Matrices Class 12 Maths 1 Exercise 2.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
Find the co-factors of the elements of the following matrices
(i) \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Here, a11 = -11, M11 = 4
∴ A11 = (-1)1+1(4) = 4
a12 = 2, M12 = -3
∴ A12 = (-1)1+2(- 3) = 3
a21 = – 3, M21 = -2
∴ A21 = (- 1)2+1(2) = -2
a22 = 4, M22 = -1
∴ A22 = (-1)2+2(-1) = -1.

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
The co-factor of aij is given by Aij = (-1)i+jMij
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 1
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 2

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

Question 2.
Find the matrix of co-factors for the following matrices
(i) \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Here, a11 = 1, M11 = -1
∴ A11 = (-1)1+1(-1) = -1
a12 = 3, M12 = 4
∴ A12 = (-1)1+2(4) = -4
a21 = 4, M21 = 3
∴ A21 = (-1)2+1(3) = -3
a22 = -1, M22 = 1
∴ A22 = (-1)2+1(1) = 1
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left(\begin{array}{rr}
-1 & -4 \\
-3 & 1
\end{array}\right)\)

(ii) \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 21
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 22
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 23
A11 = -14, A12 = -10, A13 = -6,
A21 = 6, A22 = -5, A23 = -3,
A31 = -2, A32 = -7, A33 = 1.
∴ the co-factor matrix
= \(\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]\) = \(\left[\begin{array}{rrr}
-14 & -10 & -6 \\
6 & -5 & -3 \\
-2 & -7 & 1
\end{array}\right]\)

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

Question 3.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Here, a11 = 2, M11= 5
∴ A11 = (-1)1+1(5) = 5
a12 = -3, M12 = 3
∴ A12 = (-1)1+2(3) = -3
a21 = 3, M21 = -3
∴ A A21 = (-1)2+1(-3) = 3
a22 = 5, M22 = 2
∴ A22 = (-1)2+1 = 2
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
5 & -3 \\
3 & 2
\end{array}\right]\)
∴ adj A = \(\left(\begin{array}{rr}
5 & 3 \\
-3 & 2
\end{array}\right)\)

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

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 1
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 2
A11 = -3, A12 = -12, A13 = 6,
A21 = -1, A22 = 3, A23 = 2,
A31 = -11, A32 = -9, A33 = 1
∴ the co-factor matrix = \(\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\
\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\
\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-3 & -12 & 6 \\
-1 & 3 & 2 \\
-11 & -9 & 1
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{rrr}
-3 & -1 & -11 \\
-12 & 3 & -9 \\
6 & 2 & 1
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\), verify that A (adj A) = (adj A) A = | A | ∙ I
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 3
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 4
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 5
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 6
From (1), (2) and (3), we get,
A(adj A) = (adj A)A = |A|∙I.
Note: This relation is valid for any non-singular matrix A.

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

Question 5.
Find the inverse of the following matrices by the adjoint method
(i) \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right|\) = -2 + 15 = 13 ≠ 0
∴ A-1 exists.
First we have to find the co-factor matrix
= [Aij]2×2, where Aij = (-1)i+jMij
Now, A11 = (-1)1+1M11 = 2
A12 = (-1)1+2M12 = -(-3) = 3
A21 = (-1)2+1M21 = -5
A22 = (-1)2+2M22 = -1
Hence, the co-factor matrix
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 7

(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
|A| = \(\) = 6 + 8 = 14 ≠ 0
∴ A-1 exist
First we have to find the co-factor matrix
= [Aij] 2×2 where Aij = (-1)i+jMij
Now, A11 = (-1)1+1M11 = 3
A12 = (-1)1+2M = -4
A21 = (-2)2+1M21 = (-2) = 2
A22 = (-1)2+2M22 = 2
Hence the co-factor matrix
= \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
∴ A-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A) = \(\frac{1}{14}\left(\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right)\)

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

(iii) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 8
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 9
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 10
∴ A-1 = \(\frac{1}{3}\left[\begin{array}{rrr}
3 & 0 & 0 \\
-3 & 1 & 0 \\
9 & 2 & -3
\end{array}\right]\)

(iv) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A-1 exists.
First we have to find the co-factor matrix
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 24
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 25
∴ A-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A)
= \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)
∴ A-1 = \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)

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

Question 6.
Find the inverse of the following matrices
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 11
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 12
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 13

(ii) \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 14
∴ A-1 = \(\left(\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right)\)

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

(iii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 15
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 16
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 17

(iv) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 18
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 19
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 20

Class 12 Maharashtra State Board Maths Solution 

Matrices Class 12 Maths 1 Exercise 2.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Apply the given elementary transformation on each of the following matrices.
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\), R1 ↔ R2
Solution:
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\)
By R1 ↔ R2, we get,
A ~ \(\left[\begin{array}{rr}
-1 & 3 \\
1 & 0
\end{array}\right]\)

Question 2.
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\), R1 → R1 → R2
Solution:
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\),
R1 → R1 → R2 gives,
B ~ \(\left[\begin{array}{rrr}
-1 & -6 & -1 \\
2 & 5 & 4
\end{array}\right]\)

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

Question 3.
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\), C1 ↔ C2; B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\), R1 ↔ R2. What do you observe?
Solution:
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\)
By C1 ↔ C2, we get,
A ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(1)
B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\)
By R1 ↔ R2, we get,
B ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\), 2C2
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\), -3R1
Find the addition of the two new matrices.
Solution:
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\)
By 2C2, we get,
A ~ \(\left[\begin{array}{rrr}
1 & 4 & -1 \\
0 & 2 & 3
\end{array}\right]\)
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\)
By -3R1, we get,
B ~ \(\left[\begin{array}{rrr}
-3 & 0 & -6 \\
2 & 4 & 5
\end{array}\right]\)
Now, addition of the two new matrices
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.1 1

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

Question 5.
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\), 3R3 and then C3 + 2C2.
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By 3R3, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
9 & 9 & 3
\end{array}\right]\)
By C3 + 2C2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
9 & 9 & 3+2(9)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)

Question 6.
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\), C3 + 2C2 and then 3R3. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\)
By C3 + 2C2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
3 & 3 & 1+2(3)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
3 & 3 & 7
\end{array}\right)\)
By 3R3, we get
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)
We conclude from Ex. 5 and Ex. 6 that the matrix remains same by interchanging the order of the elementary transformations. Hence, the transformations are commutative.

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

Question 7.
Use suitable transformation on \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) into an upper triangular matrix.
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
By R2 – 3R1, we get,
A ~ \(\left[\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right]\)
This is an upper triangular matrix.

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

Question 8.
Convert \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) into an identity matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
By R2 – 2R1, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]\)
By \(\left(\frac{1}{5}\right)\)R2, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]\)
By R1 + R2, we get,
A ~ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
This is an identity matrix.

Question 9.
Transform \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangular matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\)
By R2 – 2R1 and R3 – 3R1, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 5 & -2
\end{array}\right]\)
By R3 – \(\left(\frac{5}{3}\right)\)R2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 0 & -\frac{1}{3}
\end{array}\right)\)
This is an upper triangular matrix.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Miscellaneous Exercise 1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

Question 1.
Select and write the correct answer from the given alternatives in each of the following questions:
i) If p ∧ q is false and p ∨ q is true, the ________ is not true.
(A) p ∨ q
(B) p ↔ q
(C) ~p ∨ ~q
(D) q ∨ ~p
Solution:
(b) p ↔ q.

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F

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

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

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

Question 2.
Which of the following sentences are statements in logic? Justify. Write down the truth
value of the statements :
(i) 4! = 24.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos2θ – sin2θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

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

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

Question 3.
Write the truth values of the following statements :
(i) \(\sqrt {5}\) is an irrational but \(3\sqrt {5}\) is a complex number.
Solution:
Let p : \(\sqrt {5}\) is an irrational.
q : \(3\sqrt {5}\) is a complex number.
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) Ɐ n ∈ N, n2 + n is even number while n2 – n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n2 + n is an even number.
q : Ɐ n ∈ N, n2 – n is an odd number.
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].

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

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ N, satisfy n + 5 > 10).

(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
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 T. … [F ∨ T ≡ T].

(v) In ∆ ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

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

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ N do not satisfy n + 6 > 8).

Question 4.
If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement :
(i) Ǝ x ∈ A such that x + 8 = 15.
Solution:
True

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

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

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

Question 5.
Write the negations of the following :
(i) Ɐ n ∈ A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ǝ n ∈ A, such that n + 7 ≤ 6.
OR Ǝ n ∈ A, such that n + 7 ≯ 6.

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

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

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 1

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 2

(iii) ~(~p ∧ ~q) ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 3

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

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 4

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 5

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p → q) ∧ ~q)] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 6
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q)] → ~p is a tautology.

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

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 7
All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 8
All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 9
All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

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

(v) [(p ∧ (p → q)] → q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 10
All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 11
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 12
The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

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

(viii) (p → q) ∨ (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 13
All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 14
Since p ∨ q and p ∧ q both are T, from the table the truth values of both p and q are T.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 15
Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

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

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 16
Since the truth values of (p ∧ q) is F and (p ∧ q) → q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.

Question 9.
Using truth tables prove the following logical equivalences :
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 17
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

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

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 18
The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

Question 10.
Using rules in logic, prove the following :
(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
By the rules of negation of biconditional,
~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q
∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q … (Negation of disjunction)
≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

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

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

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

Question 11.
Using the rules in logic, write the negations of the following :
(i) (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 disjunction)
≡ {~ p ∧ ~q) ∨ (~q ∧ r) … (Negation of negation)
≡ (~q ∧ ~p) ∨ (~q ∧ r) … (Commutative law)
≡ (~ q) ∧ (~ p ∨ r) … (Distributive Law)

(ii) 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 disjunction)

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

(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)

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

Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 19
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is :
(p ∧ q) ∨ (~p) ∨ (p ∧ ~q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 21

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

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 20
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 22

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 23
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p: the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the given circuit in symbolic form is :
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~q)
Using the laws of logic, we have,
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~ q)
= (p ∧ ~q) ∨ [(~p ∧ q) ∨ (~p ∧ ~q) …(By Complement Law)
= (p ∧ ~q) ∨ [~p ∧ (q ∨ ~q)} (By Distributive Law)
= (p ∧ ~q) ∨ (~p ∧ T) …(By Complement Law)
= (p ∧ ~q) ∨ ~ p …(By Identity Law)
= (p ∨ ~p) ∧ (~q ∨ ~p) …(By Distributive Law)
= ~q ∨ ~p …(By Identity Law)
= ~p ∨ ~p …(By Commutative Law)
Hence, the simplified circuit for the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 24

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

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 25
Solution:
(ii) Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
s : the switch S4 is closed
t : the switch S5 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open
~ s : the switch S4‘ is closed or the switch S4 is open
~ t : the switch S5‘ is closed or the switch S5 is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 26

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 27
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
(A) The symbolic form of the given switching circuits are
p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) respectively.
By Distributive Law, p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Hence, the given switching circuits are logically equivalent.

(B)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 28
Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.

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

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 29
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed, or the switch S1 is open
~q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form Of the given circuit is :
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
≡ (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) y (p ∧ ~q ∧ r) …(By Commutative Law)
≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Complement Law)
≡ F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [(~p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) … (By Distributive Law)
≡ [T ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) = (q ∧ r) ∨ (p ∧ ~q ∧ r) …(By Complement Law)
≡ (q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [q ∨ (p ∧ ~q)] ∧ r … (By Distributive Law)
≡ [q ∨ p) ∧ ((q ∨ ~q)] ∧ r … (By Distributive Law)
≡ [(q ∨ p) ∧ T] ∧ r …(By Complement Law)
≡ (q ∨ p) ∧ r … (By Identity Law)
≡ (p ∨ q) ∧ r …(By Commutative Law)
∴ the alternative arrangement of the new circuit with minimum switches is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 30

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

Question 16.
Simplify the following so that the new circuit circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 31
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given switching circuit is :
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
Using the laws of logic, we have,
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
≡ (~p ∨ q ∨ p ∨ ~q) ∨ (p ∨ q)
≡ [(~p ∨ p) ∨ (q ∨ ~q)] ∨ (p ∨ q) … (By Commutative Law)
≡ (T ∨ T) ∨ (p ∨ q) … (By Complement Law)
≡ T ∨ (p ∨ q) … (By Identity Law)
≡ T … (By Identity Law)
∴ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
∴ the simplified form of given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 32

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

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 33
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open.
Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 34
From the table, the’ final column’ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S1.
the simplified form of the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 35

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.5 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.5 Questions And Answers Maharashtra Board

Question 1.
Express the following circuits in the symbolic form of logic and write the input-output table.
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 1
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed or the switch S1is open
~q : the switch S2‘ is closed or the switch S2 is open
~r : the switch S3‘ is closed or the switch S3 is open
l : the lamp L is on
(i) The symbolic form of the given circuit is : p ∨ (q ∧ r) = l
l is generally dropped and it can be expressed as : p ∨ (q ∧ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 7

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 2
Solution:
The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 8

(iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 3
Solution:
The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 9

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

(iv)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 4
Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 10

(v)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 5
Solution:
The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 11

(vi)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 6
Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 12

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

Question 2.
Construct the switching circuit of the following :
(i) (~p∧ q) ∨ (p∧ ~r)
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open.
Then the switching circuits corresponding to the given statement patterns are :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 13

(ii) (p∧ q) ∨ [~p ∧ (~q ∨ p ∨ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 14

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

(iii) [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 15

(iv) (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 16

(v) p ∨ (~p ) ∨ (~q) ∨ (p ∧ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 17

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

(vi) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 18

Question 3.
Give an alternative equivalent simple circuits for the following circuits :
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 19
Solution:
(i) Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch Si is open Then the symbolic form of the given circuit is :
p ∧ (~p ∨ q).
Using the laws of logic, we have,
p ∧ (~p ∨ q)
= (p ∧ ~ p) ∨ (p ∧ q) …(By Distributive Law)
= F ∨ (p ∧ q) … (By Complement Law)
= p ∧ q… (By Identity Law)
Hence, the alternative equivalent simple circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 20

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

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 21
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~q : the switch S2‘ is closed or the switch S2 is open
~r : the switch S3‘ is closed or the switch S3 is open.
Then the symbolic form of the given circuit is :
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p).
Using the laws of logic, we have
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p)
≡ [p ∧ (q ∨ r)] ∨ [ ~(r ∨ q) ∧ p] …. (By De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] … (By Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)) … (By Distributive Law)
≡ p ∧ T … (By Complement Law)
≡ p … (By Identity Law)
Hence, the alternative equivalent simple circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 22

Question 4.
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 23
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is :
(p ∨ ~q) ∨ (~p ∧ q)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 24
Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

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

ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 25
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~p : the switch S1 is closed or the switch S1 is open.
~q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is : p ∨ (~p ∧ ~q) ∨ (p ∧ q)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 26
Since the final column contains ‘0’ when p is 0 and q is ‘1’, otherwise it contains ‘1′.
Hence, the lamp will not glow when S1 is OFF and S2 is ON, otherwise the lamp will glow.

iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 27
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~q : the switch S2‘ is closed or the switch S2 is open
~r: the switch S3‘ is closed or the switch S3 is open.
Then the symbolic form of the given circuit is : [p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 28
From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S1 is ‘ON’.

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

Question 5.
Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(i) p ∨ (q ∧ ~ q)
Solution:
Using the laws of logic, we have, p ∨ (q ∧ ~q)
≡ p ∨ F … (By Complement Law)
≡ p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S1 is closed
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 29

(ii) (~p ∧ q) ∨ (~p ∧ ~q) ∨ (p ∧ ~q)]
Solution:
Using the laws of logic, we have,
(~p ∧ q) ∨ (~p ∨ ~q) ∨ (p ∧ ~q)
≡ [~p ∧ (q ∨ ~q)] ∨ (p ∧ ~ q)… (By Distributive Law)
≡ (~p ∧ T) ∨ (p ∧ ~q) … (By Complement Law)
≡ ~p ∨ (p ∧ ~q) … (By Identity Law)
≡ (~p ∨ p) ∧ (~p ∧~q) … (By Distributive Law)
≡ T ∧ (~p ∧ ~q) … (By Complement Law)
≡ ~p ∨ ~q … (By Identity Law)
Hence, the simple logical expression of the given expression is ~ p ∨ ~q.
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open,
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 30

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

(iii) [p (∨ (~q) ∨ ~r)] ∧ (p ∨ (q ∧ r)
Solution:
Using the laws of logic, we have,
[p ∨ (~ (q) ∨ (~r)] ∧ [p ∨ (q ∧ r)]
= [p ∨ { ~(q ∧ r)}] ∧ [p ∨ (q ∧ r)] … (By De Morgan’s Law)
= p ∨ [~(q ∧ r) ∧ (q ∧ r) ] … (By Distributive Law)
= p ∨ F … (By Complement Law)
= p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S1 is closed
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 31

(iv) (p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ r)
Question is Modified
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r)∨ (p ∧ q ∧ r)
Solution:
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r)
= (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Commutative Law)
= (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Complement Law)
= F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~ p ∨ p) ∧ (q ∧ r) … (By Distributive Law)
= T ∧ (q ∧ r) … (By Complement Law)
= q ∧ r … (By Identity Law)
Hence, the simple logical expression of the given expression is q ∧ r.
Let q : the switch S2 is closed
r : the switch S3 is closed.
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 32

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.4 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.4 Questions And Answers Maharashtra Board

Question 1.
Using rules of negation write the negations of the following with justification.
(i) ~q → p
Solution:
The negation of ~q → p is
~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)

(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)

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

(iv) (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 disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)

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

(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)

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

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

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

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

Question 2.
Rewrite the following statements without using if .. then.
(i) If a man is a judge then he is honest.
Solution:
Since p → ≡ ~p ∨ q, the given statements can be written as :
A man is not a judge or he is honest.

(ii) It 2 is a rational number then \(\sqrt {2}\) is irrational number.
Solution:
2 is not a rational number or \(\sqrt {2}\) is irrational number.

(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).

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

Question 3.
Without using truth table prove that :
(i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p ↔ q) ∧ (q ↔ p) … (Biconditional Law)
≡ (~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) ∨ F] ∨ [F ∨ (q ∧ p)] … (ComplementLaw)
≡ (~ p ∧ ~ q) ∨ (q ∧ p) … (Identity Law)
≡ (~ p ∧ ~ q) ∨ (p ∧ q) … (Commutative Law)
≡ (p ∧ q) ∨ (~p ∧ ~q) … (Commutative Law)
≡ RHS.

(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.

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

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

(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.3 Questions And Answers Maharashtra Board

Question 1.
If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.
(i) Ǝ x ∈ A such that x – 8 = 1
Solution:
Clearly x = 9 ∈ A satisfies x – 8 = 1. So the given statement is true, hence its truth value is T.

(ii) Ɐ x ∈ A, x2 + x is an even number
Solution:
For each x ∈ A, x2 + x is an even number. So the given statement is true, hence its truth value is T.

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

(iii) Ǝ x ∈ A such that x2 < 0
Solution:
There is no x ∈ A which satisfies x2 < 0. So the given statement is false, hence its truth value is F.

(iv) Ɐ x ∈ A, x is an even number
Solution:
x = 3 ∈ A, x = 5 ∈ A, x = 7 ∈ A, x = 9 ∈ A, x = 11 ∈ A do not satisfy x is an even number. So the given statement is false, hence its truth value is F.

(v) Ǝ x ∈ A such that 3x + 8 > 40
Solution:
Clearly x = 11 ∈ A and x = 12 ∈ A satisfies 3x + 8 > 40. So the given statement is true, hence its truth value is T.

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

(vi) Ɐ x ∈ A, 2x + 9 > 14
Solution:
For each x ∈ A, 2x + 9 > 14. So the given statement is true, hence its truth value is T.

Question 2.
Write the duals of each of the following.
(i) p ∨ (q ∧ r)
Solution:
The duals of the given statement patterns are :
p ∧ (q ∨ r)

(ii) p ∧ (q ∧ r)
Solution:
p ∨ (q ∨ r)

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

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

(iv) p ∧ ~q
Solution:
p ∨ ~q

(v) (~p ∨ q) ∧ (~r ∧ s)
Solution:
(~p ∧ q) ∨ (~r ∨ s)

(vi) ~p ∧ (~q ∧ (p ∨ q) ∧ ~r)
Solution:
~p ∨ (~q ∨ (p ∧ q) ∨ ~r)

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

(vii) [~(p ∨ q)] ∧ [p ∨ ~(q ∧ ~s)]
Solution:
[ ~(p ∧ q)] ∨ [p ∧ ~(q ∨ ~s)]

(viii) c ∨ {p ∧ (q ∨ r)}
Solution:
t ∧ {p ∧ (q Ar)}

(ix) ~p ∨ (q ∧ r) ∧ t
Solution:
~p ∧ (q ∨ r) ∨ c

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

(x) (p ∨ q) ∨ c
Solution:
(p ∧ q) ∧ t

Question 3.
Write the negations of the following.
(i) x + 8 > 11 or y – 3 = 6
Solution:
Let p : x + 8 > 11, q : y — 3 = 6.
Then the symbolic form of the given statement is p ∨ q.
Since ~(p ∨ q) ≡ ~p ∧ ~q, the negation of given statement is :
‘x + 8 > 11 and y – 3 ≠ 6’ OR
‘x + 8 ≮ 11 and y – 3 ≠ 6’

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

(ii) 11 < 15 and 25 > 20
Solution:
Let p: 11 < 15, q : 25 > 20.
Then the symbolic form of the given statement is p ∧ q.
Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of given statement is :
’11 ≮ 15 or 25 > 20.’ OR
’11 ≯ 15 or 25 ≮ 20.’

(iii) Qudrilateral is a square if and only if it is a rhombus.
Solution:
Let p : Quadrilateral is a square.
q : It is a rhombus.
Then the symbolic form of the given statement is p ↔ q.
Since ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of given statement is :
‘ Quadrilateral is a square but it is not a rhombus or quadrilateral is a rhombus but it is not a square.’

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

(iv) It is cold and raining.
Solution:
Let p : It is cold.
q : It is raining.
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 not cold or not raining.’

(v) If it is raining then we will go and play football.
Solution:
Let p : It is raining.
q : We will go.
r : We play football.
Then the symbolic form of the given statement is p → (q ∧ r).
Since ~[p → (q ∧ r)] ≡ p ∧ ~(q ∧ r) ≡ p ∧ (q ∨ ~r), the negation of the given statement is :
‘It is raining and we will not go or not play football.’

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

(vi) \(\sqrt {2}\) is a rational number.
Solution:
Let p : \(\sqrt {2}\) is a rational number.
The negation of the given statement is
‘ ~p : \(\sqrt {2}\) is not a rational number.’

(vii) All natural numbers are whole numers.
Solution:
The negation of the given statement is :
‘Some natural numbers are not whole numbers.’

(viii) Ɐ n ∈ N, n2 + n + 2 is divisible by 4.
Solution:
The negation of the given statement is :
‘Ǝ n ∈ N, such that n2 + n + 2 is not divisible by 4.’

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

(ix) Ǝ x ∈ N such that x – 17 < 20
Solution:
The negation of the given statement is :
‘Ɐ x ∈ N, x – 17 ≯ 20.’

Question 4.
Write converse, inverse and contrapositive of the following statements.
(i) If x < y then x2 < y2 (x, y ∈ R)
Solution:
Let p : x < y, q : x2 < y2.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p → q.
i.e. If x2 < y2, then x < y.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If x ≯ y, then x2 ≯ y2. OR
If x ≮ y, then x2 ≮ y2.
Contrapositive : ~q → p is the contrapositive of
p → q i.e. If x2 ≯ y2, then x ≯ y. OR
If x2 ≮ y2, then x ≮ y.

(ii) A family becomes literate if the woman in it is literate.
Solution:
Let p : The woman in the family is literate.
q : A family become literate.
Then the symbolic form of the given statement is p → q
Converse : q → p is the converse of p → q.
i.e. If a family become literate, then the woman in it is literate.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the woman in the family is not literate, then the family does not become literate.
Contrapositive : ~q → ~p is the contrapositive of p → q. i e. If a family does not become literate, then the woman in it is not literate.

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

(iii) If surface area decreases then pressure increases.
Solution:
Let p : The surface area decreases.
q : The pressure increases.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p→ q.
i.e. If the pressure increases, then the surface area decreases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the surface area does not decrease, then the pressure does not increase.
Contrapositive : ~q → ~p is the contrapositive of p → q.
i.e. If the pressure does not increase, then the surface area does not decrease.

(iv) If voltage increases then current decreases.
Solution:
Let p : Voltage increases.
q : Current decreases.
Then the symbolic form of the given statement is p → q.
Converse : q →p is the converse of p → q.
i.e. If current decreases, then voltage increases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If voltage does not increase, then current does not decrease.
Contrapositive : ~q → ~p, is the contrapositive of p → q.
i.e. If current does not decrease, then voltage doesnot increase.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.2 Questions And Answers Maharashtra Board

Question 1.
Construct the truth table for each of the following statement patterns:
(i) [(p → q) ∧ q] → p
Solution :
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 1

(ii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 2

(iii) (p ∧ q) ↔ (q ∨ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 3

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

(iv) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(v) ~p ∧ [(p ∨ ~q ) ∧ q]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 5

(vi) (~p → ~q) ∧ (~q → ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 6

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

(vii) (q → p) ∨ (~p ↔ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 7

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 8

(ix) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(x) (p ∨ ~q) → (r ∧ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 9

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

Question 2.
Using truth tables prove the following logical equivalences.
(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 10
The entries in the columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 11
The entries in the columns 3 and 7 are identical.
∴ ~(p ∨ q) ∧ (~p ∧ q) = ~p.

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 12
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

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

(iv) p → (q → p) ≡ ~p → (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 13
The entries in the columns 4 and 7 are identical.
∴ p → (q → p) ≡ ~p → (p → q).

(v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 14
The entries in the columns 5 and 8 are identical.
∴ (p ∨ q ) → r ≡ (p → r) ∧ (q → r).

(vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 15
The entries in the columns 5 and 8 are identical.
∴ p → (q ∧ r) ≡ (p → q) ∧ (p → r).

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

(vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 16
The entries in the columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

(viii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 17
The entries in the columns 3 and 7 are identical.
∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r.

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 18
The entries in the columns 6 and 9 are identical.
∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

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

Question 3.
Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.
(i) (p ∧ q) → (q ∨ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 19
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → (q ∨ p) is a tautology.

(ii) (p → q) ↔ (~p ∨ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 20
All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~p ∨ q) p is a tautology.

(iii) [~(~p ∧ ~q)] ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 21
The entries in the last column of the above truth table are neither all T nor all F.
∴ [~(~p ∧ ~q)] ∨ q is a contingency.

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

(iv) [(p → q) ∧ q)] → p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 22
The entries in the last column of the above truth table are neither all T nor all F.
∴ [(p → q) ∧ q)] → p is a contingency

(v) [(p → q) ∧ ~q] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 23
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q] → ~p is a tautology.

(vi) (p ↔ q) ∧ (p → ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 24
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.

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

(vii) ~(~q ∧ p) ∧ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 25
The entries in the last column of the above truth table are neither all T nor all F.
∴ ~(~q ∧ p) ∧ q is a contingency.

(viii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 26
All the entries in the last column of the above truth table are F.
∴ (p ∧ ~q) ↔ (p → q) is a contradiction.

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

(ix) (~p → q) ∧ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 27
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.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 28
All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
State which of the following sentences are statements. Justify your answer. In case of the statement, write down the truth value :
(i) 5 + 4 = 13.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(ii) x – 3 = 14.
Solution:
It is an open sentence, hence it is not a statement.

(iii) Close the door.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) Zero is a complex number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

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

(v) Please get me breakfast.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) Congruent triangles are also similar.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vii) x2 = x.
Solution:
It is an open sentence, hence it is not a statement,

(viii) A quadratic equation cannot have more than two roots.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

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

(ix) Do you like Mathematics ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(x) The sun sets in the west.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xi) All real numbers are whole numbers.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(xii) Can you speak in Marathi ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(xiii) x2 – 6x – 7 = 0, when x = 7.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

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

(xiv) The sum of cuberoots of unity is zero.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xv) It rains heavily.
Solution :
It is an open sentence, hence it is not a statement.

Question 2.
Write the following compound statements symbolically:
(i) Nagpur is in Maharashtra and Chennai is in Tamil Nadu.
Solution:
Let p : Nagpur is in Maharashtra.
q : Chennai is in Tamil Nadu.
Then the symbolic form of the given statement is P∧q.

(ii) Triangle is equilateral or isosceles,
Solution:
Let p : Triangle is equilateral.
q : Triangle is isosceles.
Then the symbolic form of the given statement is P∨q.

(iii) The angle is right angle if and only if it is of measure 90°.
Solution:
Let p : The angle is right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p↔q

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

(iv) Angle is neither acute nor obtuse.
Solution:
Let p : Angle is acute.
q : Angle is obtuse.
Then the symbolic form of the given statement is
~p ∧ ~q.

(v) If ∆ ABC is right angled at B, then m∠A + m∠C = 90°.
Solution:
Let p : ∆ ABC is right angled at B.
q : m∠A + m∠C = 90°.
Then the symbolic form of the given statement is p → q

(vi) Hima Das wins gold medal if and only if she runs fast.
Solution:
Let p : Hima Das wins gold medal
q : She runs fast.
Then the symbolic form of the given statement is p ↔ q.

(vii) x is not irrational number but it is a square of an integer.
Solution:
Let p : x is not irrational number
q : It is a square of an integer
Then the symbolic form of the given statement is p ∧ q
Note : If p : x is irrational number, then the symbolic form of the given statement is ~p ∧ q.

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

Question 3.
Write the truth values of the following :
(i) 4 is odd or 1 is prime.
Solution:
Let p : 4 is odd.
q : 1 is prime.
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 v q is F. … [F ∨ F = F]

(ii) 64 is a perfect square and 46 is a prime number.
Solution:
Let p : 64 is a perfect square.
q : 46 is a prime number.
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) 5 is a prime number and 7 divides 94.
Solution:
Let p : 5 is a prime number.
q : 7 divides 94.
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]

(iv) It is not true that 5 – 3i is a real number.
Solution:
Let p : 5 – 3i is a real number.
Then the symbolic form of the given statement is ~ p.
The truth values of p is F.
∴ the truth values of ~ p is T. … [~ F ≡ T]

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

(v) If 3 × 5 = 8, then 3 + 5 = 15.
Solution:
Let p : 3 × 5 = 8.
q : 3 + 5 = 15.
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]

(vi) Milk is white if and only if sky is blue.
Solution:
Let p : Milk is white.
q : Sky is blue
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]

(vii) 24 is a composite number or 17 is a prime number.
Solution :
Let p : 24 is a composite number.
q : 17 is a prime 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]

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

Question 4.
If the statements p, q are true statements and r, s are false statements, then determine the truth values of the following:
(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 ≡ T
Hence the truth value of the given statement is true.

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

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

(iv) (p → q) ∧ (~ r)
Solution:
(p → q) ∧ (~ r) ≡ (T → T) ∧ (~ F)
≡ T ∧ T ≡ T
Hence the truth value of the given statement is true.

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

(v) (~r ↔ p) → (~q)
Solution:
(~r ↔ p) → (~q) ≡ (~F ↔ T) → (~T)
≡ (T ↔ T) → F
≡ T → F ≡ F
Hence the truth value of the given statement is false.

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

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

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

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

Question 5.
Write the negations of the following :
(i) Tirupati is in Andhra Pradesh.
Solution:
The negations of the given statements are :
Tirupati is not in Andhra Pradesh.

(ii) 3 is not a root of the equation x2 + 3x – 18 = 0.
Solution:
3 is a root of the equation x2 + 3x – 18 = 0.

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

(iii) \(\sqrt {2}\) is a rational number.
Solution:
\(\sqrt {2}\) is not a rational number.

(iv) Polygon ABCDE is a pentagon.
Solution:
Polygon ABCDE is not a pentagon.

(v) 7 + 3 > 5.
Solution :
7 + 3 > 5.

Class 12 Maharashtra State Board Maths Solution 

Differentiation Class 11 Maths 2 Miscellaneous Exercise 9 Solutions Maharashtra Board

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

11th Maths Part 2 Differentiation Miscellaneous Exercise 9 Questions And Answers Maharashtra Board

(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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q1

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q2

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

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q3

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q4

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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q7

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q8

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

(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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1.1
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.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q2
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

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

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.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3.1
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.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4.1

Question 5.
Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5.1
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.

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

Question 6.
Test whether the function
f(x) = 2x – 3, for x ≥ 2
= x – 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q6

Question 7.
Test whether the function
f(x) = x2 + 1, for x ≥ 2
= 2x + 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7.1

Question 8.
Test whether the function
f(x) = 5x – 3x2, for x ≥ 1
= 3 – x, for x < 1
is differentiable at x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q8
Here, Lf'(1) = Rf'(1)
∴ f(x) is differentiable at x = 1.

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

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q9

Question 10.
If y = \(\frac{\mathbf{e}^{x}}{\sqrt{x}}\), find \(\frac{d y}{d x}\) when x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q10

Class 11 Maharashtra State Board Maths Solution 

Differentiation Class 11 Maths 2 Exercise 9.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

11th Maths Part 2 Differentiation Exercise 9.2 Questions And Answers Maharashtra Board

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q1

Question 2.
y = √x + tan x – x3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q2

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q3

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4.1

Question 5.
y = 7x + x7 – \(\frac{2}{3}\) x√x – log x + 77
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q5

Question 6.
y = 3 cot x – 5ex + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q6

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x5 tan x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q1

Question 2.
y = x3 log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 3.
y = (x2 + 2)2 sin x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q3

Question 4.
y = ex log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q4

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q5

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q6

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x2√x + x4 log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q1.1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 2.
y = ex sec x – \(x^{\frac{5}{3}}\) log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q2

Question 3.
y = x4 + x√x cos x – x2 ex
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q3

Question 4.
y = (x3 – 2) tan x – x cos x + 7x . x7
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q4

Question 5.
y = sin x log x + ex cos x – ex √x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q5

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 6.
y = ex tan x + cos x log x – √x 5x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q6
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q6.1

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q1

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q2
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q2.1

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q3

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q4

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q5

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q6

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax2 + bx + c …..(i)
∴ f(0) = a(0)2 + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x2 – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x2 – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)2 – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 V Q2
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 V Q2.1
Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = ex . tan x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q1.1

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q2.1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 3.
y = (3x2 + 5) cos x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q3.1

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q4.1

Class 11 Maharashtra State Board Maths Solution