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You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5

Question 1.
Find the value, if it exists. If not, give the reason for non-existence.
(i) sin^-1(cos π)
(ii) \(\tan ^{-1}\left(\sin \left(-\frac{5 \pi}{2}\right)\right)\)
(iii) sin^-1[sin 5]
Solution:
(i) sin^-1(cos π) = sin^-1(-1) = \(-\frac{\pi}{2}\)
(ii) \(\tan ^{-1}\left(\sin \frac{5 \pi}{2}\right)=\tan ^{-1}\left(-\sin \frac{\pi}{2}\right)=\tan ^{-1}(-1)=-\frac{\pi}{4}\)
(iii) sin^-1(sin 5) = sin^-1[sin (5 – 2π)] = 5 – 2π

Question 2.
Find the value of the expression in terms of x, with the help of a reference triangle.
(i) sin(cos^-1(1 – x))
(ii) cos(tan^-1(3x – 1))
(iii) \(\tan \left(\sin ^{-1}\left(x+\frac{1}{2}\right)\right)\)
Solution:
(i) Let cos^-1(1 – x) = θ
1 – x = cos θ
We know sin² θ = 1 – cos² θ

Question 3.
Find the value of
(i) \(\sin ^{-1}\left(\cos \left(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right.\)
(ii) \(\cot \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right)\)
(iii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Solution:
(i) \(\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3} \text { and } \cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \text { and }\)

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}\)
(ii) \(\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
Solution:

Question 5.
Prove that tan^-1 x + tan^-1 y + tan^-1 z = tan^-1 \(\left[\frac{x+y+z-x y z}{1-x y-y z-z x}\right]\)
Solution:

Question 6.
If tan^-1 x + tan^-1 y + tan^-1 z = π, show that x + y + z = xyz.
Solution:
Given tan^-1 x + tan^-1 y + tan^-1 z = π

Question 7.
Prove that \(\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1} \frac{3 x-x^{3}}{1-3 x^{2}},|x|<\frac{1}{\sqrt{3}}\)
Solution:

Question 8.
Simplify: \(\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y}\)
Solution:

Question 9.
Find the value of
(i) \(\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}\)
(ii) \(2 \tan ^{-1} x=\cos ^{-1} \frac{1-a^{2}}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}\), a > 0, b > 0
(iii) 2 tan^-1(cos x) = tarn^-1 (2 cosec x)
(iv) cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
Solution:

Question 10.
Find the number of solution of the equation tan^-1(x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x).
Solution:
tan^-1(x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x)
tan^-1(x – 1) + tan^-1 (x + 1) = tan^-1 3x – tan^-1 x

LHS = RHS
⇒ \(\tan ^{-1} \frac{2 x}{2-x^{2}}=\tan ^{-1} \frac{2 x}{1+3 x^{2}}\)
⇒ \(\frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}}\)
⇒ 2 – x² = 1 + 3x²
⇒ 4x² = 1
⇒ x² = \(\frac{1}{4}\)
⇒ x = ±\(\frac{1}{2}\)
So, the equation has 2 solutions.

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5 Additional Questions

Question 1.
Solve the following equation: sin^-1(1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:

Question 2.
Solve: tan^-1 2x + tan^-1 3x = \(\frac{\pi}{4}\)
Solution:

Question 3.

Solution:
Do it yourself

Question 4.

Solution:
Do it yourself

Question 5.

Solution:
Do it yourself

Question 6.

Solution:
Do it yourself

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequences whose n^th terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.

Solution:


It is not a G.P. or A.P. or H.P. or A.G.P.

It is not an A.P. or G.P. or H.P. or A.G.P

It is not an A.P. or G.P. or H.P. or A.G.P.


It is a A.G.P.

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below.

Solution:
a₁ = 1 + 1 = 2 ; a₂ = 2
a₃ = 3 + 1 = 4 ; a₄ = 4
a₅ = 5 + 1 = 6 ; a₆ = 6
So, the first 6 terms are 2, 2, 4, 4, 6, 6

Solution:
a₁ = 1 ; a₂ = 2, a₃ = 3
a₄ = a₃ + a₂ + a₁ = 3 + 2 + 1 = 6 ⇒ a₄ = 6
a₅ = a₄ + a₃ + a₂ = 6 + 3 + 2 = 11 ⇒ a₅ = 11
a₆ = a₅ + a₄ + a₃ = 11 + 6 + 3 = 20 ⇒ a₆ = 20
So the first 6 terms are 1, 2, 3, 5, 8, 13.

Solution:

Question 3.
Write the n_th term of the following sequences.
Solution:
(i) 2, 2, 4, 4, 6, 6……
Solution:

Solution:
Nr: 1, 2, 3, ……t_n = n
Dr: 2, 3, 4, …..t_n = n + 1


Solution:
Nr: 1, 3, 5, 7, . . .which is an A.P. a = 1, d = 3 – 1 = 2
t_n = a + (n – 1)d
t_n = 1 + (n – 1)2 = 1 + 2n – 2 = 2n – 1.
Dr : 2, 4, 6, 8, . . .
So the n^th term is 2 + (n – 1)2 = 2 + 2n – 2 = 2n.

(iv) 6, 10, 4, 12, 2, 14, 0, 16, -2,….
Solution:
t₁ = 6 ; t₂ = 10
t₃ = 4 ; t₄ = 12
t₅ = 2 ; t₆ = 14
t₇ = 0 ; t₈ = 16
When n is odd, the sequence is 6, 4, 2, 0,…
(i.e.) a = 6 and d = 4 – 6 = -2.
So, t_n = 6 + (n – 1)(-2) = 6 – 2n + 2 = 8 – 2n
When n is even, the sequence is 10, 12, 14, 16,…
Here a = 10 and d = 12 – 10 = 2
t_n = 10 + (n – 1)2 = 10 + 2n – 2 = 2n + 8 (i.e.) 8 + 2n

Question 4.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.
Solution:
The 3 numbers in a G.P. is taken as \(\frac{a}{r}\), a, ar
Their product is 5832.

6r² + 6 = 13
6r² – 13r + 6 = 0
(3r – 2)(2r – 3) = 0
r = 2/3 or 3/2

Question 5.

Solution:

Question 6.
If t_k is the k^th term of a G.P., then show that t_{n – k}, t_n, t_{n + k} also form a GP for any positive integer k.
Solution:
Let a be the first term and r be the common ratio.
We are given t_k = ar^{k – 1}
We have to Prove : t_{n – k}, t_n, t_{n + k} form a G.P.

Question 7.
If a, b, c are in geometric progression, and if \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\), then prove that x, y, z are in arithmetic progression.
Solution:
Given a, b, c are in G.P.
⇒ b² = ac
⇒ log b² = log ac
(i.e.) 2log b = log a + log c …(1)

Substituting these values in equation (1) we get 2y = x + z ⇒ x, y z are in A.P.

Question 8.
The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers.
Solution:
Let the two numbers be a and b.

So, (a + b – 20)² = 4ab
(i.e.) (a + b)² + 400 – 40(a + b) = 4ab
(a + b)² – 4ab = 40(a + b) – 400
from(3) (a + b)² – 4ab = 32(a + b)
⇒ 32(a + b) = 40(a + b) – 400
(÷ by 8) 4(a + b) = 5(a + b) – 50
4a + 4b = 5a + 5b – 50
a + b = 50
a = 50 – b
Substituting a = 50 – b in (3) we get
(50 – b – b)² = 32(50)
(50 – 2b)² = 32 × 50

When b = 5, a = 50 – 5 = 45
When b = 45, a = 50 – 45 = 5
So the two numbers are 5 and 45.

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + p – q = 0 are equal, then show that p, q and r are in AP.
Solution:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a, b, c are respectively the p^th, q^th and r^th terms of a G.P., show that (q – r) log a + (r – p) log b + (p – q) log c = 0.
Solution:
Let the G.P. be l, lk, lk²,…
We are given t_p = a, t_q = b, t_r = c

LHS = (q – r) log a + (r – p) log b + (p – q) log c

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Additional Questions Solved

Question 1.

Solution:
Here a₁ = 1
Substituting n = 2, we obtain a₂ = a₁ + 2 = 1 + 2 = 3
Substituting n = 3, 4 and 5, we obtain respectively
a₃ = a₂ + 2 = 3 + 2 = 5, a₄ = a₃ + 2 = 5 + 2 = 7
a₅ = a₄ + 2 = 7 + 2 = 9
Thus, the first five terms are 1, 3, 5, 7 and 9.

Question 2.
Find the 18^th and 25^th term of the sequence defined by

Solution:
When n = 18 (even)
a_n = n(n + 2) = 18(18 + 2) = 18(20) = 360
When n = 25(odd)

Question 3.
Write the first six terms of the sequences given by
(i) a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
(ii) a₁ = 4, a_{n + 1} = 2na_n
Solution:
(i) Here a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
Putting n = 3, a₃= a₂ + a₁ = 1 + 1 = 2
Putting n = 4, a₄ = a₃ + a₂ = 2 + 1 = 3
Putting n = 5, a₅ = a₄ + a₂ = 3 + 2 = 5
Putting n = 6, a₆ = a₅ + a₄ = 5 + 3 = 8
∴ First six terms of the sequence are 1, 1, 2, 3, 5, 8

(ii) Here a₁ = 4 and a_{n + 1} = 2na_n
Putting n = 1, a₂ = 2 × 1 × a₁ = 2 × 1 × 4 = 8
Putting n = 2, a₃ = 2 × 2 × a₂ = 4 × 8 = 32
Putting n = 3, a₄ = 8 × 192 = 1536
Putting n = 4, a₅ = 2 × 4 × a₄ = 8 × 192 = 1536
Putting n = 5, a₆ = 2 × 5 × a₅ = 10 × 1536 = 15360

Question 4.
An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the series.
Solution:
Let a₁ be the first term and d, be the common difference. Here n = 21.
∴ The three middle terms are a₁₀,a₁₁, a₁₂
Now, a₁₀ + a₁₁ + a₁₂ = 129 [Given]
∴ (a₁ + 9d) + (a₁ + 10d) + (a₁ + 11d) = 129
⇒ 3a₁ + 30d = 129 ⇒ a₁ + 10d = 43 ……(i)
The last three terms are a₁₉, a₂₀, a₂₁
a₁₉, a₂₀, a₂₁ = 237 [Given]
∴ (a₁ + 18d)+(a₁ + 19d) + (a₁ + 20d) = 237
(i.e.,) 3a₁ + 57d = 237 ⇒ a₁ + 19d = 79 … (2)
Subtracting (i) from (ii), we get 9d = 36, ⇒ d = 4
∴ From (i), a₁ + 40 = 43 ⇒ a₁ = 3
Hence, the series is 3, 7, 11, 15 …….

Question 5.
Prove that the product of the 2^nd and 3^rd terms of an arithmetic progression exceeds the product of the first and fourth by twice the square of the difference between the 1^st and 2^nd.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of A.P.
Then, a₁ = a, a₂ = a + (2 – 1)d = a + d
a₃ = a + (3 – 1)d = a + 2d, a₄ = a + (4 – 1)d = a + 3d
We have to show that a₂.a₃ – a₁.a₄ = 2(a₂ – a₁)²
LHS = a₂.a₃ – a₁.a₄ = (a + d)(a + 2d) – a(a + 3d)
= a₂ + 3ad + 2d₂ – a₂ – 3ad = 2d₂
RHS = 2(a₂ – a₁)² = 2d₂
Since LHS = RHS. Hence proved.

Question 6.
If the p^th, q^th and r^th terms of an A.P. are a, b, c respectively, prove that a(q – r) + b (r – p) + c(p – q) = 0.
Solution:
Let A be the first term and D be the common difference of A.P.
a_p = a, ∴ A + (p – 1)D = a ….. (1)
a_q = b, ∴ A + (q – 1)D = b ……. (2)
a_r = c, ∴ A + (r – 1)D = c …….. (3)
∴ a (q – r) + b (r – p) + c (p – q) = [A + (p – l) D] (q – r) + [A + (q – 1) D]
(r – p) + [A + (r – 1) D] (p – q) [Using (1), (2) and (3)]
= (q – r + r – p + p – q)A + [(p – l)(q – r) + (q – l)(r – p) + (r – l)(p – q)]D
= (0) A + (pq – pr – q + r + qr – pq – r + p + pr – p – qr + q)D
= (0)A + (0)D = 0.

Question 7.
If a, b, c are in A.P. and p is the A.M. between a and b and q is the A.M. between b and c, show that b is the A.M. between p and q.
Solution:
a, b, c are in A.P.
2b = a + c …… (1)
p is the A.M. between a and b

q is the A.M. between b and c


Hence, b is the A.M. between p and q

Question 8.
If x, y, z be respectively the p^th, q^th and r^th terms of a G.P. show that x^{q – r}, y^{r – p} ,z^{p – q} = 1
Solution:
Let A be the first term and R be the common ratio of G.P.
a_p = x ⇒ x = AR^{p – 1} ……… (1)
a_q = y ⇒ x = AR^{q – 1} ……… (2)
a_r = z ⇒ x = AR^{r – 1} ……… (3)
Raising (1), (2), (3) to the powers q – r, r – p, p – q respectively, we get

Multiplying (4), (5) and (6), we get

Hence, x^{q – r}, y^{r – p}, z^{p – q} = 1

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.7

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
A zero of x³ + 64 is _______
(a) 0
(b) 4
(c) 4i
(d) -4
Answer:
(d) -4
Hint: x³ + 64 = 0
⇒ x³ = -64
⇒ x³ = (-4)³
⇒ x = -4

Question 2.
If f and g are polynomials of degrees m and n respectively, and if h(x) = (f 0 g) (x), then the degree of h is ______
(a) mn
(b) m + n
(c) mⁿ
(d) n^m
Answer:
(a) mn

Question 3.
A polynomial equation in x of degree n always has _______
(a) n distinct roots
(b) n real roots
(c) n imaginary roots
(d) at most one root.
Answer:
(c) n imaginary roots (Every real number is also imaginary)

Question 4.
If α, β and γ are the zeros of x³ + px² + qx + r, then \(\sum \frac{1}{\alpha}\) is ______
(a) \(-\frac{q}{r}\)
(b) \(\frac{q}{p}\)
(c) \(\frac{q}{r}\)
(d) \(-\frac{q}{p}\)
Answer:
(a) \(-\frac{q}{r}\)
Hint:

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d) 5
Answer:
(c) \(\frac{4}{5}\)
Hint:
a_n = 4; a₀ = 5
Let \(\frac{p}{q}\) be the root of P (x). P must divide 5, possible values of P are ±1, ±5
q must divide 4, possible values of q are ±1, ±2, ±4
Possible roots are \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}\)

Question 6.
The polynomial x³ – kx² + 9x has three real zeros if and only if, k satisfies.
(a) |k| ≤ 6
(b) k = 0
(c) |k| > 6
(d) |k| ≥ 6
Answer:
(d) |k| ≥ 6
Hint:
x³ – kx² + 9x = 0
⇒ x (x² – kx + 9) = 0
x = 0 is one real root. If the remaining roots to be real if the
b² – 4ac ≥ 0
⇒ k² – 36 ≥ 0
⇒ k² ≥ 36
⇒ |k| ≥ 6

Question 7.
The number of real numbers in [0, 2π] satisfying sin⁴ x – 2sin² x + 1 is ______
(a) 2
(b) 4
(c) 1
(d) ∞
Answer:
(c) 1
Hint:
sin⁴ x – 2sin² x + 1 = 0
⇒ t² – 2t + 1 = 0
⇒ (t – 1)² = 0
⇒ t – 1 = 0
⇒ t = 1
⇒ sin² x = 1
⇒ \(\frac{1-\cos 2 x}{2}=1\)
⇒ 1 – cos 2x = 2
⇒ cos 2x = cos 0
⇒ 2x = 2nπ
⇒ x = nπ
n = 0, x = 0
n = 1, x = π
n = 2, x = 2π

Question 8.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if _______
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a ≤ 0
Answer:
(c) a < 0
Hint:
If a < 0, then P(x) = x³ + 12x² + 10ax + 1999 has 2 changes of sign.
∴ P (x) has atmost two positive roots. So a < 0

Question 9.
The polynomial x³ + 2x + 3 has _______
(a) one negative and two imaginary zeros
(b) one positive and two imaginary zeros
(c) three real zeros
(d) no zeros
Answer:
(a) one negative and two imaginary zeros
Hint:
P(x) = x³ + 2x + 3; No positive root.
P(-x) = -x³ – 2x + 3; Only one change in the sign.
∴ One negative root.

Question 10.
The number of positive zeros of the polynomial is ______
(a) 0
(b) n
(c) <n
(d) r
Answer:
(b) n

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 1.
Discuss the maximum possible number of positive and negative roots of the polynomial equation 9x⁹ – 4x⁸ + 4x⁷ – 3x⁶ + 2x⁵ + x³ + 7x² + 7x + 2 = 0.
Solution:
P(x) = 9x⁹ – 4x⁸ + 4x⁷ – 3x⁶ + 2x⁵ + x³ + 7x² + 7x + 2
The number of sign changes in P(x) is 4.
∴P(x) has atmost 4 positive roots.
P(-x) = -9x⁹ – 4x⁸ – 4x⁷ – 3x⁶ – 2x⁵ – x³ + 7x² – 7x + 2
The number of sign changes in P(-x) is 3.
P(x) has almost 3 negative roots. Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even.
The number of negative roots = at most 2.

Question 2.
Discuss the maximum possible number of positive and negative zeros of the polynomials x² – 5x + 6 and x² – 5x + 16. Also, draw a rough sketch of the graphs.
Solution:
P(x) = x² – 5x + 6
The number of sign changes in P(x) is 2.
P(x) has atmost 2 positive roots. P(-x) = x² + 5x + 6.
The number of sign changes in P(-x) is 0.
∴ P (x) has no negative roots. P (x) = x² – 5x + 16

Question 3.
Show that the equation x⁹ – 5x⁵ + 4x⁴ + 2x² +1 = 0 has atleast 6 imaginary solutions.
Solution:
P(x) = x⁹ – 5x⁵ + 4x⁴ + 2x² + 1
(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2.
(ii) P(-x) = -x⁹ + 5x⁵ + 4x⁴ + 2x² + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even.
P(x) has one negative root.
(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3.
∴ The number of imaginary roots at least 6.

Question 4.
Determine the number of positive and negative roots of the equation x⁹ – 5x⁸ – 14x⁷ = 0.
Solution:
x⁹ – 5x⁸ – 14x⁷ = 0
P (x) = x⁹ – 5x⁸ – 14x⁷. The number of sign changes is P(x) is 1.
The number of positive roots is 1. P (-x) = -x⁹ – 5x⁸ + 14x⁷
The number of sign changes is P(-x) is one. The number of negative zero of P(-x) is 1. It is clear that 0 is a root of the equation.
∴ The number of the imaginary roots is at least 6.

Question 5.
Find the exact number of real zeros and imaginary of the polynomial x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x.
Solution:
P(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x.
There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 Additional Problems

Question 1.
Find the maximum possible number of real roots of the equation, x⁵ – 6.x² – 4x + 5 = 0.
Solution:
Let f(x) = x⁵ – 6x² – 4x + 5
Check the terms when it changes sign
Number of changes = 2
∴ Maximum number of positive real roots = 2
f(-x) = (- x)⁵ – 6 (- x)² – 4 (-x) + 5
= -x⁵ – 6x² + 4x + 5
Check the terms when it changes signs.
Number of changes = 1
∴ Maximum number of negative real roots = 1
∴ Total maximum number of real roots = 2 + 1 = 3

Question 2.
Find the number of real roots of the equation, x² + 5 |x| + 6 = 0.
Solution:
|x²| – 5|x| + 6 = 0
(|x| -2) (|x| -3) = 0

It has four real roots. The real roots are 2, -2, 3, -3

Question 3.
Find the real roots of the equation, x² + 5 |x| + 6 = 0.
Solution:
x² + 5 |x| + 6 = 0
Case (i) If x ≥ 0
x² + 5x + 6 = 0
(x + 2) (x + 3) = 0
x = -2 and x = -3
Case (ii) If x < 0
x² – 5x + 6 = 0
(x – 2) (x – 3) = 0
x = 2 and x = 3

Question 4.
Solve x⁴ – 4x² + 8x + 35 = 0. Given (2 + \(i \sqrt{3}\)) is a root.
Solution:

⇒ x² – 4x + 7 is a factor of P (x). Dividing the polynomial P(x) = 0 by x² – 4x + 7.
We get x² + 4x + 5 = 0 is a other factor. The roots of x² + 4x + 5 = 0 are

Question 5.
Solve x⁴ – 5x³ + 4x² + 8x – 8 = 0. Given (1 – \(\sqrt{5}\)) is a root of the polynomial equation.
Solution:
Since (1 – \(\sqrt{5}\)) is a root of the polynomial P(x) = 0
(1 + \(\sqrt{5}\)) is also a root of P (x) = 0
⇒ x² – [(1 + \(\sqrt{5}\)) + (1 – \(\sqrt{5}\))] x + (1 + \(\sqrt{5}\) ) (1 – \(\sqrt{5}\) ) = 0 is a factor of P(x) = 0 ⇒ x² – 2x – 4 = 0 is a factor of P(x) = 0.
Dividing the polynomial by x² – 2x – 4 = 0
We get the other factor x² – 3x + 2 = 0
The roots of x² – 3x + 2 = 0
(x – 2) (x – 1) = 0
x = 1, 2
The roots are 1, 2, 1 ± \(\sqrt{5}\)

Question 6.
Find a polynomial equation of the lowest degree with rational co-efficient having \(\sqrt{3}\), (1 – 2i) as two of its roots.
Solution:
When \(\sqrt{3}\) is a root, – \(\sqrt{3}\) will also be a root.
Now the quadratic equation with \(\sqrt{3}\) , – \(\sqrt{3}\) are roots is x² – (\(\sqrt{3}\) – \(\sqrt{3}\))x + (\(\sqrt{3}\))(-\(\sqrt{3}\)) = 0
(i.e) x² – 3 = 0
When 1 – 2i is a root, 1 + 2i will be another root.
Now the quadratic equation with roots 1 – 2i and 1 + 2i is
x² – (1 – 2i + 1 + 2i)x + (1 – 2i)(1 + 2i) = 0
(i.e) x² – 2x + 5 = 0
∴ The equation with roots ± \(\sqrt{3}\) and 1 ± 2i is (x² – 3) (x² – 2x + 5) = 0
(i.e) x⁴ – 2x³ + 5x² – 3x² + 6x – 15 = 0 ,
(i.e) x⁴ – 2x³ + 2x² + 6x – 15 = 0

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5

Question 1.
Solve the following equations:
(i) sin²x – 5sinx + 4 = 0
(ii) 12x³ + 8x = 29x² – 4
Solution:
(i) sin²x – 5sinx + 4 = 0
Let y = sin x
(y² – 5y + 4 = 0
(y – 1) (y – 4) = 0
(y – 1) = o or (y – 4) = 0
y = 1 or y = 4
sin x = 1 or sin x = 4 [not possible since sin x ≤ 1]
sin x = sin \(\frac{\pi}{2}\)
x = nπ + (-1)ⁿ α, n ∈ z.
x = nπ + (-1)ⁿ \(\frac{\pi}{2}\)
(ii) 12x³ + 8x = 29x² – 4
12x³ – 29x² + 8x + 4 = 0 ….. (1)
By Trail and error method, (x – 2) is a factor of (1)
The other factor is 12x² – 5x – 2
The roots is 12x² – 5x – 2 = 0
(3x – 2) (4x + 1) = 0
x = \(\frac{2}{3}\), x = \(-\frac{1}{4}\)
The roots are 2, \(\frac{2}{3}\), \(-\frac{1}{4}\)

[Here a_n = 12, a₀ = 4; Let \(\frac{p}{q}\) be the root of the equation (1)
The factors of a₀ : ±1, ±2, ±4 (P must divisible by 4)
The factor of a_n : ±1, ±2, ±3, ±4, ±6, ±12
q must divide as (12)
Using these p and q we can form \(\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{2}{3}, \pm \frac{1}{6}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm 3\) are the possible roots of equation. (1)]

Question 2.
Examine for the rational roots of:
(i) 2x³ – x² – 1 = 0
(ii) x⁸ – 3x + 1 = 0
Solution:
(i) 2x³ – x² – 1 = 0
Sum of the co-efficients = 0
∴ (x – 1) is a factor
The other factor is 2x² + x + 1.
The root is (2x² + x + 1) = 0
Here ∆ = b² – 4ac = (1)² – 4(2) (1) = 1 – 8 = -7 < 0
The remaining roots are imaginary.
The only rational root is x = 1

(ii) x⁸ – 3x + 1 = 0 …. (1)
Here a_n = 1, a₀ = 1
If \(\frac{p}{q}\) is a rational root of (1)
Then q is a factor a_n, p is a factor of a₀
The possible values of p and q are ± 1.
Among the possible values 1, -1, [(p, q) = 1]
None of them satisfies the equation (1)
The above equation has no rational roots.

Question 3.
Solve: \(8 x^{\frac{3}{2 n}}-8 x^{\frac{-3}{2 n}}=63\)
Solution:

Squaring on both sides

only possible solution is x = 4ⁿ

Question 4.
Solve: \(2 \sqrt{\frac{x}{a}}+3 \sqrt{\frac{a}{x}}=\frac{b}{a}+\frac{6 a}{b}\)
Solution:

Question 5.
Solve the equations:
(i) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0
(ii) x⁴ + 3x³ – 3x – 1 = 0
Solution:
(i) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0 ….. (1)
It is a even degree reciprocal equation as p(x) = \(x^{n} p\left(\frac{1}{x}\right)\)
Dividing equation (1) by x²,


(ii) x⁴ + 3x³ – 3x – 1 = 0 …… (1)
It is an even degree reciprocal function of type II.
1, -1 are the solution of equation (1)
(x – 1), (x + 1) are the factor of (1)
(x² – 1) is a factor of (1)
Dividing (1) by (x² – 1)
we get, x² + 3x + 1 = 0 is the other factor.

Question 6.
Find all real numbers satisfying 4^x – 3(2^x+2) + 2⁵ = 0
Solution:
4^x – 3 (2^x+2) + 2⁵ = 0
⇒ (2²)^x – 3(2^x . 2²) + 2⁵ = 0
(2²)^x – 12 . 2^x+ 32 = 0
Let y = 2^x
y² – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
y – 4 = 0 or y – 8 = 0
Case (i): 2^x = 4
⇒ 2^x = (2)²
⇒ x = 2
Case (ii): 2x = 8
⇒ 2^x = (2)³
⇒ x = 3
∴ The roots are 2, 3

Question 7.
Solve the equation 6x⁴ – 5x³ – 38x² – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.
Solution:
6x⁴ – 5x³ – 38x² – 5x + 6 = 0 …… (1)
x = \(\frac{1}{3}\) is a Solution
∴ (3x – 1) is a factor of (1)
(1) is a Reciprocal equation even degree divide (1) by x².

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5 Additional Problems

Question 1.
Solve: 3^{2x + 4} + 1 = 2.3^{x + 2}
Solution:
3^{2x + 4} = 3^{x + 2} + 3^{x + 2} – 1
3^{2x + 4} – 3^{x + 2} = 3^{x + 2} – 1 ⇒ 3^{x + 2}[3^{x + 2} – 1] = [3^{x + 2} – 1]
3^{x + 2}= 1 ⇒ 3^{x + 2} = 3⁰
x + 2 = 0 ⇒ x = -2

Question 2.
Solve: 2^x – 2^{x + 3} + 2⁴ = 0.
Solution:
2^2x – (2^x.2³) + 2⁴ = 0 since 2³ = 8 = 4 + 4 = 2² + 2²
2^2x – (4 + 4)2^x + 2⁴ = 0 ⇒ 2^2x – (2² + 2²)2^x + 2⁴ = 0
(2^x – 2²)(2^x – 2²) = 0 ⇒ (2^x – 2²)² = 0
2^x = 2² ⇒ x = 2

Question 3.
Solve: (x – 4) (x + 2) (x + 3) (x – 3) + 8 = 0.
Solution:
(x – 4) (x + 3) (x + 2) (x – 3) + 8 = 0
(x² – x – 12) (x² – x – 6) + 8 = 0
Let y = x² – x
(y – 12) (y – 6) + 8 = 0 ⇒ y² – 18y + 72 + 8 = 0
y² – 18y + 80 = 0 ⇒ (y – 10)( y – 8) = 0
Case (i) y – 10 = 0
x² – x – 8 = 0

Question 4.

Solution:

Question 5.
5^{x – 1} + 5^{1 – x} = 26
Solution:

Question 6.
Solve: 12x⁴ – 56x³ + 89x² – 56x + 12 = 0
Solution:
Since the co-efficients of the equations are equal from both ends.
Divide the equation by x²

Question 7.

Solution:

Question 8.
Solve: x⁴ + 4x³ + 5x² + 4x + 1 = 0
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 1.
Show that each of the following expressions is a solution of the corresponding given differential equation.

(i) y = 2x² ; xy’ = 2y
Solution:
v = 2x² …(1)
Differential equation: xy’ = 2y
Differentiate with respect to ‘x’

On simplifying, 2y = xy’
∴ (1) is solution of the given differential equation.

(ii) y = ae^x + be^-x ; y” – y = 0
Solution:
y = ae^x + be^-x …(1) Differential equation: y” – y = 0
Differentiate with respect to ‘x’
y’ = ae^x – be^-x
Again differentiate with respect to ‘x’
y” = ae^x + be^-x
y” = y ⇒ y” – y = 0
∴ (1) is the solution of the given differential equation.

Question 2.
Find value of m so that the function y = e^mx is a solution of the given differential equation.
(i) y + 2y = 0
Solution:
Given solution y = e^mx
Differentiate with respect to ‘x’

(ii) y” – 5y’ + 6y = 0

Given differential equation is y” – 5y’ + 6y = 0
Substitute (1), (2) and (3) in this

Question 3.
The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Solution:
Slope of the tangent is the reciprocal of four times the ordinate

Question 4.

Solution:

Question 5.
Show that y = ax + \(\frac{b}{x}\), x ≠ 0, is a solution of the differential equation x²y”+ xy’ – y = 0.
Solution:

Here ‘a’ and ‘b’ are arbitrary constants

Differentiate with respect to ‘x’
xy’ + y . 1 = a (2x) = 2ax ……….. (2)
Differentiate again with respect to ‘x’ .
xy” + y’ . 1 + y = 2a ⇒ xy” + 2y’ = 2a …….. (3)
Substitute (3) in (2)
xy’ + y = (xy” + 2y’)x
xy’ + y = x²y” + 2xy’ ⇒ x²y” + xy’ – y = 0
Hence proved.

Question 6.
Show that y = ae^-3x + b, where a and b are arbitrary constants, is a solution of the differential equation

Solution:

Question 7.
Show that the differential equation representing the family of curves where a is a positive parameter, is
Solution:

Question 8.
Show that, y = a cos bx is a solution of the differential equation .
Solution:
y = a cos bx …(1) (a is an arbitrary constant)
Differentiating with respect to ‘x’

Again, differentiating with respect to ‘x’

Hence proved

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 Additional Problems

Question 1.
Verify that the function y = a cos x + b sin x is a solution of the differential equation cos \(\frac{d y}{d x}\) + y sin x = b. dx
Solution:
The given function is y = a cos x + b sin x
Differentiating both sides with respect to x, we have

Putting values of \(\frac{d y}{d x}\) and y in the given differential equation, we have
L.H.S. = cos x (- a sin x + b cos x) + {a cos x + b sin x) sin x
= – a sin x cos x + b cos2 x + a sin x cos x + b sin2 x = b (cos2 x + sin2 x)
= b × 1 = b = R.H.S
Thus, y = a cos x + b sin x is a solution of differential equation

Question 2.
Verify that the function y = 4 sin 3x is a solution of the differential equation
Solution:
The given function is y = 4 sin 3x
Differentiating both sides with respect to x, we have

Again, differentiating both sides with respect to x, we have

Putting values of \(\frac{d^{2} y}{d x^{2}}\) and y in the given differential equation, we have
L.H.S. = – 36 sin 3x + 9 (4 sin 3x) = – 36 sin 3x + 36 sin 3x = 0 = R.H.S.
Thus, y = 4 sin 3x is a solution of differential equation

Question 3.
Verify that the function y = ax² + bx + c is a solution of the differential equation   .
Solution:
The given function is y = ax² + bx + c
Differentiating both sides with respect to x, we have

Again differentiating both sides with respect to x, we have

Which is the given differential equation.
Thus, y = ax² + bx + c is a solution of differential equation \(\frac{d^{2} y}{d x^{2}}\)= 2a.

Question 4.
Verify that the function y = e^-3x is a solution of the differential equation
Solution:
The given function is y = e^-3x
Differentiating both sides with respect to x, we have

Again, differentiating both sides with respect to x, we have

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Choose the correct or more suitable answer

Question 1.
The equation of the locus of the point whose distance from y-axis is half the distance from origin is ……..
(a) x² + 3y² = 0
(b) x² – 3y² = 0
(c) 3x² + y² = 0
(d) 3x² – y² = 0
Solution:
(c) 3x² + y² = 0
Hint:
Given that PA = \([\frac{1}{2}/latex]OP
2PA = OP

4PA² = OP²
4(x)² = x² + y² ⇒ 3x² – y² = 0

Question 2.
Which of the following equation is the locus of (at², 2at) ……

Solution:
(d) y² = 4ax
Hint:
y² = 4ax ⇒ Equation that satisfies the given point (at², 2at)

Question 3.
Which of the following point lie on the locus of 3x² + 3y² – 8x – 12y + 17 = 0?
(a) (0, 0)
(b) (-2, 3)
(c) (1, 2)
(d) (0, -1)
Solution:
(c) (1, 2)
Hint:
The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0
(-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0
(1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17
32 – 32 = 0, 0 = 0

Question 4.

(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Question 5.
Straight line joining the points (2, 3) and (-1, 4) passes through the point (α, β) if
(a) α + 2β = 7
(b) 3α + β = 9
(c) α + 3β = 11
(d) 3α + 3β = 11
Solution:
(c) α + 3β = 11
Hint:
Equation joining (2, 3), (-1, 4)

3y – 12 = – x -1 ⇒ x + 3y – 11 = 0, (α, β) lies on it ⇒ α + 3β – 11 = 0.

Question 6.
The slope of the line which makes an angle 45° with the line 3x – y = – 5 are
(a) 1, -1
(b) [latex]\frac{1}{2},-2\)
(c) \(1, \frac{1}{2}\)
(d) \(2,-\frac{1}{2}\)
Solution:
(c) \(1, \frac{1}{2}\)
Hint:
Equation of line 3x – y = -5, y = 3x + 5, m₁ = 3

Question 7.
Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter \(4+2 \sqrt{2}\) is
(a)x + y + 2 = 0
(b) x + y – 2 = 0
(c) x + y – \(\sqrt{2}\) = 0
(d) x + y + \(\sqrt{2}\) = 0
Solution:
(b) x + y – 2 = 0
Hint.
Let the sides be x, x

Question 8.
The coordinates of the four vertices of a quadrilateral are (-2, 4), (-1, 2), (1, 2) and (2, 4) taken in order. The equation of the line passing through the vertex (-1, 2) and dividing the quadrilateral in the equal areas is ………
(a) x + 1 = 0
(b) x + y = 1
(c) x + y + 3 = 0
(d) x – y + 3 = 0
Solution:
(b) x + y = 1
Hint:
This equation passes through (-1, 2)
-1 + 2 = 1 ⇒ 1 = 1

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1, 2) and (3, 4) with coordinate axes are ……….
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Solution:
(b) 5, 5

Question 10.
The equation of the line with slope 2 and the length of the perpendicular from the origin equal to \(\sqrt{5}\) is ……
(a) x + 2y = \(\sqrt{5}\)
(b) 2x + y = \(\sqrt{5}\)
(c) 2x + y = 5
(d) x + 2y – 5 = 0
Solution:
(c) 2x + y = 5

The required line is y = 2x + 5 ⇒ 2x – y + 5 = 0

Question 11.
A line perpendicular to the line 5x – y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is …….

Solution:
(a) x + 5y ± 5\(\sqrt{2}\) = 0
Hint:
Equation of a line perpendicular to 5x – y = 0 is

Question 12.
Equation of the straight line perpendicular to the line x – y + 5 = o, through the point of intersection the y-axis and the given line …….
(a) x – y – 5 = 0
(b) x + y – 5 = 0
(c) x + y + 5 = 0
(d) x + y + 10 = 0
Solution:
(b) x + y – 5 = 0
Hint:
x – y + 5 = 0 ⇒ put x = 0, y = 5
The point is (0, 5)
Equation of a line perpendicular to x – y + 5 = 0 is x + y + k = 0
This passes through (0, 5)
k = -5
x + 7 – 5 = 0

Question 13.
If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2, then the length of a side is ………

Solution:
\(\sqrt{6}\)
Hint:
In an equilateral ∆ the perpendicular will bisects the base in to two equal parts. Length of the perpendicular drawn from (2, 3) to the line x + 7 – 2 = 0

Question 14.
The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the point ……

Solution:
(d) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Hint:
(p + 2 q)x + (p – 3q)y = p – q
px + 2qx + py – 3qy = p – q
P(x + y) + q (2x – 3y) = p – q
The fourth option x = 2/5, y = 3/5

= p – q = RHS

Question 15.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …
(a) (7, 3)
(b) (4, 1)
(c) (1, -1)
(d) (-2, 3)
Solution:
(b) (4, 1)
Hint:
Let (a, b) be on 2x – 3y = 5 ⇒ 2a – 3b = 5
It is equidistance from (1, 2) and (3, 4)

(a – 1)² + (b – 2)² = (a – 3)² + (6 – 4)²
a² – 2a + 1 + b² – 4b + 4 = a² – 6a + 9 + b² – 8b + 16
4a + 4b = 20
2a+ 2b = 10
2a – 3b = 5
5b = 5
b = 1 ∴ a = 4
∴ The point is (4, 1)

Question 16.
The image of the point (2, 3) in the line y = – x is ………
(a) (-3, -2)
(b) (-3, 2)
(c) (-2, -3)
(d) (3, 2)
Solution:
(a) (-3, -2)

x – 2 = -5, y – 3 = -5
x = -3, y = -2
(-3,-2)

Question 17.
The length of ⊥ from the origin to the line \(\frac{x}{3}-\frac{y}{4}=1\) is ……

Solution:
(c) \(\frac{12}{5}\)
Hint:
4x – 3y = 12 ⇒ 4x – 3y – 12 = 0

Question 18.
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is ……..

Solution:
(b) \(\frac{9}{2}\)
Hint:
Equation of a line perpendicular to 2x – 3y + 1 = 0 is 3x + 2y = k. It passes through (1, 3).
3 + 6 = k ⇒ k = 9, 3x + 2y = 9
To find y-intercept x = 0, 2y = 9, y = 9/2

Question 19.
If the two straight lines x + (2k – 7)y + 3 = 0 and 3kx + 9y – 5 = 0 are perpendicular then the value of k is ……

Solution:
(a) k = 3
Hint.

Since the lines are perpendicular m₁m₂ = – 1

Question 20.
If a vertex of a square is at the origin and its one side lies along the line 4x + 3y – 20 = 0, then the area of the square is ……..
(a) 20 sq. units
(b) 16 sq. units
(c) 25 sq. units
(d) 4 sq.units
Solution:
(b) 16 sq. units
Hint:
One side of a square = Length of the perpendicular from (0, 0) to the line.

Question 21.
If the lines represented by the equation 6x² + 41xy – 7y² = 0 make angles α and β with
x – axis, then tan α tan β =

Solution:
(a) \(-\frac{6}{7}\)
Hint.
6x² + 41xy – 7y² = 0 ⇒ 6x² – xy + 42xy – 7y² = 0 ⇒ x (6x – y) + 7y (6x – y) = 0
(x + 7y) (6x – y) = 0 ⇒ x + 7y = 0, 6x – y = 0

Question 22.
The area of the triangle formed by the lines x² – 4y² = 0 and x = a is …….

Solution:
(c) \(\frac{1}{2} a^{2}\)
Hint:
x² – 4y² = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0

Question 23.
If one of the lines given by 6x² – x + 4x² = 0 is 3x + 4y = 0, then c equals to ……
(a) -3
(b) -1
(c) 3
(d) 1
Solution:
(a) -3
Hint.
6x² – xy + 4cy² = 0, 3x + 4y = 0
The other line may be (2x + by)
(3x + 4y) (2x + by) = 6x² – xy + 4cy²
6x² + 3xby + 8xy + 4by² = 6x² – xy + 4cy²
6x² + xy (3b + 8) + 4by² = 6x² – xy + 4cy²
paring, 3b + 8 = -1
3b = -9 ⇒ b = -3
4b = 4c ⇒ 4(-3) = 4c
-12 = 4c ⇒ c = -3

Question 24.


Solution:
(c) \(\frac{5}{9}\)
Hint:

Question 25.
The equation of one the line represented by the equation x² + 2xy cot θ – y² = 0 is ………
(a) x – y cotθ = 0
(b) x + y tan θ = 0
(e) x cos θ + y(sin θ + 1) = 0
(d) x sin θ + y(cos θ + 1) = 0
Solution:
(d) x sin θ + y(cos θ + 1)=0

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.9

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ is ______
(a) 0
(b) 1
(c) -1
(d) i
Answer:
(a) 0

Question 2.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ______
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 3.
The area of the triangle formed by the complex numbers z, iz, and z + iz in the Argand’s diagrams is ______
(a) \(\frac { 1 }{ 2 }\) |z|
(b) |z|²
(c) \(\frac { 3 }{ 2 }\)|z|²
(d) 2|z|²
Answer:
(a) \(\frac { 1 }{ 2 }\) |z|
Hint: Area of triangle = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) |z| |iz|
= \(\frac { 1 }{ 2 }\) |z|²

Question 4.
The conjugate of a complex number is \(\frac{1}{i-2}\). Then, the complex number is _____
(a) \(\frac{1}{i+2}\)
(b) \(\frac{-1}{i+2}\)
(c) \(\frac{-1}{i-2}\)
(d) \(\frac{1}{i-2}\)
Answer:
(b) \(\frac{-1}{i+2}\)
Hint:

Question 5.
If \(z=\frac{(\sqrt{3}+i)^{3}(3 i+4)^{2}}{(8+6 i)^{2}}\) then |z| is equal to ______
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2
Hint:

Question 6.
If z is a non zero complex number, such that 2iz² = \(\bar{z}\) then |z| is ______
(a) \(\frac { 1 }{ 2 }\)
(b) 1
(c) 2
(d) 3
Answer:
(a) \(\frac { 1 }{ 2 }\)
Hint:

Question 7.
If |z – 2 + i| ≤ 2, then the greatest value of |z| is _______
(a) √3 – 2
(b) √3 + 2
(c) √5 – 2
(d) √5 + 2
Answer:
(d) √5 + 2
Hint:

Question 8.
If \(\left|z-\frac{3}{z}\right|\), then the least value of |z| is ______
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1
Hint:

Question 9.
If |z| = 1, then the value of \(\frac{1+z}{1+\bar{z}}\) is _______
(a) z
(b) \(\bar{z}\)
(c) \(\frac { 1 }{ z }\)
(d) 1
Answer:
(a) z
Hint:

Question 10.
The solution of the equation |z| – z = 1 + 2i is _______
(a) \(\frac { 3 }{ 2 }\) – 2i
(b) –\(\frac { 3 }{ 2 }\) + 2i
(c) 2 – \(\frac { 3 }{ 2 }\)i
(d) 2 + \(\frac { 3 }{ 2 }\)i
Answer:
(a) \(\frac { 3 }{ 2 }\) – 2i

Question 11.
If |z₁| = 1, |z₂| = 2, |z₃| = 3 and |9z₁ z₂ + 4z₁ z₃ + z₂ z₃| = 12, then the value of |z₁ + z₂ + z₃| is ______
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Hint: |z₁ + z₂ + z₃| = 2

Question 12.
If z is a complex number such that z ∈ C/R and z + \(\frac { 1 }{ z }\) ∈ R, then |z| is ______
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Hint: We have

Question 13.
z₁, z₂ and z₃ are complex numbers such that z₁ + z₂ + z₃ = 0 and |z₁| = |z₂| = |z₃| = 1 then \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}\) is ______
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(d) 0
Hint:

Question 14.
If \(\frac{z-1}{z+1}\) is purely imaginary, then |z| is _______
(a) \(\frac { 1 }{ 2 }\)
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 15.
If z = x + iy is a complex number such that |z + 2| = |z – 2|, then the locus of z is _____
(a) real axis
(b) imaginary axis
(c) ellipse
(d) circle
Answer:
(b) imaginary axis
Hint:
|z + 2| = | z – 2|
⇒ |x + iy + 2| = |x + iy – 2|
⇒ |x + 2 + iy|² = |x – 2 + iy|²
⇒ (x + 2)² + y² = (x – 2)² + y²
⇒ x² + 4 + 4x = x² + 4 – 4x
⇒ 8x = 0
⇒ x = 0

Question 16.
The principal argument of \(\frac{3}{-1+i}\) is ______
(a) \(-\frac{5 \pi}{6}\)
(b) \(-\frac{2 \pi}{3}\)
(c) \(-\frac{3 \pi}{4}\)
(d) \(\frac{-\pi}{2}\)
Answer:
(c) \(-\frac{3 \pi}{4}\)
Hint:

The complex number lies in III quadrant. \(\theta=-(\pi-\alpha)=-\left(\pi-\frac{\pi}{4}\right)=\frac{-3 \pi}{4}\)

Question 17.
The principal argument of (sin 40° + i cos 40°)⁵ is ______
(a) -110°
(b) -70°
(c) 70°
(d) 110°
Answer:
(a) -110°
Hint:
(sin 40° + i cos 40°)⁵
= (cos 50° + i sin 50°)⁵
= (cos 250° + i sin 250°)
250° lies in III quadrant.
To find the principal argument the rotation must be in a clockwise direction which coincides with 250°
θ = -110°

Question 18.
If (1 + i) (1 + 2i) (1 + 3i)…(1 + ni) = x + iy, then 2.5.10……(1 + n² ) is _____
(a) 1
(b) i
(c) x² + y²
(d) 1 + n²
Answer:
(c) x² + y²

Question 19.
If ω ≠ 1 is a cubic root of unity and (1 + ω)⁷ = A + B ω, then (A, B) equal to ______
(a) (1, 0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)
Hint:

Question 20.
The principal argument of the complex number \(\frac{(1+i \sqrt{3})^{2}}{4 i(1-i \sqrt{3})}\) is _____
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{5 \pi}{6}\)
(d) \(\frac{\pi}{2}\)
Answer:
(d) \(\frac{\pi}{2}\)
Hint:

Question 21.
If α and β are the roots of x² + x + 1 = 0, then α²⁰²⁰ + β²⁰²⁰ is ______
(a) -2
(b) -1
(c) 1
(d) 2
Answer:
(b) -1
Hint:
x² + x + 1 = 0
α and β are the roots of the equation.
There are the two roots of cube roots of unity except 1.

Question 22.
The product of all four values of \(\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{\frac{3}{4}}\) is ____
(a) -2
(b) -1
(c) 1
(d) 2
Answer:
(c) 1

Question 23.
If ω ≠ 1 is a cubic root of unity and \(\left|\begin{array}{cccc}{1} & {1} & {1} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega^{7}}\end{array}\right|\) = 3k, then k is equal to ______
(a) 1
(b) -1
(c) i√3
(d) -i√3
Answer:
(d) -i√3
Hint:

Question 24.
The value of \(\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}\) is ______
(a) cis \(\frac{2 \pi}{3}\)
(b) cis \(\frac{4 \pi}{3}\)
(c) -cis \(\frac{2 \pi}{3}\)
(d) -cis \(\frac{4 \pi}{3}\)
Answer:
(a) cis \(\frac{2 \pi}{3}\)
Hint:

Question 25.
If ω = cis \(\frac{2 \pi}{3}\), then the number of distinct roots of \(\left|\begin{array}{ccc}{z+1} & {\omega} & {\omega^{2}} \\ {\omega} & {z+\omega^{2}} & {1} \\ {\omega^{2}} & {1} & {z+\omega}\end{array}\right|=0\) are _____
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.9 Additional Problems

Question 1.

(a) (2, -1)
(b) (1, 0)
(c) (0, 1)
(d) (-1, 2)
Solution:
(b) (1, 0)

Question 2.

(a) (0, 2)
(b) (-2, 0)
(c) (0, -2)
(d) (-2, 0)
Solution:
(c) (0, -2)

Question 3.
The point represented by the complex number 2 – i is rotated about origin through an angle
of \(\frac{\pi}{2}\). in the clockwise direction, the new position of point is ……
(a) 1 + 2 i
(b) -1 – 2 i
(c) 2 + i
(d) -1 + 2 i
Solution:
(b) -1 – 2i
Hint.
Let z = 2 – i. Let the new position of point when the point represented by the complex number z = 2 – i is rotated about origin through an angle of \(\frac{\pi}{2}\) in the clockwise direction be denoted by z₁.

Question 4.
Fill in the blanks of the following:
For any two complex numbers z₁, z₂ and any real number a, b,
Solution:

Question 5.
Multiplicative inverse of 1 + i is …..
Solution:

Question 6.

Solution:

Question 7.
If |z| = 4 and arg (z) = \(\frac{5 \pi}{6}\), then z = ……
Solution:

Question 8.
State true or false for the following:
(i) The order relation is defined on the set of complex numbers.
Solution:
The given statement is false because the order relation “greater than” and “less than” are not defined on the set of complex numbers.

(ii) For any complex number 2, the minimum value of |z| + |z – 1| is 1.
Solution:
Let z = x + iy be any complex number.

When x = 0, y = 0, the value of |z| + |z – 1| is minimum and the minimum value.

Hence, the given statement is true.

(iii) 2 is not a complex number.
Solution:
It is a false statement because 2 = 2 + i(0), which is of the form α + ib, and the number of the form a + ib, where a and b are real numbers is called a complex number.

(iv) The locus represented by |z – 1| = |z – i| is a line perpendicular to the join of (1, 0) and (0, 1).
Solution:
Equation of a straight line passing through (1, 0) and (0, 1) is

Now, the lines y = -x + 1 and y = x are perpendicular to each other, so the given statement is true.

Question 9.

Solution:
1

Question 10.
If – \(\bar{z}\) lies in the third quadrant, then z lies in the ……..
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Solution:
(d) fourth quadrant
Hint:

Question 11.

(a) cos 2(α – β + γ) + i sin 2(α – β + γ)
(b) – 2cos(α – β + γ)
(c) – 2 i sin(α – β + γ)
(d) 2 cos(α – β + γ)
Solution:
(c) -2 i sin (α – β + γ)
Hint: α β γ

Question 12.


Solution:

Hint:

Question 13.

(a) 1
(b) -1
(c) 0
(d)-i
Solution:
(c) 0
Hint:

Question 14.
If -i + 2 is one root of the equation ax² – bx + c = 0, then the other root is
(a) – i – 2
(b) i – 2
(c) 2 + i
(d) 2i + i
Solution:
(c) 2 + i
Hint:
Complex roots occur in pairs when -i + 2 is one root, i.e., when 2 – i is a root, the other root is 2 + i

Question 15.
The equation having 4 – 3i and 4 + 3i as roots is …….
(a) x² + 8x + 25 = 0
(b) x² + 8x – 25 = 0
(c) x² – 8x + 25 = 0
(d) x² – 8x – 25 = 0
Solution:
(c) x² – 8x + 25 = 0
Hint.

Question 16.

(a) (1, 1)
(b) (1, -1)
(c) (0, 1)
(d) (1, 0)
Solution:
(d) (1, 0)
Hint:

Question 17.
If – i + 3 is a root of x² – 6x + k = 0, then the value of k is

Solution:
(d) 10
Hint:
α = -i + 3 = 3 – 7 ⇒ The other root is β = 3 + i
Product of the roots = αβ = (3 – i)(3 + i) = 10
Product of the roots of x² – 6x + k = 0 is k ⇒ k = 10

Question 18.
If ω is a cube root of unity, then the value of (1 – ω + ω²)⁴ + (1 + ω – ω²)⁴ is ……
(a) 0
(b) 32
(c) -16
(d) -32
Solution:
(c) -16
Hint.

Question 19.
If ω is a cube root of unity, then the value of (1 – ω) (1 – ω²) (1 – ω⁴)(1 – ω⁸) is …….
(a) 9
(b) -9
(c) 16
(d) 32
Solution:
(a) 9
Hint.

Question 20.

(a) | z | = 1
(b) | z | > 1
(c) | z | < 1
(d) None of these.
Solution:
(a) | z | = 1

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4

Question 1.
Solve:
(i) (x – 5) (x – 7) (x + 6) (x + 4) = 504
(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16
Solution:
(i) (x – 5) (x + 4) (x – 7) (x + 6) = 504
(x² – x – 20) (x² – x – 42) = 504
Let y = (x² – x)
(y – 20) (y – 42) = 504
⇒ y² – 42y – 20y + 840 = 504
⇒ y² – 62y + 336 = 0
⇒ (y – 56) (y – 6) = 0
⇒ (y – 56) = 0 or (y – 6) = 0
⇒ x² – x – 56 = 0 or x² – x – 6 = 0
⇒ (x – 8) (x + 7) = 0 or (x – 3) (x + 2) = 0
⇒ x = 8, -7 or x = 3, -2
The roots are 8, -7, 3, -2

(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16
⇒ (x – 4) (x – 2) (x – 7) (x + 1) = 16
⇒ (x² – 6x + 8) (x² – 6x – 7) = 16
Let x² – 6x = y
(y + 8)(y – 7) = 16
⇒ y² – 7y + 8y – 56 – 16 = 0
⇒ y² + y – 72 = 0
⇒ (y + 9) (y – 8) = 0
y + 9 = 0
x² – 6x + 9 = 0
(x – 3)² = 0
x = 3, 3
or
y – 8 = 0
x² – 6x – 8 = 0

The roots are 3, 3, 3 ±√17

Question 2.
Solve : (2x – 1) (x + 3) (x – 2) (2x + 3) + 20 = 0.
Solution:
(2x – 1) (2x + 3) (x + 3) (x – 2) + 20 = 0
⇒ (4x² + 6x – 2x – 3) (x² – 2x + 3x – 6) + 20 = 0
⇒ (4x² + 4x – 3) (x² + x – 6) + 20 = 0
⇒ [4(x² + x) – 3] [x² + x – 6] + 20 = 0
Let y = x² + x
⇒ (4y – 3) (y – 6) + 20 = 0
⇒ 4y² – 24y – 3y + 18 + 20 = 0
⇒ 4y² – 27y + 38 = 0
⇒ (4y – 19) (y – 2) = 0
(4y – 19) = 0
4(x² + x) – 19 = 0
4x² + 4x – 19 = 0

or
(y – 2) = 0
x² + x – 2 = 0
(x + 2) (x – 1) = 0
x = -2, +1
The roots are -2, 1, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4 Additional Questions

Questions 1.
Solve (x – 3) (x – 6) (x – 1) (x + 2) + 54 = 0.
Solution:
(x – 3) (x – 1) (x – 6) (x + 2) + 54 = 0
(x² – 4x + 3) (x² – 4x – 12) + 54 = 0
Put x² – 4x = y
(y + 3)(y – 12) + 54 = 0
y² – 9y – 36 + 54 = 0
y² – 9y + 18 = 0
(y – 3)(y – 6) = 0

Question 2.
Solve the equation (x – 4) (x – 2) (x – 1) (x + 1) + 8 = 0
Solution:
The equation can be rewritten as {(x – 4) (x + 1)} {(x – 2) (x – 1)} + 8 = 0
(x² – 3x – 4)(x² – 3x + 2) + 8 = 0

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3

Question 1.
Solve the cubic equation: 2x³ – x² – 18x + 9 = 0 if sum of two of its roots vanishes.
Solution:
The given equation is 2x³ – x² – 18x + 9 = 0
\(x^{3}-\frac{x^{2}}{2}-9 x+\frac{9}{2}=0\)
Let the roots be α, -α, β
α – α + β = \(-\left(\frac{-1}{2}\right)\)
\(\Rightarrow \beta=\frac{1}{2}\)
(α) (-α) (β) = \(\frac{-9}{2}\)
\(\Rightarrow-\alpha^{2}\left(\frac{1}{2}\right)=\frac{-9}{2}\)
α² = 9
α = ±3
The roots are 3, -3, \(\frac { 1 }{ 2 }\)

Question 2.
Solve the equation 9x³ – 36x² + 44x – 16 = 0 if the roots form an arithmetic progression.
Solution:
The given equation is 9x³ – 36x² + 44x – 16 = 0
\(x^{3}-4 x^{2}+\frac{44}{9} x-\frac{16}{9}=0\)
Let the roots be α – d, α, α + d
As they are in A.P

Question 3.
Solve the equation 3x³ – 26x² + 52x – 24 = 0 if its roots form a geometric progression.
Solution:
The given equation is 3x³– 26x² + 52x – 24 = 0
\(x^{3}-\frac{26}{3} x^{2}+\frac{52}{3} x-8=0\)
Given that the root are GP

Question 4.
Determine ft and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
Solution:
The given equation is 2x³ – 6x² + 3x + k = 0

Question 5.
Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and √3 are two of its zeros.
Solution:
The given equation is x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135 = 0
The given roots are 1 + 2i, √3
The other roots are 1 – 2i, -√3
The factors are
= {x² – x(2) + (1 + 4)}{(x + √3)(x – √3)}
= (x² – 2x + 5) (x² – 3)
= x⁴ – 3x² – 2x³ + 6x + 5x² – 15
= x⁴ – 2x³ + 2x² + 6x – 15
To find this roots,

Question 6.
Solve the cubic equations:
(i) 2x³ – 9x² + 10x = 3
(ii) 8x³ – 2x² – 7x + 3 = 0
Solution:
(i) Given equation is 2x³ – 9x² + 10x = 3
Sum of the co-efficients = 0
(x – 1) is a factor.
The other factor is 2x² – 7x + 3
2x² – 7x + 3 = 0
(x – 3)(2x – 1) = 0
The roots are 1, 3, \(\frac { 1 }{ 2 }\)

(ii) Given equation is 8x³ – 2x² – 7x + 3 = 0
Sum of odd co-efficients = Sum of even co-efficients
(x + 1) is a factor.
The other factor is 8x² – 10x + 3
8x² – 10x + 3 = 0
(4x – 3) (2x – 1) = 0
The roots are \(\frac{3}{4}, \frac{1}{2},-1\)

Question 7.
Solve the equation: x⁴ – 14x² + 45 = 0.
Solution:
The given equation is x⁴ – 14x² + 45 = 0
Let x² = y
y² – 14y + 45 = 0
(y – 9)(y – 5) = 0
y = 9, 5
x² = 9, x² = 5
x = ± 3, x = ± √5
The roots are ± 3, ± √5

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3 Additional Problems

Question 1.
If one root of x³ + 2x² + 3x + k = 0 is sum of the other two roots then find the value of k.
Solution:
Let α, β, γ be the roots of given equation.
But, α = β + γ
α + β + γ = -2 … (1) ⇒ 2α = -2 ⇒ α = -1
αβ + βγ + γα = 3 … (2) This gives β + γ = – 1
αβγ = -k …(3)

Question 2.
If sum of the roots of the equation x³ – 3x² – 16x + k = 0 is zero then find the value of k.
Solution:
Let α, β, γ be the roots of given equation.
But, α + β = 0

Question 3.
Find all zeros of the polynomial x³ – 5x² + 9x – 5 = 0, If 2 + i is a root.
Solution: