Class 11

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 10 Differentiability and Methods of Differentiation Ex 10.1

Question 1.
Find the derivatives of the following functions using first principle.
(i) f(x) = 6
Solution:
Given f(x) = 6
f(x + h) = 6

[h → 0 means h is very nears to zero from left to right but not zero]

(ii) f(x) = -4x + 7
Solution:
Given f(x) = -4x + 7
f(x + h) = -4(x + h) + 7
= -4x – 4h + 7

(iii) f(x) = -x² + 2
Given f(x) = -x² + 2
f(x + h) = -(x + h)² + 2
= -x² – h² – 2xh + 2

Question 2.
Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(i) f(x) = |x – 1|
Solution:


f'(1) does not exist
∴ ‘f’ is not differentiable at x = 1.

(ii) f(x) = \(\sqrt{1-x^{2}}\)
Solution:

∴ ‘f’ is not differentiable at x = 1.

(iii)

Solution:

‘f’ is not differentiable at x = 1

Question 3.
Determine whether the following functions is differentiable at the indicated values.
(i) f(x) = x |x| at x = 0
Solution:


Limits exists
Hence ‘f’ is differentiable at x = 0.

(ii) f(x) = |x² – 1| at x = 1
Solution:

f(x) is not differentiable at x = 1.

(iii) f(x) = |x| + |x – 1| at x = 0, 1
Solution:

∴ f(x) is not differentiable at x = 0.

∴ f(x) is not differentiable at x = 1.

(iv) f(x) = sin |x| at x = 0
Solution:

∴ f(x) is not differentiable at x = 0.

Question 4.
Show that the following functions are not differentiable at the indicated value of x.
(i)
Solution:


f(x) is not differentiable at x = 2.

(ii)
Solution:

f(x) is not differentiable at x = 0.

Question 5.
The graph off is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Solution:

(i) at x = – 1 and x = 8. The graph is not differentiable since ‘ has vertical tangent at x = -1 and x = 8 (also At x = -1. The graph has shape edge v] and at x = 8; The graph has shape peak.
(ii) At x = 4: The graph f is not differentiable, since at x =4. The graph f’ is not continuous.
(iii) At x = 11; The graph f’ is not differentiable, since at x = 11. The tangent line of the graph is perpendicular.

Question 6.
If f(x) = |x + 100| + x², test whether f’ (-100) exists.
Solution:
f(x) = |x + 100| + x²

Question 7.
Examine the differentiability of functions in R by drawing the diagrams.
(i) |sin x|
Solution:

The limit exists and continuous for all x ∈ R clearly, differentiable at R — {nπ n ∈ z) Not differentiable at x = nπ, n ∈ z.

(ii) |cos x|
Solution:

Limit exist and continuous for all x ∈ R clearly, differentiable at R {(2n + 1)π/2/n ∈ z} Not differentiable at x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z.

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.1 Additional Questions

Question 1.
Is the function f(x) = |x| differentiable at the origin. Justify your answer.
Solution:

Question 2.
Discuss the differentiability of the functions:

Solution:

∴ f(2) is not differentiable at x = 2. Similarly, it can be proved for x = 4.

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 2.13

Choose the correct or the most suitable questions.

Question 1.
If |x + 2| ≤ 9, then x belongs to
(a) (-∞, -7)
(b) [-11, 7]
(c) (-∞, -7) ∪ [11, ∞)
(d)(-11, 7)
Solution:
(b) [-11, 7]
Hint:
Given |x + 2| ≤ 9
– 9 ≤ (x + 2) ≤ 9
– 9 – 2 ≤ x ≤ 9 – 2
– 11 ≤ x ≤ 7
∴ x ∈ [-11, 7]

Question 2.
Given that x, y and b are real numbers x < y, b ≥ 0, then ……..
(a) xb < yb (b) xb > yb
(c) xb ≤ vb
(d) xlb ≥ ylb
Solution:
(a) xb < yb
Hint:

Question 3.

(a) [2, ∞]
(b) (2, ∞)
(c) (-∞, 2)
(d) (-2, ∞)
Solution:
(b) (2, ∞)
Hint:

Question 4.
The solution of 5x – 1 < 24 and 5x + 1 > -24 is …….
(a) (4, 5)
(b) (-5, -4)
(c) (-5, 5)
(d) (-5, 4)
Solution:
(c) (-5, 5)
Hint:

Question 5.
The solution set of the following inequality |x – 1| ≥ |x – 3| is …….
(a) [0, 2]
(b) (2, ∞)
(c) (0, 2)
(d) (-∞, 2)
Solution:
(b) (2, ∞)

Question 6.

(a) 16
(b) 18
(c) 9
(d) 12
Solution:
(b) 18
Hint:

Question 7.

(a) -2
(b) -8
(c) -4
(d) -9
Solution:
(c) -4
Hint:

Question 8.

(a) 0.5
(b) 2.5
(c) 1.5
(d) 1.25
Solution:
(a) 0.5
Hint:

Question 9.

(a) 2
(b) 1
(c) 3
(d) 4
Solution:
(b) 1
Hint:

Question 10.
If 3 is the logarithm of 343, then the base is ……
(a) 5
(b) 7
(c) 6
(d) 9
Solution:
(b) 7
Hint.
Given log_x 343 = 3
343 = x³
7 × 7 × 7 = x³
7³ = x³
x = 7
Base of the logarithm x = 7

Question 11.
Find a so that the sum and product of the roots of the equation 2x² + (a – 3)x + 3a – 5 = 0 are equal is ……..
(a) 1
(b) 2
(c) 0
(d) 4
Solution:
(b) 2
Hint:

Question 12.
If a and b are the roots of the equation x² – kx + 16 = 0 and satisfy a² + b² = 32, then the value of k is ……
(a) 10
(b) -8
(c) (-8, 8)
(d) 6
Solution:
(c) -8, 8
Hint:
Given a and b are the roots of x² – kx + 16 = 0 satisfying a² + b² = 32
Sum of the roots a + b = \(\frac{-(-k)}{1}\)
a + b = k
Product of the roots ab = \(\frac{16}{1}\)
ab = 16
a² + b² = (a + b)² – 2ab
32 = k² – 2 × 16
32 = k² – 32
k² = 32 + 32 = 64
k = ± 8

Question 13.
The number of solutions of x² + |x – 1| = 1 is ………
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(c) 2

We have two solutions 0, 1

Question 14.
The equations whose roots are numerically equal but opposite in sign to the roots of 3x² – 5x – 7 = 0 is ……
(a) 3x² – 5x – 7 = 0
(b) 3x² + 5x – 7 = 0
(c) 3x² – 5x + 7 = 0
(d) 3x² + x – 7 = 0
Solution:
(b) 3x² + 5x – 7 = 0
Hint:

Question 15.
If 8 and 2 are the roots of x² + ax + c = 0 and 3, 3 are the roots of x² + ax + b = 0, then the roots of the equation x² + ax + b = 0 are …….
(a) 1, 2
(b) -1, 1
(c) 9, 1
(d) -1, 2
Solution:
(c) 9, 1
Hint:
Sum = 8 + 2 = 10 = -a ⇒ a = -10
Product = 3 × 3 = 9 = b ⇒ b = 9
Now the equation x² + ax + b = 0
⇒ x² – 10x + 9 = 0
⇒ (x- 9) (x – 1) = 0
x = 1 or 9

Question 16.
If a and b are the real roots of the equation x² – kx + c = 0, then the distance
between the points (a, 0) and (b, 0) is ……..

Solution:

Hint:
a + b = k, ab = c

Question 17.

(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 18.

(a) -1/2
(b) -2/3
(c) 1/2
(d) 2/3
Solution:
(a) -1/2
Hint:

Question 19.
The number of real roots of (x + 3)⁴ + (x + 5)⁴ = 16 is ……
(a) 4
(b) 2
(c) 3
(d) 0
Solution:
(a) 4
Hint:
The given equation is (x + 3)⁴ + (x + 5)⁴ = 16
Since it is a fourth degree equation it has four roots.
∴ Number of roots = 4

Question 20.
The value of log₃ 11 . log₁₁ 13 . log₁₃ 15 . log₁₅ 27 . log₂₇ 81 is …….
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Solution:
(d) 4
Hint.
log₃ 11 . log₁₁ 13 . log ₁₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log₃ 13 . log ₁₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log ₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log ₃ 27 . log ₂₇ 81
= log ₃ 81
= log ₃3⁴
= 4 log ₃3
= 4 × 1
= 4

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 9.6

Choose the correct or the most suitable answer from the given four alternatives
Question 1.

(a) 1
(b) 0
(c) ∞
(d) -∞
Solution:
(b) 0

Question 2.

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

Question 3.

(a) 0
(b) 1
(c) 2
(d) does not exist
Solution:
(d) does not exist

Question 4.

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

Question 5.

(a) e⁴
(b) e²
(c) e³
(d) 1
Solution:
(a) e⁴

Question 6.

(a) 1
(b) 0
(c) -1
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Question 7.

(a) log ab
(b) log \(\left(\frac{a}{b}\right)\)
(c) log \(\left(\frac{b}{a}\right)\)
(d) \(\frac{a}{b}\)
Solution:
(b) log \(\left(\frac{a}{b}\right)\)

Question 8.

(a) 2 log 2
(b) 2 (log 2)²
(c) log 2
(d) 3 log 2
Solution:
(b) 2 (log 2))²

Question 9.
If f(x) = \(x(-1)^{ \left\lfloor \frac { 1 }{ x } \right\rfloor }\), x ≤ θ, then the value of is equal to …………….
(a) -1
(b) 0
(c) 2
(d) 4
Solution:
(b) 0

Question 10.

(a) 2
(b) 3
(c) does not exist
(d) 0
Solution:
(c) does not exist

Limit does not exist

Question 11.
Let the function f be defined f(x) = then ……………

Solution:

The limit does not exist

Question 12.
If f: R → R is defined by f(x) = \(\lfloor x-3\rfloor+|x-4|\) for x ∈ R, then is equal to …………..
(a) -2
(b) -1
(c) 0
(d) 1
Solution:
(c) 0

Question 13.

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

Question 14.
If then the value of p is ………….
(a) 6
(b) 9
(c) 12
(d) 4
Solution:
(c) 12

Question 15.

(a) \(\sqrt{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) 2
Solution:
(a) \(\sqrt{2}\)

Question 16.

(a) \(\frac{1}{2}\)
(b) 0
(c) 1
(d) ∞
Solution:
(a) \(\frac{1}{2}\)

Question 17.

(a) 1
(b) e
(c) \(\frac{1}{e}\)
(d) 0
Solution:
(a) 1

Question 18.

(a) 1
(b) e
(c) \(\frac{1}{2}\)
(d) 0
Solution:
(a) 1

Question 19.
The value of is ……………
(a) 1
(b) -1
(c) 0
(d) ∞
Solution:
(d) ∞
Hint:

So limit does not exist

Question 20.
The value of where k is an integer is …………..
(a) -1
(b) 1
(c) 0
(d) 2
Solution:
(b) 1

Question 21.
At x = \(\frac{3}{2}\) the function f(x) = \(\frac{|2 x-3|}{2 x-3}\) is ………….
(a) Continuous
(b) discontinuous
(c) Differentiate
(d) non-zero
Solution:
(b) discontinuous

Question 22.
Let f: R → R be defined by f(x) = then f is ……………
(a) Discontinuous at x = \(\frac{1}{2}\)
(b) Continuous at x = \(\frac{1}{2}\)
(c) Continuous everywhere
(d) Discontinuous everywhere
Solution:
(b) Continuous at x = \(\frac{1}{2}\)

Question 23.
The function f(x) = is not defined for x = -1. The value of f(-1) so that the function extended by this value is continuous is …………..
(a) \(\frac{2}{3}\)
(b) \(-\frac{2}{3}\)
(c) 1
(d) 0
Solution:
(b) \(-\frac{2}{3}\)
Hint: For the function to be continuous at x = 1

Question 24.
Let f be a continuous function on [2, 5]. If f takes only rational values for all x and f(3) = 12, then f(4.5) is equal to ……………
(a) \(\frac{f(3)+f(4.5)}{7.5}\)
(b) 12
(c) 17.5
(d) \(\frac{f(4.5)-f(3)}{1.5}\)
Solution:
(b) 12

Question 25.
Let a function f be defined by f(x) = \(\frac{x-|x|}{x}\) for x ≠ 0 and f(0) = 2. Then f is …………..
(a) Continuous nowhere
(b) Continuous everywhere
(c) Continuous for all x except x = 1
(d) Continuous for all x except x = 0
Solution:
(d) Continuous for all x except x = 0
Hint:

∴ f(x) is not continuous at x = 0
⇒ f(x) is continuous for all except x = 0

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:
Let the two irrational numbers be 2 + √5 and 4 + √5
Their difference = (2 + √5) – (4 + √5)
= 2 + √5 – 4 – √5
= 2 – 4 = – 2
which is a rational 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:
Let the two irrational numbers be 3 + √5 and 7 – √5
Their sum = 3 + √5 + 7 – √5 = 3 + 7 = 10
which is a rational number
Consider the two irrational numbers 2 + √3, 2 – √3
Their product = (2 + √3) (2 – √3)
= 2² – (√3)² = 4 – 3 = 1
which is a rational number.

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 11 Integral Calculus Ex 11.3

Integrate the following w.r.to x.
Question 1.
(x + 4)⁵ + \(\frac{5}{(2-5 x)^{4}}\) – cosec²(3x – 1)
Solution:

Question 2.
4cos(5 – 2x) + 9e^3x-6 + \(\frac{24}{6-4 x}\)
Solution:

Question 3.
sec²\(\frac{x}{5}\) + 18 cos 2x + 10 sec (5x + 3) tan (5x + 3)
Solution:

Question 4.
\(\frac{8}{\sqrt{1-(4 x)^{2}}}+\frac{27}{\sqrt{1-9 x^{2}}}-\frac{15}{1+25 x^{2}}\)
Solution:

Question 5.

Solution:

Question 6.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.3 Additional Problems

Question 1.
5x⁴ + 3(2x + 3)⁴ – 6(4 – 3x)⁵
Solution:

Question 2.
4 – \(\frac{5}{x+2}\) + 3 cos 2x
Solution:

Question 3.
9 cosec²(px – q) – 6(1 – x)⁴ + 4e^{3 – 4x}
Solution:

Question 4.
\(\frac{4}{(3+4 x)}\) + (10x + 3)⁹ – 3cosec(2x + 3) cot (2x + 3)
Solution:

Question 5.
a sec²(bx + c) + \(\frac{q}{e^{l-m x}}\)
Solution:

Question 6.

Solution:

Question 7.

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 5.5

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.
20C₁₀ > 20C_r for all possible value of r.
n = 20

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.
(38)¹⁵ = (39 – 1)¹⁵
= (39)¹⁵ – 15C₁(39 )¹⁴ + 15C₂(39)¹³ + . . . . . . . + 15C₁₄ (39) – 1
In the Binomial expansion , all the terms except the last term (- 1) are divisible by 13
∴ The remainder = 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 2 Basic Algebra Ex 2.8

Question 1.
Find all values of x for which \(\frac{x^{3}(x-1)}{(x-2)}\) > 0
Solution:

Now we have to find the signs of
x³, x – 1 and x – 2 as follows
x³ = 0; x – 1 = 0 ⇒ x = 1; x – 2 = 0 ⇒ x = 2
Plotting the points in a number line and finding intervals

So the solution set = (0, 1) ∪ (2, ∞)

Question 2.

Solution:

Plotting the points 3/2, 2 and 4 on the number line and taking the intervals.

Question 3.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.8 Additional Questions

Question 1.

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 11 Integral Calculus Ex 11.2

Integrate the following functions w.r.to ‘x’.
Question 1.
(i) (x + 5)⁶
(ii) \(\frac{1}{(2-3 x)^{4}}\)
(iii) \(\sqrt{3 x+2}\)
Solution:

Question 2.
(i) sin 3x
(ii) cos (5 – 11x)
(iii) cosec² (5x – 7)
Solution:

Question 3.
(i) e^{3x – 6}
(ii) e^{8 – 7x}
(iii) \(\frac{1}{6-4 x}\)
Solution:

Question 4.
(i) sec²\(\frac{x}{5}\)
(ii) cosec (5x + 3) cot (5x + 3)
(iii) sec (2 – 15x) tan (2 – 15x)
Solution:

Question 5.
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)
(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)
(iii) \(\frac{1}{1+36 x^{2}}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2 Additional Problems

Question 1.
(3x + 4)⁶
Solution:

Question 2.
\(\frac{1}{(x+5)^{4}}\)
Solution:

Question 3.
\(\frac{1}{p+q x}\)
Solution:

Question 4.
cos(4x + 5)
Solution:

Question 5.
cosec²(7 – 11x)
Solution:

Question 6.
sec(3 + x) tan(3 + x)
Solution:

Question 7.
cosec(3 – 2x) cot(3 – 2x)
Solution:

Question 8.
e^{3x + 2}
Solution:

Question 9.
\(\frac{1}{\sin ^{2}(l-m x)}\)
Solution:

Question 10.
(lx + m)^1/2
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 8.4

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 2 Basic Algebra Ex 2.2

Question 1.
Solve for x.

(i) |3 – x| < 7
Solution:
– 7 < 3 – x < 7
– 7 – 3 < – x < 7 – 3
– 10 < – x < 4 10 > x > – 4
– 4 < x < 10
∴ The solution set is x ∈ (-4, 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
-3 |x| ≤ – 7
3 |x| ≥ 7
|x| ≥ \(\frac{7}{3}\)

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: