Bhagya

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3

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

Question 1.
y = \(x^{x^{2 x}}\)
Solution:

Question 2.
y = \(x^{e^{x}}\)
Solution:

Question 3.
y = \(e^{x^{x}}\)
Solution:

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

Question 1.
y = \(\left(1+\frac{1}{x}\right)^{x}\)
Solution:

Question 2.
y = (2x + 5)^x
Solution:

Question 3.
y = \(\sqrt[3]{\frac{(3 x-1)}{(2 x+3)(5-x)^{2}}}\)
Solution:

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

Question 1.
y = (log x)^x + x^{log x}
Solution:

Question 2.
y = x^x + a^x
Solution:

Question 3.
y = \(10^{x^{x}}+10^{x^{10}}+10^{10^{x}}\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2

1. Find the rate of change of demand (x) of a commodity with respect to price (y) if:

Question 1.
y = 12 + 10x + 25x²
Solution:
Given y = 12 + 10x + 25x²

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{1}{10+50 x}\)

Question 2.
y = 18x + log(x – 4)
Solution:
Given y = 18x + log (x – 4)

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{x-4}{18 x-71}\)

Question 3.
y = 25x + log(1 + x²)
Solution:
Given y = 25x + log(1 + x²)

Hence, the rate of change of demand (x) with respect to price (y) \(\frac{d x}{d y}=\frac{1+x^{2}}{25 x^{2}+2 x+25}\)

2. Find the marginal demand of a commodity where demand is x and price is y.

Question 1.
y = xe^-x + 7
Solution:

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

Question 3.
y = \(\frac{5 x+9}{2 x-10}\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.9 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.9

Question 1.
Without using truth table, show that
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~(q ∨ p) …..(Conditional Law)
≡ [~p ∧ (~(q ∨ p)] ∨ [q ∧ (~q ∨ p] …..(Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] ……(Distributive Law)
≡ [(~p ∧ ~q) ∨ c] ∨ [c ∨ (q ∧ p)] …..(Complement Law)
≡ (~p ∧ ~q) ∨ (q ∧ p) ……(Identity Law)
≡ (~p ∧ ~q) ∨ (p ∧ q) ……(Commutative Law)
≡ (p ∧ q) ∨ (~p∧ q) ……(Commutative Law)
≡ RHS.

(ii) p ∧ [~p ∨ q) ∨ (~q)] ≡ p
Solution:
LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.

(iii) ~[(p ∧ q) → ~(q)] ≡ p ∧ q
Solution:
LHS = ~[(p ∧ q) → ~(~q)]
≡ (p ∧ q) ∧ ~(~q) ……(Negation of implication)
≡ (p ∧ q) ∧ q …..(Negation of negation)
≡ p ∧ (q ∧ q) …..(Associative Law)
≡ P ∧ q ……(Idempotent Law)
= RHS

(iv) ~r → ~(p ∧ q) ≡ [~(q → r)] → (~p)
Solution:
LHS = ~r → ~(p ∧ q)
≡ ~q → (~p ∨ ~q) ……(De Morgan’s Law)
≡ ~(~r) ∨ (~p ∨~q) …..(Conditional Law)
≡ r ∨ (~p ∨ ~q) …..(Involution Law)
≡ r ∨ ~q ∨ ~p …..(Commutative Law)
≡ (~q ∨ r) ∨ (~p) ……(Commutative Law)
≡ ~(q → r) ∨ (~p) ……(Conditional Law)
≡ ~(q → r) → (~p) ……(Conditional Law)
= RHS.

(v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Solution:
LHS = (p ∨ q) → r
≡ ~(p → q) ∨ r ……..(Conditional Law)
≡ (~p ∧ ~q) ∨ r ……….(De Morgan’s Law)
≡ (~p ∨ r) ∧ (~q ∨ r) ………..(Distributive Law)
≡ (p → r) ∧ (q → r) …….(Conditional Law)
= RHS.

Question 2.
Using the algebra of statement, prove that:
(i) [p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p] ≡ p
Solution:
LHS = [p ∧ (q ∨ r)] ∨ [ ~r ∧ ~q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p] ……(Associative Law)
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ……(Commutative Law)
≡ [p ∧ (q ∨ r)] ∨ [ ~ (q ∨ r) ∧ p] ……(De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] ……(Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~ (q ∨ r) ] …..(Distributive Law)
≡ p ∧ t …….(Complement Law)
≡ p ……(Identity Law)
= RHS.

(ii) (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q) ≡ p ∨ ~q
Solution:
LHS = (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~q) ∨ (~p ∧ ~q)] ……(Associative Law)
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~q ∧ ~p)] …..(CommutativeLaw)
≡ (p ∧ q) ∨ [ ~q ∧ (p ∨ ~ p)] …..(Distributive Law)
≡ (p ∧ q) ∨ (~q ∧ t) ……(Complement Law)
≡ (p ∧ q) ∨ (~q) …….(Identity Law)
≡ (p ∨ ~q) ∧ (q ∨ ~q) ……(Distributive Law)
≡ (p ∨ ~q) ∧ t …….(Complement Law)
≡ p ∨ ~q …..(Identity Law)
= RHS.

(iii) (p ∨ q) ∧ (~p ∨ ~q) ≡ (p ∧ ~q) ∨ (~p ∧ q)
Solution:
LHS = (p ∨ q) ∧ (~p ∨ ~q)
≡ [p ∧ (~p ∨ ~q)] ∨ [q ∧ (~p ∨ ~q)] ……(Distributive Law)
≡ [(p ∧ ~p) ∨ (p ∧ ~q)] ∨ [q ∧ ~p) ∨ (q ∧ ~q)] ……(Distributive Law)
≡ [c ∨ (p ∧ ~q)] ∨ [(q ∧ ~p) ∨ c] ……(Complement Law)
≡ (p ∧ ~q) ∨ (q ∧ ~p) ……..(Identity Law)
≡ (p ∧ ~q) ∨ (~p ∧ q) ………(Commutative Law)
= RHS.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.8

Question 1.
Write the negation of each of the following statements:
(i) All the stars are shining if it is night.
Solution:
The given statement can be written as:
If it is night, then all the stars are shining.
Let p : It is night.
q : All the stars are shining.
Then the symbolic form of the given statement is p → q
Since, ~(p → q) ≡ p ∧ ~q,
the negation of the given statement is ‘It is night and all the stars are not shining.’

(ii) ∀ n ∈ N, n + 1 > 0.
Solution:
The negation of the given statement is
‘∃ n ∈ N, such that n + 1 ≤ 0.’

(iii) ∃ n ∈ N, such that (n² + 2) is odd number.
Solution:
The negation of the given statement is
‘∀ n ∈ N, n² + 2 is not an odd number.’

(iv) Some continuous functions are differentiable.
Solution:
The negation of a given statement is ‘All continuous functions are not differentiable.’

Question 2.
Using the rules of negation, write the negations of the following:
(i) (p → r) ∧ q
Solution:
The negation of (p → r) ∧ q is
~[(p → r) ∧ q] ≡ ~(p → r) ∨ (~q) …..[Negation of conjunction]
≡ (p ∧ ~r) ∨ (~q) ……[Negation of implication]

(ii) ~(p ∨ q) → r
Solution:
The negation of ~(p ∨ q) → r is
~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ (~r) …..[Negation of implication]
≡ (~p ∧ ~q) ∧ (~r) ……[Negation of disjunction]

(iii) (~p ∧ q) ∧ (~q ∨ ~r)
Solution:
The negation of (~p ∧ q) ∧ (~q ∨ ~r) is
~[(~p ∧ q) ∧ (~q ∨ ~ r)] ≡ ~(~p ∧ q) ∨ ~(~q ∨ ~r) ……[Negation of conjunction]
≡ [~(~p) ∨ ~q] ∨ [~(~q) ∧ ~(~r)] … [Negation of conjunction and disjunction]
≡ (p ∨ ~q) ∨ (q ∧ r) …..[Negation of negation]

Question 3.
Write the converse, inverse, and contrapositive of the following statements:
(i) If it snows, then they do not drive the car.
Solution:
Let p : It snows.
q : They do not drive the car.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If they do not drive the car, then it snows.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If it does not snow, then they drive the car.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If they drive the car, then it does not snow.

(ii) If he studies, then he will go to college.
Solution:
Let p : He studies.
q : He will go to college.
Then two symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If he will go to college, then he studies.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If he does not study, then he will not go to college.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If he will not go to college, then he does not study.

Question 4.
With proper justification, state the negation of each of the following:
(i) (p → q) ∨ (p → r)
Solution:
The negation of (p → q) ∨ (p → r) is
~[(p → q) ∨ (p → r)] ≡ ~(p → q) ∧ ~(p → r) …..[Negation of disjunction]
≡ (p ∧ ~q) ∧ (p ∧ ~r) …[Negation of implication]

(ii) (p ↔ q) ∨ (~q → ~r)
Solution:
The negation of (p ↔ q) ∨ (~q → ~r) is
~[(p ↔ q) ∨ (~q → ~r)] ≡ ~(p ↔ q) ∧ ~(~q → ~r) …..[Negation of disjunction]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~q ∧ ~(~r)] ……[Negation of biconditional and implication]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~q ∧ r) ……[Negation of negation]

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~[(p → q) ∧ r] ≡ ~(p → q) ∨ (~r) …..[Negation of conjunction]
≡ (p ∧ ~q) ∨ (~r) …..[Negation of implication]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.7

Question 1.
Write the dual of each of the following:
(i) (p ∨ q) ∨ r
Solution:
(p ∧ q) ∧ r

(ii) ~(p ∨ q) ∧ [p ∨ ~(q ∧ ~r)]
Solution:
~(p ∧ q) ∨ [p ∧ ~(q ∨ ~r)]

(iii) p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
Solution:
p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

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

Question 2.
Write the dual statement of each of the following compound statements:
(i) 13 is prime number and India is a democratic country.
Solution:
13 is prime number or India is a democratic country.

(ii) Karina is very good or everybody likes her.
Solution:
Karina is very good and everybody likes her.

(iii) Radha and Sushmita can not read Urdu.
Solution:
Radha or Sushmita can not read Urdu.

(iv) A number is real number and the square of the number is non-negative.
Solution:
A number is real number or the square of the number is non-negative.

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 2 Matrices Ex 2.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 1.
Evaluate:
(i) \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right]\) = \(\left[\begin{array}{rrr}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]\)

(ii) \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\) = [8 – 3 + 3] = [8]

Question 2.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\). State whether AB = BA? Justify your answer.
Solution:

From (1) and (2), AB ≠ BA.

Question 3.
Show that AB = BA, where A = \(\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
Solution:

From (1) and (2), AB = BA.

Question 4.
Verify A(BC) = (AB)C, if A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
2 & 3 & 0 \\
0 & 4 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 1 \\
0 & 3
\end{array}\right]\), and C = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
2 & 0 & -2
\end{array}\right]\)
Solution:


From (1) and (2), A(BC) = (AB)C.

Question 5.
Verify that A(B + C) = AB + AC, if A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
Solution:


From (1) and (2), A(B + C) = AB + AC.

Question 6.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non-singular.
Solution:

Hence, AB is a non-singular matrix.

Question 7.
If A + I = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
5 & 4 & 2 \\
0 & 7 & -3
\end{array}\right]\), find the product (A + I)(A – I).
Solution:

Question 8.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A² – 4A is a scalar matrix.
Solution:

which is a scalar matrix.

Question 9.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A² – 8A – kI = O, where I is a 2 × 2 unit matrix and O is null matrix of order 2.
Solution:

By equality of matrices,
-k – 7 = 0
∴ k = -7.

Question 10.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A² – 5A + 7I = 0, where I is a 2 × 2 unit matrix.
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and if(A + B)² = A² + B², find values of a and b.
Solution:
(A + B)² = A² + B²
∴ (A + B)(A + B) = A² + B²
∴ A² + AB + BA + B² = A² + B²
∴ AB + BA = 0
∴ AB = -BA

By the equality of matrices, we get
0 = a – 2 ……..(1)
0 = 1 + b ……..(2)
a + 2b = 2a – 4 ……..(3)
-a – 2b = 2 + 2b ……..(4)
From equations (1) and (2), we get
a = 2 and b = -1
The values of a and b satisfy equations (3) and (4) also.
Hence, a = 2 and b = -1.

Question 12.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A² = kA – 2I.
Solution:

By equality of matrices,
1 = 3k – 2 ……..(1)
-2 = -2k ……..(2)
4 = 4k ……..(3)
-4 = -2k – 2 ……..(4)
From (2), k = 1.
k = 1 also satisfies equation (1), (3) and (4).
Hence, k = 1.

Question 13.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
x \\
y
\end{array}\right]\)
Solution:

By equality of matrices,
x = 19 and y = 12.

Question 14.
Find x, y, z, if \(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:


By equality of matrices,
-6 = x – 3, 0 = y – 1 and -2 = 2z
∴ x = -3, y = 1 and z = -1.

Question 15.
Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks. Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.
Solution:
The given data can be written in matrix form as:
Number of Pens and Notebooks

For finding the amount each one of them requires to buy the pens and notebook, we require the multiplication of the two matrices A and B.

Hence, Jay requires ₹ 104 and Ram requires ₹ 150 to buy the pens and notebooks.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

(I) Select the appropriate option from the given alternatives.

Question 1.
If y = \(\frac{x-4}{\sqrt{x+2}}\), then \(\frac{d y}{d x}\) is
(A) \(\frac{1}{x+4}\)
(B) \(\frac{\sqrt{x}}{\left(\sqrt{x+2)^{2}}\right.}\)
(C) \(\frac{1}{2 \sqrt{x}}\)
(D) \(\frac{x}{(\sqrt{x}+2)^{2}}\)
Answer:
(C) \(\frac{1}{2 \sqrt{x}}\)
Hint:

Question 2.
If y = \(\frac{a x+b}{c x+d}\),then \(\frac{d y}{d x}\) =
(A) \(\frac{a b-c d}{(c x+d)^{2}}\)
(B) \(\frac{a x-c}{(c x+d)^{2}}\)
(C) \(\frac{a c-b d}{(c x+d)^{2}}\)
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Answer:
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Hint:

Question 3.
If y = \(\frac{3 x+5}{4 x+5}\), then \(\frac{d y}{d x}\) =
(A) \(-\frac{15}{(3 x+5)^{2}}\)
(B) \(-\frac{15}{(4 x+5)^{2}}\)
(C) \(-\frac{5}{(4 x+5)^{2}}\)
(D) \(-\frac{13}{(4 x+5)^{2}}\)
Answer:
(C) \(-\frac{5}{(4 x+5)^{2}}\)
Hint:

Question 4.
If y = \(\frac{5 \sin x-2}{4 \sin x+3}\), then \(\frac{d y}{d x}\) =
(A) \(\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
(C) \(-\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(D) \(-\frac{15 \cos x}{(4 \sin x+3)^{2}}\)
Answer:
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
Hint:

Question 5.
Suppose f(x) is the derivative of g(x) and g(x) is the derivative of h(x).
If h(x) = a sin x + b cos x + c, then f(x) + h(x) =
(A) 0
(B) c
(C) -c
(D) -2(a sin x + b cos x)
Answer:
(B) c
Hint:
h(x) = a sin x + b cos x + c
Differentiating w.r.t. x, we get
h'(x) = a cos x – b sin x = g(x) …..[given]
Differentiating w.r.t. x, we get
g'(x) = -a sin x – b cos x = f(x) …..[given]
∴ f(x) + h(x) = -a sin x – b cos x + a sin x + b cos x + c
∴ f(x) + h(x) = c

Question 6.
If f(x) = 2x + 6, for 0 ≤ x ≤ 2
= ax² + bx, for 2 < x ≤ 4
is differentiable at x = 2, then the values of a and b are
(A) a = \(-\frac{3}{2}\), b = 3
(B) a = \(\frac{3}{2}\), b = 8
(C) a = \(\frac{1}{2}\), b = 8
(D) a = \(-\frac{3}{2}\), b = 8
Answer:
(D) a = \(-\frac{3}{2}\), b = 8
Hint:
f(x) = 2x + 6, 0 ≤ x ≤ 2
= ax² + bx, 2 < x ≤ 4
Lf'(2) = 2, Rf'(2) = 4a + b
Since f is differentiable at x = 2,
Lf'(2) = Rf'(2)
∴ 2 = 4a + b …..(i)
f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{+}} f(x)=f(2)=\lim _{x \rightarrow 2^{-}} f(x)\)
∴ 4a + 2b = 2(2) + 6
∴ 4a + 2b = 10
∴ 2a + b = 5 …..(ii)
Solving (i) and (ii), we get
a = \(-\frac{3}{2}\), b = 8

Question 7.
If f(x) = x² + sin x + 1, for x ≤ 0
= x² – 2x + 1, for x ≤ 0, then
(A) f is continuous at x = 0, but not differentiable at x = 0
(B) f is neither continuous nor differentiable at x = 0
(C) f is not continuous at x = 0, but differentiable at x = 0
(D) f is both continuous and differentiable at x = 0
Answer:
(A) f is continuous at x = 0, but not differentiable at x = 0
Hint:

Question 8.
If f(x) = \(\frac{x^{50}}{50}+\frac{x^{49}}{49}+\frac{x^{48}}{48}+\ldots .+\frac{x^{2}}{2}+x+1\), then f'(1) =
(A) 48
(B) 49
(C) 50
(D) 51
Answer:
(C) 50
Hint:

(II).

Question 1.
Determine whether the following function is differentiable at x = 3 where,
f(x) = x² + 2, for x ≥ 3
= 6x – 7, for x < 3.
Solution:
f(x) = x² + 2, x ≥ 3
= 6x – 7, x < 3
Differentiability at x = 3


Here, Lf'(3) = Rf'(3)
∴ f is differentiable at x = 3.

Question 2.
Find the values of p and q that make function f(x) differentiable everywhere on R.
f(x) = 3 – x, for x < 1
= px² + qx, for x ≥ 1.
Solution:
f(x) is differentiable everywhere on R.
∴ f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.

f(x) is differentiable at x = 1.
∴ Lf'(1) = Rf'(1)
∴ -1 = 2p + q …..(ii)
Subtracting (i) from (ii), we get
p = -3
Substituting p = -3 in (i), we get
p + q = 2
∴ -3 + q = 2
∴ q = 5

Question 3.
Determine the values of p and q that make the function f(x) differentiable on R where
f(x) = px³, for x < 2
= x² + q, for x ≥ 2
Solution:
f(x) is differentiable on R.
∴ f(x) is differentiable at x = 2.
∴ f(x) is continuous at x = 2.
Continuity at x = 2:
f(x) is continuous at x = 2.


f(x) is differentiable at x = 2.
∴ Lf'(2) = Rf'(2)
∴ 12p = 4
∴ p = \(\frac{1}{3}\)
Substituting p = \(\frac{1}{3}\) in (i), we get
8(\(\frac{1}{3}\) – q = 4
∴ q = \(\frac{8}{3}\) – 4 = \(\frac{-4}{3}\)

Question 4.
Determine all real values of p and q that ensure the function
f(x) = px + q, for x ≤ 1
= tan(\(\frac{\pi x}{4}\)), for 1 < x < 2
is differentiable at x = 1.
Solution:
f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Continuity at x= 1:
f(x) is continuous at x = 1.

Question 5.
Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.
Solution:


Here, Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.
∴ f is not differentiable at x = -1 and x = 1.
∴ f is not differentiable ∀ x ∈ R.

Question 6.
Test whether the function
f(x) = 2x – 3, for x ≥ 2
= x – 1, for x < 2
is differentiable at x = 2.
Solution:

Question 7.
Test whether the function
f(x) = x² + 1, for x ≥ 2
= 2x + 1, for x < 2
is differentiable at x = 2.
Solution:

Question 8.
Test whether the function
f(x) = 5x – 3x², for x ≥ 1
= 3 – x, for x < 1
is differentiable at x = 1.
Solution:

Here, Lf'(1) = Rf'(1)
∴ f(x) is differentiable at x = 1.

Question 9.
If f(2) = 4, f'(2) = 1, then find \(\lim _{x \rightarrow 2}\left[\frac{x f(2)-2 f(x)}{x-2}\right]\)
Solution:

Question 10.
If y = \(\frac{\mathbf{e}^{x}}{\sqrt{x}}\), find \(\frac{d y}{d x}\) when x = 1.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Ex 6.1

Question 1.
Check if the following relations are functions.
(a)

Solution:
Yes.
Reason: Every element of set A has been assigned a unique element in set B.

(b)

Solution:
No.
Reason: An element of set A has been assigned more than one element from set B.

(c)

Solution:
No.
Reason:
Not every element of set A has been assigned an image from set B.

Question 2.
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}
(iii) {(1, 3), (4, 1), (2, 2)}
(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}
Solution:
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1), (2, 2), show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has been assigned a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason:
3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

Question 3.
Check if the relation given by the equation represents y as function of x.
(i) 2x + 3y = 12
(ii) x + y² = 9
(iii) x² – y = 25
(iv) 2y + 10 = 0
(v) 3x – 6 = 21
Solution:
(i) 2x + 3y = 12
∴ y = \(\frac{12-2 x}{3}\)
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(ii) x + y² = 9
∴ y² = 9 – x
∴ y = ±\(\sqrt{9-x}\)
∴ For one value of x, there are two values of y.
∴ y is not a function of x.

(iii) x² – y = 25
∴ y = x² – 25
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(iv) 2y + 10 = 0
∴ y = -5
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(v) 3x – 6 = 21
∴ x = 9
∴ x = 9 represents a point on the X-axis.
There is no y involved in the equation.
So the given equation does not represent a function.

Question 4.
If f(m) = m² – 3m + 1, find
(i) f(0)
(ii) f(-3)
(iii) f(\(\frac{1}{2}\))
(iv) f(x + 1)
(v) f(-x)
(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\), h ≠ 0.
Solution:
f(m) = m² – 3m + 1
(i) f(0) = 0² – 3(0) + 1 = 1

(ii) f (-3) = (-3)² – 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) f(\(\frac{1}{2}\)) = \(\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)² – 3(x + 1) + 1
= x² + 2x + 1 – 3x – 3 + 1
= x² – x – 1

(v) f(-x) = (-x)2 – 3(-x) + 1 = x2 + 3x + 1

(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\)
= \(\frac{(2+h)^{2}-3(2+h)+1-\left(2^{2}-3(2)+1\right)}{h}\)
= \(\frac{\mathrm{h}^{2}+\mathrm{h}}{\mathrm{h}}\)
= h + 1

Question 5.
Find x, if g(x) = 0 where
(i) g(x) = \(\frac{5 x-6}{7}\)
(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
(iii) g(x) = 6x² + x – 2
(iv) g(x) = x³ – 2x² – 5x + 6
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 0
∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
g(x) = 0
\(\frac{18-2 x^{2}}{7}\) = 0
∴ 18 – 2x² = 0
∴ x² = 9
∴ x = ±3

(iii) g(x) = 6x² + x – 2
g(x) = 0
∴ 6x² + x – 2 = 0
∴ (2x – 1) (3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

(iv) g(x) = x³ – 2x² – 5x + 6
= ( x- 1) (x² – x – 6)
= (x – 1) (x + 2) (x – 3)
g(x) = 0
∴ (x – 1) (x + 2) (x – 3) = 0
∴ x – 1 = 0 or x + 2 = 0 or x – 3 = 0
∴ x = 1, -2, 3

Question 6.
Find x, if f(x) = g(x) where
(i) f(x) = x⁴ + 2x², g(x) = 11x²
(ii) f(x) = √x – 3, g(x) = 5 – x
Solution:
(i) f(x) = x⁴ + 2x², g(x) = 11x²
f(x) = g(x)
∴ x⁴ + 2x² = 11x²
∴ x⁴ – 9x² = 0
∴ x² (x² – 9) = 0
∴ x² = 0 or x² – 9 = 0
∴ x = 0 or x² = 9
∴ x = 0, ±3

(ii) f(x) = √x – 3, g(x) = 5 – x
f(x) = g(x)
∴ √x – 3 = 5 – x
∴ √x = 5 – x + 3
∴ √x = 8 – x
on squaring, we get
x = 64 + x² – 16x
∴ x² – 17x + 64 = 0
∴ x = \(\frac{17 \pm \sqrt{(-17)^{2}-4(64)}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{289-256}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{33}}{2}\)

Question 7.
If f(x) = \(\frac{a-x}{b-x}\), f(2) is undefined, and f(3) = 5, find a and b.
Solution:
f(x) = \(\frac{a-x}{b-x}\)
Given that,
f(2) is undefined
b – 2 = 0
∴ b = 2 …..(i)
f(3) = 5
∴ \(\frac{a-3}{b-3}\) = 5
∴ \(\frac{a-3}{2-3}\) = 5 ….. [From (i)]
∴ a – 3 = -5
∴ a = -2
∴ a = -2, b = 2

Question 8.
Find the domain and range of the following functions.
(i) f(x) = 7x² + 4x – 1
Solution:
f(x) = 7x² + 4x – 1
f is defined for all x.
∴ Domain of f = R (i.e., the set of real numbers)

∴ Range of f = [\(-\frac{11}{7}\), ∞)

(ii) g(x) = \(\frac{x+4}{x-2}\)
Solution:
g(x) = \(\frac{x+4}{x-2}\)
Function g is defined everywhere except at x = 2.
∴ Domain of g = R – {2}
Let y = g(x) = \(\frac{x+4}{x-2}\)
∴ (x – 2) y = x + 4
∴ x(y – 1) = 4 + 2y
∴ For every y, we can find x, except for y = 1.
∴ y = 1 ∉ range of function g
∴ Range of g = R – {1}

(iii) h(x) = \(\frac{\sqrt{x+5}}{5+x}\)
Solution:
h(x) = \(\frac{\sqrt{x+5}}{5+x}=\frac{1}{\sqrt{x+5}}\), x ≠ -5
For x = -5, function h is not defined.
∴ x + 5 > 0 for function h to be well defined.
∴ x > -5
∴ Domain of h = (-5, ∞)
Let y = \(\frac{1}{\sqrt{x+5}}\)
∴ y > 0
Range of h = (0, ∞) or R+

(iv) f(x) = \(\sqrt[3]{x+1}\)
Solution:
f(x) = \(\sqrt[3]{x+1}\)
f is defined for all real x and the values of f(x) ∈ R
∴ Domain of f = R, Range of f = R

(v) f(x) = \(\sqrt{(x-2)(5-x)}\)
Solution:
f(x) = \(\sqrt{(x-2)(5-x)}\)
For f to be defined,
(x – 2)(5 – x) ≥ 0
∴ (x – 2)(x – 5) ≤ 0
∴ 2 ≤ x ≤ 5 ……[∵ The solution of (x – a) (x – b) ≤ 0 is a ≤ x ≤ b, for a < b]
∴ Domain of f = [2, 5]
(x – 2) (5 – x) = -x² + 7x – 10
= \(-\left(x-\frac{7}{2}\right)^{2}+\frac{49}{4}-10\)
= \(\frac{9}{4}-\left(x-\frac{7}{2}\right)^{2} \leq \frac{9}{4}\)
∴ \(\sqrt{(x-2)(5-x)} \leq \sqrt{\frac{9}{4}} \leq \frac{3}{2}\)
Range of f = [0, \(\frac{3}{2}\)]

(vi) f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
Solution:
f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
For f to be defined,
\(\sqrt{\frac{x-3}{7-x}}\) ≥ 0, 7 – x ≠ 0
∴ \(\sqrt{\frac{x-3}{7-x}}\) ≤ 0 and x ≠ 7
∴ 3 ≤ x < 7
Let a < b, \(\frac{x-a}{x-b}\) ≤ 0 ⇒ a ≤ x < b
∴ Domain of f = [3, 7)
f(x) ≥ 0 … [∵ The value of square root function is non-negative]
∴ Range of f = [0, ∞)

(vii) f(x) = \(\sqrt{16-x^{2}}\)
Solution:
f(x) = \(\sqrt{16-x^{2}}\)
For f to be defined,
16 – x² ≥ 0
∴ x² ≤ 16
∴ -4 ≤ x ≤ 4
∴ Domain of f = [-4, 4]
Clearly, f(x) ≥ 0 and the value of f(x) would be maximum when the quantity subtracted from 16 is minimum i.e. x = 0
∴ Maximum value of f(x) = √16 = 4
∴ Range of f = [0, 4]

Question 9.
Express the area A of a square as a function of its
(a) side s
(b) perimeter P
Solution:
(a) area (A) = s²
(b) perimeter (P) = 4s
∴ s = \(\frac{\mathrm{P}}{4}\)
Area (A) = s² = \(\left(\frac{\mathrm{P}}{4}\right)^{2}\)
∴ A = \(\frac{\mathrm{P}^{2}}{16}\)

Question 10.
Express the area A of a circle as a function of its
(i) radius r
(ii) diameter d
(iii) circumference C
Solution:
(i) Area (A) = πr²

(ii) Diameter (d) = 2r
∴ r = \(\frac{\mathrm{d}}{2}\)
∴ Area (A) = πr² = \(\frac{\pi \mathrm{d}^{2}}{4}\)

(iii) Circumference (C) = 2πr
∴ r = \(\frac{C}{2 \pi}\)
Area (A) = πr² = \(\pi\left(\frac{\mathrm{C}}{2 \pi}\right)^{2}\)
∴ A = \(\frac{C^{2}}{4 \pi}\)

Question 11.
An open box is made from a square of cardboard of 30 cms side, by cutting squares of length x centimeters from each corner and folding the sides up. Express the volume of the box as a function of x. Also, find its domain.
Solution:

Length of the box = 30 – 2x
Breadth of the box = 30 – 2x
Height of the box = x
Volume = (30 – 2x)² x, x < 15, x ≠ 15, x > 0
= 4x(15 – x)², x ≠ 15, x > 0
Domain (0, 15)

Question 12.
Let f be a subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify?
Solution:
f = {(ab, a + b): a, b ∈ Z}
Let a = 1, b = 1. Then, ab = 1, a + b = 2
∴ (1, 2) ∈ f
Let a = -1, b = -1. Then, ab = 1, a + b = -2
∴ (1, -2) ∈ f
Since (1, 2) ∈ f and (1, -2) ∈ f,
f is not a function as element 1 does not have a unique image.

Question 13.
Check the injectivity and surjectivity of the following functions.
(i) f : N → N given by f(x) = x²
Solution:
f: N → N given by f(x) = x²

∴ f is injective.
For every y = x² ∈ N, there does not exist x ∈ N.
Example: 7 ∈ N (codomain) for which there is no x in domain N such that x² = 7
∴ f is not surjective.

(ii) f : Z → Z given by f(x) = x²
Solution:
f: Z → Z given by f(x) = x2

∴ f is not injective.
(Example: f(-2) = 4 = f(2). So, -2, 2 have the same image. So, f is not injective.)
Since x² ≥ 0,
f(x) ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iii) f : R → R given by f(x) = x²
Solution:
f : R → R given by f(x) = x²

∴ f is not injective.
f(x) = x² ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iv) f : N → N given by f(x) = x³
Solution:
f: N → N given by f(x) = x³

∴ f is injective.
Numbers from codomain which are not cubes of natural numbers are not images under f.
∴ f is not surjective.

(v) f : R → R given by f(x) = x³
Solution:
f: R → R given by f(x) = x³

∴ For every y ∈ R, there is some x ∈ R.
∴ f is surjective.

Question 14.
Show that if f : A → B and g : B → C are one-one, then gof is also one-one.
Solution:
f is a one-one function.
Let f(x₁) = f(x₂)
Then, x₁ = x₂ for all x₁, x₂ …..(i)
g is a one-one function.
Let g(y₁) = g(y₂)
Then, y₁ = y₂ for all y₁, y₂ …..(ii)
Let (gof) (x₁) = (gof) (x₂)
∴ g(f(x₁)) = g(f(x₂))
∴ g(y₁) = g(y₂),
where y₁ = f(x₁), y₂ = f(x₂) ∈ B
∴ y₁ = y₂ …..[From (ii)]
i.e., f(x₁) = f(x₂)
∴ x₁ = x₂ ….[From (i)]
∴ gof is one-one.

Question 15.
Show that if f : A → B and g : B → C are onto, then gof is also onto.
Solution:
Since g is surjective (onto),
there exists y ∈ B for every z ∈ C such that
g(y) = z …….(i)
Since f is surjective,
there exists x ∈ A for every y ∈ B such that
f(x) = y …….(ii)
(gof) x = g(f(x))
= g(y) ……[From (ii)]
= z …..[From(i)]
i.e., for every z ∈ C, there is x in A such that (gof) x = z
∴ gof is surjective (onto).

Question 16.
If f(x) = 3(4^x+1), find f(-3).
Solution:
f(x) = 3(4^x+1)
∴ f(-3) = 3(4^-3+1)
= 3(4^-2)
= \(\frac{3}{16}\)

Question 17.
Express the following exponential equations in logarithmic form:
(i) 2⁵ = 32
(ii) 54⁰ = 1
(iii) 23¹ = 23
(iv) \(9^{\frac{3}{2}}\) = 27
(v) 3^-4 = \(\frac{1}{81}\)
(vi) 10^-2 = 0.01
(vii) e² = 7.3890
(viii) \(e^{\frac{1}{2}}\) = 1.6487
(ix) e^-x = 6
Solution:

Question 18.
Express the following logarithmic equations in exponential form:
(i) log₂ 64 = 6
(ii) \(\log _{5} \frac{1}{25}\) = -2
(iii) log₁₀ 0.001 = -3
(iv) \(\log _{\frac{1}{2}}\)(8) = -3
(v) ln 1 = 0
(vi) ln e = 1
(vii) ln \(\frac{1}{2}\) = -0.693
Solution:
(i) log₂ 64 = 6
∴ 64 = 2⁶, i.e., 2⁶ = 64

Question 19.
Find the domain of
(i) f(x) = ln (x – 5)
(ii) f(x) = log10 (x² – 5x + 6)
Solution:
(i) f(x) = ln (x – 5)
f is defined, when x – 5 > 0
∴ x > 5
∴ Domain of f = (5, ∞)

(ii) f(x) = log₁₀ (x² – 5x + 6)
x² – 5x + 6 = (x – 2) (x – 3)
f is defined, when (x – 2) (x – 3) > 0
∴ x < 2 or x > 3
Solution of (x – a) (x – b) > 0 is x < a or x > b where a < b
∴ Domain of f = (-∞, 2) ∪ (3, ∞)

Question 20.
Write the following expressions as sum or difference of logarithms:
(a) \(\log \left(\frac{p q}{r s}\right)\)
Solution:

(b) \(\log (\sqrt{x} \sqrt[3]{y})\)
Solution:

(c) \(\ln \left(\frac{a^{3}(a-2)^{2}}{\sqrt{b^{2}+5}}\right)\)
Solution:

(d) \(\ln \left[\frac{\sqrt[3]{x-2}(2 x+1)^{4}}{(x+4) \sqrt{2 x+4}}\right]^{2}\)
Solution:

Question 21.
Write the following expressions as a single logarithm.
(i) 5 log x + 7 log y – log z
Solution:

(ii) \(\frac{1}{3}\) log(x – 1) + \(\frac{1}{2}\) log(x)
Solution:

(iii) ln (x + 2) + ln (x – 2) – 3 ln (x + 5)
Solution:

Question 22.
Given that log 2 = a and log 3 = b, write log √96 terms of a and b.
Solution:
log 2 = a and log 3 = b
log √96 = \(\frac{1}{2}\) log (96)
= \(\frac{1}{2}\) log (2⁵ x 3)
= \(\frac{1}{2}\) (log 2⁵ + log 3) …..[∵ log mn = log m + log n]
= \(\frac{1}{2}\) (5 log 2 + log 3) ……[∵ log mⁿ = n log m]
= \(\frac{5 a+b}{2}\)

Question 23.
Prove that:
(a) \(b^{\log _{b} a}=a\)
Solution:
We have to prove that \(b^{\log _{b} a}=a\)
i.e., to prove that (logb a) (log_b b) = log_b a
(Taking log on both sides with base b)
L.H.S. = (log_b a) (log_b b)
= log_b a …..[∵ log_b b = 1]
= R.H.S.

(b) \(\log _{b^{m}} a=\frac{1}{m} \log _{b} a\)
Solution:

(c) \(a^{\log _{c} b}=b^{\log _{c} a}\)
Solution:

Question 24.
If f(x) = ax² – bx + 6 and f(2) = 3 and f(4) = 30, find a and b.
Solulion:
f(x) = ax² – bx + 6
f(2) = 3
∴ a(2)² – b(2) + 6 = 3
∴ 4a – 2b + 6 = 3
∴ 4a – 2b + 3 = 0 …..(i)
f(4) = 30
∴ a(4)² – b(4) + 6 = 30
∴ 16a – 4b + 6 = 30
∴ 16a – 4b – 24 = 0 …..(ii)
By (ii) – 2 × (i), we get
8a – 30 = 0
∴ a = \(\frac{30}{8}=\frac{15}{4}\)
Substiting a = \(\frac{15}{4}\) in (i), we get
4(\(\frac{15}{4}\)) – 2b + 3 = 0
∴ 2b = 18
∴ b = 9
∴ a = \(\frac{15}{4}\), b = 9

Question 25.
Solve for x:
(i) log 2 + log (x + 3) – log (3x – 5) = log 3
Solution:
log 2 + log (x + 3) – log (3x – 5) = log 3
∴ log 2(x + 3) – log(3x – 5) = log 3 …..[∵ log m + log n = log mn]
∴ log \(\frac{2(x+3)}{3 x-5}\) = log 3 …..[∵ log m – log n = log \(\frac{m}{n}\)]
∴ \(\frac{2(x+3)}{3 x-5}\) = 3
∴ 2x + 6 = 9x – 15
∴ 7x = 21
∴ x = 3

Check:
If x = 3 satisfies the given condition, then our answer is correct.
L.H.S. = log 2 + log (x + 3) – log (3x – 5)
= log 2 + log (3 + 3) – log (9 – 5)
= log 2 + log 6 – log 4
= log (2 × 6) – log 4
= log \(\frac{12}{4}\)
= log 3
= R.H.S.
Thus, our answer is correct.

(ii) 2log₁₀ x = 1 + \(\log _{10}\left(x+\frac{11}{10}\right)\)
Solution:


∴ x² = 10x + 11
∴ x² – 10x – 11 = 0
∴ (x – 11)(x + 1) = 0
∴ x = 11 or x = -1
But log of a negative numbers does not exist
∴ x ≠ -1
∴ x = 11

(iii) log₂ x + log₄ x + log₁₆ x = \(\frac{21}{4}\)
Solution:

(iv) x + log₁₀ (1 + 2^x) = x log₁₀ 5 + log₁₀ 6
Solution:

∴ a + a² = 6
∴ a² + a – 6 = 0
∴ (a + 3)(a – 2) = 0
∴ a + 3 = 0 or a – 2 = 0
∴ a = -3 or a = 2
Since 2x = -3 is not possible,
2x = 2 = 21
∴ x = 1

Question 26.
If log \(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\) log x + \(\frac{1}{2}\) log y, show that \(\frac{x}{y}+\frac{y}{x}\) = 7.
Solution:

Question 27.
If log\(\left(\frac{x-y}{4}\right)\) = log√x + log√y, show that (x + y)² = 20xy.
Solution:

Question 28.
If x = log_abc, y = log_b ca, z = log_c ab, then prove that \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) = 1.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Ex 5.1

Question 1.
Describe the following sets in Roster form:
(i) A = {x/x is a letter of the word ‘MOVEMENT’}
(ii) B = {x/x is an integer, –\(\frac{3}{2}\) < x < \(\frac{9}{2}\)>
(iii) C = {x/x = 2n + 1, n ∈ N}
Solution:
(i) A = {M, O, V, E, N, T}
(ii) B = {-1, 0, 1, 2, 3, 4}
(iii) C = {3, 5, 7, 9, … }

Question 2.
Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
(iv) {0, -1, 2, -3, 4, -5, 6,…}
Solution:
(i) Let A = {0}
0 is a whole number but it is not a natural number.
∴ A = {x / x ∈ W, x ∉ N}

(ii) Let B = {0, ±1, ±2, ±3}
B is the set of elements which belongs to Z from -3 to 3.
∴ B = {x /x ∈ Z, -3 ≤ x ≤ 3}

(iii) Let C = \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
∴ C = {x / x = \(\frac{n}{n^{2}+1}\), n ∈ N, n ≤ 7}

(iv) Let D = {0, -1, 2, -3, 4, -5, 6, …}
∴ D = {x/x = (-1)n-1 × (n – 1), n ∈ N}

Question 3.
If A = {x / 6x² + x – 15 = 0}, B = {x / 2x² – 5x – 3 = 0}, C = {x / 2x² – x – 3 = 0}, then find
(i) (A ∪ B ∪ C)
(ii) (A ∩ B ∩ C)
Solution:
A = [x/6x² + x – 15 = 0)
6x² + x – 15 = 0
6x² + 10x – 9x – 15 = 0
2x(3x + 5) – 3(3x + 5) = 0
(3x + 5) (2x – 3) = 0
3x + 5 = 0 or 2x – 3 = 0
x = \(\frac{-5}{3}\) or x = \(\frac{3}{2}\)
A = {\(\frac{-5}{3}\), \(\frac{3}{2}\)}

B = {x/2x² – 5x – 3 = 0}
2x² – 5x – 3 = 0
2x² – 6x + x – 3 = 0
2x(x – 3) + 1(x – 3) = 0
(x – 3)(2x + 1) = 0
x – 3 = 0 or 2x + 1 = 0
x = 3 or x = \(\frac{-1}{2}\)
B = (\(\frac{-1}{2}\), 3)

C = {x/2x² – x – 3 = 0}
2x² – x – 3 = 0
2x² – 3x + 2x – 3 = 0
x(2x – 3) + 1(2x – 3) = 0
(2x – 3) (x + 1) = 0
2x – 3 = 0 or x + 1 = 0
x = \(\frac{3}{2}\) or x = -1
C = {-1, \(\frac{3}{2}\)}

(i) A ∪ B ∪ C = \(\left\{-\frac{5}{3}, \frac{3}{2}\right\} \cup\left\{\frac{-1}{2}, 3\right\} \cup\left\{-1, \frac{3}{2}\right\}\) = \(\left\{\frac{-5}{3},-1, \frac{-1}{2}, \frac{3}{2}, 3\right\}\)

(ii) A ∩ B ∩ C = { }

Question 4.
If A, B, C are the sets for the letters in the words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that [A – (B ∪ C)] = [(A – B) ∩ (A – C)].
Solution:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A – (B ∪ C) = {c, o}
A – B = {c, o, l}
A – C = {c, o}
∴ [(A – B) ∩ (A – C)] = {c, o} = A – (B ∪ C)
∴ [A -( B ∪ C)] = [(A – B) ∩ (A – C)]

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify the following:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(iii) (A ∪ B)’ = A’ ∩ B’
(iv) (A ∩ B)’ = A’ ∪ B’
(v) A = (A ∩ B) ∪ (A ∩ B’)
(vi) B = (A ∩ B) ∪ (A’ ∩ B)
(vii) (A ∪ B) = (A – B) ∪ (A ∩ B) ∪ (B – A)
(viii) A ∩ (B ∆ C) = (A ∩ B) ∆ (A ∩ C)
(ix) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(x) n(B) = n (A’ ∩ B) + n (A ∩ B)
Solution:
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8},
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) B ∩ C = {4, 5, 6}
∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} …..(i)
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} …….(ii)
From (i) and (ii), we get
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) B ∪ C = {3, 4, 5, 6, 7, 8}
∴ A ∩ (B ∪ C) = {3, 4} ………(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) ∪ (A∩ C) = {3, 4} ……..(ii)
From (i) and (ii), we get
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∪ B = {1, 2, 3, 4, 5, 6}
∴ (A ∪ B)’ = {7, 8, 9, 10} ………(i)
A’ = {5, 6, 7, 8, 9, 10},
B’ = {1, 2, 7, 8, 9,10}
∴ A’ ∩ B’ = {7, 8, 9, 10} …….(ii)
From (i) and (ii), we get
(A ∪ B)’ = A’ ∩ B’

(iv) A ∩ B = {3, 4}
(A ∩ B)’= {1, 2, 5, 6, 7, 8, 9, 10} …….(i)
A’ = {5, 6, 7, 8, 9, 10}
B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∪ B’ = {1, 2, 5, 6, 7, 8, 9, 10} …….(ii)
From (i) and (ii), we get
(A ∩ B)’ = A’ ∪ B’

(v) A = {1, 2, 3, 4} ……(i)
A ∩ B = {3, 4}
B’ = {1, 2, 7, 8, 9, 10}
A ∩ B’ = {1, 2}
∴ (A ∩ B) ∪ (A ∩ B’) = {1, 2, 3, 4} …..(ii)
From (i) and (ii), we get
A = (A ∩ B) ∪ (A ∩ B’)

(vi) B = {3, 4, 5, 6} …..(i)
A ∩ B = {3, 4}
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ (A ∩ B) ∪ (A’ ∩ B) = {3, 4, 5, 6} …..(ii)
From (i) and (ii), we get
B = (A ∩ B) ∪ (A’ ∩ B)

(vii) A ∪ B = {1, 2, 3, 4, 5, 6} …….(i)
A – B = {1, 2}
A ∩ B = {3, 4}
B – A = {5, 6}
∴ (A – B) ∪ (A ∩ B) ∪ (B – A) = {1, 2, 3, 4, 5, 6} ……(ii)
From (i) and (ii), we get
A ∪ B = (A – B) ∪ (A ∩ B) ∪ (B – A)

(viii) B – C = {3}
C – B = {7, 8}
B Δ C = (B – C) ∪ (C – B) = {3, 7, 8}
∴ A ∩ (B Δ C) = {3} ……(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) Δ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)] = {3} …..(ii)
From (i) and (ii), we get
A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

(ix) A = {1, 2, 3, 4}, B = {3, 4, 5, 6}
A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6}
∴ n(A) = 4, n(B) = 4,
n(A ∩ B) = 2, n(A ∪ B) = 6 ……(i)
∴ n(A) + n(B) – n(A ∩ B) = 4 + 4 – 2
∴ n(A) + n(B) – n(A ∩ B) = 6 …..(ii)
From (i) and (ii), we get
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(x) B = {3, 4, 5, 6}
∴ n(B) = 4 …..(i)
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ n(A’ ∩ B) = 2
A ∩ B = {3, 4}
∴ n(A ∩ B) = 2
∴ n(A’ ∩ B) + n(A ∩ B) = 2 + 2 = 4 …..(ii)
From (i) and (ii), we get
n(B) = n(A’ ∩ B) + n (A ∩ B)

Question 6.
If A and B are subsets of the universal set X and n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5, find
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A’ ∩ B)
(iv) n(A ∩ B’)
Solution:
n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5
(i) n(A ∪ B) = n(X) – [n(A ∪ B)’]
= n(X) – n(A’ ∩ B’)
= 50 – 5
= 45

(ii) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 35 + 20 – 45
= 10

(iii) n(A’ ∩ B) = n(B) – n(A ∩ B)
= 20 – 10
= 10

(iv) n(A ∩ B’) = n(A) – n(A ∩ B)
= 35 – 10
= 25

Question 7.
In a class of 200 students who appeared in certain examinations, 35 students faded in CET, 40 in NEET and 40 in JEE, 20 faded in CET and NEET, 17 in NEET and JEE, 15 in CET and JEE and 5 faded in ad three examinations. Find how many students
(i) did not fail in any examination.
(ii) faded in NEET or JEE entrance.
Solution:
Let A = set of students who failed in CET
B = set of students who failed in NEET
C = set of students who failed in JEE
X = set of all students
∴ n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40, n(A ∩ B) = 20, n(B ∩ C) = 17, n(A ∩ C) = 15, n(A ∩ B ∩ C) = 5

(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 35 + 40 + 40 – 20 – 17 – 15 + 5
= 68
∴ No. of students who did not fail in any exam = n(X) – n(A ∪ B ∪ C)
= 200 – 68
= 132

(ii) No. of students who failed in NEET or JEE entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63

Question 8.
From amongst 2000 Uterate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read
(i) at least one of the newspapers.
(ii) neither Marathi nor English newspaper.
(iii) only one of the newspapers.
Solution:
Let M = set of individuals who read Marathi newspapers
E = set of individuals who read English newspapers
X = set of all literate individuals
∴ n(X) = 2000,
n(M) = \(\frac{70}{100}\) × 2000 = 1400
n(E) = \(\frac{50}{100}\) × 2000 = 1000
n(M ∩ E) = \(\frac{32.5}{100}\) × 2000 = 650
(i) n(M ∪ E) = n(M) + n(E) – n(M ∩ E)
= 1400 + 1000 – 650
= 1750

No. of individuals who read at least one of the newspapers = n(M ∪ E) = 1750.

(ii) No. of individuals who read neither Marathi nor English newspaper = n(M’ ∩ E’)
= n(M ∪ E)’
= n(X) – n(M ∪ E)
= 2000 – 1750
= 250

(iii) No. of individuals who read only one of the newspapers = n(M ∩ E’) + n(M’ ∩ E)
= n(M ∪ E) – n(M ∩ E)
= 1750 – 650
= 1100

Question 9.
In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the total number of students in the hostel.
Solution:
Let T = set of students who take tea
C = set of students who take coffee
M = set of students who take milk
∴ n(T) = 25, n(C) = 20, n(M) = 15, n(T ∩ C) = 10, n(M ∩ C) = 8, n(T ∩ M) = 0, n(T ∩ M ∩ C) = 0

∴ Total number of students in the hostel = n(T ∪ C ∪ M)
= n(T) + n(C) + n(M) – n(T ∩ C) – n(M ∩ C) – n(T ∩ M) + n(T ∩ M ∩ C)
= 25 + 20 + 15 – 10 – 8 – 0 + 0
= 42

Question 10.
There are 260 persons with skin disorders. If 150 had been exposed to the chemical A, 74 to the chemical B, and 36 to both chemicals A and B, find the number of persons exposed to
(i) Chemical A but not Chemical B
(ii) Chemical B but not Chemical A
(iii) Chemical A or Chemical B
Solution:
Let A = set of persons exposed to chemical A
B = set of persons exposed to chemical B
X = set of all persons
∴ n(X) = 260, n(A) = 150, n(B) = 74, n(A ∩ B) = 36

(i) No. of persons exposed to chemical A but not to chemical B = n(A ∩ B’)
= n(A) – n(A ∩ B)
= 150 – 36
= 114

(ii) No. of persons exposed to chemical B but not to chemical A = n(A’ ∩ B)
= n(B) – n(A ∩ B)
= 74 – 36
= 38

(iii) No. of persons exposed to chemical A or chemical B = n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 150 + 74 – 36
= 188

Question 11.
Write down the power set of A = {1, 2, 3}.
Solution:
A = {1, 2, 3}
The power set of A is given by
P(A) = {{Φ}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

Question 12.
Write the following intervals in Set-Builder form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, ∞)
(iv) (-∞, 5]
(v) (2, 5]
(vi) [-3, 4)
Solution:
(i) (-3, 0) = {x / x ∈ R, -3 < x < 0}

(ii) [6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, ∞) = {x / x ∈ R, x > 6}

(iv) (-∞, 5] = {x / x ∈ R, x ≤ 5}

(v) (2, 5] = {x / x ∈ R, 2 < x ≤ 5}

(vi) [-3, 4) = {x / x ∈ R, -3 ≤ x < 4}

Question 13.
A college awarded 38 medals in volleyball, 15 in football, and 20 in basketball. The medals were awarded to a total of 58 players and only 3 players got medals in all three sports. How many received medals in exactly two of the three sports?
Solution:
Let A = Set of students who received medals in volleyball
B = Set of students who received medals in football
C = Set of students who received medals in basketball
n(A) = 38, n(B) = 15, n(C) = 20, n(A ∪ B ∪ C) = 58, n(A ∩ B ∩ C) = 3
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
58 = 38 + 15 + 20 – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + 3
∴ n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 ……(i)
Number of players who got exactly two medals = p + q + r
Here, s = n(A ∩ B ∩ C) = 3

n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 …..[From (i)]
∴ p + s + s + r + q + s = 18
∴ p + q + r + 3s = 18
∴ p + q + r + 3(3) = 18
∴ p + q + r = 18 – 9 = 9
∴ Number of players who received exactly two medals = 9.

Question 14.
Solve the following inequalities and write the solution set using interval notation.
(i) -9 < 2x + 7 ≤ 19
(ii) x² – x > 20
(iii) \(\frac{2 x}{x-4}\) ≤ 5
(iv) 6x² + 1 ≤ 5x
Solution:
(i) -9 < 2x + 7 ≤ 19
∴ -16 < 2x ≤ 12
∴ -8< x ≤ 6
∴ x ∈ (-8, 6]

(ii) x² – x > 20
∴ x² – x – 20 > 0
∴ x² – 5x + 4x – 20 > 0
∴ (x – 5) (x + 4) > 0
∴ either x – 5 > 0 and x + 4 > 0 or x – 5 < 0 and x + 4 < 0

Case I: x – 5 > 0 and x + 4 > 0
∴ x > 5 and x > -4
∴ x > 5 ….(i)

Case II:
x – 5 < 0 and x + 4 < 0
∴ x < 5 and x < -4
∴ x < -4 …..(ii)
From (i) and (ii), we get
x ∈ (-∞, – 4) ∪ (5, ∞)

(iii) \(\frac{2 x}{x-4}\) ≤ 5
∴ \(\frac{2 x}{x-4}\) – 5 ≤ 0
∴ \(\frac{2 x-5 x+20}{x-4}\) ≤ 0
∴ \(\frac{20-3 x}{x-4}\) ≤ 0
When \(\frac{a}{b}\) ≤ 0,
a ≥ 0 and b < 0 or a ≤ 0 and b > 0
∴ either 20 – 3x ≥ 0 and x – 4 < 0 or 20 – 3x ≤ 0 and x – 4 > 0
Case I:
20 – 3x ≥ 0 and x – 4 < 0
∴ x ≤ \(\frac{20}{3}\) and x < 4
∴ x < 4 ……(I)

Case II: 20 – 3x ≤ 0 and x – 4 > 0
∴ x ≥ \(\frac{20}{3}\) and x > 4
∴ x ≥ \(\frac{20}{3}\) ……(ii)
From (i) and (ii), we get
x ∈ (-∞, 4) ∪ [\(\frac{20}{3}\), ∞)

(iv) 6x² + 1 ≤ 5x
6x² – 5x + 1 ≤ 0
6x² – 3x – 2x + 1 ≤ 0
(3x – 1) (2x – 1) ≤ 0
either 3x – 1 ≤ 0 and 2x – 1 ≥ 0 or 3x – 1 ≥ 0 and 2x – 1 ≤ 0
Case I:
3x – 1 ≤ 0 and 2x – 1 ≥ 0
∴ x ≤ \(\frac{1}{3}\) and x ≥ \(\frac{1}{2}\), which is not possible.

Case II:
3x – 1 ≥ 0 and 2x – 1 ≤ 0
∴ x ≥ \(\frac{1}{3}\) and x ≤ \(\frac{1}{2}\)
∴ x ∈ [\(\frac{1}{3}\), \(\frac{1}{2}\)]

Question 15.
If A = (-7, 3], B = [2, 6] and C = [4, 9], then find
(i) A ∪ B
(ii) B ∪ C
(iii) A ∪ C
(iv) A ∩ B
(v) B ∩ C
(vi) A ∩ C
(vii) A’ ∩ B
(viii) B’ ∩ C’
(ix) B – C
(x) A – B
Solution:
A = (-7, 3], B = [2, 6], C = [4, 9]
(i) A ∪ B = (-7, 6]

(ii) B ∪ C = [2, 9]

(iii) A ∪ C = (-7, 3] ∪ [4, 9]

(iv) A ∩ B = [2, 3]

(v) B ∩ C = [4, 6]

(vi) A ∩ C = { }

(vii) A’ = (-∞, – 7] ∪ (3, ∞)
∴ A’ ∩ B = (3, 6]

(viii) B’ = (-∞, 2) ∪ (6, ∞)
C’ = (-∞, 4) ∪ (9, ∞)
∴ B’ ∩ C’ = (-∞, 2) ∪ (9, ∞)

(ix) B – C = [2, 4)

(x) A – B = (-7, 2)