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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 9 Limits and Continuity Ex 9.4

Evaluate the following limits

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

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 9 Limits and Continuity Ex 9.3

Question 1.
(a) Find the left right limits of f(x) = \(\frac{x^{2}-4}{\left(x^{2}+4 x+4\right)(x+3)}\) at x = -2
(b) f(x) = tan x at x = \(\frac{\pi}{2}\)
Solution:
(a) f(x) = \(\frac{x^{2}-4}{\left(x^{2}+4 x+4\right)(x+3)}\) at x → -2

Evaluate the following limits
Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

f(x) → ∞ as x → ∞

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or “recruits” that will return to the rivers during the reproductive period. If S is the number of spawners and R the number of recruits, “Beverton-Holt spawner recruit function” is R(S) = \(\frac{S}{(\alpha S+\beta)}\) where α and β are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.
Solution:

Question 10.
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after t minutes (in grams per litre) is C(t) = \(\frac{30 t}{200+t}\)
What happens to the concentration as t → ∞?
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.3 Additional Problems

Question 1.
Evaluate
Solution:
The given expression is of the form ∞ – ∞. So we first write it in the rational form \(\frac{f(x)}{g(x)}\). So that it reduces to either \(\frac{0}{0}\) form or \(\frac{\infty}{\infty}\) form.

Question 2.
Evaluate:
Solution:
Here the expression assumes the form ∞ – ∞ as x → ∞. So, we first reduce it to the rational form \(\frac{f(x)}{g(x)}\)

Question 3.
Evaluate
Solution:
We have

Question 4.
Evaluate:
Solution:
We have,

Question 5.
Evaluate:
Solution:

Question 6.
Evaluate:
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 8 Vector Algebra – I Ex 8.5

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

Question 1.
The value of \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{CD}}\) is ………………

Solution:
(c) \(\overrightarrow{0}\)

Question 2.
If \(\vec{a}+2 \vec{b}\) and \(3 \vec{a}+m \vec{b}\) are parallel, then the value of m is ………………
(a) 3
(b) \(\frac{1}{3}\)
(c) 6
(d) \(\frac{1}{6}\)
Solution:
(c) 6

Question 3.
The unit vector parallel to the resultant of the vectors \(\hat{i}+\hat{j}-\hat{k}\) and \(\hat{i}-2 \hat{j}+\hat{k}\) is ………………

Solution:

Question 4.
A vector \(\overrightarrow{O P}\) makes 60° and 45° with the positive direction of the x and y axes respectively. Then the angle between \(\overrightarrow{O P}\) and the z-axis is …………….
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Solution:
(b) 60°
α = 60°, β = 45°
We know cos²α + cos²β + cos²γ = 1
(i.e.,) \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\) + cos²γ = 1
cos²γ = 1 – \(\frac{1}{4}-\frac{1}{2}=\frac{1}{4}\)
cos γ = \(\frac{1}{2}\) ⇒ y = π/3 = 60°

Question 5.
If \(\overrightarrow{B A}=3 \hat{i}+2 \hat{j}+\hat{k}\) and the position vector of B is \(\hat{i}+3 \hat{j}-\hat{k}\), then the position vector A is …………………

Solution:

Question 6.
A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to …………..

Solution:

Question 7.
The vectors \(\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}\) are ……………
(a) parallel to each other
(b) unit vectors
(c) mutually perpendicular vectors
(d) coplanar vectors
Solution:
(d) coplanar vectors

Question 8.
If ABCD is a parallelogram, then \(\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}\) is equal to ……………

Solution:

Question 9.
One of the diagonals of parallelogram ABCD with \(\vec{a}\) and \(\vec{b}\) as adjacent sides is \(\vec{a}+\vec{b}\). The other diagonal \(\overrightarrow{\mathrm{BD}}\) is ……………

Solution:

Question 10.
If \(\vec{a}\), \(\vec{b}\) are the position vectors A and B, then which one of the following points whose position vector lies on AB, is ………….

Solution:
(c) \(\frac{2 \vec{a}+\vec{b}}{3}\)

Question 11.
If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of three collinear points, then which of the following is true?

Solution:
(b) \(2 \vec{a}=\vec{b}+\vec{c}\)

Question 12.
If \(\vec{r}=\frac{9 \vec{a}+7 \vec{b}}{16}\), then the point P whose position vector \(\vec{r}\) divides the line joining the points with position vectors \(\vec{a}\) and \(\vec{b}\) in the ratio ………………
(a) 7 : 9 internally
(b) 9 : 7 internally
(c) 9 : 7 externally
(d) 7 : 9 externally
Solution:

Question 13.
If \(\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}\) is a unit vector, then the value of λ is ……………..
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Solution:

Question 14.
Two vertices of a triangle have position vectors \(3 \hat{i}+4 \hat{j}-4 \hat{k}\) and \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) .If the position vector of the centroid is \(\hat{i}+2 \hat{j}+3 \hat{k}\), then the position vector of the third vertex is ………………….

Solution:

Question 15.

(a) 42
(b) 12
(c) 22
(d) 32
Solution:
(c) 22

Question 16.
If \(\vec{a}\) and \(\vec{b}\) having same magnitude and angle between them is 60° and their scalar product is \(\frac{1}{2}\) then \(|\vec{a}|\) is ……………
(a) 2
(b) 3
(c) 7
(d) 1
Solution:
(d) 1

Question 17.
The value of θ ∈ (0, \(\frac{\pi}{2}\)) for which the vectors are perpendicular, is equal to …………………

Solution:

Question 18.

(a) 15
(b) 35
(c) 45
(d) 25
Solution:
(d) 25

Question 19.
Vectors \(\vec{a}\) and \(\vec{b}\) are inclined at an angle θ = 120°
is equal to ………….
(a) 225
(b) 275
(c) 325
(d) 300
Solution:
(d) 300

Question 20.
If \(\vec{a}\) and \(\vec{b}\) are two vectors of magnitude 2 and inclined at an angle 60°, then the angle between \(\vec{a}\) and \(\vec{a}+\vec{b}\) is ………………
(a) 30°
(b)60°
(c) 45 °
(d) 90°
Solution:
(a) 30°

Question 21.
If the projection of \(5\hat{i} -\hat{j}-3 \hat{k}\) on the vector \(\hat{i}+3 \hat{j}+\lambda \hat{k}\) is same as the projection of \(\hat{i}+3 \hat{j}+\lambda \hat{k}\) on \(5\hat{i}- \hat{j}-3 \hat{k}\) then λ is equal to ………………
(a) ±4
(b) ±3
(c) ±5
(d) ±1
Solution:
(c) ±5

Question 22.
If (1, 2, 4) and (2, – 3λ, – 3) are the initial and terminal points of the vector \(\hat{i}+5 \hat{j}-7 \hat{k}\), then the value of λ is equal to ……………..
(a) \(\frac{7}{3}\)
(b) \(-\frac{7}{3}\)
(c) \(-\frac{5}{3}\)
(d) \(\frac{5}{3}\)
Solution:

Question 23.
If the points whose position vector \(10 \hat{i}+3 \hat{j}, 12 \hat{i}-5 \hat{j}\) and \(\vec{a} \hat{i}+11 \hat{j}\) are collinear then a is equal to ………………
(a) 6
(b) 2
(c) 5
(d) 8
Solution:
(d) 8

equating \(\hat{j}\) components
⇒ -8 = 8t ⇒ t = -1
equation \(\hat{i}\) components
t(a – 10) = 2
(i.e.,) (-1) (a – 10) = 2
a – 10 = -2
a = – 2 + 10 = -8

Question 24.
If then x is equal to …………..
(a) 5
(b) 7
(c) 26
(d) 10
Solution:
(c) 26

Question 25.
If \(\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k},|\vec{b}|=5\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\), then the area of the triangle formed by these two vectors as two sides, is …………….
(a) \(\frac{7}{4}\)
(b) \(\frac{15}{4}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{17}{4}\)
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 2 Basic Algebra Ex 2.1

Question 1.

Solution:

3.14 ∈ Q
0, 4 are integers and 0 ∈ Z, 4 ∈ N, Z, Q
\(\frac{22}{7} \in \mathrm{Q}\)

Question 2.
Prove that \(\sqrt{3}\) is an irrational number.
(Hint: Follow the method that we have used to prove \(\sqrt{2}\) ∉ Q.
Solution:
Suppose that \(\sqrt{3}\) is rational P

⇒ 3 is a factor of q also
so 3 is a factor ofp and q which is a contradiction.
⇒ \(\sqrt{3}\) is not a rational number
⇒ \(\sqrt{3}\) is an irrational number

Question 3.
Are there two distinct irrational numbers such that their difference is a rational number? Justify.
Solution:
Taking two irrational numbers as 3 + \(\sqrt{2}\) and 1 + \(\sqrt{2}\)
Their difference is a rational number. But if we take two irrational numbers as 2 – \(\sqrt{3}\) and 4 + \(\sqrt{7}\).
Their difference is again an irrational number. So unless we know the two irrational numbers we cannot say that their difference is a rational number or irrational number.

Question 4.
Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number.
Solution:
(i) Let the two irrational numbers as 2 + \(\sqrt{3}\) and 3 – \(\sqrt{3}\)
Their sum is 2 + \(\sqrt{3}\) + 3 – 3\(\sqrt{3}\) which is a rational number.
But the sum of 3 + \(\sqrt{5}\) and 4 – \(\sqrt{7}\) is not a rational number. So the sum of two irrational numbers is either rational or irrational.

(ii) Again taking two irrational numbers as π and \(\frac{3}{\pi}\) their product is \(\sqrt{3}\) and \(\sqrt{2}\) = \(\sqrt{3}\) × \(\sqrt{2}\) which is irrational, So the product of two irrational numbers is either rational or irrational.

Question 5.
Find a positive number smaller than \(\frac{1}{2^{1000}}\). Justify.
Solution:

There will not be a positive number smaller than 0.
So there will not be a +ve number smaller than \(\frac{1}{2^{1000}}\)

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 Additional Questions

Question 1.
Prove that \(\sqrt{5}\) is an irrational number.
Solution:
Suppose that \(\sqrt{5}\) is rational

So let p = 5c
substitute p = 5c in (1) we get
(5c)² = 5q² ⇒ 25c² = 5q²

⇒ 5 is a factor of q also
So 5 is a factor of p and q which is a contradiction.
⇒ \(\sqrt{5}\) is not a rational number
⇒ \(\sqrt{5}\) is an irrational number

Question 2.
Prove that 0.33333 = \(\frac{1}{3}\)
Solution:
Let x = 0.33333….
10x = 3.3333 ….
10x – x = 9x = 3

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 5.5

Choose the correct or the most suitable answer:

Question 1.
The value of 2 + 4 + 6 + … + 2n is …..

Solution:
(d) n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is ……….
(a) ¹⁰C₆
(b) 2⁶
(c) ¹⁰C₆2⁶
(d) ¹⁰C₆2¹⁰
Solution:
(d) ¹⁰C₆2¹⁰
Hint.
t_{r + 1} = 2¹⁰(ⁿC_r)
To find coefficient of x₆ put r = 6
∴ coefficient of x₆ = 210 [¹⁰C₆]

Question 3.
The coefficient of x⁸y¹² in the expansion of (2x + 3y)²⁰ is …….
(a) 0
(b) 2⁸3¹²
(c) 2⁸3¹² + 2¹²3⁸
(d) ²⁰C₈2⁸3¹²
Solution:
(d) ²⁰C₈2⁸3¹²

Question 4.
If ⁿC₁₀ > ⁿC_r for all possible r, then a value of n is ……..
(a) 10
(b) 21
(c) 19
(d) 20
Solution:
(d) 20
Hint.
Out of ¹⁰C₁₀, ²¹C₁₀, ¹⁹C₁₀ and ²⁰C₁₀, ²⁰C₁₀ is larger.

Question 5.
If a is the arithmetic mean and g is the geometric mean of two numbers, then ……..
(a) a ≤ g
(b) a ≥ g
(c) a = g
(d) a > g
Solution:
(b) a ≥ g
Hint. AM ≥ GM
∴ a ≥ g

Question 6.
If (1 + x²)² (1 + x)ⁿ = a₀ + a₁x + a₂x² + …. + x^{n + 4} and if a₀, a₁, a₂ are in AP, then n is ……
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b or c)n = 2 or 3

Question 7.
If a, 8, b are in A.P, a, 4, b are in G.P, if a, x, b are in HP then x is ……
(a) 2
(b) 1
(c) 4
(d) 16
Solution:
(a) 2

Question 8.

(a) AP
(b) GP
(c) HP
(d) AGP
Solution:
(c) HP

Question 9.
The HM of two positive numbers whose AM and GM are 16, 8 respectively is ………
(a) 10
(b) 6
(c) 5
(d) 4
Solution:
(d) 4
Hint.
Let the two numbers be a and b

Question 10.
If S_n denotes the sum of n terms of an AP whose common difference is d, the value of \(\mathrm{S}_{n}-2 \mathrm{S}_{n-1}+S_{n-2}\) is ……
(a) d
(b) 2d
(c) 4d
(d) d²
Solution:
(a) d

Question 11.
The remainder when 38¹⁵ is divided by 13 is …….
(a) 12
(b) 1
(c) 11
(d) 5
Solution:
(a) 12
Hint.
38¹⁵ = (39 – 1)¹⁵ = 39¹⁵ – 15C₁ 39¹⁴(1) + 15C₂ (39)¹³(1)² – 15C₃ (39)¹²(1)³ ….. + 15C₁₄ (39)¹(1) – 15C₁₅(1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.

Question 12.
The n^th term of the sequence 1, 2, 4, 7, 11, …… is

Solution:
(d) \(\frac{n^{2}-n+2}{2}\)

Question 13.

Solution:

Solution:
(d) \(\frac{\sqrt{2 n+1}-1}{2}\)

Question 14.


Solution:

Question 15.


Solution:
(c) \(\frac{n(n+1)}{\sqrt{2}}\)

Question 16.

(a) 14
(b) 7
(c) 4
(d) 6
Solution:
(a) 14

Question 17.
The sum of an infinite GP is 18. If the first term is 6, the common ratio is ………
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{3}{4}\)
Solution:
(b) \(\frac{2}{3}\)
Hint:

18r = 18 – 6 = 12
r = 12/18 = 2/3

Question 18.
The coefficient of x⁵ in the series e^-2x is ………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{-4}{15}\)
(d) \(\frac{4}{15}\)
Answer:
(c) \(\frac{-4}{15}\)

Question 19.

Answer:
(c) \(\frac{(e-1)^{2}}{2 e}\)

Question 20.


Solution:
(b) \(\frac{3}{2} \log \left(\frac{5}{3}\right)\)

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 8 Vector Algebra – I Ex 8.4

Question 1.
Find the magnitude of
Solution:

Question 2.
Show that
Solution:

Question 3.
Find the vectors of magnitude \(10 \sqrt{3}\) that are perpendicular to the plane which contains \(\hat{i}+2 \hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+4 \hat{k}\)
Solution:

Question 4.
Find the unit vectors perpendicular to each of the vectors

Solution:

Question 5.
Find the area of the parallelogram whose two adjacent sides are determined by the vectors \(\hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:

Question 6.
Find the area of the triangle whose vertices are A(3, -1, 2), B(1, -1, -3) and C(4, -3, 1)
Solution:
A = (3, -1, 2); B = (1, -1, -3) and C = (4, -3, 1)

Question 7.
If \(\vec{a}, \vec{b}, \vec{c}\) are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is \(\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|\). Also deduce the condition for collinearity of the points A, B, C
Solution:

If the points A, B, C are collinear, then the area of ∆ABC = 0.

Question 8.
For any vector \(\vec{a}\) prove that
Solution:

Question 9.
Let \(\vec{a}, \vec{b}, \vec{c}\) be unit vectors such that \(\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}=\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{c}}=\mathbf{0}\) and the angle between \(\vec{b} \text { and } \vec{c} \text { is } \frac{\pi}{3}\). Prove that \(\vec{a}=\pm \frac{2}{\sqrt{3}}(\vec{b} \times \vec{c})\)
Solution:
Given \(|\vec{a}|=|\vec{b}|=|\vec{c}|\) = 1

Question 10.
Find the angle between the vectors using vector product
Solution:
The angle between \(\vec{a}\) and \(\vec{b}\) using vector product is given by

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.4 Additional Problems

Question 1.

Solution:

Question 2.
If \(\vec{a}\), \(\vec{b}\) are any two vectors, then prove that
Solution:

Question 3.
Find the angle between the vectors by using cross product.
Solution:

Question 4.
Find the vector of magnitude 6 which are perpendicular to both the vectors
Solution:

Question 5.
Find the vectors whose length 5 which are perpendicular to the vectors
Solution:

Question 6.

Solution:
Given \(\vec{a} \times \vec{b}=\vec{c} \times \vec{d}\) and \(\vec{a} \times \vec{c}=\vec{b} \times \vec{d}\)

Question 7.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) if \(|\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}|=\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}\)
Solution:

Question 8.

Solution:

Question 9.
If find the angle between \(\vec{a}\) and \(\vec{b}\)
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 8 Vector Algebra – I Ex 8.3

Question 1.
Find \(\vec{a} \cdot \vec{b}\) when

Solution:

Question 2.
Find the value of λ for which the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular, where

Solution:
When \(\vec{a}\) and \(\vec{b}\) are ⊥^r then \(\vec{a} \cdot \vec{b}\) = 0
\(\vec{a}\) ⊥^r \(\vec{b}\) ⇒ \(\vec{a} \cdot \vec{b}\) = 0
(i) (2) (1) + (λ) (-2) + (1) (3) = 0 ⇒ λ = 5/2
(ii) (2) (3) + (4) (-2) + (-1) (λ) = 0
6 – 8 – λ = 0
-λ – 2 = 0 ⇒ -λ = 2 ⇒ λ = -2

Question 3.
If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 15 and \(\vec{a} \cdot \vec{b}\) = 75\(\sqrt{2}\) , find the angle between \(\vec{a}\) and \(\vec{a}\) .
Solution:

Question 4.
Find the angle between the vectors

Solution:

Question 5.
If \(\overrightarrow{\boldsymbol{a}}, \overrightarrow{\boldsymbol{b}}, \overrightarrow{\boldsymbol{c}}\) are three vectors such that \(\vec{a}+2 \vec{b}+\vec{c}=\overrightarrow{0}\) and \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=7\) find the angle between \(\vec{a}\) and \(\vec{b}\)
Solution:

Question 6.
Show that the vectors are mutually orthogonal.
Solution:

Question 7.
Show that the vectors form a right angled triangle.
Solution:

So, the given vectors form the sides of a right angled triangle

Question 8.

Solution:

Question 9.
Show that the points (2, -1, 3) (4, 3, 1) and (3, 1, 2) are collinear
Solution:
Let the given points be A, B, C

Question 10.
If \(\vec{a}, \vec{b}\) are unit vectors and θ is the angle between them, show that

Solution:

Question 11.
Let \(\vec{a}, \vec{b}, \vec{c}\) be the three vectors such that \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5\) and each one of them being perpendicular to the sum of the other two, find \(|\vec{a}+\vec{b}+\vec{c}|\)
Solution:

Question 12.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(2 \hat{i}+6 \hat{j}+3 \hat{k}\)
Solution:

Question 13.
Find λ, when the projection of is 5 units.
Solution:

Question 14.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3 Additional Problems

Question 1.
Find λ so that the vectors are perpendicular to each other.
Solution:

Question 2.

Solution:

Question 3.
If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is \(\sqrt{3}\)
Solution:

Question 4.
Show that the vectors form a right angled triangle.
Solution:

⇒ The given vectors form the sides of a right of a right angled triangle.

Question 5.
Find the projection of

Solution:

Question 6.
Show that the vector \(\hat{i}+\hat{j}+\hat{k}\) is equally inclined with the coordinate axes.
Solution:

Question 7.
If \(\vec{a}, \vec{b}, \vec{c}\) are three mutually perpendicular unit vectors, then prove that \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}\)
Solution:

Question 8.
Show that the points whose positions vectors from a right angled triangle.
Solution:

⇒ The given points form a right angled triangle.

Question 9.

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 5.4

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
Solution:

Question 2.
Find \(\sqrt[3]{1001}\) approximately (two decimal places).
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
(ii) e^-2x
(iii) \(e^{\frac{x}{2}}\)
Solution:

Question 6.
Write the first 4 terms of the logarithmic series
(i) log(1 + 4x),
(ii) log(1 – 2x),
(iii) \(\log \left(\frac{1+3 x}{1-3 x}\right)\)
(iv) \(\log \left(\frac{1-2 x}{1+2 x}\right)\).
Find the intervals on which the expansions are valid.
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Additional Questions

Question 1.

Solution:

Question 2.
Write the four terms in the expansions of the following:

Solution:

Question 3.
Evaluate the following:

Solution:

Question 4.

Solution:

Question 5.

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 5.3

Question 1.
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

Question 2.

Solution:

Question 3.
Compute the sum of first n terms of the following series:
(i) 8 + 88 + 888 + 8888 + ……
(i) 6 + 66 + 666 + 6666 + …..
Solution:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + ……..
Solution:

Question 5.
Find the general term and sum to n terms of the sequence \(1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \ldots \ldots \ldots\)
Solution:

Denominator 1, 3, 9, 27, which is a G.P. a = 1, r = 3

It is an arithmetico Geometric series. Here the n^th term is t_n = [a + (n – 1)d]r^{n – 1} where a = 1, d = 3 and r = 1/3
Now the sum to n terms is

Question 6.

Solution:

Question 7.
Show that the sum of (m + n)^th and (m – n)^th term of an A.P. is equal to twice the m^th term.
Solution:
Let the A.P. be a, a + d, a + 2d, ……..
t_{m + n} = a + (m + n – 1)d
t_{m – n} = a + (m – n – 1)d
t_m = a + (m – 1)d
2t_m = 2[a + (m – 1)d]
To prove t_{m + n} + t_{m – n} = 2t_m
LHS t_{m + n} + t_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d

= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t_m = RHS

Question 8.
A man repays an amount of ₹ 3250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 per month. How long will it take him to clear the amount?
Solution:
a = 20, d = 15, S_n = 3250 to find n.

3n² + 5n – 1300 = 0
3n² – 60n + 65n – 1300 = 0
3n (n – 20) + 65 (n – 20) = 0
(3n + 65) (n – 20) = 0
n = – 65/3 or 20
n = – 65/3 is not possible
∴ n = 20
So he will take 20 months to clear the amount.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Solution:
t₁ = 24 × 2 = 48, t₂ =48 + 8 = 56 or (24 + 4)2, t₃ =(28 + 4)2 = 64 which is an A.P.
Here a = 48,
d = 56 – 48 = 8

The contestant has to run 2480 m to bring all the balls.

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?
Solution:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2² = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n^th hour = 30 × 2ⁿ

Question 11.
What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:

Question 12.
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Solution:
a = 5, r = 2, t_n >150000, To find ‘n’

On the 15^th day it will grow over 150000 units.

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Additional Questions

Question 1.
If the ratio of the sums of m terms and n terms of an A.P. be m² : n², prove that the ratio of its m^th and n^th terms is (2m – 1) : (2n – 1).
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

⇒ 2an + (mn – n)d = 2am + (mn – m)d
⇒ 2an – 2am = (mn – m – mn + n)d
⇒ 2a(n – m) = (n – m)d ⇒ d = 2a, [n – m ≠ 0, as n ≠ m]

Hence, the required ratio is (2m – 1) : (2n – 1)

Question 2.
Determine the number of terms of geometric progression {a_n} if a₁ = 3, a_n = 96, S_n = 189.
Solution:

Question 3.
The sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. Find the common ratio of the G.P.
Solution:

Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

Question 4.
Find the sum to n terms the series: (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …
Solution:
(x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + … to n terms

Question 5.
Sum the series: (1 + x) + (1 + x + x²) + (1 + x + x² + x³) + … up to n terms
Solution:

Question 6.
Sum up to n terms the series: 7 + 77 + 777 + 7777 + …
Solution:
S = 1 + 77 + 777 + 7777 + … to n terms

Question 7.
If S₁, S₂, S₃ be respectively the sums of n, 2n, 3n, terms of a G.P. , then prove that S₁(S₃ – S₂) = (S₂ – S₁)₂.
Solution:
Let a be the first term and r be the common ratio of G.P.

Question 8.
If sum of the n terms of a G.P be S, their product P and the sum of their reciprocals R, the prove that \(P^{2}=\left(\frac{S}{R}\right)^{n}\)
Solution:
Let a be the first term and r be the common ratio of the G.P.

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 8 Vector Algebra – I Ex 8.1

Question 1.
Represent graphically the displacement of
(i) 45 cm 30 ° north of east
(ii) 80 km, 60° south of west
Solution:
(i) 45 cm 30 0 north of east

(ii) 80 km 60° south of west

Question 2.
Prove that the relation R defined on the set V of all vectors by \(\vec{a}\) R \(\vec{b}\) if \(\vec{a}=\vec{b}\) is an equivalence relation on V.
Solution:
\(\vec{a}\) R \(\vec{b}\) is given as \(\vec{a}=\vec{b}\).
(i) \(\vec{a}\) = \(\vec{a}\) ⇒ \(\vec{a}\) R \(\vec{a}\)
(i.e.,) the relation is reflexive.

(ii) \(\vec{a}=\vec{b}\) ⇒ \(\vec{b}\) = \(\vec{a}\)
(i.e.,) \(\vec{a}\) R \(\vec{b}\) – \(\vec{b}\) R \(\vec{a}\)
So, the relation is symmetric.

(iii) \(\vec{a}=\vec{b} ; \vec{b}=\vec{c} \Rightarrow \vec{a}=\vec{c}\)
(i.e„) \(\vec{a}\) R \(\vec{b}\) ; \(\vec{b}\) R \(\vec{c}\) ⇒ \(\vec{a}\) R \(\vec{c}\)
So the given relation is transitive
So, it is an equivalence relation.

Question 3.
Let \(\vec{a}\) and \(\vec{a}\) be the position vectors of the points A and B. Prove that the position vectors of the points which trisects the line segment AB are
Solution:

Question 4.
If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that

Solution:

Question 5.
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:

Question 6.
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:

In a quadrilateral when opposite sides are equal and parallel it is a parallelogram So, PQRS is a parallelogram, from (1) and (2).

Question 7.
If \(\vec{a}\) and \(\vec{b}\) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Solution:
OABC is a parallelogram where

Question 8.
If \(\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}\), prove that the points P, Q, R are collinear.
Solution:

But Q is a common point.
⇒ P, Q, R are collinear.

Question 9.
If D is the midpoint of the side BC of a triangle ABC, prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A C}}=2 \overrightarrow{\mathbf{A D}}\)
Solution:
D is the midpoint of ∆ ABC.

Question 10.
If G is the centroid of a triangle ABC, prove that \(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Solution:
For any triangle ABC,
\(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Now G is the centroid of ∆ABC, which divides the medians (AD, BE and CF) in the ratio 2 : 1.

Question 11.
Let A, B and C be the vertices of a triangle. Let D, E and F be the midpoints of the sides BC, CA, and AB respectively. Show that \(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}=\overrightarrow{0}\)
Solution:
In ∆ABC, D, E, F are the midpoints of BC, CA and AB respectively.

Question 12.
If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathrm{EF}}\)
Solution:
ABCD is a quadrilateral in which E and F are the midpoints of AC and BD respectively.

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1 Additional Problems

Question 1.
Shown that the points with position vectors are collinear.
Solution:
To prove the points P, Q, R are collinear we have to prove that
\(\overrightarrow{\mathrm{PQ}}\) = t \(\overrightarrow{\mathrm{PR}}\) where t is a scalar.
Let the given points be P, Q, R.

So, the points P, Q, R are collinear (i.e,) the given points are collinear.

Question 2.
If ABC and A’B’C’ are two triangles and G, G’ be their corresponding centroids, prove that \(\overrightarrow{\mathrm{AA}^{\prime}}+\overrightarrow{\mathrm{BB}^{\prime}}+\overrightarrow{\mathrm{CC}^{\prime}}=3 \overrightarrow{\mathrm{GG}}\)
Solution:
Let O be the origin.
We know when G is the centroid of ∆ ABC,

Question 3.
Prove using vectors the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Solution:
ABCD is a quadrilateral with position vectors
OA = \(\vec{a}\), OB = \(\vec{b}\), OC = \(\vec{c}\) and OD = \(\vec{d}\)
P is the midpoint of BC and R is the midpoint of AD.
Q is the midpoint of AC and S is the midpoint of BD.
To prove PQRS is a parallelogram. We have to prove that \(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}\)