ କଲେ ଏହା ଅନୁକ୍ରମ ହେବ ଓଏହା S₁₅ ହେବ ।
= (1 – 2) + (3 – 4) + (5 – 6) + ……… (29 – 30)
= (- 1) + (- 1) + (- 1) + ………. = -1 × 15 = -15

(e) 1 – 2 + 3 – 4 + ……… ରେ S₄₁
ପଦସଂଖା 41 ହେତୁ ଅନୁକ୍ରମଟି
(1 – 2) + (3 – 4) + ……. (39 – 40) + 41
= ( – 1 ) + ( – 1) + (- 1) + ……. S₂₀ + 41 = -20 + 41 = 21

(f) 1 + 1 + 2 + 2 + 3 + 3 ……… ରେ S₁₇ ନିର୍ଣ୍ଣୟ କରାଯିବ ।
= (1 + 2 + 3 + ……. + 9) + (1 + 2 + 3 + ……. + 8)
= \(\frac{9×10}{2}+\frac{8×9}{2}\) (∵ Sn = \(\frac{n(n-1)}{2}\))
= 9 × 5 + 4 × 9 = 45 + 36 = 81

(g) 1 + 2 + 3 + 2 + 3 + 4 + 3 + 4 + 5 ….. ରେ S₃₉ କେତେ ?
ତିନୋଟି ପଦକୁ ଗୋଟିଏ ପଦରେ ପରିଣତ କଲେ, 39ଟି ପଦ \(\frac{39}{3}\) = 13ଟି ପଦ ହେବ ।
(1 + 2 + 3) + (2 + 3 + 4) + (3 + 4 +5) ………. ରେ S₁₃
6 + 9 + 12 + ……. ରେ S₁₃
= \(\frac{13}{2}\) {2 × 6 + (13 – 1) 3} = \(\frac{13}{2}\) {12 + 36} = \(\frac{13}{2}\) × 48 = 13 × 24 = 312

(h) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = – 7, d = -10 – (-7) = – 10 + 7 = – 3, n = 21
S₂₁ = \(\frac{21}{2}\) {2 · (- 7) + (21 – 1) (- 3)}
= \(\frac{21}{2}\) { – 14 – 60} = = \(\frac{21}{2}\) { -74) = 21 × (-37) =-777

(i) ଏଠାରେ a = 10, d = 6 – 10 = 2 – 6 = – 4, n = 15
S₁₅ = \(\frac{15}{2}\) {2 × 10 + (15 – 1) (- 4)}
= \(\frac{15}{2}\) {20+ (-56)} = \(\frac{15}{2}\) × -36 = 15 × (-18) = -270

(j) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 20, d = 9 – 20 = -2 – 9 = – 11, n = 25
S₂₅ = \(\frac{25}{2}\) {2 × 20+ (25 – 1) (- 11)}
= \(\frac{25}{2}\) {40 + 24 × (-11)} = \(\frac{25}{2}\) (40 – 264)
= \(\frac{25}{2}\) × – 224 = 25 × (- 112) = – 2800

(k) ଏଠାରେ a = n, d = (n – 1) – n = -1, n = n
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{n}{2}\) {2n + (n – 1)(-1)}
=\(\frac{n}{2}\) (2n – n + 1) = \(\frac{n}{2}\) (n + 1)

(l) ଏଠାରେ a = 5, d = \(4 \frac{1}{3}-5=\frac{13}{3}-5=\frac{13-15}{3}=-\frac{2}{3}\), n = 20
S₂₀ = \(\frac{20}{2}\) {2 × 5 + (20 – 1) (-3)} [∵ S_n = \(\frac{n}{2}\) {2a + (n – 1) d}]
= 10 {10 – \(\frac{38}{3}\)} = 10 (\(\frac{30-38}{3}\)) = 10 × \(\frac{-8}{3}\) = \(\frac{-80}{3}\) = \(-26 \frac{2}{3}\)

Question 9.
(a) ଯଦି a = 3, d = 4, n = 10, ତେବେ S_n କେତେ ?
(b) ଯଦି a = – 5, d = – 3, ତେବେ S₁₇ କେତେ ?
(c) ଯଦି t_n = 2n – 1, ତେବେ ପ୍ରଥମ 5ଟି ପଦ ଲେଖ ।
(d) ଯଦି t_n = 3n + 2, S₆₁ ନିର୍ଣ୍ଣୟ କର ।
(e) ଯଦି t_n = 3n – 5, ତେବେ S₅₀ ନିର୍ଣ୍ଣୟ କର ।
(f) ଯଦି t_n = 2 – 3n, ତେବେ S_n ନିର୍ଣ୍ଣୟ କର ।
(g) ଯଦି S_n = n², ତେବେ t₁₅ କେତେ ?
(h) ଏକ A.P.ର a = 3, d = 4, S_n = 903, ତେବେ n କେତେ?
(i) ଏକ A.P. ର d = 2, S₁₅ = 285, ତେବେ a କେତେ?
(j) ଏକ A.P. ର t₁₅ = 30, t₂₀ = 50, ତେବେ S₁₇ କେତେ?
ସମାଧାନ :
MBD
(a) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 5, d = 4, n = 10
S₁₀ = \(\frac{10}{2}\) {2 × 3 + (10 – 1) 4} = 5 {6 + 36} = 5 × 42 = 210

(b) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = -5, d = -3, n = 17
S₁₇ = \(\frac{17}{2}\) {2 × (-5) + (17 – 1) (-3)} = \(\frac{17}{2}\) {-10 – 48} = \(\frac{17}{2}\) × (-58) = 17 × (-29) = – 493

(c) t_n = 2n – 1,
t₁ = a = 2 × 1 – 1 = 2 – 1 = 1
t₂ = a = 2 × 2 – 1 = 4 – 1 = 3,
t₃ = a = 2 × 3 – 1 = 6 – 1 = 5,
t₄ = a = 2 × 4 – 1 = 8 – 1 = 7,
t₅ = a = 2 × 5 – 1 = 10 – 1 = 9,

(d) t_n = 3n + 2
t₁ = a = 3 × 1 + 2 = 5
t₂ = a = 3 × 2 + 2 = 8
t₃ = a = 3 × 3 + 2 = 11
ଏଠାରେ a = 5, d = 8 – 5 = 11 – 8 = 3, n = 61
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
S₆₁ = \(\frac{61}{2}\) {2 × 5+ (61 – 1) 3} = \(\frac{61}{2}\) {10 + 180} = 61 × 95 = 5795

(e) t_n = 3n – 5
t₁ = 3 × 1 – 5 = -2
t₂ = 3 × 2 – 5 = 1
t₂ = 3 × 3 – 5 = 4
a = -2, d = 1 – (- 2) = 4 – 1 = 3, n = 50
S₅₀ = \(\frac{50}{2}\) {2 × (-2) + (50 – 1) 3} = 25 × (- 4 + 147) = 25 × 143 = 3575

(f) t_n = 2 – 3n
t₁ = 2 – 3 × 1 = – 1
t₂ = 2 – 3 × 2 = -4
t₃ = 2 – 3 × 3 = -7 ଇତ୍ୟାଦି
ଏଠାରେ a = -1, d= – 4 – (- 1) = -7 – (- 4) = -3
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{n}{2}\) (2 × (- 1) + (n – 1) (- 3)}
= \(\frac{n}{2}\) (-2 – 3n + 3} = \(\frac{n}{2}\) (1 – 3n)

(g) t_n = S_n – S_n-1
S_n = n², S₁₅ = 15², S₁₄ = 14²
t₁₅ = S₁₅ – S₁₄ = 15² – 14² = 225 – 196 = 29

(h) ଏକ A.P.ର a = 3, d = 4, S_n = 903

∴ n = 2

(i) A.P.ର d = 2, S₁₅ = 285, n = 15

(j) ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a ଓ ସାଧାରଣ ଅନ୍ତର = d
t₁₅ = 30 ⇒ a + (15 – 1) d = 30 ⇒ a + 14d = 30 …(i)
t₂₀ = 50 ⇒ a + (20 – 1) d = 50 ⇒ a + 19d = 50 …(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ

d ର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ
a + 14d = 30 = a + 14 × 4 = 30
= a + 56 = 30 = a = 30 – 56 = -26
S₁₇ = \(\frac{17}{2}\) {-52 + (17 – 1)4} = \(\frac{17}{2}\) {-52 + 64}= \(\frac{17}{2}\) × 12 = 17 × 6 = 102

Question 10.
(i) ‘ଓଲଟାଇ ମିଶାଇବା କୌଶଳରେ’ ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
(a) 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।
(b) 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।
(c) 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।

(ii) 1, 2, 3, …….. ଅନୁକ୍ରମର।
(a) S₂₀ ନିଶ୍ଚୟ କର (b) S₅₀ ନିର୍ଣ୍ଣୟ କର ।
(iii) 32 ଠାରୁ 85 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
(iv) 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
(v) 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(i) (a) 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 105
ମନେକର 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

∴ 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ଯୋଗଫଳ = 5565

(b) 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 93 – 24 = 69
ମନେକର 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

∴ 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି = 4071

(c) 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ସଂଖ୍ୟା = 222 – 110 = 112
ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

⇒ 2S = 333 × 112 ⇒ S = \(\frac{333×112}{2}\) = 333 × 56 = 18648
∴ ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି = 18648

(ii) S_n = \(\frac{n(n-1)}{2}\)

(a) 1, 2, 3 ଅନୁକ୍ରମର S₂₀ = \(\frac{20×21}{2}\) = 10 × 21 = 210

(b) S₅₀ = \(\frac{50×51}{2}\) = 25 × 51 = 1275

(iii) S = 32 +33 +34 …….. + 85
ଏଠାରେ a = 32, d = 33 – 32 = 1, n = 85 – 31 = 54
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{54}{2}\) {2 × 32 + (54 – 1) × 1}
= 27 × {64 +53} = 27 × 117 = 3159
∴ ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି 3159।

(iv) S_n = \(\frac{n(n-1)}{2}\)
100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି
S = 2 + 4 + 6 + 8 …… + 98
= 2 (1 + 2 + 3 + 4 ……. + 49)
= 2 × \(\frac{49 (49 + 1)}{2}\) = \(\frac{49×50}{2}\) = 49 × 50 = 2450
∴ 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି = 2450

(v) 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି
S = 1 + 3 + 5 + 7 ……. + 149
ଏଠାରେ a = 1, d = 3 – 1 = 5 – 3 = 2
t_n = 149
⇒ a + (n – 1) d = 149 ⇒ 1 + (n – 1) 2 = 149
⇒ n – 1 = \(\frac{149 – 1}{2}=\frac{148}{2}=74\) ⇒ n = 74 + 1 = 75
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{75}{2}\) {2 × 1 + (75 – 1) 2}
= \(\frac{75}{2}\) {2 + 148} = \(\frac{75×150}{2}\)
= 75 × 75 = 5625
∴ 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି 5625 ।

Question 11.
ଯେଉଁ ସମାନ୍ତର ଅନୁକ୍ରମର ପ୍ରଥମ ପଦ 17 ଓ ସାଧାରଣ ଅନ୍ତର – 2 ତାହାର କେତୋଟି ପଦର ସମଷ୍ଟି 72 ହେବ ? ଏହାର ଦୁଇଟି ଉତ୍ତର ମିଳିବାର କାରଣ ଲେଖ ।
ସମାଧାନ :
ଏକ ଅନୁକ୍ରମର ପ୍ରଥମ ପଦ (a) = 17, ସାଧାରଣ ଅନ୍ତର (d) = – 2
ମନେକର nଟି ପଦର ସମଷ୍ଟି 72 ।
⇒ S_n = 72 ⇒ \(\frac{n}{2}\) {2a + (n – 1) d} = 72
⇒ \(\frac{n}{2}\) {2 × 17 + (n – 1) × (- 2)} = 72 ⇒ \(\frac{n}{2}\) {34 – 2n + 2} = 72
⇒ \(\frac{n}{2}\) {36 – 2n } = 72 ⇒ \(\frac{n}{2}\) × 2 (18 – 2n) = 72
⇒ 18n – 2n² – 72 = 0 ⇒ -2 (n² – 18n + 72) = 0
⇒ n² – 18n + 72 = 0 ⇒ n² – 12n – 6n + 72 = 0
⇒ n (n – 12) – 6 (n – 12) = 0 ⇒ (n – 6) (n – 12) = 0
n = 6 ବା n = 12
t₇ = a + (7 – 1) d = 17 + 6 (- 2) = 17 – 12 = 5
t₈ = a + (8 – 1) d = 17 + 7 (- 2) = 17 – 14 = 3
t₉ = a + (9 – 1) d = 17 + 8 (- 2) = 17 – 16 = 1
t₁₀ = a + (10 – 1) d = 17 + 9 (- 2) = 17 – 18 = – 1
t₁₁ = a + (11 – 1) d = 17 + 10 (- 2) = 17 – 20 = -3
t₁₂ = a + (12 – 1) d = 17 + 11 (- 2) = 17 – 22 = -5
∴ t₇ + t₁₁ ………. + t₁₂ = (5 + 3 + 1) – (1 + 3 + 5) = 0
∴ ସପ୍ତମ ପଦରୁ ଦ୍ଵାଦଶ ତମ ପଦ ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକର ଯୋଗଫଳ 0 ହୋଇଥିବାରୁ
ପ୍ରଥମ 6ଟି ପଦର ସମଷ୍ଟି = ପ୍ରଥମ 12 ଟି ପଦର ସମଷ୍ଟି
ତେଣୁ ଆମେ ଦୁଇଟି ଉତ୍ତର ପାଇଲୁ ।

Question 12.
(i) ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ତିନୋଟି ରାଶିର ଯୋଗଫଳ 18 ଏବଂ ଗୁଣଫଳ 192 ହେଲେ, ସଂଖ୍ୟାଗୁଡ଼ିକ ସ୍ଥିର କର ।
(ii) ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ଛଅଟି ପଦ ମଧ୍ୟରୁ ପ୍ରାନ୍ତ ପଦଦ୍ୱୟର ଯୋଗଫଳ 16 ଏବଂ ମଧ୍ୟ ପଦଦ୍ୱୟର ଗୁଣଫଳ 63 ହେଲେ, ପଦଗୁଡ଼ିକ ସ୍ଥିର କର ।
ସମାଧାନ :
(i) ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ପଦତ୍ରୟ a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ ପଦତ୍ରୟର ଯୋଗଫଳ = 18
⇒ a – d + a + a + d = 18 ⇒ 3a = 18 ⇒ a = \(\frac{18}{3}\) ⇒ a = 6.
ପୁନଶ୍ଚ (a – d) × a (a + d) = 192 ⇒ a (a² – d²) = 192
⇒ 6 {(6)² – d²} = 192 ⇒ 36 – d² = \(\frac{192}{6}\) = 32
⇒ d² = 32 – 36 = -4 ⇒ d² = 4 ⇒ d= ±√4⇒ d = ±2
a = 6 ଓ d = 2 ହେଲେ A.P. ର ପଦତ୍ରୟ a – d = 6 – 2 = 4
a = 6 ଏବଂ a + d = 6 + 2 = 8
a = 6 ଓ d = – 2 ହେଲେ A.P.ର ପଦତ୍ରୟ a – d = 6 – (-2) = 6 + 2 = 8
a = 6, a – d = 6 + (- 2) = 6 – 2 = 4
∴ A.P.ର ପଦତ୍ରୟ 4, 6, 8 ବା 8, 6, 4 ।

(ii) ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ଛଅଟି ପଦ a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d ।
ପ୍ରଶ୍ନନୁସାରେ, a – 5d + a + 5d = 16 ⇒ a = \(\frac{16}{2}\) = 8
ପୁନଶ୍ଚ (a – d) × (a + d) = 63
⇒ (a² – d²) = 63 ⇒ 64 – d² = 63
⇒ d² = 1 ⇒ d = ± √1 = ±1
a = 8 ଓ d = 1 ହେଲେ ପଦଗୁଡ଼ିକ a – 5d = 8 – 5 = 3, a – 3d = 8 – 3 = 5,
a – d = 8 – 1 = 7, a + d = 8 + 1 = 9, a + 3d = 8 + 3 = 11, a + 5d = 8+ 5 = 13,
a = 8 ଓ d = -1 ହେଲେ ପଦଗୁଡ଼ିକ a – 5d = 8 – 5 (-1) = 8 + 5 = 13, a – 3d = 8 – 3 (- 1)
= 8 + 3 = 11, a – d = 8 – (- 1) = 9
a + d = 8+ (-1) = 7, a + 3d = 8 + 3 (- 1) = 8 – 3 = 5
a + 5d = 8 +5(-1) = 8 – 5 = 3
∴ A.P.ର ପଦତ୍ରୟ 3, 5, 7, 9, 11, 13 ବା 13, 11, 9, 7, 5, 3 ।

Question 13.
ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ତିନୋଟି ପଦର ଯୋଗଫଳ 21 ଏବଂ ସେମାନଙ୍କ ବର୍ଗର ଯୋଗଫଳ 155; ପଦଗୁଡ଼ିକ କେତେ ?
ସମାଧାନ :
ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ତିନୋଟି ପଦ a + d, a, a + 3d ।
ପ୍ରଶ୍ନନୁସାରେ, a – d + a + a + d = 21 ⇒ 3a = 21 ⇒ a = \(\frac{21}{3}\) = 7
ପୁନଶ୍ଚ a (a – d)² + a² + (a + d)² = 155 ⇒ a² + (a – d)² + (a + d)² = 155
⇒ a² + 2 (a² + d²) = 155 ⇒ 7² + 2 (7² + d²) = 155
⇒ 2 (49 + d²) = 155 – 49 = 106 ⇒ 98 + 2d² = 106
⇒2d² = 106 – 98 ⇒ d² = \(\frac{8}{2}\) = 4 ⇒ d = = ±√4 = ±2
a = 7 ଓ d = 2 ହେଲେ ପଦଗୁଡ଼ିକ a – d = 7 – 2 = 5, a = 7, a + d = 7 + 2 = 9
a = 7 ଓ d = -2 ହେଲେ ପଦଗୁଡ଼ିକ a – d = 7 – (- 2) = 7 + 2 = 9,
a = 7, a + d = 7 + (- 2) = 7 – 2 = 5
∴ ସମାନ୍ତର ଅନୁକ୍ରମର ପଦତ୍ରୟ 5, 7, 9 ବା 9, 7, 5 ।

Question 14.
ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଥିଲେ ପ୍ରମାଣ କର ଯେ, ସେମାନଙ୍କର ଅନୁପାତ 3 : 4 : 5 ହେବ ।
ସମାଧାନ :
ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ।
ମନେକର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ, (a – d)² + a² = (a + d)² ⇒ (a + d)² – (a – d)² = a²
⇒ (a + d + a + d) (a + d – a + d) = a² ⇒ 2 × 2d = a² ⇒ a = 4d
⇒ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ, a – d = 4d – d = 3d, a = 4d,
a + d = 4d + d = 5d
ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ଅନୁପାତ = 3d : 4d : 5d = 3 : 4 : 5 (ପ୍ରମାଣିତ)

Question 15.
100 ରୁ କ୍ଷୁଦ୍ରତର ଏବଂ 5 ଦ୍ୱାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ଏବଂ 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ = S
S = 5 + 10 + 15 ……… + 95
ଏଠାରେ a = 5, d = 5, n = \(\frac{95}{5}\) = 19
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
= \(\frac{19}{2}\) {2 × 5+ (19 – 1) 5} = \(\frac{19}{2}\) (10 + 90) = \(\frac{19}{2}\) × 100 = 19 × 50 = 950
∴ ନିର୍ଦେୟ ଯୋଗଫଳ = 950

Question 16.
200 ରୁ କ୍ଷୁଦ୍ରତର ଓ 3 ଦ୍ଵାରା ଅବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ବସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର 1 ଠାରୁ ଆରମ୍ଭ କରି 200ରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ = S₁
S₁ = 1 + 2 + 3 + 4 ……. + 199 = \(\frac{199×200}{2}\) = 199 × 100 = 19900
ପୁନଶ୍ଚ 200 ଠାରୁ କ୍ଷୁଦ୍ରତର 3 ଦ୍ଵାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ
S₂ = 3 + 6 + 9 ………. + 198
ଏଠାରେ a = 3, d = 6-3 = 3, n = \(\frac{198}{3}\) = 66
S₂ = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{66}{2}\) {2 × 3 + (66 – 1) × 3}
= 33 {6 + 65 × 3) = 33 × (6 + 195) = 33 × 201 = 6633
200 ଠାରୁ କ୍ଷୁଦ୍ରତର 3 ଦ୍ଵାରା ଅବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ
= S = S₁ – S₂ = 19900 – 6633 = 13267

Question 17.
15 କୁ ଏପରି 3 ଭାଗରେ ବିଭକ୍ତ କର ଯେପରିକି ସେମାନେ ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ରହିବେ ଓ ସେମାନଙ୍କର ଗୁଣଫଳ 120 ହେବ ।
ସମାଧାନ :
15କୁ ଏପରି ତିନି ଭାଗରେ ବିଭକ୍ତ କରାଯିବ ଯେପରି ସେମାନେ A.P.ରେ ରହିବେ ।
ମନେକର ସଂଖ୍ୟା ତ୍ରୟ a – d, a, a + d ।
ପ୍ରଶାନୁସାରେ, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = 5
ପୁନଶ୍ଚ (a – d) a (a + d) = 120
⇒ 5 (5² – d²) = 120 ⇒ 25 – d² = \(\frac{120}{5}\) ⇒ d² = 25 – 24
⇒ d² = 1 = d = +1
a=5 ଓ d = 1 ହେଲେ a – d = 5 – 1 = 4, a = 5, a + d = 5 + 1 = 6
a = 5 ଓ d = -1 ହେଲେ a – d = 5 – (- 1) = 5 + 1 = 6, a= 5,
a + d = 5 + (- 1) = 5 – 1= 4
∴ ସଂଖ୍ୟାତ୍ରୟ 4, 5, 6 ବା 6, 5,4 ।

Question 18.
A.P. ରେ ଥ‌ିବା ତିନୋଟି ପଦର ଯୋଗଫଳ 15 ଏବଂ ପ୍ରାନ୍ତ ପଦଦ୍ୱୟର ବର୍ଗର ଯୋଗଫଳ 58 ହେଲେ ପଦତ୍ରୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର A.P. ରେ ଥ‌ିବା ପଦ ତିନୋଟି a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = \(\frac{15}{3}\) = 5
ପୁନଶ୍ଚ (a – d)² + (a + d)² = 58 ⇒ 2 (a² + d²) = 58
⇒ 5²+ d² = \(\frac{58}{2}\) ⇒ d² = 29 – 25 = 4 ⇒ d = ± √4 = ±2
a = 5 ଓ d = 2 ହେଲେ ପଦତ୍ରିୟ a – d = 5 – 2 = 3, a = 5, a + d = 5 + 2 = 7 ̧
a = 5 ଓ d = -2 ହେଲେ ପଦତ୍ରିୟ a – d = 5 – (- 2) = 5 + 2 = 7, a = 5,
a + d = 5 + (-2) = 3 ।
∴ A.P.ର ପଦତ୍ରୟ 3, 5, 7 ବା 7, 5, 3 ।

Question 19.
A.P. ରେ ଥିବା ଚାରୋଟି ପଦ ମଧ୍ୟରୁ ପ୍ରାନ୍ତ ପଦ ଦ୍ଵୟର ଯୋଗଫଳ 8 ଏବଂ ମଧ୍ୟ ପଦ ଦ୍ଵୟର ଗୁଣଫଳ 15 ହେଲେ ପଦଗୁଡ଼ିକ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର A.P.ରେ ଥ‌ିବା ଚାରୋଟି ପଦ a – 3d, a – d, a + d ଓ a + 3d ।
ପ୍ରଶ୍ନନୁସାରେ, a – 3d + a + 3d = 8⇒ 2a = 8 ⇒ a = \(\frac{8}{2}\) = 4
ପୁନଶ୍ଚ (a – d) (a + d) = 15 ⇒ a² – d² = 15 ⇒ 4² – 15 = d²
⇒ d² = 16 – 15 = 1 ⇒ d = ±√1 = ±1
a = 4 ଓ d = 1 ହେଲେ ପଦତ୍ରିୟ a – 3d = 4 – 3 (1) = 4 – 3 = 1
a – d = 4 – 1 = 3, a + d = 4 + 1 = 5
a + 3d = 4 + 3 · 1 = 4 + 3 = 7
a =4 ଓ d = 1 ହେଲେ ପଦତ୍ରିୟ a – 3d = 4 – 3 (-1) = 4 + 3 = 7,
a – d = 4 – (-1) = 4 + 1 = 5,
a + d = 4 + (-1) = 3 ଏବଂ a + 3d = 4 + 3 (- 1) = 4 – 3 = 1
∴ A.P.ର ଥ‌ିବା ପଦତ୍ରୟ 1, 3, 5, 7 ବା 7, 5, 3, 1 ।

Question 20.
A.P. ରେ ଥ‌ିବା ତିନୋଟି ରାଶିମାଳାର n ସଂଖ୍ୟକ ପଦମାନଙ୍କର ସମଷ୍ଟି S₁ S₂ ଏବଂ S₃ । ପ୍ରତ୍ୟେକ ରାଶିମାଳାର ପ୍ରଥମ ପଦ 1 ଏବଂ ସାଧାରଣ ଅନ୍ତର ଯଥାକ୍ରମେ 1, 2, 3 ହେଲେ ପ୍ରମାଣ କର ଯେ, S₁ + S₃ = 2S₂ ।
ସମାଧାନ :
A.P. ରେ ଥିବା ତିନୋଟି ରାଶିମାଳାର n ସଂଖ୍ୟକ ପଦମାନଙ୍କର ସମଷ୍ଟି S₁ S₂ ଏବଂ S₃ ।
ପ୍ରତ୍ୟେକ ରାଶିମାଳାର ପ୍ରଥମ ପଦ = a
ପ୍ରଥମ ରାଶିମାଳାର d = 1, ଦ୍ୱିତୀୟ ରାଶିମାଳାର d = 2, ତୃତୀୟ ରାଶିମାଳାର d = 3
S₁ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 1} = \(\frac{n}{2}\) (n + 1)
S₂ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 2} = n²
ଏବଂ S₃ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 3} = \(\frac{n}{2}\) (3n – 1)
L.H.S. = S₁ + S₃ = \(\frac{n}{2}\) (n + 1) + \(\frac{n}{2}\) (3n – 1)
= \(\frac{n}{2}\) {n + 1 + 3n – 1}
R.H.S. = 2S₂ = 2 × n² = 2n²
∴ L.H.S = R.H.S. (ପ୍ରମାଣିତ)

Question 21.
ଏକ A.P. ର p-ତମ, ୟୁ-ତମ ଏବଂ r-ତମ ପଦଗୁଡ଼ିକର ମାନ ଯଥାକ୍ରମେ a, b ଏବଂ c ହେଲେ ପ୍ରମାଣ 6, a (q – r) + b (r -p) + c (p – q) = 0 |
ସମାଧାନ :
ମନେକର A.P. ର ପ୍ରଥମ ପଦ = x, ସାଧାରଣ ଅନ୍ତର = d
t = x + (p – 1) d = a ……… (i)
t = x + (q – 1) d = b ……… (ii)
t = x + (r – 1) d = c …….(iii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,

⇒ d (p – q) = a – b ⇒ d = \(\frac{a – b}{p – q}\) …… (iv)
ସେହିପରି (ii)ରୁ (iii)କୁ ବିୟୋଗ କଲେ ଆମେ ପାଇବା d = \(\frac{b – c}{q – r}\) …… (v)
ସମୀକରଣ (iv) ଓ (v) ରୁ ⇒ \(\frac{a – b}{p – q}=\frac{b – c}{q – r}\) ⇒ (a – b) (q- r) = (b -c) (p – q)
⇒ a (q – r) – b(q – r) = b(p – q) – c (p – q)
⇒ a (q – r) – b(q – r) = b (p – q) + c (p – q) = 0
⇒ a (q – r) – b(q – r + p – q) + c (p – q) = 0
⇒ a (q – r) – b(p – r) + c (p – q) = 0
⇒ a (q – r) – b(r – p) + c (p – q) = 0 (ପ୍ରମାଣିତ)
ବି.ଦ୍ର. : (i) ରୁ a = x + (p – 1)d
∴ a(q – r) = x (q – r) + (p – 1) (q – r)d
ସେହିପରି b(r- p) = x (r – p) + (q – 1) (r – p)d ଏବଂ
c (p – q) = x (p – q) + (r – 1) (p – q) d
ଯୋଗକଲେ a (q – r) + b (r- p) + c (p -q) = 0 ପାଇବା

Question 22.
ତିନୋଟି ସଂଖ୍ୟା a, b, ୯ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିଲେ ପ୍ରମାଣ କର ଯେ ନିମ୍ନରେ ପ୍ରଦତ୍ତ ସଂଖ୍ୟା ତ୍ରୟ ମଧ୍ୟ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ ।
(i) \(\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\)
(ii) b+c, c + a, a + b
(iii) b+c-a, c + a-b, a + b-c
(iv) \(\frac{1}{a}(\frac{1}{b}+\frac{1}{c}),\frac{1}{b}(\frac{1}{c}+\frac{1}{a}),\frac{1}{c}(\frac{1}{a}+\frac{1}{b})\)
(v) a² (b+c), b² (c + a), c²(a + b)
ସମାଧାନ :
(i) a, b, c ସମାନ୍ତର ପ୍ରଗତିରେ ଅବସ୍ଥିତ ।
ପ୍ରତ୍ୟେକ ପଦକୁ abc ଦ୍ବାରା ଭାଗକଲେ \(\frac{1}{abc},\frac{1}{abc},\frac{1}{abc}\) ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ ।
⇒ \(\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\) (ପ୍ରମାଣିତ)

(ii) a, b, c ସମାନ୍ତର ପ୍ରଗତିରେ ଅବସ୍ଥିତ ।
(a + b + c) ବିୟୋଗ କଲେ ବିୟୋଗଫଳ A.P.ରେ ରହିବ ।
ଅର୍ଥାତ୍ (a+b+c), b – (a + b + c), c (a + b + c) A.P.ରେ ରହିବ ।
⇒ – (b + c), -(c + a), -(a + b) A.P.ରେ ରହିବ ।
⇒ b + c, c + a, a + b ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ । (ପ୍ରତ୍ୟେକ ପଦକୁ –1 ଦ୍ଵାରା ଗୁଣିଲେ)

(iii) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
⇒ 2a, 2b, 2c ମଧ୍ଯ A.P. ରେ ରହିବେ (ପ୍ରତ୍ୟେକ ପଦରେ 2 ଗୁଣନ କଲେ ।)
ପ୍ରତ୍ୟେକରୁ (a + b + c) ପଦକୁ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ A.P ରେ ରହିବ ।
⇒ 2a – (a+b+c), 2b – (a + b + c), 2c – (a + b + c) A.P.ରେ ରହିବ ।
⇒ (b + c – a), (c + a – b), (a + b – c) A.P.ରେ ରହିବ ।
⇒ b+c-a, c+a-b, a+b-c A.P.ରେ ରହିବ ।

(iv) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ (-1) ଗୁଣନ କଲେ, -a, -b, -c A.P.ରେ ରହିବ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ a + b + c ଯୋଗକଲେ
a + b + c – a, a + b + c – b, a + b + c – c A.P.ରେ ରହିବ ।
∴ b + c, c + a, a + b A.P.ରେ ରହିବ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦକୁ abc ଦ୍ଵାରା ଭାଗକଲେ \(\frac{b + c}{abc},\frac{c + a}{abc},\frac{a + b}{abc}\) A.P.ରେ ରହିବ ।
⇒ \(\frac{1}{a}(\frac{b + c}{bc}),\frac{1}{b}(\frac{c + a}{ac}),\frac{1}{c}(\frac{a + b}{ab})\) A.P.ରେ ରହିବ ।
⇒ \(\frac{1}{a}(\frac{1}{b}+\frac{1}{c}),\frac{1}{b}(\frac{1}{c}+\frac{1}{a}),\frac{1}{c}(\frac{1}{a}+\frac{1}{b})\) A.P.ରେ ରହିବ ।

(v) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ ab + bc + ca ଗୁଣନ କଲେ ଗୁଣଫଳ A.P.ରେ ରହିବ ।
a (ab + bc + ca), b (ab + bc + ca), c (ab + bc + ca) A.P.ରେ ରହିବେ ।
a²b+ abc + ca², ab²+ b²c + abc, abc + bc² + c²a A.P.ରେ ରହିବ ।
= A.P.ର ପ୍ରତ୍ୟେକ ପଦରୁ abc ବିୟୋଗ କଲେ ଲବ୍‌ଧ ଅନୁକ୍ରମ A.P.ରେ ରହିବ ।
a²b + ca², ab² + b²c, bc² + c²a A.P.ରେ ରହିବ ।
a²(b+c), b²(c + a), c²(a + b), A.P.ରେ ରହିବ ।

Question 23.
(i) \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) ରେ ରହିଲେ ଏବଂ a + b + c ≠ 0 ହେଲେ, ପ୍ରମାଣ କର ଯେ \(\frac{b + c}{a},\frac{c + a}{b},\frac{a + b}{c}\) ମଧ୍ୟ A.P. ରେ ରହିବେ ।
(ii) \(\frac{a}{b + c},\frac{b}{c + a},\frac{c}{a + b}\) ଅନୁକ୍ରମ A.P.ରେ ରହିଲେ ଏବଂ a + b + c ≠ 0 ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{1}{b + c},\frac{1}{c + a},\frac{1}{a + b}\) A.P.ରେ ରହିବେ ।
ସମାଧାନ :
(i) \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) A.P.ରେ ଅବସ୍ଥିତ । (a + b + c ≠ 0)
⇒ A.P. ର ପ୍ରତ୍ୟେକ ପଦରେ a + b + c ଗୁଣନ କଲେ,
\(\frac{a+b+c}{a},\frac{a+b+c}{b},\frac{a+b+c}{c}\) A.P.ରେ ରହିବ ।
⇒ \(\frac{a}{a}+\frac{b+c}{a}, \frac{c+a}{b}+\frac{b}{b}, \frac{a+b}{c}+\frac{c}{c}\)
⇒ \(1+\frac{b+c}{a}, \frac{c+a}{b}+1, \frac{a+b}{c}+1\) A.P.ରେ ରହିବ ।
⇒ \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) A.P.ରେ ରହିବ । (ପ୍ରତ୍ୟେକ ପଦରୁ 1 ବିୟୋଗ କଲେ ।) (ପ୍ରମାଣିତ)

Question 24.
ଯଦି କୌଣସି A.P.ର ପ୍ରଥମ ପଦ a ଏବଂ ଶେଷ ପଦ l ହୁଏ ପ୍ରମାଣ କର ଯେ, ଅନୁକ୍ରମର ପ୍ରଥମରୁ r ତମ ପଦ ଏବଂ ଶେଷରୁ r ତମ ପଦର ସମଷ୍ଟି, ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି ସହିତ ସମାନ ।
ସମାଧାନ :
A.P.ର ପ୍ରଥମପଦ = a, ଶେଷ ପଦ = l
ମନେକର ସାଧାରଣ ଅନ୍ତର = d ।
∴ A.P. ଟି a, a + d, a + 2d, a + 3d, …….. l + 2d, l + d, l
ପ୍ରଥମରୁ rତମ ପଦ = a + (r – 1) d
ଶେଷ r ତମ ପଦର a = l, d = -d
t_r = l + (r – 1 ) (-d)
∴ ପ୍ରଥମରୁ r ତମ ପଦ + ଶେଷରୁ r ତମ ପଦ = [a + (r – 1) d] + [l + (r – 1) (- d)]
= a + (r – 1) d + l – (r – 1) d = a + l
∴ ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି = a + l
∴ ପ୍ରଥମରୁ r ତମ ପଦ ଓ ଶେଷରୁ 1 ତମ ପଦର ସମଷ୍ଟି = ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ : ପ୍ରଥମରୁ 1 ତମ ପଦ (t) = a + (r – 1)d
ଶେଷରୁ r ତମ ପଦ = ପ୍ରଥମରୁ (n – r + 1) ତମ ପଦ
= t_n-r+1 = a + {(n – r+1) – 1}d = a + (n – r)d
∴ t_r + _n-r + 1 = a = (r – 1 )d + a + (n – r) d = 2a + (n – 1)d = a + {a + (n – 1)d} = a + 1

Question 25.
ଗୋଟିଏ ସମାନ୍ତର ପ୍ରଗତିର ପ୍ରଥମ p ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି r, ପ୍ରଥମ q ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି s ଏବଂ ସାଧାରଣ ଅନ୍ତର । ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{r}{p}-\frac{s}{q}\) = (p – q) \(\frac{d}{2}\) ହେବ ।
ସମାଧାନ :
ଗୋଟିଏ ସମାନ୍ତର ପ୍ରଗତିର S_p = x ଏବଂ S_q = s
ସାଧାରଣ ଅନ୍ତର = d । ମନେକର ପ୍ରଥମ ପଦ = a
.. ପ୍ରଥମ p-ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି = S_p = \(\frac{p}{2}\) {2a + (p – 1) d}
ପ୍ରଶାନୁସାରେ, \(\frac{p}{2}\) {2a + (p – 1) d} = r ⇒ 2a + (p – 1) d = \(\frac{2r}{p}\) …….(i)
.. ପ୍ରଥମ q-ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି = S_q = \(\frac{q}{2}\) {2a + (q – 1) d}
ପୁନଶ୍ଚ, \(\frac{q}{2}\) {2a + (q – 1) d} = s ⇒ 2a + (q – 1) d = \(\frac{2s}{q}\) …….(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,

Question 26.
ଗୋଟିଏ ସମାନ୍ତର ଶ୍ରେଣୀର ପ୍ରଥମ p, q, r ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି a, b, c ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{a}{p}\) (q – r) + \(\frac{b}{q}\) (r – p) + \(\frac{c}{r}\) (p – q) = 0 ହେବ ।
ସମାଧାନ :
ଗୋଟିଏ ସମାନ୍ତର ଶ୍ରେଣୀର ପ୍ରଥମ p, q, r ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି a, b, c ।
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = x ଓ ସାଧାରଣ ଅନ୍ତର = d
S_p = \(\frac{p}{2}\) {2a + (p – 1) d} = a ⇒ 2x + (p – 1) d = \(\frac{2a}{p}\) ……(i)
S_q = \(\frac{q}{2}\) {2a + (q – 1) d} = b ⇒ 2x + (q – 1) d = \(\frac{2b}{q}\) ……(ii)
S_r = \(\frac{r}{2}\) {2a + (r – 1) d} = c ⇒ 2x + (r – 1) d = \(\frac{2c}{r}\) ……(iii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,


ବି.ଦ୍ର. : (i) ରୁ 2 =x + \(\frac{1}{2}\)(p – 1)d ⇒ \(\frac{a}{p}\) (q – r) = x(q – r) + \(\frac{1}{2}\) (p – 1)(q – r)d
ସେହିପର \(\frac{b}{q}\)(r – p) = x(r – p) + \(\frac{1}{2}\) (q – 1)(r – p)d
c(p – q) = x(p – q) + 2(r – 1)(p – q)d
ଯୋଗକଲେ \(\frac{a}{p}\) (q – r) + \(\frac{b}{q}\) (r – p) + \(\frac{c}{r}\) (p – q) = 0 ହେବ ।

Question 27.
କୌଣସି A.P.ର t = q, t = p ହେଲେ ପ୍ରମାଣ କର ଯେ t = p + q – m ।
ସମାଧାନ :
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a ଓ ସାଧାରଣ ଅନ୍ତର = d
ଏକ A.P.ର t_p = q ⇒ a + (p – 1)d = q … (i)
⇒ t_q = p ⇒ a + (q – 1)d = p … (ii)
t_p – t_q = q – p ⇒ a + (p – 1)d – a + (q – 1)d = q – p
⇒ d(p – 1 – q + 1) = q – p ⇒ d(p – q) = q – p
⇒ d = \(\frac{q-p}{p-q}=\frac{-(p-q)}{p-q}=-1\)
‘d’ର ମାନ (i) ରେ ବସାଇଲେ a + (p – 1) d = q
⇒ a + (p – 1) (-1) = q ⇒ a – p + 1 = q
⇒ a = p + q – 1
:. t = a + (m – 1)d = p + q – 1 +(m – 1)(-1)
= p + q – 1 – m + 1 = p + q – m

Question 28.
କୌଣସି A.P.ର S_m = n, S_n = m ହେଲେ, ପ୍ରମାଣ କର ଯେ S_m+n = -(m + n) ହେବ ।
ସମାଧାନ :
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a, ସାଧାରଣ ଅନ୍ତର = d
S_m = n ⇒ \(\frac{m}{2}\) {2a + (m – 1) d} = n ⇒ am + (m² – m) \(\frac{d}{2}\) = n ……(i)
S_n = m ⇒ \(\frac{n}{2}\) {2a + (n – 1) d} = m ⇒ an + (n² – n) \(\frac{d}{2}\) = m ……(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(c)

Question 1.

Evaluate the following limits :
(i) \(\lim _{x \rightarrow 0} \frac{x}{\sin 2 x}\)
Solution:

(ii) \(\lim _{x \rightarrow 0} \frac{\sin 3 x}{\sin 5 x}\)
Solution:

(iii) \(\lim _{x \rightarrow 0} \frac{\sin m x}{\sin n x}\)
Solution:

(iv) \(\lim _{x \rightarrow 0} \frac{\tan \alpha x}{x}\)
Solution:

(v) \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}\)
Solution:

(vi) \(\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{x}\)
Solution:

(vii) \(\lim _{x \rightarrow \pi} \frac{\sin x}{\pi-x}\)
Solution:

(viii) \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\left(\frac{\pi}{2}-x\right)^2}\)
Solution:

(ix) \(\lim _{x \rightarrow 0} \frac{1-\cos ^3 x}{x \sin 2 x}\)
Solution:

(x) \(\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x}{1-\sin x-\cos x}\)
Solution:

(xi) \(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}\)
Solution:

(xii) \(\lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{\tan ^3 x-\sin ^3 x}\)
Solution:

(xiii) \(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\pi}{2}-x\right) \tan x\)
Solution:

(xiv) \(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos 2 x}\)
Solution:

(xv) \(\lim _{x \rightarrow 0} \frac{x-x \cos 2 x}{\sin ^3 2 x}\)
Solution:

(xvi) \(\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x}\)
Solution:

(xvii) \(\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}\)
Solution:

(xviii) \(\lim _{x \rightarrow 0} \frac{\cos x-\cos 5 x}{\cos 2 x-\cos 6 x}\)
Solution:

(xix) \(\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}\)
Solution:

Question 2.
Evaluate
(i) \(\lim _{x \rightarrow \alpha} \frac{x \sin \alpha-\alpha \sin x}{x-\alpha}\)
Solution:

(ii) \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}\)
Solution:

Question 3.
Evaluate the following limits :
(i) \(\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}\)
Solution:

(ii) \(\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}\)
Solution:

(iii) \(\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan x}{h}\)
Solution:

(iv) \(\lim _{h \rightarrow 0} \frac{{cosec}(x+h)-{cosec} x}{h}\)
Solution:

(v) \(\lim _{h \rightarrow 0} \frac{\sec (x+h)-\sec x}{h}\)
Solution:

(vi) \(\lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h}\)
Solution:

(vii) \(\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\)
Solution:

(viii) \(\lim _{h \rightarrow 0} \frac{\log _{\mathrm{a}}(x+h)-\log _a x}{h}\)
Solution:

(ix) \(\lim _{h \rightarrow 0} \frac{\ln (x+h)-\ln x}{h}\)
Solution:

(x) \(\lim _{h \rightarrow 0} \frac{a^{x+h}-e^x}{h}\)
Solution:

(xi) \(\lim _{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}\)
Solution:

(xii) \(\lim _{h \rightarrow 0}\left\{\frac{1}{(x+h)^3}-\frac{1}{x^3}\right\}\)
Solution:

(xiii) \(\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}\)
Solution:

(xiv) \(\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}-\frac{1}{\sqrt{x}}}\right\}\)
Solution:

Question 4.
Evaluate the following :
(i) \(\lim _{x \rightarrow 0} \frac{\log _e\left(1+\frac{x}{2}\right)}{x}\)
Solution:

(ii) \(\lim _{x \rightarrow 1} \frac{x-1}{\log _e x}\)
Solution:

(iii) \(\lim _{x \rightarrow 1} \frac{\log _e(2 x-1)}{x-1}\)
Solution:

(iv) \(\lim _{x \rightarrow 0} \frac{\log _e(x+1)}{\sqrt{x+1}-1}\)
Solution:

(v) \(\lim _{x \rightarrow 2} \frac{\log _e(x-1)}{x^2-3 x+2}\)
Solution:

(vi) \(\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x}\)
Solution:

(vii) \(\lim _{x \rightarrow 0} \frac{e^{a x}-e^{-a x}}{x}\)
Solution:

(viii) \(\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{e^{4 x}-e^{3 x}}\)
Solution:

(ix) \(\lim _{x \rightarrow 0} \frac{a^{2 x}-1}{x}\)
Solution:

(x) \(\lim _{x \rightarrow 0} \frac{a^x-b^x}{x}\)
Solution:

(xi) \(\lim _{x \rightarrow 1} \frac{2^{x-1}-1}{x-1}\)
Solution:

(xii) \(\lim _{x \rightarrow 0} \frac{a^x-a^{-x}}{x}\)
Solution:

(xiii) \(\lim _{x \rightarrow 1} \frac{3^x-3}{x-1}\)
Solution:

(xiv) \(\lim _{x \rightarrow 0} \frac{3^x-2^x}{4^x-3^x}\)
Solution:

(xv) \(\lim _{x \rightarrow 1} \frac{2^{x-1}-1}{\sqrt{x}-1}\)
Solution:

Question 5.
Evaluate the following :
(i) \(\lim _{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}\)
Solution:

(ii) \(\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{x}\)
Solution:

(iii) \(\lim _{x \rightarrow 0} \frac{\sqrt{x}-\sqrt{5}}{x-5}\)
Solution:

(iv) \(\lim _{x \rightarrow 0} \frac{\sqrt{3-2 x}-\sqrt{3}}{x}\)
Solution:

(v) \(\lim _{x \rightarrow 5} \frac{\sqrt{x-1}-2}{x-5}\)
Solution:

(vi) \(\lim _{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}\)
Solution:

(vii) \(\lim _{x \rightarrow a} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}\), (a > b)
Solution:

(viii) \(\lim _{x \rightarrow 1} \frac{x^{\frac{1}{m}}-1}{x^{\frac{1}{n}}-1}\) (m, n are integers)
Solution:

(ix) \(\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+4}-2}\)
Solution:

= \(\frac{2+2}{1+1}=\frac{4}{2}\) = 2

(x) \(\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\)
Solution:

(xi) \(\lim _{x \rightarrow \infty}\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right)\)
Solution:

(xii) \(\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\)
Solution:

(xiii) \(\lim _{x \rightarrow 0} \frac{(x+9)^{\frac{3}{2}}-27}{x}\)
Solution:

(xiv) \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\)
Solution:

(xv) \(\lim _{x \rightarrow \infty} \frac{a_0+a_1 x+a_2 x^2+\ldots+a_m x^m}{b_0+b_1 x+b_2 x^2+\ldots+b_n x^n}\)
Solution:
\(\lim _{x \rightarrow \infty} \frac{a_0+a_1 x+a_2 x^2+\ldots+a_m x^m}{b_0+b_1 x+b_2 x^2+\ldots+b_n x^n}\)
= \(\left\{\begin{array}{lll}
\infty & \text { if } & m>n \\
0 & \text { if } & m<n \\
\frac{a_m}{b_n} & \text { if } & m=n
\end{array}\right.\)

Question 6.
Evaluate the following :
(i) \(\lim _{x \rightarrow \infty} \frac{\sin x}{x}\)
Solution:

(ii) \(\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}-1\right)\), a > 0
Solution:

(iii) \(\lim _{x \rightarrow 0} \frac{x^{\frac{1}{2}}+2 x+3 x^{\frac{3}{2}}}{2 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}+4 x^{\frac{7}{2}}}\)
Solution:
\(\lim _{x \rightarrow 0} \frac{x^{\frac{1}{2}}+2 x+3 x^{\frac{3}{2}}}{2 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}+4 x^{\frac{7}{2}}}\)
= \(\lim _{x \rightarrow 0} \frac{1+2 \sqrt{x}+3 x}{2-2 x^2+4 x^3}=\frac{1}{2}\)

(iv) \(\lim _{x \rightarrow \infty} \sqrt{x}\{\sqrt{x+1}-\sqrt{x}\}\)
Solution:

(v) \(\lim _{x \rightarrow \infty} x^2\left\{\sqrt{x^4+a^2}-\sqrt{x^4-a^2}\right\}\)
Solution:

(vi) \(\lim _{x \rightarrow 0} \cos (\sin x)\)
Solution:
\(\lim _{x \rightarrow 0} \cos (\sin x)\)
= cos (sin 0) = cos 0 = 1

(vii) \(\lim _{x \rightarrow 0} \log _e \frac{\sqrt{1+x}-1}{x}\)
Solution:

(viii) \(\lim _{x \rightarrow 2} \log _e \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}}\)
Solution:

(ix) \(\lim _{x \rightarrow \infty} \log _e\left(1+\frac{a}{x}\right)^x\)
Solution:

(x) \(\lim _{x \rightarrow 0} \log _e(1+b x)^{\frac{1}{x}}\)
Solution:

(xi) \(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(\frac{1-\tan x}{1+\tan x}\right)}{\frac{\pi}{4}-x}\)
Solution:

(xii) \(\lim _{x \rightarrow \frac{\pi}{2}} \log \frac{1-\sin ^3 x}{\cos ^2 x}\)
Solution:

(xiii) \(\lim _{x \rightarrow \infty} e^x\left(a^{\frac{1}{x}}-1\right)\)
Solution:

(xiv) \(\lim _{x \rightarrow 0} \frac{x\left(e^{\frac{\sqrt{1+x^2+x^4-1}}{x}-1}\right)}{\sqrt{1+x^2+x^4}-1}\)
Solution:

(xv) \(\lim _{x \rightarrow 0+} \frac{b \tan x\left(e^{\sin \frac{a x}{b x}-\frac{a}{b}}\right)}{b \sin a x-a \tan b x}\)
Solution:

Question 7.
Examine the existence of the following limits :
(i) \(\lim _{x \rightarrow 0+} \log _a x\)
Solution:
\(\lim _{x \rightarrow 0+} \log _a x\)
= \(\lim _{h \rightarrow 0} \log _a h=-\infty\)
∴ The limit exists

(ii) \(\lim _{x \rightarrow \frac{\pi}{2}} \tan x\)
Solution:

(iii) \(\lim _{x \rightarrow 0}{cosec} x\)
Solution:

(iv) \(\lim _{x \rightarrow 0-} \frac{1}{e^x}\)
Solution:
\(\lim _{x \rightarrow 0-} \frac{1}{e^x}\) = 0 because as
x → 0, \(\frac{1}{x}\) → ∞
So \(e^{\frac{1}{x}}\) → 0
∴ The limit exists.

(v) \(\lim _{x \rightarrow 0+} \frac{1}{e^x}\)
Solution:
\(\lim _{x \rightarrow 0+} \frac{1}{e^x}\) = \(\lim _{h \rightarrow 0} e^{\frac{1}{h}}=e^{\infty}\) = ∞
The limit exists.

(vi) \(\lim _{x \rightarrow 0} \frac{1}{e^{\frac{1}{x}}-1}\)
Solution:

Question 8.
(i) \(\lim _{x \rightarrow \alpha} \frac{\tan a(x-\alpha)}{x-\alpha}=\frac{1}{2}\)
Solution:

(ii) \(\lim _{x \rightarrow \alpha} \frac{\tan a x}{\sin 2 x}=1\)
Solution:

(iii) \(\lim _{x \rightarrow 0} \frac{e^{a x}-e^x}{x}\) = 2
Solution:

(iv) \(\lim _{x \rightarrow 1} \frac{5^x-5}{(x-1) \log _e a}\) = 5
Solution:

(v) \(\lim _{x \rightarrow 2} \frac{\log _e(2 x-3)}{a(x-2)}\) = 1
Solution:

Odisha State Board CHSE Odisha Class 12 Optional Odia Solutions Chapter 21 ଉପସର୍ଗ Textbook Exercise Questions and Answers.

+2 2nd Year Odia Optional Chapter 21 ଉପସର୍ଗ Question Answer

(କ) ୧ ନମ୍ବର ବିଶିଷ୍ଟ ଗୋଟିଏ ବାକ୍ୟ ବା ଶବ୍ଦରେ ଉତ୍ତରମୂଳକ ପ୍ରଶ୍ନବଳୀ ।

Question ୧।
ଉପସର୍ଗ କାହାକୁ କୁହାଯାଏ ?
ଉ –
ପ୍ର, ପରା, ଅପ୍, ସମ୍ ପ୍ରଭୃତି କେତେକ ପ୍ରତ୍ୟୟ (ଅବ୍ୟୟ) ଧାତୁର ପୂର୍ବରେ ବ୍ୟବହୃତ ହୋଇ ଏହାର ଅର୍ଥରେ ବିବିଧ ପରିବର୍ତ୍ତନ ସାଧନ କରିଥା’ନ୍ତି । ଏମାନଙ୍କୁ ଉପସର୍ଗ କୁହାଯାଏ ।

Question ୨।
ଉପସର୍ଗକୁ କେଉଁ ପ୍ରତ୍ୟୟ କୁହାଯାଏ ?
ଉ –
ଉପସର୍ଗକୁ ‘ପୂର୍ବ ପ୍ରତ୍ୟୟ’ ଓ ‘ପର ପ୍ରତ୍ୟୟ’ ଭାବରେ କୁହାଯାଇଥାଏ ।

Question ୩।
ସଂସ୍କୃତ ଉପସର୍ଗର ସଂଖ୍ୟା କେତେ ?
ଉ –
ସଂସ୍କୃତ ଉପସର୍ଗର ସଂଖ୍ୟା ୨୦ ଗୋଟି ।

Question ୪।
ନିମ୍ନୋକ୍ତ ଉପସର୍ଗଗୁଡ଼ିକୁ ଲଗାଇ ଦୁଇଟି ଲେଖାଏଁ ନୂତନ ଶବ୍ଦ ଲେଖ ।
ପ୍ର, ଅନୁ, ଅବ, ଆ, ଉପ, ପରି, ପ୍ରତି, ଅଧ୍, ଅପି, ଅଭି ।
ଉ –
ପ୍ର – ପ୍ରଚଣ୍ଡ, ପ୍ରହାର
ଅବ – ଅବଧାନ, ଅବଧାନ
ପ୍ରତି – ପ୍ରତିଧ୍ଵନି, ପ୍ରତିଫଳନ
ଅନୁ – ଅନୁରାଗ, ଅନୁରୂପ
ଆ – ଆଗମ, ଆକର୍ଷଣ
ପରି – ପରିଣତ, ପରିଚୟ
ଅଧୂ – ଅଧୂବେଶନ, ଅଧ୍ୟାସୀ
ଅପି – ଅପିହିତ. ଅପିନିହିତ
ଅଭି – ଅଭିଭାଷଣ, ଅଭିନବ

Question ୫।
ନିମ୍ନୋକ୍ତ ଶବ୍ଦ ଗୁଡ଼ିକରେ ଉପସର୍ଗ ବ୍ୟବହାର କରି ଦୁଇଟି ଲେଖାଏଁ ନୂତନ ଶବ୍ଦ ଗଢ଼ ।
ଯୋଗ, କାର, ଦାନ, ଭାବ, ହାର, କାଶ, ଚାର, ବନ, କଥା, ବାଦ ।
ଉ –

Question ୬।
ନିମ୍ନୋକ୍ତ ଉପସର୍ଗଯୁକ୍ତ ଶବ୍ଦମାନଙ୍କରୁ ଉପସର୍ଗକୁ ପୃଥକ କର ।
ପ୍ରକ୍ରିୟା, ପରିପନ୍ଥୀ, ଉପକାର, ଅନୁଭବ, ଉତ୍କଳ, ପରାଭବ, ଅବତାର, ଅବତୀର୍ଣ୍ଣ, ଅଧିକାର, ଉପକରଣ, ଅନୁରାଗ, ଆକର୍ଷଣ, ପ୍ରଭାବ, ଅଭିନେତା, ଆଜୀବନ, ପରିଧାନ, ଅଭିଶାପ ।
ଉ –
ପ୍ରକ୍ରିୟା – ପ୍ର + କ୍ରିୟା ।
ପରିପନ୍ଥୀ – ପରି + ପନ୍ଥୀ ।
ଉପକାର – ଉପ + କାର ।
ଅନୁଭବ – ଅନୁ + ଭବ ।
ଉତ୍କଳ – ଉତ୍ + କଳ ।
ପରାଭବ – ପରା + ଭବ ।
ଅବତାର – ଅବ + ତାର।
ଅଧିକାର – ଅଧ୍ + କାର ।
ଅନୁରାଗ – ଅନୁ + ରାଗ ।
ଆଜୀବନ – ଆ + ଜୀବନ ।
ଅବତୀର୍ଣ୍ଣ – ଅବ + ତୀର୍ଣ୍ଣ ।
ଉପକରଣ – ଉପ + କରଣ ।
ଆକର୍ଷଣ – ଆ + କର୍ଷଣ ।
ଅଭିନେତା – ଅଭି + ନେତା ।
ପରିଧାନ – ପରି + ଧାନ ।
ଅଭିଶାପ – ଅଭି + ଶାପ ।

Question ୭।
ଓଡ଼ିଆ ଭାଷାରେ ପ୍ରଚଳିତ ଅଣସଂସ୍କୃତ ଉପସର୍ଗ ବ୍ୟବହାର କରି ୧୦ଟି ଶବ୍ଦ ଲେଖ ।
ଉ –
ବଦ୍ + ଅଭ୍ୟାସ = ବଦଭ୍ୟାସ
ବେ = ବେ + ହିସାବୀ = ବେହିସାବୀ
ଦର = ଦର + ବୁଢ଼ା = ଦରବୁଢ଼ା
ଅଣ = ଅଣ + ଫୁଟା = ଅଣଫୁଟା
ଦର = ଦର + ପାଚିଲା
ଅ = ଅ + ଶୁଣା = ଅଶୁଣା
ଅଣ = ଅଣ + ନାତି = ଅଣନାତି
ବେ + ଆଇନ = ବେଆଇନ

ଉପସର୍ଗ :

ନୂଆ ନୂଆ ଶବ୍ଦ ଗଠନରେ ପ୍ରତ୍ୟୟଗୁଡ଼ିକର ଆବଶ୍ୟକତା ରହିଛି । ଉଦାହରଣ ସ୍ଵରୂପ
ଅ + କ୍ଷମ = ଅକ୍ଷମ
ଭାରତ + ଈୟ = ଭାରତୀୟ
ବି + ଜ୍ଞାନ = ବିଜ୍ଞାନ
ସତ୍ୟ + ତା = ସତ୍ୟତା

ଏଠାରେ କ୍ଷମ, ଜ୍ଞାନ, ଭାରତ, ସତ୍ୟ ମୂଳ ଶବ୍ଦ । ଏଗୁଡ଼ିକରେ ଅ, ବି, ଈୟ, ତା ଯଥାକ୍ରମେ ପ୍ରତ୍ୟୟ ଭାବରେ ବ୍ୟବହୃତ । ଏହି ପ୍ରତ୍ୟୟଗୁଡ଼ିକ ଯେତେବେଳେ ଶବ୍ଦର ପୂର୍ବରେ ପ୍ରୟୋଗ କରାଯାଏ, ତାହାକୁ ପୂର୍ବସର୍ଗ କହନ୍ତି; ପରେ ବ୍ୟବହୃତ ହେଲେ ତାହା ପରସର୍ଗ ନାମରେ ନାମିତ ହୋଇଥାଏ ।

ଯେଉଁ ପ୍ରତ୍ୟୟଗୁଡ଼ିକ ଶବ୍ଦର ପୂର୍ବରେ ଯୁକ୍ତ ହେବାଦ୍ଵାରା ଅର୍ଥରେ ପରିବର୍ତ୍ତନ ଘଟେ ସେଗୁଡ଼ିକୁ ଉପସର୍ଗ କହନ୍ତି । ସଂସ୍କୃତ ଭାଷାରେ ପ୍ରଚଳିତ ପ୍ର, ପରା, ଅପ, ସମ୍ ଇତ୍ୟାଦି ୨୦ ଗୋଟି ପୂର୍ବ ପ୍ରତ୍ୟୟ ଧର୍ମୀ ଉପସର୍ଗ ଓଡ଼ିଆ ଭାଷାରେ ମଧ୍ୟ ପ୍ରଚଳିତ । ଉପସର୍ଗଗୁଡ଼ିକ ମୁଖ୍ୟତଃ କ୍ରିୟା, କ୍ରିୟାମୂଳ ବା ଧାତୁ ପୂର୍ବରେ ସଂଯୁକ୍ତ ହୋଇ ଅର୍ଥ ବୈଚିତ୍ର ସୃଷ୍ଟି କରିଥା’ନ୍ତି ।
ଯଥା –
ସମ୍ + ଭାଷଣ = ସମ୍ଭାଷଣ
ପ୍ର + ତାପ = ପ୍ରତାପ
ସୁ + ଗମ = ସୁଗମ

କ୍ରିୟା ବା କ୍ରିୟାଜାତ ଶବ୍ଦ ବ୍ୟତୀତ ଅନ୍ୟ ଶବ୍ଦରେ ମଧ୍ୟ ଉପସର୍ଗ ଯୋଗ କରାଯାଏ ।
ଯଥା-
ପ୍ରତି + ମୂର୍ତ୍ତି = ପ୍ରତିମୂର୍ତ୍ତି
ଦୁର୍ + ଭାଗ୍ୟ = ଦୁର୍ଭାଗ୍ୟ

ଉପସର୍ଗର ପ୍ରକାରଭେଦ :

ଯଥା – (କ) ସଂସ୍କୃତ ଉପସର୍ଗ (ଖ) ଅଣ ସଂସ୍କୃତ ଉପସର୍ଗ

(କ) ଓଡ଼ିଆରେ ଅବିକଳ ପ୍ରଚଳିତ ୨୦ଟି ସଂସ୍କୃତ ଉପସର୍ଗ, ଯଥା –

ପ୍ର – ପ୍ରଶଂସା, ପ୍ରକାଶ, ପ୍ରହାର ।
ପରା – ପରାଜୟ, ପରାଭବ, ପରାକ୍ରମ ।
ଅପ – ଅପଯଶ, ଅପମାନ, ଅପଳାପ ।
ସମ୍ – ସଂହାର, ସଂରକ୍ଷଣ, ସଂଯୋଜନା ।
ନି – ନିବୃତ୍ତି, ନିଯୁକ୍ତ, ନିକ୍ଷେପ ।
ଅଧ୍ – ଅଧ୍ୟାର, ଅଧ୍ୟାତି, ଅଧ୍ଯକ୍ଷ ।
ସୁ – ସୁଦର୍ଶନ, ସୁଗନ୍ଧ, ସୁସମ୍ବାଦ ।
ନିର୍ – ନିରାକାର, ନିରାକରଣ, ନିରସନ, ନିରଳସ ।
ଦୁର – (ଦୁଃ) ଦୁର୍ଭାଗ୍ୟ, ଦୁର୍ଗମ ।
ଉତ୍ – ଉତପ୍ତ, ଉତ୍କର୍ଷ, ଉନ୍ମୋଚ ।
ଅତି – ଅତିଶୟ, ଅତିକ୍ରମ, ଅତିକାୟ ।
ପରି – ପରିଚ୍ଛେଦ, ପର୍ଯ୍ୟାୟ, ପରିଚୟ ।
ପ୍ରତି – ପ୍ରତିଦିନ, ପ୍ରତିମୂର୍ତ୍ତି, ପ୍ରତିଧ୍ବନି, ପ୍ରତିଫଳନ ।
ଅବ – ଅବକାଶ, ଅବଗତି, ଅବଧାନ,
ଅନୁ – ଅନୁ ମାନ, ଅନୁ ଜ୍ଞା, ଅନୁ ପାନ, ଅନୁରାଗ ।
ବି – ବିଖ୍ୟାତ, ବିଜ୍ଞାନ, ବିଶୁଦ୍ଧ ।
ଅଭି – ଅଭିସାର, ଅଭିଯାନ, ଅଭିଜାତ ।
ଉପ – ଉପକୂଳ, ଉପବନ, ଉପସାଗର ।
ଅପି – ଅପିଧାନ, ଅପିହିତ, ଅପିନଦ୍ଧ ।
ଆ – ଆଗମନ, ଆନୟନ, ଆରକ୍ତ ।

(ଖ) ଅଣସଂସ୍କୃତ ଉପସର୍ଗ :
ସଂସ୍କୃତ ଅନ୍ୟ କେତେକ ଉପସର୍ଗର ବ୍ୟବହାର ରହିଛି । କେତୋଟିର ଦୃଷ୍ଟାନ୍ତ ପ୍ରଦାନ କରାଗଲା । ଯଥା
ଅ – ଅକଟା, ଅବଟା, ଅଶୁଣା ।
ଅଣ – ଅଣବାବୁଆ, ଅଣବାଟୁଆ, ଅଣବାହୁଡ଼ା !
ଦର – ଦରଫୁଟା, ଦରହସା, ଦରଖୁ।
ବଦ୍‌ – ବଦ୍‌ନାମ, ବଦ୍‌ରାଗୀ, ବଦ୍‌ହଜମ, ବଦ୍‌ଖର୍ଚ୍ଚ, ବଦ୍ୟାଭ୍ୟାସ, ବଦ୍‌ଗନ୍ଧ ।
ବେ – ବେଆଇନ, ବେରସିକ, ବେକାର ।