Category: Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.2

Question 1.
Find the Cartesian co-ordinates of the point whose polar co-ordinates are :
(i) \(\left(\sqrt{2}, \frac{\pi}{4}\right)\)
Solution:
Here, r = \(\sqrt {2}\) and θ = \(\frac{\pi}{4}\)
Let the cartesian coordinates be (x, y)
Then, x = rcosθ = \(\sqrt {2}\)cos\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
y = rsinθ = \(\sqrt {2}\)sin\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
∴ the cartesian coordinates of the given point are (1, 1).

(ii) \(\left(4, \frac{\pi}{2}\right)\)
Solution:

(iii) \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\)
Solution:
Here, r = \(\frac{3}{4}\) and θ = \(\frac{3 \pi}{4}\)
Let the cartesian coordinates be (x, y)

(iv) \(\left(\frac{1}{2}, \frac{7 \pi}{3}\right)\)
Solution:
Here, r = \(\frac{1}{2}\) and θ = \(\frac{7 \pi}{4}\)
Let the cartesian coordinates be (x, y)


∴ the cartesian coordinates of the given point are \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)

Question 2.
Find the of the polar co-ordinates point whose Cartesian co-ordinates are.
(i) \((\sqrt{2}, \sqrt{2})\)
Solution:
Here x = \(\sqrt {2}\) and y = \(\sqrt {2}\)
∴ the point lies in the first quadrant.
Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (\(\sqrt {2}\) )² + (\(\sqrt {2}\) )² = 2 + 2 = 4
∴ r = 2 … [∵ r > 0]
cos θ = \(\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
and sin θ = \(\frac{y}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
∴ tan θ = 1
Since the point lies in the first quadrant and
0 ≤ θ ≤ 2π, tan θ = 1 = tan\(\frac{\pi}{4}\)
∴ θ = \(\frac{\pi}{4}\)
∴ the polar coordinates of the given point are \(\left(2, \frac{\pi}{4}\right)\).

(ii) \(\left(0, \frac{1}{2}\right)\)
Solution:
Here x = 0 and y = \(\frac{1}{2}\)
the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (0)² + \(\left(\frac{1}{2}\right)^{2}=0+\frac{1}{4}=\frac{1}{4}\)
∴ r = \(\frac{1}{2}\) …[∵ r > 0]
cosθ = \(\frac{x}{r}=\frac{0}{(1 / 2)}\) = 0
and sin θ = \(\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}\) = 1
Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π
cosθ = 0 = cos\(\frac{\pi}{2}\) and sinθ = 1 = sin\(\frac{\pi}{2}\)
∴ θ = \(\frac{\pi}{2}\)
∴ the polar coordinates of the given point are \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\).

(iii) \((1,-\sqrt{3})\)
Solution:
Here x = 1 and y = \(-\sqrt{3}\)
∴ the point lies in the fourth quadrant.
Let the polar coordinates be (r, θ).
Then, r² = x² + y² = (1)² + (\(-\sqrt {3}\) )² = 1 + 3 = 4
∴ r = 2 … [∵ r > 0]

∴ the polar coordinates of the given point are \(\left(2, \frac{5 \pi}{3}\right)\).

(iv) \(\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\)
Solution:

Question 3.
In ∆ABC, if ∠A = 45º, ∠B = 60º then find the ratio of its sides.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\)
∴ \(\frac{a}{b}=\frac{\sin A}{\sin B}\) and \(\frac{b}{c}=\frac{\sin B}{\sin C}\)
∴ a : b : c = sinA : sinB : sinC
Given ∠A = 45° and ∠B = 60°
∵ ∠A + ∠B + ∠C = 180°
∴ 45° + 60° + ∠C = 180°
∴ ∠C = 180° – 105° = 75°

Question 4.
In ∆ABC, prove that sin \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)=\left(\frac{\boldsymbol{b}-\boldsymbol{c}}{a}\right)\) cos \(\frac{A}{2}\).
Solution:
By the sine rule,

Question 5.
With usual notations prove that 2 \(\left\{a \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{A}{2}\right\}\) = a – b + c.
Solution:

Question 6.
In ∆ABC, prove that a³sin(B – C) + b³sin(C – A) + c³sin(A – B) = 0
Solution:
By the sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a³sin (B – C) + b³sin (C – A) + c³sin (A – B)
= a³(sin B cos C – cos B sin C) + b³(sinCcos A – cos C sin A) + c³(sinAcosB – cos A sin B)

= \(\frac{1}{2 k}\) [a²(a² + b² – c²) – a²(a² + c² – b²) + b²(b² + c² – a²) – b²(a² + b² – c²) + c²(c² + a² – b²) – c²(b² + c² – a²)]
= \(\frac{1}{2 k}\) [a⁴ + a²b² – a²c² – a⁴ – a²c² + a²b² + b⁴ + b²c² – a²b² – a²b² – b⁴ + b²c² + c⁴ + a²c² – b²c² – b²c² – c⁴ + a²c²]
= \(\frac{1}{2 k}\)(0) = 0 = RHS.

Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a², b², c² are also in A.P
Solution:
By the sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) =\(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb, sin C = kc …(1)
Now, cot A, cotB, cotC are in A.P.
∴ cotC – cotB = cotB – cot A
∴ cotA + cotC = 2cotB

Question 8.
In ∆ABC, if a cos A = b cos B then prove that the triangle is right angled or an isosceles traingle.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = k
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sin A cos A = k sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B ∴ sin 2A – sin 2B = 0
∴ 2 cos (A + B)∙sin (A -B) = 0
∴ 2cos (π – C)∙sin(A – B) = 0 … [∵ A + B + C = π]
∴ -2 cos C∙sin (A – B) = 0
∴ cos C = 0 OR sin(A -B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.

Question 9.
With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a² + b² + c².
Solution:
LHS = 2 (bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
= 2bc \(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\) + 2ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) + 2ab\(\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\) …(By cosine rule]
= b² + c² – a² + c² + a² – b² + a² + b² – c² = a² + b² + c² = RHS.

Question 10.
In △ABC, if a = 18, b = 24, c = 30 then find the values of
(i) cos A
Solution:
Given : a = 18, b = 24 and c = 30
∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36

(ii) sin\(\frac{A}{2}\)
Solution:

(iii) cos\(\frac{A}{2}\)
Solution:

(iv) tan\(\frac{A}{2}\)
Solution:

(v) A(△ABC)
Solution:

(iv) sin A.
Solution:

Question 11.
In △ABC prove that (b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan\(\frac{B}{2}\) = (a + b – c) tan\(\frac{C}{2}\).
Solution:
(b + c – a) tan \(\frac{A}{2}\)

Question 12.
In △ABC prove that sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\) = \(\frac{[A(\triangle A B C)]^{2}}{a b c s}\)
Solution:
LHS = sin \(\frac{A}{2}\)∙sin \(\frac{B}{2}\)∙sin \(\frac{C}{2}\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.1

Question 1.
Find the principal solutions of the following equations :
(i) cos θ= \(\frac{1}{2}\)
Solution:
We know that, cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) and cos (2π – θ) = cos θ
∴ cos\(\frac{\pi}{3}\) = cos(2π – \(\frac{\pi}{3}\)) = cos\(\frac{5 \pi}{3}\)
∴ cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\) = \(\frac{1}{2}\), where
0 < \(\frac{\pi}{3}\) < 2π and 0 < \(\frac{5 \pi}{3}\) < 2π
∴ cos θ = \(\frac{1}{2}\) gives cos θ = cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\)
∴ θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)
Hence, the required principal solutions are
θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

(ii) sec θ = \(\frac{2}{\sqrt{3}}\)
Solution:

(iii) cot θ = \(\sqrt {3}\)
Solution:
The given equation is cot θ = \(\sqrt {3}\) which is same as tan θ = \(\frac{1}{\sqrt{3}}\).
We know that,

Hence, the required principal solution are
θ = \(\frac{\pi}{6}\) and θ = \(\frac{7 \pi}{6}\).

(iv) cot θ = 0.
Solution:

Question 2.
Find the principal solutions of the following equations:
(i) sinθ = \(-\frac{1}{2}\)
Solution:
We know that,
sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\) and sin (π + θ) = -sinθ,
sin(2π – θ) = -sinθ

Hence, the required principal solutions are
θ = \(\frac{7\pi}{6}\) and θ = \(\frac{11 \pi}{6}\).

(ii) tanθ = -1
Solution:
We know that,
tan\(\frac{\pi}{4}\) = 1 and tan(π – θ) = -tanθ,
tan(2π – θ) = -tanθ

Hence, the required principal solutions are
θ = \(\frac{3\pi}{4}\) and θ = \(\frac{7 \pi}{4}\).

(iii) \(\sqrt {3}\) cosecθ + 2 = 0.
Solution:

Question 3.
Find the general solutions of the following equations :
(i) sinθ = \(\frac{1}{2}\)
Solution:
(i) The general solution of sin θ = sin ∝ is
θ = nπ + (-1 )ⁿ∝, n ∈ Z
Now, sinθ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\) …[∵ sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\)]
∴ the required general solution is
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z.

(ii) cosθ = \(\frac{\sqrt{3}}{2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
Now, cosθ = \(\frac{\sqrt{3}}{2}\) = cos\(\frac{\pi}{6}\) …[∵ cos\(\frac{\pi}{6}\) = \(\frac{\sqrt{3}}{2}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z.

(iii) tanθ = \(\frac{1}{\sqrt{3}}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, tan θ = \(\frac{1}{\sqrt{3}}\) = tan\(\frac{\pi}{6}\) …[tan\(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z.

(iv) cotθ = 0.
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, cot θ = 0 ∴ tan θ does not exist
∴ tanθ = tan\(\frac{\pi}{2}\) [∵ tan\(\frac{\pi}{2}\) does not exist]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{2}\), n ∈ Z.

Question 4.
Find the general solutions of the following equations:
(i) secθ = \(\sqrt {2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = nπ ± ∝, n ∈ Z.
Now, secθ = \(\sqrt {2}\) ∴ cosθ = \(\frac{1}{\sqrt{2}}\)
∴ cosθ = cos\(\frac{\pi}{4}\) ….[cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{4}\), n ∈ Z.

(ii) cosecθ = –\(\sqrt {2}\)
Solution:
The general solution of sinθ = sin∝ is

(iii) tanθ = -1
Solution:
The general solution of tanθ = tan∝ is

Question 5.
Find the general solutions of the following equations :
(i) sin 2θ = \(\frac{1}{2}\)
Solution:
The general solution of sin θ = sin ∝ is
θ = nπ + (-1)ⁿ∝, n ∈ Z

(ii) tan \(\frac{2 \theta}{3}\) = \(\sqrt {3}\)
Solution:
The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1
Solution:
The general solution of tan θ = tan ∝ is

Question 6.
Find the general solutions of the following equations :
(i) 4 cos²θ = 3
Solution:
The general solution of cos²θ = cos² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 cos²θ = 3

(ii) 4 sin²θ = 1
Solution:
The general solution of sin²θ = sin² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 sin²θ = 3

(iii) cos 4θ = cos 2θ
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z
Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z
Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = \(\frac{n \pi}{3}\), n ∈ Z
Hence, the required general solution is
θ = \(\frac{n \pi}{3}\), n ∈ Z or ∴ θ = nπ, n ∈ Z.
Alternative Method:
cos 4θ = cos 2θ
∴ cos4θ – cos 20 = 0
∴ -2sin\(\left(\frac{4 \theta+2 \theta}{2}\right)\)∙sin\(\left(\frac{4 \theta-2 \theta}{2}\right)\) = 0
∴ sin3θ∙sinθ = 0
∴ either sin3θ = 0 or sin θ = 0
The general solution of sin θ = 0 is
θ = nπ, n ∈ Z.
∴ the required general solution is given by
3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z
i.e. θ = \(\frac{n \pi}{3}\), n ∈ Z or θ = nπ, n ∈ Z.

Question 7.
Find the general solutions of the following equations :
(i) sinθ = tanθ
Solution:
sin θ = tan θ
∴ sin θ = \(\frac{\sin \theta}{\cos \theta}\)
∴ sin θ cos θ = sin θ
∴ sin θ cos θ – sinθ = 0
∴ sin θ (cos θ – 1) = θ
∴ either sinθ = 0 or cosθ – 1 = 0
∴ either sin θ = 0 or cos θ = 1
∴ either sinθ = 0 or cosθ = cosθ …[∵ cos0 = 1]
The general solution of sinθ = 0 is θ = nπ, n ∈ Z and cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = 2nπ ± 0, n ∈ Z
∴ θ = nπ, n ∈ Z or θ = 2nπ, n ∈ Z.

(ii) tan³θ = 3tanθ
Solution:
tan³θ = 3tanθ
∴ tan³θ – 3tanθ = 0
∴ tan θ (tan²θ – 3) = 0
∴ either tan θ = 0 or tan²θ – 3 = 0
∴ either tanθ = 0 or tan²θ = 3
∴ either tan θ = 0 or tan²θ = (\(\sqrt {3}\) )³
∴ either tan θ = 0 or tan²θ = (tan\(\frac{\pi}{3}\))³ …[tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ either tanθ = 0 or tan²θ = tan²\(\frac{\pi}{3}\)
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan²θ = tan²∝ is θ = nπ ± ∝, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) cosθ + sinθ = 1.
Solution:
cosθ + sinθ = 1

Question 8.
Which of the following equations have solutions ?
(i) cos 2θ = -1
Solution:
cos 2θ = -1
Since -1 ≤ cos θ ≤ 1 for any θ,
cos 2θ = -1 has solution.

(ii) cos²θ = -1
Solution:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5
Solution:
3tanθ = 5 ∴ tanθ = \(\frac{5}{3}\)
This is possible because tan θ is any real number.
∴ 3tanθ = 5 has solution.

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1

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

(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.

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’.

(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].

(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].

(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

(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.

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:

(iii) ~(~p ∧ ~q) ∨ q
Solution:

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

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

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(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.

(ii) [(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.

(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 → r)] ↔ [(p ∧ q) → r]
Solution:

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

(v) [(p ∧ (p → q)] → q
Solution:

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:

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:

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

(viii) (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.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:

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:

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.

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:

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:

The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:

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).

(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.

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)

(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)

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).

(ii)

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).

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.

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 :

(ii)

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 :

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)

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)

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.

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.

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 :

Question 16.
Simplify the following so that the new circuit circuit.

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 :

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.

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)]

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 :

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

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

Question 1.
Express the following circuits in the symbolic form of logic and writ the input-output table.
(i)

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).

(ii)

Solution:
The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).

(iii)

Solution:
The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).

(iv)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).

(v)

Solution:
The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).

(vi)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)

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 :

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

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

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

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

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

Question 3.
Give an alternative equivalent simple circuits for the following circuits :
(i)

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 :

(ii)

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 :

Question 4.
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
i)

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)

Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

ii)

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)

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)

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)]

From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S₁ is ‘ON’.

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 :

(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 :

(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 :

(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 :

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

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

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)

(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)

(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)

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).

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.

(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.

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

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

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.

(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.

(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)

(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)

(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

(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’

(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.’

(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.’

(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.’

(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.

(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 does not increase.

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

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

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.

(ii) (p ∧ ~q) ↔ (p → q)
Solution:

(iii) (p ∧ q) ↔ (q ∨ r)
Solution:

(iv) p → [~(q ∧ r)]
Solution:

(v) ~p ∧ [(p ∨ ~q ) ∧ q]
Solution:

(vi) (~p → ~q) ∧ (~q → ~p)
Solution:

(vii) (q → p) ∨ (~p ↔ q)
Solution:

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:

(ix) p → [~(q ∧ r)]
Solution:

(x) (p ∨ ~q) → (r ∧ p)
Solution:

Question 2.
Using truth tables prove the following logical equivalences.
(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:

The entries in the columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

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

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

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]
Solution:

The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

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

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:

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:

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

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

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:

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

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
Solution:

The entries in the columns 6 and 9 are identical.
∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

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:

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:

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:

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

(iv) [(p → q) ∧ q)] → p
Solution:

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:

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:

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.

(vii) ~(~q ∧ p) ∧ q
Solution:

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:

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

(ix) (~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.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:

All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction

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

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

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’.

(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’.

(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’.

(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

(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.

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]

(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]

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.

(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.

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.

(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.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.1

Question 1.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to a machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B and the number of man-hours available for the firm is as follows.

Profit on the sale of A is ₹ 30 and B ₹ 20 Per unit Formulate the LPP to have maximum profit.
Solution:
Let the manufacturing firm produce x gadgets of type A and y gadgets of type B.
On selling x gadgets of type A the firm gets ₹ 30 and that on type B is ₹ 20.
∴ Total profit is z = ₹ 30x + 20y.
Since x and y are the numbers of gadgets, x ≥ 0, y ≥ 0
From the given table, the availability of man-hours of labour required in each shop and for the firm is given as 60 and 35.
∴ The inequation are 10x + 6y ≤ 60 and 5x + 4y ≤ 35.
Hence the given LPP can be formulated as Maximize z = 30x + 20y
Subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.

Question 2.
In a cattle breeding farm, it is prescribed that the food ratio for one animal must contain 14, 22, and 1 unit of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit weight of these two contains the following:

The cost of fodder 1 is ₹ 3 per unit and that of fodder 2 is ₹ 2 per unit. Formulate the LPP to minimize the cost.
Solution:
Let x unit of fodder 1 and y unit of fodder 2 be included in the ration of an animal
The cost of 1 unit of fodder 1 is ₹ 3 and the cost of 1 unit of fodder 2 is ₹ 2.
∴ The total cost is ₹ 3x + 2y.
The minimum requirement of the nutrients A, B, and C is given as 14 units, 22 units, and 1 unit.
∴ From the given table, the daily food ration will include (2x + 2y) unit of Nutrient A, (2x + 3y) unit of Nutrient B, and (x + y) of Nutrient C.
The total cost is 2 = ₹ 3x + 2y
Hence the given LPP can be formulated as Minimize z = 3x + 2y
subject to 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Question 3.
A Company manufactures two types of chemicals A and B. Each chemical requires two types of raw materials P and Q. The table below shows a number of units of P and Q required to manufacture one unit of A and one unit of B.

The company gets profits of ₹ 350/- and ₹ 400/- by selling one unit of A and one unit of B respectively. Formulate the problem as LPP to maximize the profit.
Solution:
∴ Let the company manufactures x unit of chemical A and y unit of chemical B.
The availability of the raw materials for the production of chemicals A and B are given as 120 and 160 units.
The company gets ₹ 350 as profit on selling one unit of chemical A and ₹ 400 as profit on selling one unit of chemical B.
∴ Total profit is ₹ (350x + 400y).
The inequation can be written as.
3x + 2y ≤ 120
2x + 5y ≤ 160
and x & y cannot be negative
Hence the LPP can be formulated as follows,
Maximize z = 350x + 400y
Subject to 3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.

Question 4.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on the machine I, 5 hours on machine II, and 2 hours on machine III. Magazine B requires 3 hours on machine 1, 2 hours on machine II and 6 hours on machine III. Machines I, II, III are available for 36, 50, 60 hours per week respective. Formulate the linear programming problem to maximize the profit.
Solution:
Let the company print x magazine of type A and y magazines of type B.
Then the total earnings of the company are ₹ 10x + 15y.
The given problem can be tabulated as follows.

From the table, the total time required for Machine I is (2x + 3y) hours, for machine II is (5x + 2y) hours, and for machine III is (2x + 6y) hours.
The machine I, II, and III are available for 36, 50, and 60 hours per work.
∴ The constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50 and 2x + 6y ≤ 60.
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.
Hence the given LPP can be formulated as
Maximize z = 10x + 15y
Subject to 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

Question 5.
Manufacture produces bulbs and tubes. Each of these must be processed through two machines M₁ and M₂. A package of bulbs requires 1 hour of work on machine M₁ and 3 hours of work on M₂. A package of tubes requires 2 hours on machine M₁ and 4 hours on machine M₂. He earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. Formulate the LPP to maximize the profit. He operates M₁ for at most 10 hours and M₂ for at most 12 hours a day.
Solution:
Let the manufacturer produce x packages of bulbs and y packages of tubes.
He earns a profit of ₹ 13.5 per packages of bulbs and ₹ 55 per package of tubes.
∴ His total profit = ₹ (13.5x + 55y).
The given problem can be tabulated as follows.

From the above table, the total time required for M₁ is (x + 2y), and that of M₂ is (3x + 4y).
M₁ and M₂ are available for at most 10 hrs per day and 12 hours per day.
∴ The constraint for the objective function is x + 2y ≤ 10, 3x + 4y ≤ 12
Hence the give LPP can be formulated as
Maximize z = 13.5x + 55y
Subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.

Question 6.
A Company manufactures two types of fertilizers F₁ and F₂. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F₁ and F₂ and availability of the raw materials A and B per day are given in the table below

By selling one unit of F₁ and one unit of F₂, the company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as LPP to maximize the profit.
Solution:
Let the company manufacture x units of Fertilizers F₁ and y units of fertilizer F₂.
The company gets a profit of ₹ 500 and ₹ 750 by selling a unit of F₁ and F₂.
∴ Total profit = ₹ (500x + 750y)
The availability of raw materials A and B per day is given as 40 and 70.
∴ From the given table the constraints can be written as 2x + 3y ≤ 40 and x + 4y ≤ 70.
Since x & y cannot be negative, x ≥ 0, y ≥ 0
Hence the given LPP can be formulated as
Maximize z = 500x + 750y
Subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.

Question 7.
A doctor has prescribed two different kinds of feeds A and B to form a weekly diet for a sick person. The minimum requirement of fats, carbohydrates, and proteins are 18, 28,14 units respectively. One unit of food A has 4 units of fat, 14 units of carbohydrates, and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is ₹ 4.5 per unit and that of food B is ₹ 3.5 per unit. Form the LPP so that the sick person’s diet meets the requirements at minimum cost.
Solution:
Let x unit of food A and y unit of food B be consumed by a sick person.
The cost of food A in ₹ 4.5 per unit and food B is ₹ 3.5 per unit.
∴ Total cost = ₹ (4.5x + 3.5y)
The given conditions can be tabulated as follows.

∴ The given LPP can be formulated as
Minimise z = 4.5x + 3.5y
Subject to 4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.

Question 8.
If John drives a car at a speed of 60 kms/hour he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 km/ hour, the cost of petrol increases to ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as LPP.
Solution:
Let John drive x km at a speed of 60 km/hr and y km at a speed of 90 km/hr.
∴ Time required to drive a distance of x km is \(\frac{x}{60}\) hours and the time require to drive at a distance of y km is \(\frac{y}{90}\) hours.
∴ Total time required \(\left(\frac{x}{60}+\frac{y}{90}\right)\) hours.
Since he wishes to drive maximum distance within an hour,
\(\frac{x}{60}+\frac{y}{90} \leq 1\)
He has to spend ₹ 5 per km at a speed of 60 km/hr and ₹ 8 per km at a speed of 90 km/hr.
He has ₹ 600 on petrol to spend, 5x + 8y ≤ 600
The total distance he wishes to travel is (x + y) hours.
∴ The given LPP can be formulated as
Maximize z = x + y
Subject to \(\frac{x}{60}+\frac{y}{90}\) ≤ 1, 5x + 8y ≤ 600, x ≥ 0, y ≥ 0.

Question 9.
The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs ₹ 20 per kg. and sand costs ₹ 6 per kg. Strength considerations dictate that a concrete brick should contain a minimum of 4 kg of cement and not more than 2 kg of sand. Formulate the LPP for the cost to be minimum.
Solution:
Let the concrete brick contain x kg of cement and y kg of sand.
The cost of cement is ₹ 20 per kg and sand is ₹ 6 per kg.
∴ The total cost = ₹ (20x + 6y)
Since the weight of the concrete brick has to be at least 5 kg, therefore, x + y ≥ 5
Also, the concrete brick should contain a minimum of 4 kg of cement, i.e. x ≥ 4, and not more than 2 kg of sand, i.e, y ≤ 2.
∴ The LPP can be formulated as
Minimize z = 20x + 6y
Subject to x + y ≥ 5, x ≥ 4, y ≤ 2, x ≥ 0, y ≥ 0.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Question 1.
Determine the maximum and minimum values of the following functions:
(i) f(x) = 2x³ – 21x² + 36x – 20
Solution:
f(x) = 2x³ – 21x² + 36x – 20
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 21x² + 36x – 20)
= 2 × 3x² – 21 × 2x + 36 × 1 – 0
= 6x² – 42x + 36
and f”(x) = \(\frac{d}{d x}\)(6x² – 42x + 36)
= 6 × 2x – 42 × 1 + 0
= 12x – 42
f'(x) = 0 gives 6x² – 42x + 36 = 0.
∴ x² – 7x + 6 = 0
∴ (x – 1)(x – 6) = 0
∴ the roots of f'(x) = 0 are x₁ = 1 and x₂ = 6.
For x = 1, f”(1) = 12(1) – 42 = -30 < 0
∴ by the second derivative test,
f has maximum at x = 1 and maximum value of f at x = 1
f(1) = 2(1)³ – 21(1)² + 36(1) – 20
= 2 – 21 + 36 – 20
= -3
For x = 6, f”(6) = 12(6) – 42 = 30 > 0
∴ by the second derivative test,
f has minimum at x = 6 and minimum value of f at x = 6
f(6) = 2(6)³ – 21(6)² + 36(6) – 20
= 432 – 756 + 216 – 20
= -128
Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

(ii) f(x) = x . log x
Solution:
f(x) = x . log x
f'(x) = \(\frac{d}{d x}\)(x.log x)
= x.\(\frac{d}{d x}\)(log x) + log x.\(\frac{d}{d x}\)(x)
= x × \(\frac{1}{x}\) + (logx) × 1
= 1 + log x
and f”(x) = \(\frac{d}{d x}\)(1 + logx)
= 0 + \(\frac{1}{x}\)
= \(\frac{1}{x}\)
Now, f'(x) = 0, if 1 + log x = 0
i.e. if log x = -1 = -log e
i.e. if log x = log(e^-1) = log \(\frac{1}{e}\)
i.e. if x = \(\frac{1}{e}\)
When x = \(\frac{1}{e}\), f”(x) = \(\frac{1}{(1 / e)}\) = e > 0
∴ by the second derivative test,
f is minimum at x = \(\frac{1}{e}\)
Minimum value of f at x = \(\frac{1}{e}\)
= \(\frac{1}{e}\) log(\(\frac{1}{e}\))
= \(\frac{1}{e}\) log(e^-1)
= \(\frac{1}{e}\) (-1) log e
= \(\frac{-1}{e}\) ……..[∵ log e = 1]
Hence, the function f has minimum at x = \(\frac{1}{e}\) and minimum value is \(\frac{-1}{e}\).

(iii) f(x) = x² + \(\frac{16}{x}\)
Solution:

f'(x) = 0 gives 2x – \(\frac{16}{x^{2}}\) = 0
∴ 2x³ – 16 = 0
∴ x³ = 8
∴ x = 2
For x = 2, f”(2) = 2 + \(\frac{32}{(2)^{3}}\) = 6 > 0
∴ by the second derivative test, f has minimum at x = 2 and minimum value of f at x = 2
f(2) = (2)² + \(\frac{16}{2}\)
= 4 + 8
= 12
Hence, the function f has a minimum at x = 2 and a minimum value is 12.

Question 2.
Divide the number 20 into two parts such that their product is maximum.
Solution:
Let the first part of 20 be x.
Then the second part is 20 – x.
∴ their product = x(20 – x) = 20x – x² = f(x) …..(Say)
∴ f'(x) = \(\frac{d}{d x}\)(20x – x²)
= 20 × 1 – 2x
= 20 – 2x
and f”(x) = \(\frac{d}{d x}\)(20 – 2x)
= 0 – 2 × 1
= -2
The root of the equation f'(x) = 0
i.e. 20 – 2x = 0 is x = 10
and f”(10) = -2 < 0
∴ by the second derivative test, f is maximum at x = 10.
Hence, the required parts of 20 are 10 and 10.

Question 3.
A metal wire of 36 cm long is bent to form a rectangle. Find its dimensions where its area is maximum.
Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Then its perimeter is 2(x + y) = 36
∴ x + y = 18
∴ y = 18 – x
Area of the rectangle = xy = x(18 – x)
Let f(x) = x(18 – x) = 18x – x²
Then f'(x) = \(\frac{d}{d x}\)(18x – x²)
= 18 × 1 – 2x
= 18 – 2x
and f”(x) = \(\frac{d}{d x}\)(18 – 2x)
= 0 – 2 × 1
= -2
Now, f(x) = 0, if 18 – 2x = 0
i.e. if x = 9
and f”(9) = -2 < 0
∴ by the second derivative test, f has maximum value at x = 9
When x = 9, y = 18 – 9 = 9
Hence, the rectangle is a square of side 9 cm.

Question 4.
The total cost of producing x units is ₹(x² + 60x + 50) and the price is ₹(180 – x) per unit. For what units is the profit maximum?
Solution:
Let the number of units sold be x.
Then profit = S.P. – C.P.
∴ P(x) = (180 – x)x – (x² + 60x + 50)
∴ P(x) = 180x – x² – x² – 60x – 50
∴ P(x) = 120x – 2x² – 50
P'(x) = \(\frac{d}{d x}\)(120x – 2x² – 50)
= 120 × 1 – 2 × 2x – 0
= 120 – 4x
and P”(x) = \(\frac{d}{d x}\)(120 – 4x)
= 0 – 4 × 1
= -4
P'(x) = 0 if 120 – 4x = 0
i.e. if x = 30 and P”(30) = -4 < 0
∴ by the second derivative test, P(x) is maximum when x = 30.
Hence, the number of units sold for maximum profit is 30.