Bhagya

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.6

Evaluate:

Question 1.
\(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Let \(\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\)
∴ 2x + 1 = A(x – 2) + B(x + 1)
Put x + 1 = 0, i.e. x = -1, we get
2(-1) + 1 = A(-3) + B(0)
∴ A = \(\frac{1}{3}\)
Put x – 2 = 0, i.e. x = 2, we get
2(2) + 1 = A(0) + B(3)
∴ B = \(\frac{5}{3}\)

Question 2.
\(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Let \(\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}\)
∴ 2x + 1 = A(x – 1)(x – 4) + Bx(x – 4) + Cx(x – 1)
Put x = 0, we get
2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1)
∴ 1 = 4A
∴ A = \(\frac{1}{4}\)
Put x – 1 = 0, i.e. x = 1, we get
2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0)
∴ 3 = -3B
∴ B = -1
Put x – 4 = 0, i.e x = 4, we get
2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3)
∴ 9 = 12C
∴ C = \(\frac{3}{4}\)

Question 3.
\(\int \frac{x^{2}+x-1}{x^{2}+x-6} d x\)
Solution:

∴ 1 = A(x – 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
1 = A(-5) + B (0)
∴ A = \(\frac{-1}{5}\)
Put x – 2 = 0, i.e. x = 2, we get
1 = A(0) + B(5)
∴ B = \(\frac{1}{5}\)

Question 4.
\(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Solution:
Let I = \(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Let \(\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}\)
∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²
Put x – 1 = 0, i.e. x = 1, we get
1 = A(0)(3) + B(3) + C(0)
∴ B = \(\frac{1}{3}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 = A (-3)(0) + B(0) + C(9)
∴ C = \(-\frac{2}{9}\)
Put x = -1, we get,
-1 = A(-2)(1) + B(1) + C(4)
But B = \(\frac{1}{3}\) and C = \(-\frac{2}{9}\)

Question 5.
\(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Solution:
Let I = \(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Let \(\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}\)
∴ 3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)²
Put x + 1 = 0, i.e. x = -1, we get
3(-1) – 2 = A(0)(2) + B(2) + C(0)
∴ -5 = 2B
∴ B = \(-\frac{5}{2}\)
Put x + 3 = 0, i.e. x = -3, we get
3(-3) – 2 = A(-2)(0) + B(0) + C(4)
∴ -11 = 4C
∴ C = \(-\frac{11}{4}\)
Put x = 0, we get
3(0) – 2 = A(1)(3) + B(3) + C(1)
∴ -2 = 3A + 3B + C
But B = \(-\frac{5}{2}\) and C = \(-\frac{11}{4}\)

Question 6.
\(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
Solution:
Let I = \(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
= \(\int \frac{x^{4}}{x^{5}\left(x^{5}+1\right)} d x\)

Question 7.
\(\int \frac{1}{x\left(x^{n}+1\right)} d x\)
Solution:

Question 8.
\(\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x\)
Solution:

Let \(\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\)
∴ 5x² + 20x + 6 = A(x + 1)² + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i.e. x = -1, we get
5(1) + 20(-1) + 6 = A(0) + B(-1)(0) + C(-1)
∴ -9 = -C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = -1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.5

Evaluate the following.

Question 1.
∫x log x
Solution:

Question 2.
∫x² e^4x dx
Solution:

Question 3.
∫x² e^3x dx
Solution:

Question 4.
\(\int x^{3} e^{x^{2}} d x\)
Solution:

Question 5.
\(\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x\)
Solution:

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

Question 7.
\(\int e^{x} \frac{x-1}{(x+1)^{3}} d x\)
Solution:

Question 8.
\(\int e^{x}\left[(\log x)^{2}+\frac{2 \log x}{x}\right] d x\)
Solution:

Question 9.
\(\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] d x\)
Solution:

Question 10.
\(\int \frac{\log x}{(1+\log x)^{2}} d x\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.5

Question 1.
Use qualifiers to convert each of the following open sentences defined on N, into a true statement:
(i) x² + 3x – 10 = 0
Solution:
∃ x ∈ N, such that x² + 3x – 10 = 0 is a true statement
(x = 2 ∈ N satisfy x² + 3x – 10 = 0)

(ii) 3x – 4 < 9
Solution:
∃ x ∈ N, such that 3x – 4 < 9 is a true statement.
(x = 1, 2, 3, 4 ∈ N satisfy 3x – 4 < 9)

(iii) n² ≥ 1
Solution:
∀ n ∈ N, n² ≥ 1 is a true statement.
(All n ∈ N satisfy n² ≥ 1)

(iv) 2n – 1 = 5
Solution:
∃ x ∈ N, such that 2n – 1 = 5 is a true statement.
(n = 3 ∈ N satisfy 2n – 1 = 5)

(v) y + 4 > 6
Solution:
∃ y ∈ N, such that y + 4 > 6 is a true statement.
(y = 3, 4, 5, … ∈ N satisfy y + 4 > 6

(vi) 3y – 2 ≤ 9
Solution:
∃ y ∈ N, such that 2y ≤ 9 is a true statement.
(y = 1, 2, 3 ∈ N satisfy 3y – 2 ≤ 9).

Question 2.
If B = {2, 3, 5, 6, 7}, determine the truth value of each of the following:
(i) ∀ x ∈ B, x is a prime number.
Solution:
(i) x = 6 ∈ B does not satisfy x is a prime number.
So, the given statement is false, hence its truth value is F.

(ii) ∃ n ∈ B, such that n + 6 > 12.
Solution:
Clearly n = 7 ∈ B satisfies n + 6 > 12.
So, the given statement is true, hence its truth value is T.

(iii) ∃ n ∈ B, such that 2n + 2 < 4.
Solution:
No element n ∈ B satisfy 2n + 2 < 4.
So, the given statement is false, hence its truth value is F.

(iv) ∀ y ∈ B, y² is negative.
Solution:
No element y ∈ B satisfy y² is negative.
So, the given statement is false, hence its truth value is F.

(v) ∀ y ∈ B, (y – 5) ∈ N.
Solution:
y = 2 ∈ B, y = 3 ∈ B and y = 5 ∈ B do not satisfy (y – 5) ∈ N.
So, the given statement is false, hence its truth value is F.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.4

Question 1.
Write the following statements in symbolic form:
(i) If the triangle is equilateral, then it is equiangular.
Solution:
Let p : Triangle is equilateral.
q : It is equiangular.
Then the symbolic form of the given statement is p → q.

(ii) It is not true that ‘i’ is a real number.
Solution:
Let p : ‘i’ is a real number.
Then the symbolic form of the given statement is ~p.

(iii) Even though it is not cloudy, it is still raining.
Solution:
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is ~p ∧ q.

(iv) Milk is white if and only if the sky is not blue.
Solution:
Let p : Milk is white.
q : Sky is blue.
Then the symbolic form of the given statement is p ↔ (~q).

(v) Stock prices are high if and only if stocks are rising.
Solution:
Let p : Stock prices are high.
q : stocks are rising.
Then the symbolic form of the given statement is p ↔ q

(vi) If Kutub-Minar is in Delhi, then Taj Mahal is in Agra.
Solution:
Let p : Kutub-Minar is in Delhi.
q : Taj Mahal is in Agra.
Then the symbolic form of the given statement is p → q

Question 2.
Find the truth value of each of the following statements:
(i) It is not true that 3 – 7i is a real number.
Solution:
Let p : 3 – 7i be a real number.
Then the symbolic form of the given statement is ~p.
The truth value of p is F.
∴ the truth value of ~p is T. ….[~F ≡ T]

(ii) If a joint venture is a temporary partnership, then a discount on purchase is credited to the supplier.
Solution:
Let p : Joint venture is a temporary partnership.
q : Discount on purchases is credited to the supplier.
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) Every accountant is free to apply his own accounting rules if and only if machinery is an asset.
Solution:
Let p : Every accountant is free to apply his own accounting rules.
q : Machinery is an asset.
Then the symbolic form of the given statement is p ↔ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ↔ q is F. ….[F ↔ T ≡ F]

(iv) Neither 27 is a prime number nor divisible by 4.
Solution:
Let p : 27 is a prime number.
q : 27 is divisible by 4.
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 T. …..[~F ∧ ~F ≡ T ∧ T ≡ T]

(v) 3 is a prime number and an odd number.
Solution:
Let p : 3 be a prime number.
q : 3 is an odd number.
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]

Question 3.
If p and q are true and r and s are false, find the true value of each of the following statements:
(i) p ∧ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∧ (q ∧ r) ≡ T ∧ (T ∧ F)
≡ T ∧ F
≡ F
Hence, the truth value of the given statement is false.

(ii) (p → q) ∨ (r ∧ s)
Solution:
(p → q) ∨ (r ∧ s) ≡ (T → T) ∨ (F ∧ F)
≡ T ∨ F
≡ T
Hence, the truth value of the given statement is true.

(iii) ~[(~p ∨ s) ∧ (~q ∧ r)]
Solution:
~[(~p ∨ s) ∧ (~q ∧ r)] ≡ ~[(~ T ∨ F) ∧ (~T ∧ F)]
≡ ~[(F ∨ F) ∧ (F ∧ F)]
≡ ~(F ∧ F)
≡ ~F
≡ T
Hence, the truth value of the given statement is true.

(iv) (p → q) ↔ ~(p ∨ q)
Solution:
(p → q) ↔ ~(p ∨ q) = (T → T) ↔ ~(T ∨ T)
≡ T ↔ ~ (T)
≡ T ↔ F
≡ F
Hence, the truth value of the given statement is false.

(v) [(p ∨ s) → r] ∨ [~(p → q) ∨ s]
Solution:
[(p ∨ s) → r] ∨ ~[~(p → q) ∨ s]
≡ [(T ∨ F) → F] ∨ ~[ ~(T → T) ∨ F]
≡ (T → F) ∨ ~(~T ∨ F)
≡ F ∨ ~ (F ∨ F)
≡ F ∨ ~F
≡ F ∨ T
≡ T
Hence, the truth value of the given statement is true.

(vi) ~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
Solution:
~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
≡ ~[T ∨ (F ∧ F)] ∧ ~[(F ∧ ~F) ∧ T]
≡ ~[T ∨ F] ∧ ~[(F ∧ T) ∧ T]
≡ ~T ∧ ~(F ∧ T)
≡ F ∧ ~F
≡ F ∧ T
≡ F
Hence, the truth value of the given statement is false.

Question 4.
Assuming that the following statements are true:
p : Sunday is a holiday.
q : Ram does not study on holiday.
Find the truth values of the following statements:
(i) Sunday is not holiday or Ram studies on holiday.
Solution:
The symbolic form of the statement is ~p ∨ ~q.

∴ the truth value of the given statement is F.

(ii) If Sunday is not a holiday, then Ram studies on holiday.
Solution:
The symbolic form of the given statement is ~p → ~q.

∴ the truth value of the given statement is T.

(iii) Sunday is a holiday and Ram studies on holiday.
Solution:
The symbolic form of the given statement is p ∧ q.

∴ the truth value of the given statement is F.

Question 5.
If p : He swims.
q : Water is warm.
Give the verbal statements for the following symbolic statements:
(i) p ↔ ~q
Solution:
p ↔ ~ q
He swims if and only if the water is not warm.

(ii) ~(p ∨ q)
Solution:
~(p ∨ q)
It is not true that he swims or water is warm.

(iii) q → p
Solution:
q → p
If water is warm, then he swims.

(iv) q ∧ ~p
Solution:
q ∧ ~p
The water is warm and he does not swim.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Question 1.
Write the negation of each of the following statements:
(i) All men are animals.
Solution:
Some men are not animals.

(ii) 3 is a natural number.
Solution:
-3 is not a natural number.

(iii) It is false that Nagpur is the capital of Maharashtra.
Solution:
Nagpur is the capital of Maharashtra.

(iv) 2 + 3 ≠ 5.
Solution:
2 + 3 = 5.

Question 2.
Write the truth value of the negation of each of the following statements:
(i) √5 is an irrational number.
Solution:
Let p : √5 is an irrational number.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(ii) London is in England.
Solution:
Let p : London is in England.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(iii) For every x ∈ N, x + 3 < 8.
Solution:
Let p : For every x ∈ N, x + 3 < 8.
The truth value of p is F.
Therefore, the truth value of ~p is T.

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