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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 9.2

Evaluate the following limits:
Question 1.

Solution:

Question 2.
m and n are integers.
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:

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

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

∴ Limit does not exist

Question 14.

Solution:

Question 15.

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 9.1

In problems 1-6, complete the table using calculate and use the result to estimate the limit.
Question 1.

Solution:

Question 2.

Solution:

∴ Limit is 0.25

Question 3.

Solution:

∴ Limit is 0.288

Question 4.

Solution:

∴ Limit is -0.25

Question 5.

Solution:

∴ Limit is 1

Question 6.

Solution:

∴ Limit is 0

In exercise problems 7-15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
Question 7.
\(\lim _{x \rightarrow 3}\)(4 – x)
Solution:

Limit exist and is equal to 1

Question 8.
\(\lim _{x \rightarrow 1}\)(x² + 2)
Solution:

Limit exist and is equal to = 3

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Limit does not exist

Question 12.

Solution:
When x → 5, (x – 5) = -(x – 5)
∴ \(\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)}\) = -1
When x → 5⁺, (x – 5) = (x – 5)

Question 13.
\(\lim _{x \rightarrow 1}\) sin(πx)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\) (sec x)
Solution:

Question 15.
\(\lim _{x \rightarrow \frac{\pi}{2}}\) tan x
Solution:

Limit does not exist

Sketch the graph of f, then identify the values of x₀ for which \(\lim _{x \rightarrow x_{0}}\) f(x) exists.
Question 16.

Solution:

Question 17.

Solution:

Limit exists except at x₀ = π

Question 18.
Sketch the graph of a function f that satisfies the given values:

Solution:

Question 19.
Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25
Solution:

Question 20.
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Solution:
No, f(x) = 4, It is the value of the function at x = 2
This limit doesn’t exists at x = 2
Since f(2) = 4
It need not imply that \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)\)
∴ we cannot conclude at x = 2

Question 21.
If the limit of f(x) as z approaches 2 is 4, can you conclude anything about f(2)?
Explain reasoning.
Solution:
\(\lim _{x \rightarrow 2}\) f(x) 4 , \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
When x approaches 2 from the left or from the right f(x) approaches 4.
Given that \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
The existence or non existence at x =2 has no leaving on the existence of the limit of f(x) as x approaches to 2.
∴ We cannot conclude the value of f(2)

Question 22.
Evaluate: \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) if it exists by finding f(3^–) and f(3⁺).
Solution:

Question 23.
Verify the existence of
Solution:

Limit does not exist

Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1 Additional Questions

Question 1.
Suppose . What are possible values of a and b?
Solution:

Question 2.

Solution:
We have,

Question 3.

Solution:
We have,

Question 4.

Solution:

Question 5.
Let a₁, a₂ …………… a_n be fixed real numbers such that f(x) = (x – a₁) , (x – a₂), ………. (x – a_n) what \(\lim _{x \rightarrow a}\) f(x) For a ≠ a₁, a₂, ………… a_n compute \(\lim _{x \rightarrow a}\) f(x)
Solution:
We have,

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 2.2

Question 1.
Solve for x.

(i) |3 – x| < 7
Solution:
⇒ -7 < 3 – x < 7 3 – x > -7
-x > -7 -3 (= -10)
-x > -10 ⇒ x < 10
3 – x < 7
– x < 7 – 3 (= 4)
– x < 4x > -4 … .(2)
From (1) and (2) ⇒ x > -4 and x < 10
⇒ -4 < x < 10

(ii) |4x – 5| ≥ -2
Solution:

(iii)

Solution:

(iv) |x| – 10 < -3
Solution:
|x| < -3 + 10 (= 7)
|x| < 7 ⇒ -7 < x < 7

Question 2.
Solve \(\frac{1}{|2 x-1|}<6\) and express the solution using the interval notation.
Solution:

Question 3.
Solve -3|x| + 5 ≤ – 2 and graph the solution set in a number line.
Solution:
-3|x| + 5 ≤ – 2
⇒ -3 |x| ≤ – 2 – 5 (= -7)
-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Question 4.
Solve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.
Solution:

Question 5.

Solution:

Question 6.
Solve |5x – 12| < -2
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

⇒ x – 2< – 1 (or) x – 2 > 1 and – 2 < x – 2 < 2
⇒ x < 1 (or) x > 3 and -2 + 2 < x < 2 + 2
⇒ x < 1 (or) x > 3 and 0 < x < 4 Hence, the required solution is (0, 1) ∪ (3, 4)

Question 3.

Solution:

Question 4.
Solve: |x – 1| ≤ 5, |x| ≥ 2
Solution:
|x – 1| ≤ 5 and |x| ≥ 2
⇒ -5 ≤ x – 1 ≤ 5 and x ≤ -2 (or) x > 2
⇒ – 5 + 1 ≤ x ≤ 5 + 1
⇒ -4 ≤ x ≤ 6 and x ≤ -2 (or) x ≥ 2
Hence x < [-4, -2] ∪ [2, 6]

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 9 Limits and Continuity Ex 9.5

Question 1.
Prove that f(x) = 2x² + 3x – 5 is continuous at all points in R.
Solution:
Polynomial functions are continuous at every points of R.

Question 2.
Examine the continuity of the following:
(i) x + sin x
Solution:
f(x) = x + sin x
The Domain of the function (-∞, ∞)
∴ f(x) is continuous in (-∞, ∞)(i.e.,) for all x ∈ R

(ii) x² cos x
Solution:
f(x) = x² cos x
The Domain of the function (-∞, ∞)

f(x) is continuous in R

(iii) e^x tan x
Solution:
The Domain of the function in R – {(2n + 1) π/2}
∴ The functions is continuous for all x ∈ R – (2n + 1) \(\frac{\pi}{2}\), n ∈ Z

(iv) e^2x + x²
f(x) = e^2x + x² = 1 + 2x + \(\frac{(2 x)^{2}}{2 !}\) + …………. + x²
Solution:
∴ The functions is continuous for all x ∈ R

(v) x.ln x
Solution:
Thus f(x) is continuous for (0, ∞)

(vi) \(\frac{\sin x}{x^{2}}\)
Solution:
Thus f(x) is continuous for all x ∈ R – {0}

(vii) \(\frac{x^{2}-16}{x+4}\)
Solution:
f(x) = \(\frac{x^{2}-16}{x+4}=\frac{(x-4)(x+4)}{x+4}\)
The function f(x) is continuous for all x ∈ R – {-4}

(viii) |x + 2| + |x – 1|
Solution:
f(x) is continuous for x ∈ R

(ix) \(\frac{|x-2|}{|x+1|}\)
Solution:
The function is continuous for all x ∈ R – {-1}

(x) cot x + tan x
Solution:
The function is continuous for all x ∈ R – \(\frac{n \pi}{2}\), n ∈ z.

Question 3.
Find the points of discontinuity of the function f, where,
(i)

Solution:
f(3) = 12 + 5 = 17

∴ f(x) is discontinuous at x = 3

(ii)

Solution:
f(x) = 4

∴ f(x) is continuous for all x ∈ R

(iii)

Solution:
f(x) = 8 – 3 = 5

∴ f(x) is continuous for all x ∈ R

(iv)

Solution:

∴ f(x) is continuous for all x ∈ [0, π/2]

Question 4.
At the given points x₀ discover whether the given function is continuous or discontinuous citing the reasons for your answer.
(i)

Solution:
Given f(x₀) = 1

∴ f(x) is continuous at x₀ = 1

(ii)

Solution:

∴ f(x) is not continuous at x₀ = 3

Question 5.
Show that the function is continuous on (-∞, ∞)
Solution:

Given that f(1) = 3
∴ f(x) is continuous for all x ∈ R

Question 6.
For what value of α is this function f(x) = continuous at x = 1?
Solution:

∵ f(x) is continuous at x = 1, α = 4

Question 7.
Let Graph the function. Show that f(x) continuous on (-∞, ∞)
Solution:


∴ f(x) is continuous in (-∞, ∞)

Question 8.
If f and g are continuous function with f(3) = 5 and find g(3).
Solution:
Since f and g are continuous

2f(3) – g(3) = 4
2(5) – g(3) = 4
10 – 4 = g(3)
g(3) = 6

Question 9.
Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.
(i)

∴ f(x) is not continuous at x = 1
Solution:

f(x) is not continuous at x = 1

(ii)

Solution:

∴ f(x) is not continuous at x = 0

Question 10.
A function f is defined as follows:

Is the function continuous?
Solution:

From (i), (ii) and (iii)
f(x) is continuous at x = 0, 1, 3

Question 11.
Which of the following functions f has removable discontinuity at x = x₀? If the discontinuity is removable, find a function g that agrees with f for x ≠ x₀ and is continuous on R.
(i) f(x) = \(\frac{x^{2}-2 x-8}{x+2}\), x₀ = -2
Solution:

(ii) f(x) = \(\frac{x^{3}+64}{x+4}\), x₀ = -4
Solution:

(iii) f(x) = \(\frac{3-\sqrt{x}}{9-x}\), x₀ = 9
Solution:

Question 12.
Find the constant b that makes g continuous on (-∞, ∞)

Solution:
Since g(x) is continuous,

Question 13.
Consider the function f(x) = x sin \(\frac{\pi}{x}\). What value must we give f(0) in order to make the function continuous everywhere?
Solution:

so to make the function f(x) is continuous at f(0) = 0

Question 14.
The function f(x) = \(\frac{x^{2}-1}{x^{3}-1}\) is not defined at x = 1. What value must we give f(1) in order to make f(x) continuous at x = 1?
Solution:

Question 15.
State how continuity is destroyed at x = x₀ for each of the following graphs.

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 9.4

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: