Bhagya

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.
Express the following statements in symbolic form:
(i) e is a vowel or 2 + 3 = 5.
Solution:
Let p : e is a vowel.
q: 2 + 3 = 5.
Then the symbolic form of the given statement is p ∨ q.

(ii) Mango is a fruit but potato is a vegetable.
Solution:
Let p : Mango is a fruit.
q : Potato is a vegetable.
Then the symbolic form of the given statement is p ∧ q.

(iii) Milk is white or grass is green.
Solution:
Let p : Milk is white.
q : Grass is green.
Then the symbolic form of the given statement is p ∨ q.

(iv) I like playing but not singing.
Solution:
Let p : I like playing.
q : I am not singing.
Then the symbolic form of the given statement is p ∧ q.

(v) Even though it is cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is cloudy and it is still raining.
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

Question 2.
Write the truth values of the following statements:
(I) Earth is a planet and Moon is a star.
Solution:
Let p : Earth is a planet.
q : Moon is a star.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. …[T ∧ F ≡ F]

(ii) 16 is an even number and 8 is a perfect square.
Solution:
Let p : 16 is an even number.
q : 8 is a perfect square.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. ….[T ∧ F ≡ F]

(iii) A quadratic equation has two distinct roots or 6 has three prime factors.
Solution:
Let p : A quadratic equation has two distinct roots.
q : 6 has three prime factors.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …..[F ∨ F ≡ F]

(iv) The Himalayas are the highest mountains but they are part of India in the northeast.
Solution:
Let p : the Himalayas are the highest mountains.
q : They are part of India in the northeast.
Then the symbolic form of the given statement is p ∧ q.
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Question 1.
Show that C₀ + C₁ + C₂ + ….. + C₈ = 256
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + ….. + C₈ = 2⁸
∴ C₀ + C₁ + C₂ + ….. + C₈ = 256

Question 2.
Show that C₀ + C₁ + C₂ + …… + C₉ = 512
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 9, we get
C₀ + C₁ + C₂ + ….. + C₉ = 2⁹
∴ C₀ + C₁ + C₂ + …… + C₉ = 512

Question 3.
Show that C₁ + C₂ + C₃ + ….. + C₇ = 127
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 7, we get
C₀ + C₁ + C₂ + ….. + C₇ = 2⁷
∴ C₀ + C₁ + C₂ +….. + C₇ = 128
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₇ = 128
∴ C₁ + C₂ + ….. + C₇ = 128 – 1 = 127

Question 4.
Show that C₁ + C₂ + C₃ + ….. + C₆ = 63
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 6, we get
C₀ + C₁ + C₂ + ….. + C₆ = 2⁶
∴ C₀ + C₁ + C₂ + …… + C₆ = 64
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₆ = 64
∴ C₁ + C₂ + ….. + C₆ = 64 – 1 = 63

Question 5.
Show that C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128
Solution:
Since C₀ + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + C₃ + …… + C₈ = 2⁸
But, sum of even coefficients = sum of odd coefficients
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇
Let C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = k
Now, C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ + C₇ + C₈ = 256
∴ (C₀ + C₂ + C₄ + C₆ + C₈) + (C₁ + C₃ + C₅ + C₇) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128

Question 6.
Show that C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
But, C₀ = 1
∴ 1 + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
∴ C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1

Question 7.
Show that C₀ + 2C₁ + 3C₂ + 4C₃ + ….. + (n + 1)C_n = (n + 2) 2^n-1
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)^-4
Solution:

(ii) (1 – x)^1/3
Solution:

(iii) (1 – x²)^-3
Solution:

(iv) (1 + x)^-1/5
Solution:

(v) (1 + x²)^-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)^-3
Solution:
(a – b)^-3 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{-3}\)

(ii) (a + b)^-4
Solution:

(iii) (a + b)^1/4
Solution:

(iv) (a – b)^-1/4
Solution:
(a – b)^-1/4 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{\frac{-1}{4}}\)

(v) (a + b)^-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)^-4
Solution:

(ii) (1 + 3x)^-1/2
Solution:

(iii) (2 – 3x)^1/3
Solution:

(iv) (5 + 4x)^-1/2
Solution:

(v) (5 – 3x)^-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) \(\sqrt[3]{126}\)
Solution:

(iii) \(\sqrt[4]{16.08}\)
Solution:

(iv) (1.02)^-5
Solution:

(v) (0.98)^-3
Solution:

Maharashtra State Board HSC 12th Commerce Maths & Statistics Digest Pdf, 12th Commerce Maharashtra State Board Maths Solution Book Pdf Part 1 & 2 free download in English Medium and Marathi Medium 2021-2022.

Maharashtra State Board 12th Commerce Maths Digest Pdf

12th Commerce Maths Digest Pdf Part 1

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions Chapter 1 Mathematical Logic

• Chapter 1 Mathematical Logic Ex 1.1
• Chapter 1 Mathematical Logic Ex 1.2
• Chapter 1 Mathematical Logic Ex 1.3
• Chapter 1 Mathematical Logic Ex 1.4
• Chapter 1 Mathematical Logic Ex 1.5
• Chapter 1 Mathematical Logic Ex 1.6
• Chapter 1 Mathematical Logic Ex 1.7
• Chapter 1 Mathematical Logic Ex 1.8
• Chapter 1 Mathematical Logic Ex 1.9
• Chapter 1 Mathematical Logic Ex 1.10
• Chapter 1 Mathematical Logic Miscellaneous Exercise 1

12th Commerce Maths Book Pdf Chapter 2 Matrices

• Chapter 2 Matrices Ex 2.1
• Chapter 2 Matrices Ex 2.2
• Chapter 2 Matrices Ex 2.3
• Chapter 2 Matrices Ex 2.4
• Chapter 2 Matrices Ex 2.5
• Chapter 2 Matrices Ex 2.6
• Chapter 2 Matrices Miscellaneous Exercise 2

Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 3 Differentiation

• Chapter 3 Differentiation Ex 3.1
• Chapter 3 Differentiation Ex 3.2
• Chapter 3 Differentiation Ex 3.3
• Chapter 3 Differentiation Ex 3.4
• Chapter 3 Differentiation Ex 3.5
• Chapter 3 Differentiation Ex 3.6
• Chapter 3 Differentiation Miscellaneous Exercise 3

12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives

• Chapter 4 Applications of Derivatives Ex 4.1
• Chapter 4 Applications of Derivatives Ex 4.2
• Chapter 4 Applications of Derivatives Ex 4.3
• Chapter 4 Applications of Derivatives Ex 4.4
• Chapter 4 Applications of Derivatives Miscellaneous Exercise 4

12th Commerce Maths Notes Chapter 5 Integration

• Chapter 5 Integration Ex 5.1
• Chapter 5 Integration Ex 5.2
• Chapter 5 Integration Ex 5.3
• Chapter 5 Integration Ex 5.4
• Chapter 5 Integration Ex 5.5
• Chapter 5 Integration Ex 5.6
• Chapter 5 Integration Miscellaneous Exercise 5

12th Commerce Maths Digest Pdf Chapter 6 Definite Integration

• Chapter 6 Definite Integration Ex 6.1
• Chapter 6 Definite Integration Ex 6.2
• Chapter 6 Definite Integration Miscellaneous Exercise 6

12th Maharashtra State Board Maths Solution Book Pdf Part 1 Chapter 7 Application of Definite Integration

• Chapter 7 Application of Definite Integration Ex 7.1
• Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

12th Maths 1 Digest Pdf Chapter 8 Differential Equation and Applications

• Chapter 8 Differential Equation and Applications Ex 8.1
• Chapter 8 Differential Equation and Applications Ex 8.2
• Chapter 8 Differential Equation and Applications Ex 8.3
• Chapter 8 Differential Equation and Applications Ex 8.4
• Chapter 8 Differential Equation and Applications Ex 8.5
• Chapter 8 Differential Equation and Applications Ex 8.6
• Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8

12th Commerce Maths Solution Book Pdf Part 2

HSC Maths Textbook Pdf Chapter 1 Commission, Brokerage and Discount

• Chapter 1 Commission, Brokerage and Discount Ex 1.1
• Chapter 1 Commission, Brokerage and Discount Ex 1.2
• Chapter 1 Commission, Brokerage and Discount Miscellaneous Exercise 1

12th Maths Digest Pdf Part 2 Chapter 2 Insurance and Annuity

• Chapter 2 Insurance and Annuity Ex 2.1
• Chapter 2 Insurance and Annuity Ex 2.2
• Chapter 2 Insurance and Annuity Miscellaneous Exercise 2

12 Maths Book Pdf State Board Chapter 3 Linear Regression

• Chapter 3 Linear Regression Ex 3.1
• Chapter 3 Linear Regression Ex 3.2
• Chapter 3 Linear Regression Ex 3.3
• Chapter 3 Linear Regression Miscellaneous Exercise 3

12th Maths 2 Digest Pdf Chapter 4 Time Series

• Chapter 4 Time Series Ex 4.1
• Chapter 4 Time Series Miscellaneous Exercise 4

12th Maths Part 2 Digest Pdf Chapter 5 Index Numbers

• Chapter 5 Index Numbers Ex 5.1
• Chapter 5 Index Numbers Ex 5.2
• Chapter 5 Index Numbers Ex 5.3
• Chapter 5 Index Numbers Miscellaneous Exercise 5

12th Statistics Book Chapter 6 Linear Programming

• Chapter 6 Linear Programming Ex 6.1
• Chapter 6 Linear Programming Ex 6.2
• Chapter 6 Linear Programming Miscellaneous Exercise 6

12th Commerce Maths Book Pdf Chapter 7 Assignment Problem and Sequencing

• Chapter 7 Assignment Problem and Sequencing Ex 7.1
• Chapter 7 Assignment Problem and Sequencing Ex 7.2
• Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions

• Chapter 8 Probability Distributions Ex 8.1
• Chapter 8 Probability Distributions Ex 8.2
• Chapter 8 Probability Distributions Ex 8.3
• Chapter 8 Probability Distributions Ex 8.4
• Chapter 8 Probability Distributions Miscellaneous Exercise 8

Maharashtra State Board Class 12 Textbook Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Question 1.
In the following expansions, find the indicated term.
(i) \(\left(2 x^{2}+\frac{3}{2 x}\right)^{8}\), 3rd term
Solution:

(ii) \(\left(x^{2}-\frac{4}{x^{3}}\right)^{11}\), 5th term
Solution:

(iii) \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\), 7th term
Solution:

(iv) In \(\left(\frac{1}{3}+a^{2}\right)^{12}\), 9th term
Solution:

(v) In \(\left(3 a+\frac{4}{a}\right)^{13}\), 10th term
Solution:

Question 2.
In the following expansions, find the indicated coefficients.
(i) x³ in \(\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}\)
Solution:

(ii) x⁸ in \(\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}\)
Solution:

(iii) x⁹ in \(\left(\frac{1}{x}+x^{2}\right)^{18}\)
Solution:

(iv) x^-3 in \(\left(x-\frac{1}{2 x}\right)^{5}\)
Solution:

(v) x^-20 in \(\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}\)
Solution:

Question 3.
Find the constant term (term independent of x) in the expansion of
(i) \(\left(2 x+\frac{1}{3 x^{2}}\right)^{9}\)
Solution:

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
Solution:

(iii) \(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}\)
Solution:

(iv) \(\left(x^{2}-\frac{1}{x}\right)^{9}\)
Solution:

(v) \(\left(2 x^{2}-\frac{5}{x}\right)^{9}\)
Solution:

Question 4.
Find the middle terms in the expansion of
(i) \(\left(\frac{x}{y}+\frac{y}{x}\right)^{12}\)
Solution:

(ii) \(\left(x^{2}+\frac{1}{x}\right)^{7}\)
Solution:

(iii) \(\left(x^{2}-\frac{2}{x}\right)^{8}\)
Solution:

(iv) \(\left(\frac{x}{a}-\frac{a}{x}\right)^{10}\)
Solution:

(v) \(\left(x^{4}-\frac{1}{x^{3}}\right)^{11}\)
Solution:

Question 5.
In the expansion of (k + x)⁸, the coefficient of x⁵ is 10 times the coefficient of x⁶. Find the value of k.
Solution:

Question 6.
Find the term containing x⁶ in the expansion of (2 – x) (3x + 1)⁹.
Solution:

Question 7.
The coefficient of x² in the expansion of (1 + 2x)^m is 112. Find m.
Solution:
The coefficient of x² in (1 + 2x)^m = ^mC₂ (2²)
Given that the coefficient of x² = 112
∴ ^mC₂ (4) = 112
∴ ^mC₂ = 28
∴ \(\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28\)
∴ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28\)
∴ m(m – 1) = 56
∴ m² – m – 56 = 0
∴ (m – 8) (m + 7) = 0
As m cannot be negative.
∴ m = 8

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2

Question 1.
Expand:
(i) (√3 + √2)⁴
Solution:
Here, a = √3, b = √2 and n = 4.
Using binomial theorem,

∴ (√3 + √2)⁴ = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)
= 9 + 12√6 + 36 + 8√6 + 4
= 49 + 20√6

(ii) (√5 – √2)⁵
Solution:
Here, a = √5, b = √2 and n = 5.
Using binomial theorem,

Question 2.
Expand:
(i) (2x² + 3)⁴
Solution:
Here, a = 2x², b = 3 and n = 4.
Using binomial theorem,

(ii) \(\left(2 x-\frac{1}{x}\right)^{6}\)
Solution:
Here, a = 2x, b = \(\frac{1}{x}\) and n = 6.
Using binomial theorem,

Question 3.
Find the value of
(i) (√3 + 1)⁴ – (√3 – 1)⁴
Solution:

(ii) (2 + √5)⁵ + (2 – √5)⁵
Solution:


Adding (i) and (ii), we get
∴ (2 + √5 )⁵ + (2 – √5)⁵ = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )
= 64 + 800 + 500
= 1364

Question 4.
Prove that:
(i) (√3 + √2)⁶ + (√3 – √2)⁶ = 970
Solution:

(ii) (√5 + 1)⁵ – (√5 – 1)⁵ = 352
Solution:

Question 5.
Using binomial theorem, find the value of
(i) (102)⁴
Solution:

(ii) (1.1)⁵
Solution:

Question 6.
Using binomial theorem, find the value of
(i) (9.9)³
Solution:

(ii) (0.9)⁴
Solution:

Question 7.
Without expanding, find the value of
(i) (x + 1)⁴ – 4(x + 1)³ (x – 1) + 6(x + 1)² (x – 1)² – 4(x + 1) (x – 1)³ + (x – 1)⁴
Solution:

(ii) (2x – 1)⁴ + 4(2x – 1)³ (3 – 2x) + 6(2x – 1)² (3 – 2x)² + 4(2x – 1)¹ (3 – 2x)³ + (3 – 2x)⁴
Solution:

Question 8.
Find the value of (1.02)⁶, correct upto four places of decimals.
Solution:

Question 9.
Find the value of (1.01)⁵, correct upto three places of decimals.
Solution:

Question 10.
Find the value of (0.9)⁶, correct upto four places of decimals.
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6

1. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = √x
Solution:

Question 2.
y = x⁵
Solution:

Question 3.
y = x^-7
Solution:

2. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = e^x
Solution:

Question 2.
y = e^(2x+1)
Solution:

Question 3.
y = e^{log x}
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5

1. Find \(\frac{d y}{d x}\) if:

Question 1.
x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 2.
x = 2at², y = at⁴
Solution:

Question 3.
x = e^3t, y = e^(4t+5)
Solution:
x = e^3t, y = e^(4t+5)
Differentiating x and y w.r.t. t, we get

2. Find \(\frac{d y}{d x}\) if:

Question 1.
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\)
Solution:
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 2.
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²)
Solution:
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 3.
Differentiate 5^x with respect to log x.
Solution:
Let u = 5^x and v = log x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(5^{x}\right)=5^{x} \cdot \log 5\)

3. Solve the following:

Question 1.
If x = \(a\left(1-\frac{1}{t}\right)\), y = \(a\left(1+\frac{1}{t}\right)\), then show that \(\frac{d y}{d x}\) = -1
Solution:

Question 2.
If x = \(\frac{4 t}{1+t^{2}}\), y = \(3\left(\frac{1-t^{2}}{1+t^{2}}\right)\), then show that \(\frac{d y}{d x}=-\frac{9 x}{4 y}\)
Solution:

Question 3.
If x = t . log t, y = t^t, then show that \(\frac{d y}{d x}\) – y = 0.
Solution:
x = t log t
Differentiating w.r.t. t, we get

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.5

Question 1.
Find Sn of the following arithmetico-geometric sequences.
(i) 2, 4x, 6x², 8x³, 10x⁴, ……
Solution:
2, 4x, 6x², 8x³, 10x⁴, ……
Here, 2, 4, 6, 8, 10,… are in A.P.
∴ a = 2, d = 2
∴ nth term = a + (n – 1)d
= 2 + (n – 1)(2)
= 2n

(ii) 1, 4x, 7x², 10x³, 13x⁴, ……
Solution:
1, 4x, 7x², 10x³, 13x⁴, ……
Here, 1, 4, 7, 10, 13,… are in A.P.
a = 1, d = 3
∴ nth term = a + (n – 1)d
= 1 + (n – 1)(3)
= 3n – 2
Also, 1, x, x², x³,… are in G.P.
∴ a = 1, r = x,
nth term = ar^n-1 = x^n-1
nth term of arithmetico-geometric sequence is

(iii) 1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, ……
Solution:
1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, …..
Here, 1, 2, 3, 4, 5, … are in A.P.
∴ a = 1, d = 1
∴ nth term = a + (n – 1)d
= 1 + (n – 1)1
= n

(iv) 3, 12, 36, 96, 240, ……
Solution:
3, 12, 36, 96, 240, ……
i.e., 1 × 3, 2 × 6, 3 × 12, 4 × 24, 5 × 48, …….
Here, 1, 2, 3, 4, 5, ….. are in A.P.
∴ nth term = n
Also, 3, 6, 12, 24, 48, ….. are in G.P.
∴ a = 3, r = 2
∴ nth term = ar^n-1 = 3 . (2^n-1)
∴ nth term of arithmetico-geometric sequence is

Question 2.
Find the sum to infinity of the following arithmetico-geometric sequence.
(i) \(1, \frac{2}{4}, \frac{3}{16}, \frac{4}{64}, \ldots\)
Solution:

(ii) \(3, \frac{6}{5}, \frac{9}{25}, \frac{12}{125}, \frac{15}{625}, \ldots\)
Solution:

∴ \(\frac{4}{5} S=\frac{15}{4}\)
∴ S = \(\frac{75}{16}\)

(iii) \(1, \frac{-4}{3}, \frac{7}{9}, \frac{-10}{27} \ldots\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4

1. Find \(\frac{d y}{d x}\) if:

Question 1.
√x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get

Question 2.
x³ + y³ + 4x³y = 0
Solution:
x³ + y³ + 4x³y = 0
Differentiating both sides w.r.t. x, we get

Question 3.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y.e^x + x.e^y = 1
Solution:
y.e^x + x.e^y = 1
Differentiating both sides w.r.t. x, we get

Question 2.
x^y = e^(x-y)
Solution:
x^y = e^(x-y)
∴ log x^y = log e^(x-y)
∴ y log x = (x – y) log e
∴ y log x = x – y …..[∵ log e = 1]
∴ y + y log x = x
∴ y(1 + log x) = x
∴ y = \(\frac{x}{1+\log x}\)

Question 3.
xy = log(xy)
Solution:
xy = log (xy)
∴ xy = log x + log y
Differentiating both sides w.r.t. x, we get

3. Solve the following:

Question 1.
If x⁵ . y⁷ = (x + y)¹², then show that \(\frac{d y}{d x}=\frac{y}{x}\)
Solution:
x⁵ . y⁷ = (x + y)¹²
∴ log(x⁵ . y⁷) = log(x + y)¹²
∴ log x⁵ + log y⁷ = log(x + y)¹²
∴ 5 log x + 7 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get

Question 2.
If log(x + y) = log(xy) + a, then show that \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}}\)
Solution:
log (x + y) = log (xy) + a
∴ log(x + y) = log x + log y + a
Differentiating both sides w.r.t. x, we get

Question 3.
If e^x + e^y = e^(x+y), then show that \(\frac{d y}{d x}=-e^{y-x}\).
Solution:
e^x + e^y = e^(x+y) ……….(1)
Differentiating both sides w.r.t. x, we get