Category: 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 11 Integral Calculus Ex 11.11

Integrate the following with respect to x:
Question 1.

Solution:

Question 2.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.11 Additional Problems

Question 1.
\(\frac{2 x-1}{2 x^{2}+x+3}\)
Solution:

Question 2.
\(\frac{4 x+1}{x^{2}+3 x+1}\)
Solution:

Question 3.
\(\frac{2 x-3}{\sqrt{10-7 x-x^{2}}}\)
Solution:

Question 4.
\(\frac{6 x+7}{\sqrt{(x-4)(x-5)}}\)
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 11 Integral Calculus Ex 11.10

Find the integrals of the following
Question 1.
(i) \(\frac{1}{4-x^{2}}\)
(ii) \(\frac{1}{25-4 x^{2}}\)
(iii) \(\frac{1}{9 x^{2}-4}\)
Solution:

Question 2.

Solution:
(i) Let I = \(\int \frac{1}{6 x-7-x^{2}} d x\)
Consider, -x² + 6x – 7 = -[x² – 6x + 4]
= -[(x – 3)² – 9 + 7]

Question 3.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.10 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.

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 2.13

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:
-x – 2 ≤ 9 x + 2 ≤ 9
-x < 9 + 2 = 11 x ≤ 9 – 2 = 7
⇒ x ≥ -11
so 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.
⇒ log_x343 = 3 ⇒ 343 = x³
(.i.e.,) 7³ = x³ ⇒ x = 7
⇒ 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:
a + b = k ….(1) ab = 16 ….(2)
a² + b² = (a + b)² – 2ab = 32 .
k² – 32 = 32 ⇒ k² = 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 equation is (x + 3)⁴ + (x + 5)⁴ = 16
(x + 3)⁴ + (x + 5)⁴ = 2⁴
This is biquadratic equation. It has 4 roots.

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.

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 11.9

Integrate the following with respect to x
Question 1.
e^x (tan x + log sec x)
Solution:

Question 2.
e^x\(\left(\frac{x-1}{2 x^{2}}\right)\)
Solution:

Question 3.
e^x sec x (1 + tan x)
Solution:
Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)
Take f(x) = sec x
f ‘ (x) = sec x tan x
This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e^x f(x) + c
= e^x sec x + c

Question 4.
e^x \(\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right)\)
Solution:

Question 5.

Solution:

Question 6.

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 11.8

Integrate the following with respect to x.
Question 1.
(i) e^ax cos bx
(ii) e^2x sin x
(iii) e^-x cos 2x
Solution:

Question 2.
(i) e^-3x sin 2x
(ii) e^-4x sin 2x
(iii) e^-3x cos x
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.8 Additional Problems

Question 1.
e^2x sin 3x dx
Solution:

Question 2.
e^x cos 2x
Solution:

Substituting (2) in (1) we get,
I = e^x cos 2x + [e^x + sin 2x – 2I]
I = e^x cos 2x + 2e^x sin 2x – 4I
5I = e^x(cos 2x + 2 sin 2x)
∴ I = \(\frac{e^{x}}{5}\)[cos 2x + 2 sin 2x] + c

Question 3.
e^3x sin 2x
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 2.12

Question 1.
Let b > 0 and b ≠ 1. Express y = b^x in logarithmic form. Also state the domain and range of the logarithmic function.
Solution:
Given y = b^x ⇒ log_by = x, x ∈ R with range (0, ∞) (-∞, ∞)

Question 2.
Compute log₉27 – log₂₇9.
Solution:
Let log₉27 = x ⇒ 27 = 9^x ⇒ 3³ = (3²)^x = 3^2x
⇒ 2x = 3 ⇒ x = 3/2
Let log₂₇9 = x

Question 3.
Solve log₈x + log₄x + log₂x = 11
Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Another methods

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.
Solve log_{5 – x}(x² – 6x + 65) = 2
Solution:
log_{5 – x}(x² – 6x + 65) = 2
⇒ x² – 6x + 65 = (5 – x)²
x² – 6x + 65 = 25 + x² – 10x
⇒ x² – 6x + 65 – 25 – x² + 10x = 0
4x + 40 = 0 ⇒ 4x = -40
x = -10

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 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 2.11

Question 1.

Solution:
(i)

(ii)

(iii)

(iv)

(v)

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Simplify and hence find the value of n : 3²ⁿ9²m^-n/3³ⁿ = 27
Solution:

Question 5.
Find the radius of the spherical tank whose volume is 32π/3 units.
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.11 Additional Questions Solved

Question 1.
Simplify (343)^2/3
Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

squaring on bothsides

Question 5.

Solution:

L.C.M. of 2 and 3 is 6 & Raising to the power 6

since constant term is – 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x – 7) is a factor

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 11.7

Integrate the following with respect to x
Question 1.
(i) 9xe^3x
(ii) x sin 3x
(iii) 25xe^-5x
(iv) x sec x tan x
Solution:

Question 2.
(i) x log x
(ii) 27 x² e^3x
(iii) x² cos x
(iv) x³ sin x
Solution:

Question 3.
(i) \(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\)
(ii) x² e^x2
(iii) \(\tan ^{-1}\left(\frac{8 x}{1-16 x^{2}}\right)\)
(iv) \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.7 Additional Problems

Question 1.
x cosec² x
Solution:

Question 2.
x cos 5x cos 2x
Solution:

Question 3.
x² e^2x
Solution:

Question 4.

Solution:

Question 5.
cosec³ x
Solution:

Question 6.
sec³ 2x
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 11.6

Integrate the following with respect to x.
Question 1.
\(\frac{x}{\sqrt{1+x^{2}}}\)
Solution:

Question 2.
\(\frac{x^{2}}{1+x^{6}}\)
Solution:

Question 3.
\(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)
Solution:

Question 4.

Solution:

Question 5.
\(\frac{\sin \sqrt{x}}{\sqrt{x}}\)
Solution:

Question 6.
\(\frac{\cot x}{\log (\sin x)}\)
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
\(\frac{\sqrt{x}}{1+\sqrt{x}}\)
Solution:

Question 11.
\(\frac{1}{x \log x \log (\log x)}\)
Solution:

Question 12.
αβx^α-1e^-βxα
Solution:

Question 13.
\(\tan x \sqrt{\sec x}\)
Solution:

Question 14.
x(1 – x)¹⁷
Solution:

Question 15.
sin⁵ x cos³ x
Solution:

Question 16.
\(\frac{\cos x}{\cos (x-a)}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6 Additional Problems

Question 1.
(2x + 3)\(\sqrt{x^{2}+3 x-5}\)
Solution:

Question 2.
\(\frac{x \sin ^{-1}\left(x^{2}\right)}{\sqrt{1-x^{4}}}\)
Solution:

Question 3.
sec4 x tan x
Solution:

Question 4.
\(\frac{\sin x}{\sin (x+a)}\)
Solution:

Question 5.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution:

Question 6.
x(l – x)¹⁶
Solution:

Question 7.
x²(2 – x)¹⁵
Solution:

Question 8.
(x + 1)\(\sqrt{2 x+3}\)
Solution:

Question 9.
(x²)\(\sqrt{x+1}\)
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 2 Basic Algebra Ex 2.10

Determine the region in the plane determined by the inequalities:

Question 1.
x ≤ 3y, x ≥ y
Solution:
Given in equation are x ≤ 3y,x ≥ y
Suppose x = 3y ⇒ \(\frac{x}{3}=\) = y

If x = y

Question 2.
y ≥ 2x, -2x + 3y ≤ 6
Solution:
Suppose y = 2x

-2x + 3y = 6
-2x = 6 – 3y

Question 3.
3x + 5y ≥ 45, x ≥ 0, y ≥ 0.
Solution:
If 3x + 5y = 45


x ≥ 0 is nothing but the positive portion of Y-axis and y ≥ 0 is the positive portion of X-axis.
Shaded region is the required portions.

Question 4.
2x + 3y ≤ 35, y ≥ 2, x ≥ 5
Solution:
If 2x + 3y = 35 then

y = 2 is a line parallel to X-axis at a distance 2 units
x = 5 is a line parallel to Y-axis at a distance of 5 units

The required region is below 2x + 3y = 35, above y = 2 and to the right of x = 5

Question 5.
2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0.
Solution:
If 2x + 3y = 6

x + 4y = 4

x ≥ 0, y ≥ 0 represents the area in the 1 quadrant.

The required region is below 2x + 3y = 6 and below x + 4y = 4 bounded by x-axis and y-axis.

Question 6.
x – 2y ≥ 0, 2x – y ≤ -2, x ≥ 0, y ≥ 0
Solution:
If x – 2y = 0

2x – y = -2


x ≥ 0, y ≥ 0 represents the portion in the 1 quadrant only.

Question 7.
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6.
Solution:
2x + y = 8

x + 2y = 8

x + y = 6

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10 Additional Questions

Question 1.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
The given inequality is 3x + 2y ≤ 12.
Draw the graph of the line 3x + 2y ≤ 12
Table of values satisfying the equation
3x + 2y ≤ 12

Putting (0, 0) in the given inequation, we have 3 × 0 + 2 × 0 ≤ 12
∴ Half plane of 3x + 2y ≤ 12 is towards origin
Also the given inequality is x ≥ 1.
Draw the graph of the line x = 1.

Putting (0, 0) in the given inequation, we have 0 ≥ 1 which is false.
∴ Half plane of x ≥ 1 is away from origin.
The given inequality is y ≥ 2.
Putting (0, 0) in the given inequation, we have 0 ≥ 2 which is false.
∴ Half plane of y ≥ 2 is away from origin.

Question 2.
x + y ≥ 4, 2x – y > 0
Solution:
The given inequality is x + y ≥ 4.
Draw the graph of the line x + y = 4.

Table of values satisfying the equation
x + y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 ≥ 4 ⇒ 0 ≥ 4, which is false.
∴ Half plane of x + y ≥ 4 is away from origin.
Also the given inequality is 2x – y > 0.
Draw the graph of the line 2x – y = 0.
Table of values satisfying the equation
2x – y = 0

Putting (3, 0) in the given inequation, we have 2 × 3 – 0 > 0 ⇒ 6 > 0, which is true.
∴ Half plane of 2x – y > 0 containing (3, 0)

Question 3.
x + y ≤ 9, y > x, x ≥ 0
Solution:
The given inequality is x + y ≤ 9.
Draw the graph of the line x + y = 9.

Table of values satisfying the equation
x + y = 9

Putting (0, 0) in the given inequation, we have 0 + 0 ≤ 9⇒ 0 ≤ 9, is towards is origin.
∴ Half plane of x + y ≤ 9 is away from origin.
Also the given inequality is x – y < 0.
Draw the graph of the line x -y = 0.
Table of values satisfying the equation
x – y = 0

Putting (0, 3) in the given inequation, we have 0 – 3 – 0 < 0 ⇒ -3 < 0, which is true.
∴ Half plane of x – y < 0 containing the points (0, 3).

Question 4.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
The given inequality is 5x + 4y ≤ 20.
Draw the graph of the line 5x + 4y = 20.

Table of values satisfying the equation
5x + 4y = 20

Putting (0, 0) in the given inequation, we have 5 × 0 + 4 × 0 ≤ 20 ⇒ 0 ≤ 20, which is true.
∴ Half plane of 5x + 4y ≤ 20 is away from origin.
Also the given inequality is x ≥ 1.
Draw the graph of the line x = 1.
Putting (0, 0) in the given inequation, we have 0 ≥ 1, which is false.
∴ Half plane of x ≥ 1 is y ≥ 2.
Draw the graph of the line y = 2.
Putting (0, 0) in the given inequation, we have 0 ≥ 2, which is false.
∴ Half plane of y ≥ 2 is away from origin.

Question 5.
3x +4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
The given inequality is 3x + 4y ≤ 60.
Draw the graph of the line 3x + 4y = 60.
Table of values satisfying the equation
3x + 4y = 60

Putting (0, 0) in the given inequation, we have 3 × 0 + 4 × 0 ≤ 60 ⇒ 0 ≤ 60, which is true.
∴ Half plane of 3x + 4y ≤ 60 is towards origin.
Also the given inequality is x + 3y ≤ 30.
Draw the graph of the line x + 3y = 30.
Table of values satisfying the equation .
x + 3y = 30


Putting (0, 0) in the given inequation, we have 0 + 3 × 0 ≤ 30 ⇒ 0 ≤ 30, which is true.
∴ Half plane of x + 3y ≤ 30 is towards origin.