Class 11

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

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

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

(ii) cos x
Solution:

(iii) sin x
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1st

2nd

3rd

4th

5th

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

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

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

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

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

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

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

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

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

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

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

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

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

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(c)

Question 1.

Fill in the blanks choosing the correct answer from the brackets.

(i) The number of solutions of  2 sin θ – 1 = 0 is__________. (one, two, infinite)
Solution:
Infinite

(ii) If cos α = cos β, then α + β = ____________. (0, π, 2π)
Solution:
2π

(iii) The number of solution(s) of 2 sin θ + 1 = 0 is__________. (zero, two, infinite)
Solution:
Zero

(iv) If tan θ = tan α and 90° < α < 180°, then θ can be in ____________quadrant. (1st, 3rd, 4th)
Solution:
4th

(v) If tan x. tan 2x. tan 7x = tan x + tan 2x + tan 7x, then x = _____________. (\(\frac{\pi}{4}, \frac{\pi}{5}, \frac{\pi}{10}\))
Solution:
\(\frac{\pi}{10}\)

(vi) For_____________value of θ, sin θ + cos θ = √2. (\(\frac{\pi}{4}, \frac{\pi}{2}, \frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(vii) The number of values of x for which cos² x = 1 and x² ≤ 4 is______________. (1, 2, 3)
Solution:
1

(viii) In the 1st quadrant the solution of tan² θ = 3 is_____________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}\))
Solution:
\(\frac{\pi}{3}\)

(ix) The least positive value of θ for which 1 + tan θ = 0 and √2 cos θ + 1 = 0 is___________. (\(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}\))
Solution:
\(\frac{3 \pi}{4}\)

(x) the least positive value of x for which tan 3x = tan x is______________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \pi\))
Solution:
\(\frac{\pi}{2}\)

Question 2.
Find the principal solution of the following equations:
(i) sin θ = sin 2θ
Solution:
sin θ = sin 2θ
or, 2θ = nπ + (-1)ⁿ θ
or, 2π – (-1)ⁿ θ = nπ
or, θ = \(\frac{n \pi}{2-(-1)^n}\)
when n = 0, θ = 0
when n = 1, θ = \(\frac{\pi}{3}\)
when n = 2, θ = 2π
when n = 3, θ = π
when n = 4, θ = 4π
when n = 5, θ = \(\frac{5 \pi}{3}\)
∴ The principal solution are 0, \(\frac{\pi}{3}\), π, \(\frac{5 \pi}{3}\)

(ii) √3 sin θ – cos θ = 2
Solution:
√3 sin θ – cos θ = 2
or, \(\frac{\sqrt{3}}{2}\) sin θ – 1/2 cos θ = 1

which is the only principal solution.

(iii) cos² θ + sin θ + 1 = 0
Solution:
cos² θ + sin θ + 1 = 0
or, 1 – sin² θ + sin θ + 1 = 0
or,  sin² θ – sin θ + 2 = 0
or, sin² θ – 2 sin θ + sin θ – 2 = 0
or, sinθ(sinθ – 2) + (sinθ – 2) = 0
or, (sinθ – 2) (sinθ + 1) = 0
∴ sinθ = 2, sinθ = – 1
= sin \(\left(-\frac{3 \pi}{2}\right)\) or, θ = – \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
∴ The principal solution is \(\frac{3 \pi}{2}\).

(iv) sin 4x + sin 2x = 0
Solution:

(v) sin x + cos x = \(\frac{1}{\sqrt{2}}\)
Solution:

Question 3.
Find the general solution of the following equations:
(i) cos 2x = θ
Solution:
cos 2x = θ
or, 2x = (2n + 1)\(\frac{\pi}{2}\)
or, x = (2n + 1)\(\frac{\pi}{4}\), n∈Z

(ii) sin(x° + 40°) = \(\frac{1}{\sqrt{2}}\)
Solution:

(iii) sin 5θ = sin 3θ
Solution:
sin 5θ = sin 3θ
or, 5θ = nπ + (-1)ⁿ 3θ
or, 5θ – (-1)ⁿ 3θ = nπ
or, θ[5 – (-1)ⁿ3] = nπ
or, θ = \(\frac{n \pi}{5-(-1)^n 3}\)

(iv) tan ax = cot bx
Solution:

(v) tan² 3θ = 3
Solution:

Question 4.
Solve the following:
(Hints : cos x ≠ 0 and sin² x- sin x + 1/2 = 0)
(i) tan² x + sec² x = 3
Solution:
tan² x + sec² x = 3
or, tan² x + 1 + tan2 x = 3
or, 2tan² x = 2
or, tan² x = 1
or, tan x = ± 1 = tan \(\left(\pm \frac{\pi}{4}\right)\)
∴ x = nπ ± \(\frac{\pi}{4}\)

(ii) 4 sin² x + 6 cos² x = 5
Solution:
4 sin² x + 6 cos² x = 5
or, 4 sin² x + 6(1 – sin² x) = 5
or, 4 sin² x + 6 – 6 sin² x = 5
or, 6- 2 sin² x = 5
or, 2 sin² x = 1
or, sin² x = 1/2
or, sin x = ± \(\frac{1}{\sqrt{2}}\) = sin \(\left(\pm \frac{\pi}{4}\right)\)
or, x = nπ + (-1)ⁿ \(\left(\pm \frac{\pi}{4}\right)\)

(iii) 3 sin x + 4 cos x = 5
Solution:

(iv) 3 tan x + cot x = 5 cosec x
Solution:

(v) cos x + √3 sin x = √2
Solution:

(vi) sin 3x – 2 cos² x = 0
Solution :
sin 2x – 2 cos² x = 0
or, 2 sin x cos x – 2 cos² x = 0
or, 2 cos x(sin x – cos x) = 0
∴ cos x = 0, sin x = cos x
∴ x = (2n + 1)\(\frac{\pi}{2}\), tan x = 1 = tan \(\frac{\pi}{4}\)
or, x = nπ + \(\frac{\pi}{4}\)

(vii) sec θ + tan θ = √3
Solution:

(viii) cos 2θ – cos θ = sin θ – sin 20
Solution:

(ix) sin θ + sin 2θ + sin 3θ + sin 4θ = 0
Solution:

(x) cos 2x° + cos x° – 2 = 0
Solution:
cos 2x° + cos x° – 2 = 0
or, 2 cos² x° – 1 + cos x° – 2 = 0
or, 2 cos² x° + cos x° – 3 = 0
or 2 cos² + 3cos x° – 2cos x°- 3 = 0
or, cos x°(2 cos x° + 3) – 1(2 cos x° + 3) = 0
or, (2 cos x° + 3)(cos x° – 1) = 0
∴ cos x° = 1 = cos 0°
∴ x° = 2nπ ± 0 = 2nπ
or, \(\frac{\pi x}{180}\) = 2nπ
or, x = 360 n
Again 2 cos x° + 3 = 0
⇒ cos x° = – 3/2 which has no solution.
Hence x = 360 n.

(xi) tan θ + tan 2θ = tan 3θ
Solution:

(xii) tan θ + tan (\(\theta+\frac{\pi}{3}\)) + tan (\(\theta+\frac{2\pi}{3}\)) = 3
Solution:

(xiii) cot² θ – tan² θ = 4 cot 2θ
Solution:

(xiv) cos 2θ = \((\sqrt{2}+1)\left(\cos \theta-\frac{1}{\sqrt{2}}\right)\)
Solution:

(xv) sec θ – 1 = \((\sqrt{2}-1)\) tan θ
Solution:

⇒ θ = 2nπ + \(\frac{\pi}{4}\)

(xvi) 3cot² θ – 2 sin θ = 0
Solution:

(xvii) 4 cos x. cos 2x . cos 3x = 1
Solution:
4 cos x cos 2x cos 3x = 1
⇒ 2 cos x cos 2x. 2 cos 3x = 1
⇒ (cos 3x + cos x) 2 cos 3x = 1
⇒ 2 cos² 3x + 2 cos 3x cos x = 1
⇒ 2 cos² 3x – 1 + cos 4x + cos 2x = 0
⇒ cos 6x + cos 4x + cos 2x = 0
⇒ cos 6x + cos 2x + cos 4x = 0
⇒ 2 cos 4x cos 2x + cos 4x = 0
⇒ cos 4x (2 cos 2x + 1) = 0
⇒ cos 4x = 0, cos 2x = – 1/2
cos 4x = 0 ⇒ 4x = (2n + 1) \(\frac{\pi}{2}\)

(xviii) cos 3x – cos 2x = sin 3x
Solution:

⇒ 1 – 2 sin x cos x = y²
∴ Equation (1) reduces to
1 – 2(1 – y²) + y = 0
⇒ 2y² + y – 1 = 0
⇒ (2y- 1) (y + 1) = 0

(xix) cos x + sin x = cos 2x + sin 2x
Solution:
cos x + sin x – cos 2x + sin 2x
[Refer (viii)]

(xx) tan x + tan 4x + tan 7x = tan x. tan 4x. tan 7x
Solution:
tan x + tan 4x + tan 7x = tan x tan 4x tan 7x
or, tan x + tan 4x
= – tan 7x + tan x tan 4x tan 7x
= – tan 7x (1 – tan x tan 4x)
or, \(\frac{\tan x+\tan 4 x}{1-\tan x \tan 4 x}\) = – tan 7x
or, tan (x + 4x) = tan (π – 7x)
or, tan 5x = tan (π – 7x)
or, 5x = nπ + π – 7x
or, 12x = π(n + 1)
or, x = \(\frac{\pi(n+1)}{12}\), n∈Z

(xxi) 2(sec² θ + sin² θ) = 5
Solution:

(xxii) \((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=1\)
Solution:
\((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=0\)
As cos x ≠ 0.
we have sin² – \(\frac{3}{2}\) sin x + \(\frac{1}{2}\) = 0
∴ 2 sin² x – 3 sin x + 1 = 0
or, 2 sin² x – 2 sin x – sin x + 1 = 0
or, (2sin x – 1)(sin x – 1) = 0
∴ sin x = \(\frac{1}{2}\) or, sin x = 1
But as cos x ≠ 0, we have sin x ≠ 1
∴ sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ x = nπ + (-1)ⁿ \(\frac{\pi}{6}\), n∈Z

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

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

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

X	0	1
y	0	2

x – 2y = 0

X	0	2
y	0	1

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

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

X	1	0
y	0	-1

y – x = 1 ⇒ y = x + 1

X	0	-1
y	1	0

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

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

X	-2	0
y	0	1

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

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

X	0	-1
y	1	0

3x + 4y = 12

X	4	0
y	0	3

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

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

X	1	0
y	0	1

3x – y = 3
⇒ y = 3x – 3

X	1	0
y	0	-3

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

X	-3	0
y	0	1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(iii) – 47 + 8 √-1
Solution:

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

(iv) – 8 + √-1
Solution:

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

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

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

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

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

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

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

(ii) z³ = i
Solution:

(iii) z⁶ = – i
Solution:

(iv) z³ = 1 + i
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solve graphically
Question 1.
x < y
Solution:
x < y
Step – 1: Let us draw the dotted line x = y

X	0	1
y	0	1

Step – 2: Let us take a point say (1, 0) which is not on the line. Putting x = 1, y = 0 in the equation we get 1 < 0 (false).
⇒ (1, 0) does not satisfy the inequality.
⇒ The solution is the half-plane that does not contain (1, 0)
Step – 3: The shaded region is the solution region.

Question 2.
3x + 4y ≥ 12
Solution:
3x + 4y ≥ 12
Step – 1: Let us draw the line 3x + 4y = 12

X	4	0
y	0	3

Step – 2: Let us consider the point (0, 0) which is not on the line. Putting x = 0, y = 0 in the inequality we have 0 ≥ 12 (false).
∴ (0, 0) does not satisfy the inequality.
⇒ The half-plane that does not contain (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

Question 3.
x – y > 0
Solution:
x – y > 0
Step – 1: Let us draw the dotted line x = y.

X	0	1
y	0	1

Step – 2: Let us consider (1, 0) which is not on the line.
Putting x = 1, y = 0 in the inequation we get 1 > 0 (True)
⇒ (1, 0) satisfies the inequation.
⇒ The half-plane containing (1, 0) is the solution region.

Question 4.
x + 2y – 5 ≤ 0
Solution:
x + 2y – 5 ≤ 0
Step – 1: Let us draw the line x + 2y – 5 = 0
⇒ y = \(\frac{5-x}{2}\)

X	5	1
y	0	2

Step – 2: Let us consider the point (0, 0) which does not lie on the line putting x = 0, y = 0 in the inequation we get – 5 < 0 (True).
⇒ The point satisfies the inequation.
⇒ The half-plane containing (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

Question 5.
7x – 4y < 14
Solution:
7x – 4y < 14
Step – 1: Let us draw the dotted line 7x – 4y = 14

X	2	6
y	0	7

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequality we get 0 < 14 (True).
⇒ (0, 0) satisfies the inequation.
⇒ The half-plane including (0, 0) is the solution region.
Step – 3: The solution region is the shaded region.

Question 6.
x + 8y + 10 > 0
Solution:
x + 8y + 10 > 0
Step – 1: Let us draw the dotted line x + 8y + 10 = 0

X	-10	-2
y	0	-1

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequation we get 10 > 0 (True)
⇒ (0, 0) satisfies the inequality.
⇒ The half-plane containing origin is the solution region.
Step – 3: The solution region’ is the shaded region.

Question 7.
5x + 6y < 12
Solution:
5x + 6y < 12
Step – 1: Let us draw the dotted line 5x + 6y = 12

x	-6	0	6
y	7	2	-3

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 < 12 (True)
⇒ The (0, 0) satisfies the equation.
Step – 3: The shaded region is the required solution.

Question 8.
– 3x + y > 0
Solution:
Step – 1: Let us draw the dotted line – 3x + y = 0

x	0	1	-1
y	0	3	-3

Step – 2: Putting x – 1, y = 0 in the equation we get, – 3 > 0 (false)
∴ Point (1, 0) does not satisfy the in the equation.
Step – 3: The shaded half-plane is the solution.

Question 9.
3x + 8y > 24
Solution:
Step- 1: Let us draw the dotted graph of 3x + 8y = 24

x	8	0
y	0	3

Step- 2: Putting x = 0, y- 0 we get, 0 > 24 (false)
∴ 0, (0, 0) does not satisfy the in equality.

Question 10.
x + y > 1
Solution:
Step – 1: Let us draw the graph x + y = 1

x	1	0
y	0	1

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 > 1 (false)
∴ 0 (0, 0) does not satisfy the in equation
Step – 3: The shaded region is the solution.

Question 11.
x ≤ 0
Solution:
Step – 1: Let us draw the graph x = 0

Step – 2: Putting x = – 1 we get, – 1 ≤ 0 (True)
Thus, the shaded region is the solution.

Question 12.
y > 5
Solution:
Step – 1: Let us draw the dotted graph of y = 5

x	0	1	-1
y	5	5	5

Step- 2: Putting x = 0, y = 0 we have 0 > 5 (false)
we 0(0, 0) does not satisfy. the inequality.
Step – 3: The shaded region is the solution.

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

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

Question 1.
Determine whether the solution set is finite or infinite or empty:
(i) x < 1000, x ∈ N
Solution:
Finite

(ii) x < 1, x ∈ Z (set of integers)
Solution:
Infinite

(iii) x < 2, x is a positive integer.
Solution:
Finite

(iv) x < 1, x is a positive integer.
Solution:
Empty

Question 2.
Solve as directed:
(i) 5x ≤ 20 in positive integers, in integers.
Solution:
5x ≤ 20
⇒ \(\frac{5 x}{5} \leq \frac{20}{5}\)
⇒ x ≤ 4
If x is a positive integer, then the solution set is {1, 2, 3, 4}
If x is an integer, then the solution set is:
S = {x : x ∈ Z and x ≤ 4}
= { ….. -3, -2, -1, 0, 1, 2, 3, 4}

(ii) 2x + 3 > 15 in integers, in natural numbers.
Do you mark any difference in the solution sets?
Solution:
2x + 3 > 15
⇒ 2x + 3 – 3 > 15 – 3
⇒ 2x > 12
⇒ \(\frac{2 x}{2}>\frac{12}{2}\)
⇒ x > 6
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x > 6}
= {7, 8, 9…… }
If x ∈ N. then the solution set is S = {x : x ∈ N and x > 6}
= {7, 8, 9…… }
Two solution sets are the same.

(iii) 5x + 7 < 32 in integers, in non-negative integers.
Solution:
5x + 7 < 32
⇒ 5x + 7 – 7 < 32 – 7
⇒ 5x < 25
⇒ \(\frac{5 x}{5}<\frac{25}{5}\)
⇒ x < 5
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x < 5 }
= {…..-3, -2, -1, 0, 1, 2, 3, 4}
If x is a non-negative solution then the solution set is S = {x : x is a non-negative integer < 5}
= (0, 1,2, 3,4}

(iv) -3x – 8 > 19, in integers, in real numbers.
Solution:
– 3x – 8 > 19
⇒ – 3 x – 8 + 8 > 19 + 8
⇒ – 3x > 27
⇒ \(\frac{-3 x}{-3}<\frac{27}{-3}\)
⇒ x < – 9
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x < – 9}
= { ……..- 11, – 10}
If x ∈ R then the solution set is S = {x : x ∈ R and x < – 9}
= (∞, – 9)

(v) |x – 3| < 11, in N and in R.
Solution:
|x – 3| < 11
⇒ – 1 < x – 3 < 11
⇒ – 11 + 3 < x – 3 + 3 < 11+3
⇒ – 8 < x < 14
If x ∈ N the solution set is S = {1, 2, 3, 4, 5……..12, 13}
If x ∈ R then the solution set is: S = {x : x ∈ R and – 8 < x < 14}
= (- 8, 14)

Question 3.
Solve as directed:
(i) 2x + 3 > x – 7 in R
Solution:
2x + 3 > x – 7
⇒ 2x – x > – 7 – 3
⇒  x > – 10
x ∈ R, the solution set is S = (x : x ∈ R and x > – 10} = (-10, ∞)

(ii) \(\frac{x}{2}+\frac{7}{3}\) <  3x – 1 in R
Solution:
\(\frac{x}{2}+\frac{7}{3}\) <  3x – 1
\(\frac{3 x+14}{6}\) <  3x – 1
⇒ 3x + 14 < 18x – 6
⇒ 3x – 18x < – 6 – 14
⇒ – 15x < – 20
⇒ \(\frac{-15 x}{-15}>\frac{-20}{-15}\)
⇒ x > \(\frac{4}{3}\)
If x ∈ R, the solution set is S = \(\left(\frac{4}{3}, \infty\right)=\left\{x: x \in R \text { and } x>\frac{4}{3}\right\}\)

(iii) \(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\) for non-negative real numbers.
Solution:
\(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\)
⇒ \(\frac{15 x-10 x+6 x}{30}\) ≤ \(\frac{11}{3}\)
⇒ 11x ≤ \(\frac{11}{3}\) × 30
⇒ 11x ≤ 110
⇒ x ≤ 10
If x is a non-negative real number then the solution set is S = {x : x ∈ R and 0 ≤ x ≤ 10}
= {0, 10}

(iv) 2(3x – 1) < 7x + 1 < 3 (2x + 1) for real values.
Solution:
2(3x – 1) < 7x + 1 < 3(2x + 1)
⇒ 6x – 2 < 7x + 1< 6x + 3
⇒ – 2 < x + 1 < 3
⇒ – 3 < x < 2
If x ∈ R, the solution set is S = (x : x ∈ R and -3 < x < 2}
= {-3, 2}

(v) 7(x – 3) ≤ 4 (x + 6), for non-negative integral values.
Solution:
7(x – 3) ≤ 4(x + 6)
⇒ 7x – 21 ≤ 4x + 24
⇒ 7x – 4x ≤ 24 + 21
⇒ 3x ≤ 45
⇒ x ≤ 9
If x is a non-negative integer the solution set is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(vi) Convert to linear inequality and solve for natural numbers: (x – 2) (x – 3) < (x + 3) (x – 1)
Solution:
(x – 2) (x – 3) < (x + 3) (x – 1)
⇒ x² – 5x + 6  <  x² + 2x – 3
⇒ – 5x + 6 < 2x – 3
⇒ – 5x – 2x < – 3 – 6
⇒ – 7x < – 9
⇒ x > \(\frac{9}{7}\)
If x ∈ N, the solution set is S = {2, 3, 4 }

(vii) Solve in R, \(\frac{x}{2}\) + 1 ≤ 2x – 5 < x. Also, find its solution in N.
Solution:
\(\frac{x}{2}\) + 1 ≤ 2x – 5 < x
⇒ \(\frac{x}{2}\) +1 ≤ 2x – 5 and 2x – 5 < x
⇒ \(\frac{x}{2}\) – 2x ≤ – 5 – 1 and x < 5
⇒ \(\frac{-3x}{2}\) ≤ – 6 and x < 5
⇒ – 3x ≤ – 12 and x < 5
⇒ x ≥ 4 and x < 5
⇒ 4 ≤ x < 5
If x ∈ R, the solution set is S = {x : x ∈ R and 4 < x < 5}
= {4, 5}
If x ∈ N, the solution set is S = { 4 }

(viii) Solve in R and also in Z: \(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
Solution:
\(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
⇒ \(\frac{3 x+1}{5} \geq \frac{5 x+10-15+9 x}{15}\)
⇒ 3x + 1 ≥ \(\frac{14 x-5}{3}\)
⇒ 9x + 3 ≥ 14x – 5
⇒ 9x – 14x ≥ – 5 – 3
⇒ – 5x ≥ – 8
⇒ x ≤ \(\frac{8}{5}\)
If x ∈ R, then the solution set is S = (x : x ∈ R and x ≤ \(\frac{8}{5}\)}
= (- ∞, \(\frac{8}{5}\))
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x ≤ \(\frac{8}{5}\)}
= {……. -3, -2, -1, 0, 1}

Question 4.
Solve |x – 1| >1 and represent the solution on the number line.
[Exhaustive hints: By definition of modulus function
For x – 1 ≥ 0 or x ≥ 1, |x – 1| > 1
⇔ x – 1 > 1 ⇔ x > 2 ⇔ x ∈ (2, ∞)
For x- 1 < 0 or x < 1, |x – 1| > 1
⇔ – (x – 1) > 1
⇔ x – 1 < -1 (multiplication by -1 reverses the inequality)
⇔ x < 0 ⇔ x ∈ ( -∞, 0)
∴ The solution set is the Union,
(-∞, 0) ∪ (2, ∞) Show this as two disjoint open intervals on the number line, i.e., real line.]
Solution:
|x – 1| > 1
⇒ – 1 > x – 1 > 1
⇒ 0 > x > 2
⇒ x < 0 and x > 2
∴ The solution set is S = {x : x ∈ R, x < 0 and x > 2}
= (-∞, 0) ∪ (2, ∞)
We can show this solution in the number line as

Question 5.
Solve in R and represent the solution on the number line.
(i) |x – 5| < 1
Solution:
|x – 5| < 1
⇒ – 1< x – 5 < 1
⇒ 4 < x < 6
If x ∈ R, then the solution set is S = (4, 6)
We can represent the solution on the number line as

(ii) \(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
Solution:
\(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{4 x+2+1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{x+3}{6}\)
⇒ 6x < 5x + 15
⇒ x < 15
If x ∈ R, the solution set is S = (-∞, 5)
We can represent the solution on the number line as

(iii) 2x + 1 ≥ 0
Solution:
2x + 1 ≥ 0
⇒ 2x ≥ -1
⇒ x ≥ -1/2
If x ∈ R, then the solution set is S = [\(-\frac{1}{2}\), ∞]
We can represent the solution on the number line as

(iv) \(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
Solution:
\(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
⇒ 3x – 3 ≤ 2x + 2 < 3x – 1
⇒ 3x – 3 ≤ 2x + 2 and 2x + 2 < 3x – 1
⇒ x ≤ 5 and – x < – 3
⇒ x ≤ 5 and x > 3
⇒ 3 < x ≤ 5
If x ∈ R, the solution set is S = {3, 5}
We can represent the solution on the number line as

Question 6.
In a triangle, ABC; AB, BC, and CA are x, 3x + 2, and x + 4 units respectively where x ∈ N. Find the length of its sides. (Hint: Apply triangle-inequality).
Solution:
Given AB = x
BC = 3x + 2
and CA = x + 4
Now AB + AC > BC (Triangle inequality)
⇒ x + x + 4 > 3x + 2
⇒ 2x + 4 > 3x + 2
⇒ – x > – 2
⇒ x < 2
As x ∈ N we have x = 1
The sides of triangle ABC are
AB = 1 unit
BC = 5 units
and CA = 5 units

Question 7.
The length of one side of a parallelogram is 1 cm. shorter than that of its adjacent side. If its perimeter is at least 26 c.m., find the minimum possible lengths of its sides.
Solution:
Let the longer side = x cm
∴ The smaller side = (x – 1) cm
Perimeter = 2(x + (x – 1)) = 4x – 2 cm
According to the question
4x – 2 ≥ 26
⇒ 4x ≥ 28
⇒ x > 7
The minimum value of x = 7.
∴ The minimum length of the sides is 7cm and 6 cm.

Question 8.
The length of the largest side of a quadrilateral is three times that of its smallest side. Out of the other two sides, the length of one is twice that of the smallest and the other is 1 cm. longer than the smallest. If the perimeter of the quadrilateral is at most 36 c.m., then find the maximum possible lengths of its sides.
Solution:
Let the smallest side = x cm.
Largest side = 3 times x = 3x cm.
The other two sides are 2x cm and x + 1 cm.
⇒ The perimeter = x + 3x + 2x + x + 1
= 7x + 1 cm
According to the question:
7x + 1 ≤ 36
⇒ 7x ≤ 35
⇒ x ≤ 5
Maximum value of x = 5
∴ The maximum possible length of sides are x = 5 cm, 3x = 15 cm, 2x = 10 cm, and x + 1 = 6 cm.

Question 9.
Find all pairs of consecutive odd numbers each greater than 20, such that their sum is less than 60.
Solution:
Let two consecutive odd numbers are
2n – 1 and 2n + 1
Now 2n – 1 > 20 and 2n + 1 > 20
But their sum = 2n – 1 + 2n + 1
= 4n < 60
⇒ n < 15
for n = 14 two numbers are 27, 29
for n = 13 two numbers are 25, 27
for n = 12 two numbers are 23, 25
for n = 11 two numbers are 21, 23
∴ All pairs are 21, 23; 23, 25; 25, 27 and 27, 29

Question 10.
Find all pairs of even numbers each less than 35, such that their sum is at least 50.
Solution:
Let two even numbers be x and y.
According to the question
x < 35, y < 35 and x + y ≥ 50
⇒ x ≤ 34, y ≤ 34 and x + y ≥ 50
⇒ x + y ≤ 70, x + y ≥ 50
⇒ 50 ≤ x + y ≤ 70
If x + y = 50 the numbers are {34, 16}, {32, 18}, {30, 20}, {28, 22}, {26, 24}
If x + y = 52 the numbers are {34, 18}, {32, 20}, {30, 22}, {28, 24}, {26, 26}
If x + y = 34 the numbers are {34, 20}, {32, 22}, {30, 24}
If x + y = 56 the numbers are {34, 22}, {32, 24}, {30, 26}, {28, 28}
If x + y = 58 the numbers are {34, 24}, {32, 26}, {30, 28}
If x + y = 60 the numbers are {34, 26}, {32, 28}, {30, 30}
If x + y = 62 the numbers are {34, 28}, {32, 30}
If x + y = 64 the numbers are {34, 30}, {32, 32}
If x + y = 68 the numbers are {34, 34}

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(a)

Question 1.
Compute the product A × B when
(i) A = {0} = B
(ii) A = {a, b}, B = {a, b, c}
(iii) A = Z, B = Φ
Solution:
(i) A = {0} = B
∴ A × B = {(0, 0)}

(ii) A = {a, b}, B = {a, b, c}
∴ A × B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}

(iii) A = Z, B = Φ ∴ AxB = Φ

Question 2.
If |A| = m, |B| = n, what can you say about
(i) |A × B| (ii) |P(A) × P(B)|
Solution:
If |A| = m. |B| = n then

(i) lA × B| = mn.

(ii) |P(A)| = 2^m . |P(B)| = 2ⁿ
∴  |P(A) × P(B)| =2^m × 2ⁿ = 2^m+n

Question 3.
Find x, y if
(i) (x, y) = (-3, 2)
(ii) {x + y, 1) = (1, x – y)
(iii) (2x + y, 1) = (x, 2x + 3y)
Solution:

(i) ∴ x = – 3, y = 2

(ii) ∴ x + y = 1, x – y = 1
∴ 2x = 2 or, x = 1
∴ y=0

(iii) ∴ 2x + y = x, 1 = 2x + 3y
∴ {x + y = 0} × 2
2x + 3y = 1
–      –      –
∴ – y = – 1 or, y = 1
∴ x = – 1

Question 4.
If, A × B = B × A then what can you
Solution:
If A × B = B × A then A = B

Question 5.
|A × B| = 6. If ( -1, y ), (1, x), (0, y) are in A × B. Write other elements in A × B, where x ≠ y.
Solution:
Let |A × B| = 6 and (-1, y) (1, .x) (0, y) ∈ A × B
⇒ -1, 1, 0 ∈ A and x, y ∈ B
As |A × B| = 6 and 3 × 2 = 6
We have A = {-1, 1, 0} and B = {x, y}
Thus other elements of A × B is (-1, x) , (1, y), (0, x)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(d)

Question 1.
Fill in the blanks choosing the correct answer from the brackets.

(i) In Δ ABC, b =____________. (b cos B + c cos C, a cos A + c cos C, c cos A + a cos C)
Solution:
c cos A + a cos C

(ii) If a cot A = b cot B then Δ ABC is__________. (isosceles, right-angled, equilateral)
Solution:
isosceles

(iii) In Δ ABC if b sin C = c sin B = 2 then b sin C = ___________. (0, 1, 2)
Solution:
1

(iv) In Δ ABC if \(\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}=\frac{\cos \mathrm{C}}{c}\) then the tringle is_________ (equilateral, isosceles, scalene)
Solution:
equilateral

(v) If sin A = sin B and b = 1/2, then a = _______________. (2, 1/2, 1)
Solution:
a = 1/2

(vi) In Δ ABC if A = 60°, B = 45° a : b = __________. ( √2 : √3, √6 : 2, √3 : 2)
Solution:
√6 : 2

(vii) In Δ ABC if b² + c² < a² , then _________ angle is obtuse. (A, B, C)
Solution:
A

(viii) If a cos B = b cos A. then cos B = _____________. \(\left(\frac{c}{a}, \frac{a}{2 c}, \frac{c}{2 a}\right)\)
Solution:
cos B = \(\frac{c}{2 a}\)

(ix) If a – b cos C, then __________ angle is a right angle. (A, B, C)
Solution:
∠B is a right angle

(x) If a = 12, b = 7, C = 30°, then Δ = ______________. (42, 84, 21)
Solution:
Δ = 21

Question 2.
Prove that
(i) a sin A – b sin B = c sin (A – B)
Solution:
R.H.S. = c sin (A – B)
= 2R sin C sin (A – B)
= 2R sin (A + B) sin (A – B)
[∴ A + B + C = π or, A + B = π – C
or sin (A + B) = sin (π – C) sin C]
= 2R (sin² A – sin²  B)
= 2R sin A sin A – 2R sin B sin B
= a sin A – b sin B = L.H.S.

(ii) b cos B + c cos C = a cos (B – C)
Solution:
R.H.S. = a cos (B – C)
= 2R sin A cos (B – C)
= 2R sin (B + C) cos (B – C)
= R sin (B + C + B – C) + sin (B + C – B + C)}
= R [(sin 2B + sin 2C)
= R(2sin B cos B + 2 sin C cos C)
= 2R sin B cos B + 2R sin C cos C
= b cos B + c cos C = L.H.S.

(iii) If (a + b + c)(b + c – a) = 3bc, then A = 60°
Solution:

(iv) If \(\frac{b+c}{5}=\frac{c+a}{6}=\frac{a+b}{7}\) then sin A : sin B : sin C = 4 : 3 : 2
Solution:

(v) If A: B: C = 1 : 2 : 3 then sin A: sin B: sin C = 1 : 2
Solution:

(vi) If b² + c² – a² = bc, then A = 60°
Solution:
If b² + c² – a² = bc, then A = 60°
or, \(\frac{b^2+c^2-a^2}{2 b c}\) = 1/2 or, cos A = 1/2
or, A = 60°

(vii) If A : B: C = 1 : 2 : 7, then c: a = (√5 + 1) : (√5 – 1)
Solution:

But we know that \(\frac{\sin C}{\sin A}=\frac{c}{a}\)
∴ \(\frac{c}{a}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

Question 3.
(i) If cos A = \(\frac{12}{13}\), cos B = \(\frac{5}{13}\) then find a : b.
Solution:

(ii) If a = 7, b = 3, c = 5 then find A.
Solution:

(iii) If a = 8, b = 6, c = 4 find tan \(\frac{B}{2}\)
Solution:

(iv) If \(\frac{a}{\sec A}=\frac{b}{\sec B}\) and a ≠ b then find C.
Solution:

(v) If a = 48, b = 35, ∠C = 60° then find c.
Solution:

In Δ ABC prove that (Q. 4 to Q. 26)

Question 4.
a sin (B – C) + b sin (C – A) + c sin (A – B) = 0
Solution:
a sin (B – C) + b sin (C – A) + c sin (A – B)
= 2R sin A sin (B – C) + 2R sin B sin (C – A) + 2R sin C sin (A – B)
= 2R [sin (B + C) sin (B – C) + sin (C + A) sin (C – A) + sin (A + B) sin (A – B)]
= 2R[sin² B – sin² C + sin² C – sin² A + sin² A – sin² A]
= 2R x 0 = 0

Question 5.
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b \cos C-c \cos B}{b \cos C+c \cos B}\)
Solution:

Question 6.
\(\sum \frac{a^2 \sin (B-C)}{\sin (B+C)}=0\)
Solution:

Question 7.
a²(cos² B – cos² C) + b²(cos² C – cos² A) + c²(cos² A – cos² B) = 0
Solution:

Question 8.
\(\frac{b^2-c^2}{a^2} \sin 2 A+\frac{c^2-a^2}{b^2} \sin 2 B\) \(+\frac{a^2-b^2}{c^2} \sin 2 \mathrm{C}=0\)
Solution:

Question 9.
\(\frac{a^2\left(b^2+c^2-a^2\right)}{\sin 2 \mathrm{~A}}=\frac{b^2\left(c^2+a^2-b^2\right)}{\sin 2 \mathrm{~B}}\) \(=\frac{c^2\left(a^2+b^2-c^2\right)}{\sin 2 C}\)
Solution:

Question 10.
\(\Sigma \frac{\cos A}{\sin B \cdot \sin C}=2\)
Solution:

Question 11.
(a² – b² + c²) tan B = (a² + b² – c²) tan C
Solution:

Question 12.
(b² – c²) cot A + (c² – a²) cot B + (a² – b²) cot C = 0
Solution:

Question 13.
\(\frac{b+c}{a}=\frac{\cos \mathbf{B}+\cos C}{1-\cos A}\)
Solution:

Question 14.
\(\sum a^3 \sin (B-C)=0\)
Solution:

Question 15.
(b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S. = (b + c) cos A + (c + a) cos B + (a + b) cos C
= b cos A + c cos A + a cos B + c cos B + a cos C + b cos C
= (b cos C+ c cos B) +(c cos A + a cos C) + (a cos B + b cos A)
= a + b + c = R.H.S.

Question 16.
2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²
Solution:
2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²
\(=2\left(b c \times \frac{\left(b^2+c^2-a^2\right)}{2 b c}+c a \times \frac{c^2+a^2-b^2}{2 c a}\right.\) \(\left.+a b \times \frac{a^2+b^2-c^2}{2 a b}\right)\)
= b² + c² – a² + c² + a² – b² + a² + b²– c²
= a² + b² + c²

Question 17.
a (b² + c²) cos A + b (c² + a²) cos B + c(a² + b²) cos C = 3 abc.
Solution:
a (b² + c²) cos A + b (c² + a²) cos B + c(a² + b²) cos C
= ab² cos A + ac² cos A + bc² cos B + ba² cos B + ca² cos C + cb² cos C
= ab² cos A + ba² cos B + ac² cos A + ca² cos C + bc² cos B + cb² cos C
= ab (b cos A + a cos B) + ac (c cos A + a cos C) bc (c cos B + b cos C)
= abc = abc + abc = 3abc

Question 18.
a³ cos (B – C) + b³ cos (C – A) + c³ cos (A – B) = 3 abc
Solution:
1st term of L.H.S. = a³ cos (B – C)
= a² a cos (B – C)
= a² . 2R sin A cos (B – C)
= 2a²R sin (B + C) cos (B- C)
= a²R [sin (B + C + B – C) + sin (B + C – B + C)]
= a² R (sin 2B + sin 2C)
= a²R [2 sin B cos B + 2 sin C cos C]
= a² [2R sin B cos B + 2R sin C cos C]
= a² (b cos B + c cos C)
Similarly, 2nd term
= b² (c cos C + a cos A) and
3rd term = c² (a cos A + b cos B)
∴ L.H.S.= a²b cos B+a²c cos C+b²c cos C + b²a cos A + c²a cos A + c²b cos B
= ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C)
= abc + bca + cab = 3abc = R.H.S.

Question 19.
a (cos B + cos C) = 2(b + c) sin² \(\frac{A}{2}\)
Solution:
Refer to Q. N. 13.

Question 20.
(b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan \(\frac{B}{2}\) = (a + b – c) tan \(\frac{C}{2}\)
Solution:

Question 21.
\((b+c-a)\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)\) \(=2 a \cot \frac{A}{2}\)
Solution:

Question 22.
(a – b)² cos² \(\frac{C}{2}\) + (a + b)² sin² \(\frac{C}{2}\) = c²
Solution:
L.H.S = (a – b)² cos² \(\frac{C}{2}\) + (a + b)² sin² \(\frac{C}{2}\)

Question 23.
1 – tan \(\frac{A}{2}\) tan \(\frac{B}{2}\) = \(\frac{c}{2}\)
Solution:

Question 24.
(b – c) cot \(\frac{A}{2}\) + (c – a) cot \(\frac{B}{2}\) + (a – b) cot \(\frac{C}{2}\) = 0
Solution:

Question 25.
cot A + cot B + cot C = \(\frac{a^2+b^2+c^2}{4 \Delta}\)
Solution:

Question 26.
a² cot A + b² cot B + c² cot C = 4Δ
Solution:

Question 27.
If \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\) then prove C = 60°.
Solution:

or, cos C = 1/2 or, ∠C = 60°

Question 28.
If a = 2b and A = 3B, find the measures of the angles of the triangle.
Solution:
If a = 2b and A = 3B, we have \(\frac{a}{b}\) = 2
or, \(\frac{\sin A}{\sin B}\) = 2
or, sin A = 2 sin B         …..(1)
Also  sin A = sin 3B (as a = 3B)    …..(2)
∴ From (1) and (2), we have
2 sin B = sin 3B = 3 sin B – 4 sin³ B
or, 4 sin³ B – sin B = 0
or, sin B(4 sin² B – 1) = 0
or, sin B = 0, 4 sin² B = 1
Now sin B = 0 ⇒ B = 0 (Impossible)
∴ sin² B = \(\frac{1}{2}\) or, sin B = ± \(\frac{1}{2}\)
If sin B = \(\frac{1}{2}\) then ∠B = 30°
∴ A = 3B = 3 x 30° = 90°
∠C = 60°

Question 29.
If a⁴ + b⁴ + c⁴ = 2c² (A² + b²), prove that m∠ACB = 45° or 135°.
Solution:
a⁴ + b⁴ + c⁴ = 2c² (a² + b²)
or, a⁴ + b⁴ + c⁴ + 2a²b² – 2b²c² – 2c²a² = 2a²b²
or, (a² + b² – c²)² = 2a²b²
or, a² + b² – c² = ± √2 ab
or, \(\frac{a^2+b^2-c^2}{2 a b}=\pm \frac{1}{\sqrt{2}}\)
or, cos C = ± \(\frac{1}{\sqrt{2}}\)
∴ ∠C = 45° or 135°.

Question 30.
If x² + x + 1, 2x + 1, and x² – 1 are lengths of sides of a triangle, then prove that the measure of the greatest angle is 120°.
Solution:

Question 31.
if cos B = \(\frac{\sin A}{2 \sin C}\), prove that the triangle is isosceles.
Solution:
cos B = \(\frac{\sin A}{2 \sin C}\)
⇒ \(\frac{c^2+a^2-b^2}{2 c a}=\frac{a}{2 c}\) ⇒ c² + a² – b² = a²
or, c² = b² or, c = b
∴ The triangle is isosceles.

Question 32.
If a tan A + b tan B = (a + b)tan \(\frac{1}{2}\) (A + B) prove that the triangle is isosceles.
Solution:

Question 33.
If (cos A + 2 cos C) : (cos A + 2 cos B) = sin B : sin C prove that the triangles are either isosceles or right-angled.
Solution:
\(\frac{\cos A+2 \cos C}{\cos A+2 \cos B}=\frac{\sin B}{\sin C}\)
⇒ cos A sin C = cos A sin B + 2 cos B sin B
⇒ cos A (sin B – sin C) + (sin 2B – sin 2c) = 0
⇒ cos A (sin B – sin C) + 2 cos (B + C) sin (B – C) = 0
⇒ cos A (sin B – sin C) – 2 cos A sin (B – C) = 0
(∴ cos (B + C) = cos (π – A) = – cos A)
⇒ cos A = 0 or sin B – sin C – 2 sin (B – C) = 0
cos A = 0 ⇒ A = 90°
i.e. the triangle is right-angled. sin B sin C – 2 sin (B – C) = 0

Question 34.
If cos A = sin B – cos C, prove that the triangle is right-angled.
Solution:
cos A = sin B – cos C
or, cos C + cos A = sin B

or, 2C = π
or, C = \(\frac{\pi}{2}\)

Question 35.
If a², b², c²  be in A.P., prove that cot A, cot B, cot C are also in A.P.
Solution:
If a², b², c²  be in A.P.
then b² – a² = c² – b²
or, 2b² = c² + a²
or, \(b^2=\frac{c^2+a^2}{2}\)
or, 2b² = c² + a² …..(1)
We have to prove that cot A, cot B, cot C are in A.P.
i.e. to prove cot B – cot A = cot C – cot B
i.e. 2 Cot B = cot C + cot A
∴ R.H.S. = cot C + cot A

Question 36.
If sin A: sin C = sin (A – B) : sin (B – C) prove that a², b², c² are in A.P.
Solution:
\(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
or, sin A sin (B – C) = sin C sin (A – B)
or, sin (B + C) sin (B – C) = sin (A + B) sin (A – B)
or, sin² B – sin² C = sin² A – sin² B
or, 2 sin² B = sin² C + sin² A
or, \(2 \frac{b^2}{4 \mathrm{R}^2}=\frac{c^2}{4 \mathrm{R}^2}+\frac{a^2}{4 \mathrm{R}^2}\)
or, 2b² = c² + a²
or, b² – a² = c² – b²
∴ a², b², c² are in A.P

Question 37.
If the side lengths a, b, and c are in A.P., then prove that cos \(\frac{1}{2}\) (A – C) = 2 sin \(\frac{1}{2}\) B.
Solution:
If a,b, and c are in A.P. then b – a – c – b or, 2b = c + a
We have to prove that

Question 38.
If the side lengths a, b, and c are in A.P., prove that cot \(\frac{1}{2}\) A, cot \(\frac{1}{2}\) B, cot \(\frac{1}{2}\) C are in A.P.
Solution: