Category: Class 12

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

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.6

Question 1.
Prepare the truth tables for the following statement patterns:
(i) p → (~p ∨ q)
Solution:
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (~p ∨ q) ∧ (~p ∨ ~q)
Solution:

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
Here are three statements and 4 connectives.
∴ there are 2 × 2 × 2 = 8 rows and 3 + 4 = 7 columns in the truth table.

(iv) (p ∧ q) ∨ ~r
Solution:

Question 2.
Examine, whether each of the following statement patterns is a tautology or a contradiction or a contingency:
(i) q ∨ [~(p ∧ q)]
Solution:

All the entries in the last column of the above truth table are T.
∴ q ∨ [~(p ∧ q)] is a tautology.

(ii) (~q ∧ p) ∧ (p ∧ ~p)
Solution:

All the entries in the last column of the above truth table are F.
∴ (~q ∧ p) ∧ (p ∧ ~p) is a contradiction.

(iii) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(iv) ~p → (p → ~q)
Solution:

All the entries in the last column of the truth table are T.
∴ p → (p → ~q) is a tautology.

Question 3.
Prove that each of the following statement pattern is a tautology:
(i) (p ∧ q) → q
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → q is a tautology.

(ii) (p → q) ↔ (~q → ~p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~q → ~p) is a tautology.

(iii) (~p ∧ ~q) → (p → q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∧ ~q) → (p → q) is a tautology.

(iv) (~p ∨ ~q) ↔ ~(p ∧ q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∨ ~q) ↔ ~(p ∧ q) is a tautology.

Question 4.
Prove that each of the following statement pattern is a contradiction:
(i) (p ∨ q) ∧ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

(ii) (p ∧ q) ∧ ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∧ ~p is a contradiction.

(iii) (p ∧ q) ∧ (~p ∨ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∧ q) ∧ (~p ∨ ~q) is a contradiction.

(iv) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

Question 5.
Show that each of the following statement pattern is a contingency:
(i) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(ii) (p → q) ↔ (~p ∧ q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ↔ (~p ∧ q) is a contingency.

(iii) p ∧ [(p → ~q) → q]
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ p ∧ [(p → ~q) → q] is a contingency.

(iv) (p → q) ∧ (p → r)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ∧ (p → r) is a contingency.

Question 6.
Using the truth table, verify:
(i) p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

(ii) p → (p → q) ≡ ~q → (p → q)
Solution:

The entries in columns 5 and 6 are identical.
∴ p → (p → q) ≡ ~q → (p → q)

(iii) ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q
Solution:

The entries in columns 5, 7 and 8 are identical.
∴ ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q.

(iv) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:

The entries in columns 3 and 7 are identical.
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 7.
Prove that the following pairs of statement patterns are equivalent:
(i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(ii) p ↔ q and (p → q) ∧ (q → p)
Solution:

The entries in columns 3 and 6 are identical.
∴ p ↔ q ≡ (p → q) ∧ (q → p)

(iii) p → q and ~q → ~p and ~p ∨ q
Solution:

The entries in columns 5, 6 and 7 are identical.
∴ p → q ≡ ~q → ~p ≡ ~p ∨ q.

(iv) ~(p ∧ q) and ~p ∨ ~q
Solution:

The entries in columns 6 and 7 are identical.
∴ ~(p ∧ q) ≡ ~p ∨ ~q.

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]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

State which of the following sentences are statements. Justify your answer. In case of statements, write down the truth value:

Question (i).
A triangle has ‘ n’ sides.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (ii).
The sum of interior angles of a triangle is 180°.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (iii).
You are amazing!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question (iv).
Please grant me a loan.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (v).
√-4 is an irrational number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vi).
x² – 6x + 8 = 0 implies x = -4 or x = -2.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vii).
He is an actor.
Solution:
It is an open sentence, hence it is not a statement.

Question (viii).
Did you eat lunch yet?
Solution:
It is an interrogative sentence, hence it is not a statement.

Question (ix).
Have a cup of cappuccino.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (x).
(x + y)² = x² + 2xy + y² for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xi).
Every real number is a complex number.
Solution:
It is a statement that is true, hence its truth value is ‘T.

Question (xii).
1 is a prime number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xiii).
With the sunset, the day ends.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xiv).
1! = 0.
Solution:
It is a statement that is false, hence its truth value is

Question (xv).
3 + 5 > 11.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xvi).
The number π is an irrational number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xvii).
x² – y² = (x + y)(x – y) for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xviii).
The number 2 is only even a prime number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xix).
Two coplanar lines are either parallel or intersecting.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xx).
The number of arrangements of 7 girls in a row for a photograph is 7!
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxi).
Give me a compass box.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxii).
Bring the motor car here.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxiii).
It may rain today.
Solution:
It is an open sentence, hence it is not a statement.

Question (xxiv).
If a + b < 7, where a ≥ 0 and b ≥ 0, then a < 7 and b < 7.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxv).
Can you speak English?
Solution:
It is an interrogative sentence, hence it is not a statement.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:
When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.

Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.

Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤ p_i ≤ 1
(b) Σp_i = 1.

(i)

(a) Here 0 ≤ p_i ≤ 1
(b) Σp_i = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)

P(X = 3) = -0.1, i.e. P_i < 0 which does not satisfy 0 ≤ P_i ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)

Here ∑p_i = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

(v)

Here ∑p_i = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.

(vi)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution:
(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
∴ P[X = 0] = P(0) = \(\frac{1}{4}\)
P[X = 1] = P(1) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P[X = 2] = P(2) = \(\frac{1}{4}\)
∴ the required probability distribution is

(ii) When three coins are tossed simultaneously, then the sample space is
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X denotes the number of tails.
Then X can take the value 0, 1, 2, 3.
∴ P[X = 0] = P(0) = \(\frac{1}{8}\)
P[X = 1] = P(1) = \(\frac{3}{8}\)
P[X = 2] = P(2) = \(\frac{3}{8}\)
P[X = 3] = P(3) = \(\frac{1}{8}\)
∴ the required probability distribution is

(iii) When a fair coin is tossed 4 times, then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
Let X denotes the number of heads.
Then X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n(X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
∴ the probability distribution of X is as follows:

Question 5.
Find the probability distribution of a number of successes in two tosses of a die, where success is defined as a number greater than 4 appearing on at least one die.
Solution:
When a die is tossed twice, the sample space s has 6 × 6 = 36 sample points.
∴ n(S) = 36
The trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2.
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ n(X) = 16.
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)
∴ the required probability distribution is

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P(X = 0) = P(0) = P(all 4 non-defective bulbs)
= \(\frac{24}{30} \times \frac{24}{30} \times \frac{24}{30} \times \frac{24}{30}\)
= \(\frac{256}{625}\)
P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)

P(X = 2) = P(2) = P(2 defective and 2 non-defective)

P(X = 3) = P(3) = P(3 defectives and 1 non-defective)

P(X = 4) = P(4) = P(all 4 defectives)
= \(\frac{6}{30} \times \frac{6}{30} \times \frac{6}{30} \times \frac{6}{30}\)
= \(\frac{1}{625}\)
∴ the required probability distribution is

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. If the coin is tossed twice. Find the probability distribution of a number of tails.
Solution:
Given a biased coin such that heads is 3 times as likely as tails.
∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)
The coin is tossed twice.
Let X can be the random variable for the number of tails.
Then X can take the value 0, 1, 2.
∴ P(X = 0) = P(HH) = \(\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)
P(X = 1) = P(HT, TH) = \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}=\frac{6}{16}=\frac{3}{8}\)
P(X = 2) = P(TT) = \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
∴ the required probability distribution is

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)
Solution:
(i) Since P (x) is a probability distribution of x,
\(\sum_{x=0}^{7} P(x)=1\)
⇒ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(10k – 1) = 0
⇒ 10k – 1 = 0 ……..[∵ k ≠ -1]
⇒ k = \(\frac{1}{10}\)

(ii) P(X< 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = P (1) + P (2)
= k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

Question 9.
Find expected value and variance of X for the following p.m.f.:

Solution:
We construct the following table to calculate E(X) and V(X):

From the table,
Σx_ip_i = -0.05 and \(\Sigma x_{i}^{2} \cdot p_{i}\) = 2.25
∴ E(X) = Σx_ip_i = -0.05
and V(X) = \(\Sigma x_{i}^{2}+p_{i}-\left(\sum x_{i}+p_{i}\right)^{2}\)
= 2.25 – (-0.05)²
= 2.25 – 0.0025
= 2.2475
Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10.
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution:
If a die is tossed, then the sample space for the random variable X is
S = {1, 2, 3, 4, 5, 6}
∴ P(X) = \(\frac{1}{6}\); X = 1, 2, 3, 4, 5, 6.

Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11.
Find the mean number of heads in three tosses of a fair coin.
Solution:
When three coins are tossed the sample space is {HHH, HHT, THH, HTH, HTT, THT, TTH, TTT}
∴ n(S) = 8
Let X denote the number of heads when three coins are tossed.
Then X can take values 0, 1, 2, 3
P(X = 0) = P(0) = \(\frac{1}{8}\)
P(X = 1) = P(1) = \(\frac{3}{8}\)
P(X = 2) = P(2) = \(\frac{3}{8}\)
P(X = 3) = P(3) = \(\frac{1}{8}\)
∴ mean = E(X) = Σx_iP(x_i)
= \(0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}\)
= \(0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\)
= \(\frac{12}{8}\)
= 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
When two dice are thrown, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Let X denote the number of sixes when two dice are thrown.
Then X can take values 0, 1, 2
When X = 0, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(X) = 25
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{25}{36}\)
When X = 1, then
X = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(X) = 10
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{10}{36}\)
When X = 2, then X = {(6, 6)}
∴ n(X) = 1
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{1}{36}\)
∴ E(X) = ΣxiP(xi)
= \(0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}\)
= \(0+\frac{10}{36}+\frac{2}{36}\)
= \(\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
Two numbers are chosen from the first 6 positive integers.
∴ n(S) = \({ }^{6} C_{2}=\frac{6 \times 5}{1 \times 2}\) = 15
Let X denote the larger of the two numbers.
Then X can take values 2, 3, 4, 5, 6.
When X = 2, the other positive number which is less than 2 is 1.
∴ n(X) = 1
∴ P(X = 2) = P(2) = \(\frac{n(X)}{n(S)}=\frac{1}{15}\)
When X = 3, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways.
∴ n(X) = 2
P(X = 3) = P(3) = \(\frac{n(X)}{n(S)}=\frac{2}{15}\)
Similarly, P(X = 4) = P(4) = \(\frac{3}{15}\)
P(X = 5) = P(5) = \(\frac{4}{15}\)
P(X = 6) = P(6) = \(\frac{5}{15}\)
∴ E(X) = Σx_iP(x_i)
= \(2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}\)
= \(\frac{2+6+12+20+30}{15}\)
= \(\frac{70}{15}\)
= \(\frac{14}{3}\)

Question 14.
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution:
If two fair dice are rolled then the sample space S of this experiment is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

∴ the probability distribution of X is given by

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the student is recorded. What is the probability distribution of the random variable X? Find mean, variance, and standard deviation of X.
Solution:
Let X denote the age of the chosen student. Then X can take values 14, 15, 16, 17, 18, 19, 20, 21.
We make a frequency table to find the number of students with age X:

The chances of any student selected are equally likely.
If there are m students with age X, then P(X) = \(\frac{m}{15}\)
Using this, the following is the probability distribution of X:


Variance = V(X) = \(\Sigma x_{i}^{2}\) . P(x_i) – [E(X)]²
= 312.2 – (17.53)²
= 312.2 – 307.3
= 4.9
Standard deviation = √V(X) = √4.9 = 2.21
Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Question 16.
In a meeting, 70% of the member’s favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
X takes values 0 and 1.
It is given that
P(X = 0) = P(0) = 30% = \(\frac{30}{100}\) = 0.3
P(X = 1) = P(1) = 70% = \(\frac{70}{100}\) = 0.7
∴ E(X) = Σx_i . P(x_i) = 0 × 0.3 + 1 × 0.7 = 0.7
Also, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 0 × 0.3 + 1 × 0.7 = 0.7
∴ Variance = V(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)-[E(X)]^{2}\)
= 0.7 – (0.7)²
= 0.7 – 0.49
= 0.21
Hence, E(X) = 0.7 and Var(X) = 0.21.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.
The differential equation of y = c² + \(\frac{c}{x}\) is…….
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)
(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)
(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
Answer:
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Question 3.
x² + y² = a² is a solution of ………
(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)
Answer:
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Answer:
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Question 5.
The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………
(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
y \(\frac{d y}{d x}\) + x = 0
∴ ∫y dy + ∫x dx = c
∴ \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=c\)
∴ x² + y² = 2c which is a circle.

Question 6.
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
(b) x tan^-1x + c = 0
(c) x – tan^-1x = c
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Question 7.
The solution of (x + y)² \(\frac{d y}{d x}\) = 1 is…….
(a) x = tan^-1(x + y) + c
(b) y tan^-1(\(\frac{x}{y}\)) = c
(c) y = tan^-1(x + y) + c
(d) y + tan^-1(x + y) = c
Answer:
(c) y = tan^-1(x + y) + c

Question 8.
The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……
(a) sin^-1(\(\frac{y}{x}\)) = 2 log |x| + c
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c
(c) sin(\(\frac{x}{y}\)) = log |x| + c
(d) sin(\(\frac{y}{x}\)) = log |y| + c
Answer:
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c

Question 9.
The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……
(a) y e^x = cos x + c
(b) y e^x + e^x cos x = c
(c) y e^x = e^x cos x + c
(d) y² e^x = e^x cos x + c
Answer:
(c) y e^x = e^x cos x + c
Hint:
\(\frac{d y}{d x}\) + y = cos x – sin x
I.F. = \(e^{\int 1 d x}=e^{x}\)
∴ the solution is y . e^x = ∫(cos x – sin x) e^x + c
∴ y . e^x = e^x cos x + c

Question 10.
The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x² log x is……..
(a) \(\frac{1}{x}\)
(b) k
(c) \(\frac{1}{n^{2}}\)
(d) x²
Answer:
(d) x²
Hint:
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e^{2 log x}
= x²

Question 11.
The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….
(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
\(\frac{d y}{d x}\) = sec x – y tan x
∴ \(\frac{d y}{d x}\) + y tan x = sec x
I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x
∴ the solution is
y . sec x = ∫sec x . sec x dx + c
∴ y sec x = tan x + c

Question 12.
The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……
(a) e^x-y = x + 1
(b) e^x+y = x + 1
(c) e^x + e^y = x + 1
(d) e^y-x = x – 1
Answer:
(a) e^x-y = x + 1

Question 13.
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..
(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)
(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)
Answer:
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……
(a) 5\(\frac{2}{3}\) grams
(b) 5\(\frac{1}{3}\) grams
(c) 5.1 grams
(d) 5 grams
Answer:
(b) 5\(\frac{1}{3}\) grams

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..
(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.
Determine the order and degree of the following differential equations:
(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.
∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)
Solution:
The given D.E. is

This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.
∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).
∴ order = 4
Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.
In each of the following examples verify that the given function is a solution of the differential equation.
(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)
Solution:
x² + y² = r² ……. (1)
Differentiating both sides w.r.t. x, we get

Hence, x² + y² = r² is a solution of the D.E.
\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = e^ax sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)
Solution:

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
Solution:
y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(iv) xy = ae^x + be^-x + x²; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)
Solution:

(v) x² = 2y² log y, x² + y² = xy \(\frac{d x}{d y}\)
Solution:
x² = 2y² log y ……(1)
Differentiating both sides w.r.t. y, we get

∴ x² + y² = xy \(\frac{d x}{d y}\)
Hence, x² = 2y² log y is a solution of the D.E.
x² + y² = xy \(\frac{d x}{d y}\)

Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) y² = a(b – x)(b + x)
Solution:
y² = a(b – x)(b + x) = a(b² – x²)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax
∴ y \(\frac{d y}{d x}\) = -ax …….(1)
Differentiating again w.r.t. x, we get

This is the required D.E.

(ii) y = a sin(x + b)
Solution:
y = a sin(x + b)

This is the required D.E.

(iii) (y – a)² = b(x + 4)
Solution:
(y – a)² = b(x + 4) …….(1)
Differentiating both sides w.r.t. x, we get
\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
Solution:
y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
∴ y² = a cos (log x) + b sin (log x) …….(1)
Differentiating both sides w.r.t. x, we get

(v) y = Ae^3x+1 + Be^-3x+1
Solution:
y = Ae^3x+1 + Be^-3x+1 …… (1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:

Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
∴ equation of the circle is (x – h)² + (y – 0)² = h²
∴ x² – 2hx + h² + y² = h²
∴ x² + y² = 2hx ……..(1)
Differentiating both sides w.r.t. x, we get
2x + 2y \(\frac{d y}{d x}\) = 2h
Substituting the value of 2h in equation (1), we get
x² + y² = (2x + 2y \(\frac{d y}{d x}\)) x
∴ x² + y² = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) + x² – y² = 0
This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:
Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h)² = 4b(y – k) ……. (1)
where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get
2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)
∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)
∴ (x – h) = 2b \(\frac{d y}{d x}\)
Differentiating again w.r.t. x, we get
1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)
∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.
Solution:
Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).
∴ slope of normal to this line is \(-\frac{2}{3}\)
Then the equation of the normal is
y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)
∴ 3\(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).
Solution:
The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)
i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16k, b² = 36k
∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)
i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)
∴ 9x² – 4y² = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0
∴ 9x – 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Question 5.
Solve the following differential equations:
(i) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(ii) \(\frac{d y}{d x}\) = x²y + y
Solution:

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)
Solution:

(iv) x dy = (x + y + 1) dx
Solution:

(v) \(\frac{d y}{d x}\) + y cot x = x² cot x + 2x
Solution:
\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x² cot x + 2x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \cot x d x}\)
= \(e^{\log (\sin x)}\)
= sin x
∴ the solution of (1) is given by
y(I.F.) = ∫Q . (I.F.) dx + c
∴ y sin x = ∫(x² cot x + 2x) sin x dx + c
∴ y sinx = ∫(x² cot x . sin x + 2x sin x) dx + c
∴ y sinx = ∫x² cos x dx + 2∫x sin x dx + c
∴ y sinx = x² ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c
∴ y sin x = x² (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c
∴ y sin x = x² sin x – 2∫x sin x dx + 2∫x sin x dx + c
∴ y sin x = x² sin x + c
∴ y = x² + c cosec x
This is the general solution.

(vi) y log y = (log y² – x) \(\frac{d y}{d x}\)
Solution:

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e^-3y
Solution:

Question 6.
Find the particular solution of the following differential equations:
(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e²
Solution:

(ii) (x + 2y²) \(\frac{d y}{d x}\) = y, when x = 2, y = 1
Solution:


This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)² + c(1)
∴ c = 0
∴ the particular solution is x = 2y².

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x
\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)
This is the linear differential equation of the form

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:

(v) \(2 e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0\), when y(0) = 1
Solution:

Question 7.
Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).
Solution:

Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)
∴ equation of the normal is

This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
2² + 3² = 4(2) + c
∴ c = 5
∴ equation of the required curve is x² + y² = 4x + 5.

Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.


Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.
Solution:
Let x be the assets of the presort at time t years.
Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.
∴ \(\frac{d x}{d t}\) ∝ √x
∴ \(\frac{d x}{d t}\) = -k√x, where k > 0
∴ \(\frac{d x}{\sqrt{x}}\) = -k dt
Integrating both sides, we get
\(\int x^{-\frac{1}{2}} d x\) = -k∫dt
∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c
∴ 2√x = -kt + c
At the beginning, i.e. at t = 0, x = 10,00,000
2√10,00,000 = -k(0) + c
∴ c = 2000
∴ 2√x = -kt + 2000 ……..(1)
Also, when t = 2, x = 10,000
∴ 2√10000 = -k × 2 + 2000
∴ 2k = 1800
∴ k = 900
∴ (1) becomes,
∴ 2√x = -900t + 2000
When the person will be bankrupt, x = 0
∴ 0 = -900t + 2000
∴ 900t = 2000
∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)
Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

I. Evaluate the following:

Question 1.
∫x² log x dx
Solution:

Question 2.
∫x² sin 3x dx
Solution:

Question 3.
∫x tan^-1 x dx
Solution:

Question 4.
∫x² tan^-1 x dx
Solution:

Question 5.
∫x³ tan^-1 x dx
Solution:
Let I = ∫x³ tan^-1 x dx
= ∫(tan^-1 x) . x³ dx

Question 6.
∫(log x)² dx
Solution:
Let I = ∫(log x)² dx
Put log x = t

Question 7.
∫sec³ x dx
Solution:
Let I = ∫sec³ x dx
= ∫sec x sec² x dx
= sec x ∫sec² x dx – ∫[\(\frac{d}{d x}\)(sec x) ∫sec² x dx] dx
= sec x tan x – ∫(sec x tan x)(tan x) dx
= sec x tan x – ∫sec x tan² x dx
= sec x tan x – ∫sec x (sec² x – 1) dx
= sec x tan x – ∫sec³ x dx + ∫sec x dx
∴ I = sec x tan x – I + log|sec x + tan x|
∴ 2I = sec x tan x + log|sec x + tan x|
∴ I = \(\frac{1}{2}\) [sec x tan x + log|sec x + tan x|] + c.

Question 8.
∫x . sin² x dx
Solution:

Question 9.
∫x³ log x dx
Solution:

Question 10.
∫e^2x cos 3x dx
Solution:

Question 11.
∫x sin^-1 x dx
Solution:

Question 12.
∫x² cos^-1 x dx
Solution:

Question 13.
\(\int \frac{\log (\log x)}{x} d x\)
Solution:

= t(log t – 1) + c
= (log x) . [log(log x) – 1] + c.

Question 14.
\(\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Solution:

Question 15.
∫cos√x dx
Solution:
Let I = ∫cos√x dx
Put √x = t
∴ x = t²
∴ dx = 2t dt
∴ I = ∫(cos t) 2t dt
= ∫2t cos t dt
= 2t ∫cos t dt – ∫[\(\frac{d}{d t}\)(2t) ∫cos t dt]dt
= 2t sin t – ∫2 sin t dt
= 2t sin t + 2 cos t + c
= 2[√x sin√x + cos√x] + c.

Question 16.
∫sin θ . log(cos θ) dθ
Solution:
Let I = ∫sin θ . log (cos θ) dθ
= ∫log(cos θ) . sin θ dθ
Put cos θ = t
∴ -sin θ dθ = dt
∴ sin θ dθ = -dt

= -t log t + t + c
= -cos θ . log(cos θ) + cos θ + c
= -cos θ [log(cos θ) – 1] + c.

Question 17.
∫x cos³ x dx
Solution:
cos 3x = 4 cos³ x – 3 cos x
∴ cos³ x + 3 cos x = 4cos3x
∴ cos³ x = \(\frac{1}{4}\) cos 3x + \(\frac{3}{4}\) cos x

Question 18.
\(\int \frac{\sin (\log x)^{2}}{x} \cdot \log x d x\)
Solution:

Question 19.
\(\int \frac{\log x}{x} d x\)
Solution:
Let I = \(\int \frac{\log x}{x} d x\)
Put log x = t
\(\frac{1}{x}\) dx = dt
∴ I = ∫t dt
= \(\frac{1}{2}\) t² + c
= \(\frac{1}{2}\) (log x)² + c

Question 20.
∫x sin 2x cos 5x dx.
Solution:
Let I = ∫x sin 2x cos 5x dx
sin 2x cos 5x = \(\frac{1}{2}\)[2 sin 2x cos 5x]
= \(\frac{1}{2}\) [sin(2x + 5x) + sin(2x – 5x)]
= \(\frac{1}{2}\) [sin 7x – sin 3x]
∴ ∫sin 2x cos 5x dx = \(\frac{1}{2}\) [∫sin 7x dx – ∫sin 3x dx]

Question 21.
\(\int \cos (\sqrt[3]{x}) d x\)
Solution:
Let I = \(\int \cos (\sqrt[3]{x}) d x\)

II. Integrate the following functions w.r.t. x:

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

Question 2.
e^-x cos 2x
Solution:

Question 3.
sin(log x)
Solution:

Question 4.
\(\sqrt{5 x^{2}+3}\)
Solution:
Let I = \(\sqrt{5 x^{2}+3}\) dx

Question 5.
\(x^{2} \sqrt{a^{2}-x^{6}}\)
Solution:

Question 6.
\(\sqrt{(x-3)(7-x)}\)
Solution:

Question 7.
\(\sqrt{4^{x}\left(4^{x}+4\right)}\)
Solution:

Question 8.
(x + 1) \(\sqrt{2 x^{2}+3}\)
Solution:
Let I = ∫(x + 1) \(\sqrt{2 x^{2}+3}\) dx
Let x + 1 = A[\(\frac{d}{d x}\)(2x² + 3)] + B
= A(4x) + B
= 4Ax + B
Comparing the coefficients of x and constant term on both the sides, we get
4A = 1, B = 1
∴ A = \(\frac{1}{4}\), B = 1

Question 9.
\(x \sqrt{5-4 x-x^{2}}\)
Solution:
Let I = ∫\(x \sqrt{5-4 x-x^{2}}\) dx
Let x = A[\(\frac{d}{d x}\)(5 – 4x – x²)] + B
= A[-4 – 2x] + B
= -2Ax + (B – 4A)
Comparing the coefficients of x and the constant term on both sides, we get
-2A = 1, B – 4A = 0

Question 10.
\(\sec ^{2} x \sqrt{\tan ^{2} x+\tan x-7}\)
Solution:

Question 11.
\(\sqrt{x^{2}+2 x+5}\)
Solution:

Question 12.
\(\sqrt{2 x^{2}+3 x+4}\)
Solution:

III. Integrate the following functions w.r.t. x:

Question 1.
[2 + cot x – cosec² x] e^x
Solution:
Let I = ∫e^x [2 + cot x – cosec² x] dx
Put f(x) = 2 + cot x
∴ f'(x) = \(\frac{d}{d x}\)(2 + cot x)
= \(\frac{d}{d x}\)(2) + \(\frac{d}{d x}\)(cot x)
= 0 – cosec² x
= -cosec² x
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x (2 + cot x) + c.

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

Question 3.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution:
Let I = ∫\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Let f(x) = \(\frac{1}{x}\), f'(x) = \(-\frac{1}{x^{2}}\)
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x . \(\frac{1}{x}\) + c

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

Question 5.
\(\frac{e^{x}}{x}\) . [x(log x)² + 2 log x]
Solution:

Question 6.
\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Solution:
Let I = ∫\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)

Question 7.
\(e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right]\)
Solution:

Question 8.
log(1 + x)^(1+x)
Solution :
Let I = ∫log(1 + x)^(1+x) dx

Question 9.
cosec (log x)[1 – cot(log x)]
Solution:
Let I = ∫cosec (log x)[1 – cot(log x)] dx
Put log x = t
x = e^t
dx = e^t dt
I = ∫cosec t (1 – cot t). e^t dt
= ∫e^t [cosec t – cosec t cot t] dt
= ∫e^t [cosec t + \(\frac{d}{d t}\) (cosec t)] dt
= e^t cosec t + c ….. [∵ e^t [f(t) +f'(t) ] dt = e^t f(t) + c ]
= x . cosec(log x) + c.