Class 12

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

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Miscellaneous Exercise 4

(I) Choose the correct alternative:

Question 1.
The equation of tangent to the curve y = x² + 4x + 1 at (-1, -2) is
(a) 2x – y = 0
(b) 2x + y – 5 = 0
(c) 2x – y – 1 = 0
(d) x + y – 1 = 0
Answer:
(a) 2x – y = 0

Question 2.
The equation of tangent to the curve x² + y² = 5, where the tangent is parallel to the line 2x – y + 1 = 0 are
(a) 2x – y + 5 = 0; 2x – y – 5 = 0
(b) 2x + y + 5 = 0; 2x + y – 5 = 0
(c) x – 2y + 5 = 0; x – 2y – 5 = 0
(d) x + 2y + 5; x + 2y – 5 = 0
Answer:
(a) 2x – y + 5 = 0; 2x – y – 5 = 0

Question 3.
If the elasticity of demand η = 1, then demand is
(a) constant
(b) inelastic
(c) unitary elastic
(d) elastic
Answer:
(c) unitary elastic

Question 4.
If 0 < η < 1, then the demand is
(a) constant
(b) inelastic
(c) unitary elastic
(d) elastic
Answer:
(b) inelastic

Question 5.
The function f(x) = x³ – 3x² + 3x – 100, x ∈ R is
(a) increasing for all x ∈ R, x ≠ 1
(b) decreasing
(c) neither increasing nor decreasing
(d) decreasing for all x ∈ R, x ≠ 1
Answer:
(a) increasing for all x ∈ R, x ≠ 1

Question 6.
If f(x) = 3x³ – 9x² – 27x + 15, then
(a) f has maximum value 66
(b) f has minimum value 30
(c) f has maxima at x = -1
(d) f has minima at x = -1
Answer:
(c) f has maxima at x = -1

(II) Fill in the blanks:

Question 1.
The slope of tangent at any point (a, b) is called as ___________
Answer:
gradient

Question 2.
If f(x) = x³ – 3x² + 3x – 100, x ∈ R, then f”(x) is ___________
Answer:
6x – 6 = 6(x – 1)

Question 3.
If f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0, then f”(x) is ___________
Answer:
14x^-3

Question 4.
A rod of 108 m in length is bent to form a rectangle. If area j at the rectangle is maximum, then its dimensions are ___________
Answer:
27 and 27

Question 5.
If f(x) = x . log x, then its maximum value is ___________
Answer:
\(-\frac{1}{e}\)

(III) State whether each of the following is True or False:

Question 1.
The equation of tangent to the curve y = 4xe^x at (-1, \(\frac{-4}{e}\)) is y.e + 4 = 0.
Answer:
True

Question 2.
x + 10y + 21 = 0 is the equation of normal to the curve y = 3x² + 4x – 5 at (1, 2).
Answer:
False

Question 3.
An absolute maximum must occur at a critical point or at an endpoint.
Answer:
True

Question 4.
The function f(x) = x.e^x(1-x) is increasing on (\(\frac{-1}{2}\), 1).
Answer:
True.
Hint:


Hence, function f(x) is increasing on (\(\frac{-1}{2}\), 1).

(IV) Solve the following:

Question 1.
Find the equations of tangent and normal to the following curves:
(i) xy = c² at (ct, \(\frac{c}{t}\)), where t is a parameter.
Solution:
xy = c²
Differentiating both sides w.r.t. x, we get


Hence, equations of tangent and normal are x + t²y – 2ct = 0 and t³x – ty – c(t⁴ + 1) = 0 respectively.

(ii) y = x² + 4x at the point whose ordinate is -3.
Solution:
Let P(x₁, y₁) be the point on the curve
y = x² + 4x, where y₁ = -3



Hence, the equations of tangent and normal at
(i) (-3, -3) are 2x + y + 9 = 0 and x – 2y – 3 = 0
(ii) (-1, -3) are 2x – y – 1 = 0 and x + 2y + 7 = 0

(iii) x = \(\frac{1}{t}\), y = t – \(\frac{1}{t}\), at t = 2.
Solution:
When t = 2, x = \(\frac{1}{2}\) and y = 2 – \(\frac{1}{2}\) = \(\frac{3}{2}\)
Hence, the point P at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{3}{2}\))



Hence, the equations of tangent and normal are 5x + y – 4 = 0 and x – 5y + 7 = 0 respectively.

(iv) y = x³ – x² – 1 at the point whose abscissa is -2.
Solution:
y = x³ – x² – 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\)(x³ – x² – 1)
= 3x² – 2x – 0
= 3x² – 2x
∴ \(\left(\frac{d y}{d x}\right)_{\text {at } x=-2}\) = 3(-2)² – 2(-2) = 16
= slope of the tangent at x = -2
When x = -2, y = (-2)³ – (-2)² – 1 = -13
∴ the point P is (-2, -13)
∴ the equation of the tangent at (-2, -13) is
y – (-13) = 16[x – (-2)]
∴ y + 13 = 16x + 32
∴ 16x – y + 19 = 0
The slope of the normal at x = -2
= \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at } x=-2}}=\frac{-1}{16}\)
∴ the equation of the normal at (-2, -13) is
y – (-13) = \(-\frac{1}{16}\)[x – (-2)]
∴ 16y + 208 = -x – 2
∴ x + 16y + 210 = 0
Hence, equations of tangent and normal are 16x – y + 19 = 0 and x + 16y + 210 = 0 respectively.

Question 2.
Find the equation of the normal to the curve y = \(\sqrt{x-3}\) which is perpendicular to the line 6x + 3y – 4 = 0.
Solution:
Let P(x₁, y₁) be the foot of the required normal to the curve y = \(\sqrt{x-3}\)
Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get



∴ x – 2y – \(\frac{57}{16}\) = 0
i.e. 16x – 32y – 57 = 0
Hence, the equation of the normals are 16x – 32y – 41 = 0 and 16x – 32y – 57 = 0.

Question 3.
Show that the function f(x) = \(\frac{x-2}{x+1}\), x ≠ -1 is increasing.
Solution:
f(x) = \(\frac{x-2}{x+1}\)

∴ f'(x) > 0, for all x ∈ R, x ≠ -1
Hence, the function f is increasing for all x ∈ R, where x ≠ -1.

Question 4.
Show that the function f(x) = \(\frac{3}{x}\) + 10, x ≠ 0 is decreasing.
Solution:
f(x) = \(\frac{3}{x}\) + 10

∴ f'(x) < 0 for all x ∈ R, x ≠ 0
Hence, the function f is decreasing for all x ∈ R, where x ≠ 0.

Question 5.
If x + y = 3, show that the maximum value of x²y is 4.
Solution:
x + y = 3
∴ y = 3 – x
∴ x²y = x²(3 – x) = 3x² – x³
Let f(x) = 3x² – x³
Then f'(x) = \(\frac{d}{d x}\)(3x² – x³)
= 3 × 2x – 3x²
= 6x – 3x²
and f”(x) = \(\frac{d}{d x}\)(6x – 3x²)
= 6 × 1 – 3 × 2x
= 6 – 6x
Now, f'(x) = 0 gives 6x – 3x² = 0
∴ 3x(2 – x) = 0
∴ x = 0 or x = 2
f”(0) = 6 – 0 = 6 > 0
∴ f has minimum value at x = 0
Also, f”(2) = 6 – 12 = -6 < 0
∴ f has maximum value at x = 2
When x = 2, y = 3 – 2 = 1
∴ maximum value of x²y = (2)²(1) = 4.

Question 6.
Examine the function f for maxima and minima, where f(x) = x³ – 9x² + 24x.
Solution:
f(x) = x³ – 9x² + 24x
∴ f'(x) = \(\frac{d}{d x}\)(x³ – 9x² + 24x)
= 3x² – 9 × 2x + 24 × 1
= 3x² – 18x + 24
and f”(x) = \(\frac{d}{d x}\)(3x² – 18x + 24)
= 3 × 2x – 18 × 1 + 0
= 6x – 18
f'(x) = 0 gives 3x² – 18x + 24 = 0
∴ x² – 6x + 8 = 0
∴ (x – 2)(x – 4) = 0
∴ the roots of f'(x) = 0 are x₁ = 2 and x₂ = 4.
(a) f”(2) = 6(2) – 18 = -6 < 0
∴ by the second derivative test,
f has maximum at x = 2 and maximum value of f at x = 2
f(2) = (2) – 9(2)² + 24(2)
= 8 – 36 + 48
= 20
(b) f”(4) = 6(4) – 18 = 6 > 0
∴ by the second derivative test, f has minimum at x = 4
and minimum value of f at x = 4
f(4) = (4)³ – 9(4)² + 24(4)
= 64 – 144 + 96
= 16.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.4

Question 1.
The demand function of a commodity at price P is given as D = 40 – \(\frac{5 P}{8}\). Check whether it is an increasing or decreasing function.
Solution:
D = 40 – \(\frac{5 P}{8}\)
∴ \(\frac{d D}{d P}=\frac{d}{d P}\left(40-\frac{5 P}{8}\right)\)
= 0 – \(\frac{5}{8}\) × 1
= \(\frac{-5}{8}\)
Hence, the given function is decreasing function.

Question 2.
The price P for demand D is given as P = 183 + 120D – 3D², find D for which price is increasing.
Solution:
P = 183 + 120D – 3D²
∴ \(\frac{d P}{d D}=\frac{d}{d D}\)(183 + 120D – 3D²)
= 0 + 120 × 1 – 3 × 2D
= 120 – 6D
If price P is increasing, then \(\frac{d P}{d D}\) > 0
∴ 120 – 6D > 0
∴ 120 > 6D
∴ D < 20
Hence, the price is increasing when D < 20.

Question 3.
The total cost function for production of x articles is given as C = 100 + 600x – 3x². Find the values of x for which the total cost is decreasing.
Solution:
The cost function is given as
C = 100 + 600x – 3x²
∴ \(\frac{d C}{d D}=\frac{d}{d D}\)(100 + 600x – 3x²)
= 0 + 600 × 1 – 3 × 2x
= 600 – 6x
If the total cost is decreasing, then \(\frac{d C}{d x}\) < 0
∴ 600 – 6x < 0
∴ 600 < 6x
∴ x > 100
Hence, the total cost is decreasing for x > 100.

Question 4.
The manufacturing company produces x items at the total cost of ₹(180 + 4x). The demand function for this product is P = (240 – x). Find x for which
(i) revenue is increasing
(ii) profit is increasing.
Solution:
(i) Let R be the total revenue.
Then R = P.x = (240 – x)x
∴ R = 240x – x²
∴ \(\frac{d R}{d D}=\frac{d}{d D}\)(240x – x²)
= 240 × 1 – 2x
= 240 – 2x
R is increasing, if \(\frac{d R}{d x}\) > 0
i.e. if 240 – 2x > 0
i.e. if 240 > 2x
i.e. if x < 120
Hence, the revenue is increasing, if x < 120.

(ii) Profit π = R – C
∴ π = (240x – x²) – (180 + 4x)
= 240x – x² – 180 – 4x
= 236x – x² – 180
∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(236x – x² – 180)
= 236 × 1 – 2x – 0
= 236 – 2x
Profit is increasing, if \(\frac{d \pi}{d x}\) > 0
i.e. if 236 – 2x > 0
i.e. if 236 > 2x
i.e. if x < 118
Hence, the profit is increasing, if x < 118.

Question 5.
For manufacturing x units, labour cost is 150 – 54x and processing cost is x². Price of each unit is p = 10800 – 4x². Find the values of x for which
(i) total cost is decreasing
(ii) revenue is increasing.
Solution:
(i) Total cost C = labour cost + processing cost
∴ C = 150 – 54x + x²
∴ \(\frac{d C}{d x}=\frac{d}{d x}\)(150 – 54x + x²)
= 0 – 54 × 1 + 2x
= -54 + 2x
The total cost is decreasing, if \(\frac{d C}{d x}\) < 0
i.e. if -54 + 2x < 0
i.e. if 2x < 54
i.e. if x < 27
Hence, the total cost is decreasing, if x < 27.

(ii) The total revenue R is given as
R = p.x
R = (10800 – 4x²) x
R = 10800x – 4x³
∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(10800x – 4x³)
= 10800 × 1 – 4 × 3x²
= 10800 – 12x²
The revenue is increasing, if \(\frac{d R}{d x}\) > 0
i.e. if 10800 – 12x² > 0
i.e. if 10800 > 12x²
i.e. if x² < 900
i.e. if x < 30 ……[∵ x > 0]
Hence, the revenue is increasing, if x < 30.

Question 6.
The total cost of manufacturing x articles is C = 47x + 300x² – x⁴. Find x, for which average cost is
(i) increasing
(ii) decreasing.
Solution:
The total cost is given as C = 47x + 300x² – x⁴
∴ the average cost is given by

(i) CA is increasing, if \(\frac{d C_{A}}{d x}\) > 0
i.e. if 300 – 3x² > 0
i.e. if 300 > 3x²
i.e. if x² < 100
i.e. if x < 10 …..[∵ x > 0]
Hence, the average cost is increasing, if x < 10.

(ii) CA is decreasing, if \(\frac{d C_{A}}{d x}\) < 0
i.e. if 300 – 3x² < 0
i.e. if 300 < 3x²
i.e. if x² > 100
i.e. if x > 10 ……[∵ x > 0]
Hence, the average cost is decreasing, if x > 10.

Question 7.
(i) Find the marginal revenue, if the average revenue is 45 and the elasticity of demand is 5.
Solution:
Given R_A = 45 and η = 5
Now, R_m = \(R_{A}\left(1-\frac{1}{\eta}\right)\)
= 45(1 – \(\frac{1}{5}\))
= 45(\(\frac{4}{5}\))
= 36
Hence, the marginal revenue = 36.

(ii) Find the price, if the marginal revenue is 28 and elasticity of demand is 3.
Solution:
Given R_m = 28 and η = 3

Hence, the price = 42.

(iii) Find the elasticity of demand, if the marginal revenue is 50 and price is ₹ 75.
Solution:
Given R_m = 50 and R_A = 75

Hence, the elasticity of demand = 3.

Question 8.
If the demand function is D = \(\frac{p+6}{p-3}\), find the elasticity of demand at p = 4.
Solution:
The demand function is

Elasticity of demand is given by

Hence, the elasticity of demand at p = 4 is 3.6

Question 9.
Find the price for the demand function D = \(\frac{2 p+3}{3 p-1}\), when elasticity of demand is \(\frac{11}{14}\).
Solution:
The demand function is

Elasticity of demand is given by

Question 10.
If the demand function is D = 50 – 3p – p² elasticity of demand at (i) p = 5 (ii) p = 2. Comment on the result.
Solution:
The demand function is D = 50 – 3p – p²
∴ \(\frac{d D}{d p}=\frac{d}{d p}\)(50 – 3p – p²)
= 0 – 3 × 1 – 2p
= -3 – 2p
Elasticity of demand is given by

(i) When p = 5, then

Since, η >1, the demand is elastic.
(ii) When p = 2, then

Since, 0 < η < 1, the demand is inelastic.

Question 11.
For the demand function D = 100 – \(\frac{p^{2}}{2}\), find the elasticity of demand at (i) p = 10 (ii) p = 6 and comment on the results.
Solution:
The demand function is

The elasticity of demand is given by

(i) When p = 10, then

Since, η > 1, the demand is elastic.
(ii) When p = 6, then

Since, 0 < η < 1, the demand is inelastic.

Question 12.
A manufacturing company produces, x items at a total cost of ₹(40 + 2x). Their price is given as p = 120 – x. Find the value of x for which
(i) revenue is increasing
(ii) profit is increasing
(iii) Also find an elasticity of demand for price 80.
Solution:
(i) The total revenue R is given by
R = p.x = (120 – x)x
∴ R = 120x – x²
∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(120x – x²)
= 120 × 1 – 2x
= 120 – 2x
If the revenue is increasing, then \(\frac{d R}{d x}\) > 0
∴ 120 – 2x > 0
∴ 120 > 2x
∴ x < 60
Hence, the revenue is increasing when x < 60.

(ii) Profit π = R – C
= (120x – x²) – (40 + 2x)
= 120x – x² – 40 – 2x
= 118x – x² – 40
∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(118x – x² – 40)
= 118 × 1 – 2x – 0
= 118 – 2x
If the profit is increasing, then \(\frac{d \pi}{d x}\) > 0
∴ 118 – 2x > 0
∴ 118 > 2x
∴ x < 59
Hence, the profit is increasing when x < 59.

(iii) p = 120 – x
∴ x = 120 – p
∴ \(\frac{d x}{d p}=\frac{d}{d p}\)(120 – p)
= 0 – 1
= -1
Elasticity of demand is given by

Question 13.
Find MPC, MPS, APC and APS, if the expenditure E_c of a person with income I is given as
E_c = (0.0003)I² + (0.075)I, when I = 1000.
Solution:
E_c = (0.0003)I² + (0.075)I
MPC = \(\frac{d E_{c}}{d I}=\frac{d}{d I}\)[(0.0003)I2 + (0.075)I]
= (0.0003)(2I) + (0.075)(1)
= (0.0006)I + 0.075
When I = 1000, then
MPC = (0.0006)(1000) + 0.075
= 0.6 + 0.075
= 0.675.
∴ MPC + MPS = 1
∴ 0.675 + MPS = 1
∴ MPS = 1 – 0.675 = 0.325
Now, APC = \(\frac{E_{c}}{I}=\frac{(0.0003) I^{2}+(0.075) I}{I}\)
= (0.0003)I + (0.075)
When I = 1000, then
APC = (0.0003)(1000) + 0.075
= 0.3 + 0.075
= 0.375
∵ APC + APS = 1
∴ 0.375 + APS = 1
∴ APS = 1 – 0.375 = 0.625
Hence, MPC = 0.675, MPS = 0.325, APC = 0.375, APS = 0.625.

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.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

Question 1.
Test whether the following functions are increasing and decreasing:
(i) f(x) = x³ – 6x² + 12x – 16, x ∈ R
Solution:
f(x) = x³ – 6x² + 12x – 16
∴ f'(x) = \(\frac{d}{d x}\)(x³ – 6x² + 12x – 16)
= 3x² – 6 × 2x + 12 × 1 – 0
= 3x² – 12x + 12
= 3(x² – 4x + 4)
= 3(x – 2)² > 0 for all x ∈ R, x ≠ 2
∴ f'(x) > 0 for all x ∈ R – {2}
∴ f is increasing for all x ∈ R – {2}.

(ii) f(x) = x – \(\frac{1}{x}\), x ∈ R, x ≠ 0
Solution:
f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{d}{d x}\left(x-\frac{1}{x}\right)\)
= 1 – \(\left(-\frac{1}{x^{2}}\right)\)
= 1 + \(\frac{1}{x^{2}}\) > 0 for all x ∈ R, x ≠ 0
∴ f'(x) > 0 for all x ∈ R, where x ≠ 0
∴ f is increasing for all x > R, where x ≠ 0.

(iii) f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0
Solution:
f(x) = \(\frac{7}{x}\) – 3
∴ f'(x) = \(\frac{d}{d x}\left(\frac{7}{x}-3\right)=7\left(-\frac{1}{x^{2}}\right)-0\)
= \(-\frac{7}{x^{2}}\) < 0 for all x ∈ R, x ≠ 0
∴ f'(x) < 0 for all x ∈ R, where x ≠ 0.
∴ f is decreasing for all x ∈ R, where x ≠ 0.

Question 2.
Find the values of x, such that f(x) is increasing function:
(i) f(x) = 2x³ – 15x² + 36x + 1
Solution:
f(x) = 2x³ – 15x² + 36x + 1
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² + 36x + 1)
= 2 × 3x² – 15 × 2x + 36 × 1 + 0
= 6x² – 30x + 36
= 6(x² – 5x + 6)
f is increasing, if f'(x) > 0
i.e. if 6(x² – 5x + 6) > 0
i.e. if x² – 5x + 6 > 0
i.e. if x² – 5x > -6
i.e. if x – 5x + \(\frac{25}{4}\) > -6 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{1}{4}\)
i.e. if x – \(\frac{5}{2}\) > \(\frac{1}{2}\) or x – \(\frac{5}{2}\) < –\(\frac{1}{2}\)
i.e. if x > 3 or x < 2
i.e. if x ∈ (-∞, 2) ∪ (3, ∞)
∴ f is increasing, if x ∈ (-∞, 2) ∪ (3, ∞).

(ii) f(x) = x² + 2x – 5
Solution:
f(x) = x² + 2x – 5
∴ f'(x) = \(\frac{d}{d x}\)(x² + 2x – 5)
= 2x + 2 × 1 – 0
= 2x + 2
f is increasing, if f'(x) > 0
i.e. if 2x + 2 > 0
i.e. if 2x > -2
i.e. if x > -1, i.e. x ∈ (-1, ∞)
∴ f is increasing, if x > -1, i.e. x ∈ (-1, ∞)

(iii) f(x) = 2x³ – 15x² – 144x – 7
Solution:
f(x) = 2x³ – 15x² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\) (2x³ – 15x² – 144x – 7)
= 2 × 3x² – 15 × 2x – 144 × 1 – 0
= 6x² – 30x – 144
= 6(x² – 5x – 24)
f is increasing if, f'(x) > 0
i.e. if 6(x² – 5x – 24) > 0
i.e. if x² – 5x – 24 > 0
i.e. if x² – 5x > 24
i.e. if x² – 5x + \(\frac{25}{4}\) > 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{121}{4}\)
i.e. if \(x-\frac{5}{2}>\frac{11}{2} \text { or } x-\frac{5}{2}<-\frac{11}{2}\)
i.e. if x > 8 or x < -3
i.e. if x ∈ (-∞, -3) ∪ (8, ∞)
∴ f is increasing, if x ∈ (-∞, -3) ∪ (8, ∞).

Question 3.
Find the values of x such that f(x) is decreasing function:
(i) f(x) = 2x³ – 15x² – 144x – 7
Solution:
f(x) = 2x³ – 15x² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² – 144x – 7)
= 2 × 3x² – 15 × 2x – 144 × 1 – 0
= 6x² – 30x – 144
= 6(x² – 5x – 24)
f is decreasing, if f'(x) < 0
i.e. if 6(x² – 5x – 24) < 0
i.e. if x² – 5x – 24 < 0
i.e. if x² – 5x < 24
i.e. if x² – 5x + \(\frac{25}{4}\) < \(\frac{121}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{121}{4}\)
i.e. if \(-\frac{11}{2}<x-\frac{5}{2}<\frac{11}{2}\)
i.e. if \(-\frac{11}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{11}{2}+\frac{5}{2}\)
i.e. if -3 < x < 8
∴ f is decreasing, if -3 < x < 8.

(ii) f(x) = x⁴ – 2x³ + 1
Solution:
f(x) = x⁴ – 2x³ + 1
∴ f'(x) = \(\frac{d}{d x}\)(x⁴ – 2x³ + 1)
= 4x³ – 2 × 3x² + 0
= 4x³ – 6x²
f is decreasing, if f'(x) < 0
i.e. if 4x³ – 6x² < 0
i.e. if x²(4x – 6) < 0
i.e. if 4x – 6 < 0 …….[∵ x² > 0]
i.e. if x < \(\frac{3}{2}\)
i.e. -∞ < x < \(\frac{3}{2}\)
∴ f is decreasing, if -∞ < x < \(\frac{3}{2}\).

(iii) f(x) = 2x³ – 15x² – 84x – 7
Solution:
f(x) = 2x³ – 15x² – 84x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² – 84x – 7)
= 2 × 3x² – 15 × 2x – 84 × 1 – 0
= 6x² – 30x – 84
= 6(x² – 5x – 14)
f is decreasing, if f'(x) < 0
i.e. if 6(x² – 5x – 14) < 0
i.e. if x² – 5x – 14 < 0
i.e. if x² – 5x < 14
i.e. if x – 5x + \(\frac{25}{4}\) < 14 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{81}{4}\)
i.e. if \(-\frac{9}{2}<x-\frac{5}{2}<\frac{9}{2}\)
i.e. if \(-\frac{9}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{9}{2}+\frac{5}{2}\)
i.e. if -2 < x < 7
∴ f is decreasing, if -2 < x < 7.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 1.
Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x² – x + 1 at (1, 3)
Solution:
y = 3x² – x + 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\) (3x² – x + 1)
= 3 × 2x – 1 + 0
= 6x – 1
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}\) = 6(1) – 1
= 5
= slope of the tangent at (1, 3).
∴ the equation of the tangent at (1, 3) is
y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0.
The slope of the normal at (1, 3) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}\)
∴ the equation of the normal at (1, 3) is
y – 3 = \(-\frac{1}{5}\)(x – 1)
∴ 5y – 15 = -x + 1
∴ x + 5y – 16 = 0
Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.

(ii) 2x² + 3y² = 5 at (1, 1)
Solution:
2x² + 3y² = 5
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = \(\frac{-2}{3}\)(x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}\)
∴ the equation of the normal at (1, 1) is
y – 1 = \(\frac{3}{2}\)(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.

(iii) x² + y² + xy = 3 at (1, 1)
Solution:
x² + y² + xy = 3
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}\)
= \(\frac{-1}{-1}\)
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.

Question 2.
Find the equations of the tangent and normal to the curve y = x² + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Solution:
Let P(x₁, y₁) be the point on the curve y = x² + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Differentiating y = x² + 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(x² + 5) = 2x + 0 = 2x
\(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}\)
= slope of the tangent at (x₁, y₁)
Let m₁ = 2x₁
The slope of the line 4x – y + 1 = 0 is
m₂ = \(\frac{-4}{-1}\) = 4
Since, the tangent at P(x₁, y₁) is parallel to the line 4x – y + 1 = 0,
m₁ = m₂
∴ 2x₁ = 4
∴ x₁ = 2
Since, (x₁, y₁) lies on the curve y = x² + 5, y₁ = \(x_{1}^{2}\) + 5
∴ y₁ = (2)² + 5 = 9 ……[x₁ = 2]
∴ the coordinates of the point are (2, 9) and the slope of the tangent = m₁ = m₂ = 4.
∴ the equation of the tangent at (2, 9) is
y – 9 = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x – y + 1 = 0
Slope of the normal = \(\frac{-1}{m_{1}}=-\frac{1}{4}\)
∴ the equation of the normal at (2, 9) is
y – 9 = \(-\frac{1}{4}\)(x – 2)
∴ 4y – 36 = -x + 2
∴ x + 4y – 38 = 0
Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

Question 3.
Find the equations of the tangent and normal to the curve y = 3x² – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Solution:
Let P(x₁, y₁) be the point on the curve y = 3x² – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Differentiating y = 3x² – 3x – 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(3x² – 3x – 5)
= 3 × 2x – 3 × 1 – 0
= 6x – 3
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3\)
= slope of the tangent at (x₁, y₁)
Let m₁ = 6x₁ – 3
The slope of the line 3x – y + 1 = 0
m₂ = \(\frac{-3}{-1}\) = 3
Since, the tangent at P(x₁, y₁) is parallel to the line 3x – y + 1 = 0,
m₁ = m₂
∴ 6x₁ – 3 = 3
∴ 6x₁ = 6
∴ x₁ = 1
Since, (x₁, y₁) lies on the curve y = 3x² – 3x – 5,
\(y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5\), where x₁ = 1
= 3(1)² – 3(1) – 5
= 3 – 3 – 5
= -5
∴ the coordinates of the point are (1, -5) and the slope of the tangent = m₁ = m₂ = 3.
∴ the equation of the tangent at (1, -5) is
y – (-5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x – y – 8 = 0
Slope of the normal = \(-\frac{1}{m_{1}}=-\frac{1}{3}\)
∴ the equation of the normal at (1, -5) is
y – (-5) = \(-\frac{1}{3}\)(x – 1)
∴ 3y + 15 = -x + 1
∴ x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.

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

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

(I) Choose the correct alternative:

Question 1.
If y = (5x³ – 4x² – 8x)⁹, then \(\frac{d y}{d x}\) = ___________
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)
(b) 9(5x³ – 4x² – 8x)⁹ (15x² – 8x – 8)
(c) 9(5x³ – 4x² – 8x)⁸ (5x² – 8x – 8)
(d) 9(5x³ – 4x² – 8x)⁹ (5x² – 8x – 8)
Answer:
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)

Question 2.
If y = \(\sqrt{x+\frac{1}{x}}\), then \(\frac{d y}{d x}\) = ?
(a) \(\frac{x^{2}-1}{2 x^{2} \sqrt{x^{2}+1}}\)
(b) \(\frac{1-x^{2}}{2 x^{2} \sqrt{x^{2}+1}}\)
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
(d) \(\frac{1-x^{2}}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Answer:
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Hint:

Question 3.
If y = \(e^{\log x}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{e^{\log x}}{x}\)
(b) \(\frac{1}{x}\)
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{e^{\log x}}{x}\)

Question 4.
If y = 2x² + 2² + a², then \(\frac{d y}{d x}\) = ?
(a) x
(b) 4x
(c) 2x
(d) -2x
Answer:
(b) 4x

Question 5.
If y = 5^x . x⁵, then \(\frac{d y}{d x}\) = ?
(a) 5^x . x⁴(5 + log 5)
(b) 5^x . x⁵(5 + log 5)
(c) 5^x . x⁴(5 + x log 5)
(d) 5^x . x⁵(5 + x log 5)
Answer:
(c) 5^x . x⁴(5 + x log 5)

Question 6.
If y = \(\log \left(\frac{e^{x}}{x^{2}}\right)\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{2-x}{x}\)
(b) \(\frac{x-2}{x}\)
(c) \(\frac{e-x}{ex}\)
(d) \(\frac{x-e}{ex}\)
Answer:
(b) \(\frac{x-2}{x}\)
Hint:

Question 7.
If ax² + 2hxy + by² = 0, then \(\frac{d y}{d x}\) = ?
(a) \(\frac{(a x+h y)}{(h x+b y)}\)
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)
(c) \(\frac{(a x-h y)}{(h x+b y)}\)
(d) \(\frac{(2 a x+h y)}{(h x+3 b y)}\)
Answer:
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)

Question 8.
If x⁴ . y⁵ = (x + y)^(m+1) and \(\frac{d y}{d x}=\frac{y}{x}\) then m = ?
(a) 8
(b) 4
(c) 5
(d) 20
Answer:
(a) 8
Hint:
If x^p . y^q = (x + y)^p+q, then \(\frac{d y}{d x}=\frac{y}{x}\)
∴ m + 1 = 4 + 5 = 9
∴ m = 8.

Question 9.
If x = \(\frac{e^{t}+e^{-t}}{2}\), y = \(\frac{e^{t}-e^{-t}}{2}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{-y}{x}\)
(b) \(\frac{y}{x}\)
(c) \(\frac{-x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(d) \(\frac{x}{y}\)
Hint:

Question 10.
If x = 2at², y = 4at, then \(\frac{d y}{d x}\) = ?
(a) \(-\frac{1}{2 a t^{2}}\)
(b) \(\frac{1}{2 a t^{3}}\)
(c) \(\frac{1}{t}\)
(d) \(\frac{1}{4 a t^{3}}\)
Answer:
(c) \(\frac{1}{t}\)

(II) Fill in the blanks:

Question 1.
If 3x²y + 3xy² = 0 then \(\frac{d y}{d x}\) = …………
Answer:
-1
Hint:
3x²y + 3xy² = 0
∴ 3xy(x + y) = 0
∴ x + y = 0
∴ y = -x
∴ \(\frac{d y}{d x}\) = -1

Question 2.
If x^m . yⁿ = (x+y)^(m+n) then \(\frac{d y}{d x}=\frac{\ldots \ldots}{x}\)
Answer:
y

Question 3.
If 0 = log(xy) + a then \(\frac{d y}{d x}=\frac{-y}{\ldots . .}\)
Answer:
x

Question 4.
If x = t log t and y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y
Hint:
x = t log t = log t^t = log y
∴ 1 = \(\frac{1}{y} \cdot \frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = y

Question 5.
If y = x . log x then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{1}{x}\)

Question 6.
If y = [log(x)]² then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{2(1-\log x)}{x^{2}}\)
Hint:

Question 7.
If x = y + \(\frac{1}{y}\) then \(\frac{d y}{d x}\) = …………
Answer:
\(\frac{y^{2}}{y^{2}-1}\)
Hint:

Question 8.
If y = e^ax, then x.\(\frac{d y}{d x}\) = …………
Answer:
axy

Question 9.
If x = t . log t, y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y

Question 10.
If y = \(\left(x+\sqrt{x^{2}-1}\right)^{m}\) then \(\sqrt{\left(x^{2}-1\right)} \frac{d y}{d x}\) = ………
Answer:
my
Hint:

(III) State whether each of the following is True or False:

Question 1.
If f’ is the derivative of f, then the derivative of the inverse of f is the inverse of f’.
Answer:
False

Question 2.
The derivative of log_a x, where a is constant is \(\frac{1}{x \cdot \log a}\).
Answer:
True

Question 3.
The derivative of f(x) = a^x, where a is constant is x . a^x-1
Answer:
False

Question 4.
The derivative of a polynomial is polynomial.
Answer:
True

Question 5.
\(\frac{d}{d x}\left(10^{x}\right)=x \cdot 10^{x-1}\)
Answer:
False

Question 6.
If y = log x, then \(\frac{d y}{d x}=\frac{1}{x}\).
Answer:
True

Question 7.
If y = e², then \(\frac{d y}{d x}\) = 2e.
Answer:
False

Question 8.
The derivative of a^x is a^x. log a.
Answer:
True

Question 9.
The derivative of x^m . yⁿ = (x + y)^(m+n) is \(\frac{x}{y}\)
Answer:
False

(IV) Solve the following:

Question 1.
If y = (6x³ – 3x² – 9x)¹⁰, find \(\frac{d y}{d x}\)
Solution:
Given y = (6x³ – 3x² – 9x)¹⁰

Question 2.
If y = \(\sqrt[5]{\left(3 x^{2}+8 x+5\right)^{4}}\), find \(\frac{d y}{d x}\).
Solution:

Question 3.
If y = [log(log(log x))]², find \(\frac{d y}{d x}\).
Solution:

Question 4.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = 25 + 30x – x².
Solution:

Question 5.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = \(\frac{5 x+7}{2 x-13}\)
Solution:

Question 6.
Find \(\frac{d y}{d x}\) if y = x^x.
Solution:
y = x^x
∴ log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 7.
Find \(\frac{d y}{d x}\) if y = \(2^{x^{x}}\).
Solution:

Question 8.
Find \(\frac{d y}{d x}\), if y = \(\sqrt{\frac{(3 x-4)^{3}}{(x+1)^{4}(x+2)}}\)
Solution:

Question 9.
Find \(\frac{d y}{d x}\) if y = x^x + (7x – 1)^x
Solution:

Question 10.
If y = x³ + 3xy² + 3x²y, find \(\frac{d y}{d x}\).
Solution:
y = x³ + 3xy² + 3x²y
Differentiating both sides w.r.t. x, we get

Question 11.
If x³ + y² + xy = 7, find \(\frac{d y}{d x}\).
Solution:
x³ + y² + xy = 7
Differentiating both sides w.r.t. x, we get

Question 12.
If x³y³ = x² – y², find \(\frac{d y}{d x}\).
Solution:
x³y³ = x² – y²
Differentiating both sides w.r.t. x, we get

Question 13.
If x⁷ . y⁹ = (x + y)¹⁶, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 14.
If x^a . y^b = (x + y)^a+b, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 15.
Find \(\frac{d y}{d x}\) if x = 5t², y = 10t.
Solution:
x = 5t², y = 10t
Differentiating x and y w.r.t. t, we get

Question 16.
Find \(\frac{d y}{d x}\) if x = e^3t, y = \(e^{\sqrt{t}}\).
Solution:
x = e^3t, y = \(e^{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get

Question 17.
Differentiate log(1 + x²) with respect to a^x.
Solution:
Let u = log(1 + x²) and v = a^x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 18.
Differentiate e^(4x+5) with resepct to 10^4x.
Solution:
Let u = e^(4x+5) and v = 10^4x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 19.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = log x.
Solution:
y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\)

Question 20.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = 2at, x = at².
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 21.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = x² . e^x
Solution:
y = x² . e^x
Differentiating w.r.t. x, we get

= e^x (2x + 2 + x² + 2x)
= e^x (x² + 4x + 2).

Question 22.
If x² + 6xy + y² = 10, then show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\).
Solution:
x² + 6xy + y² = 10 ……..(1)
Differentiating both sides w.r.t. a, we get

Question 23.
If ax² + 2hxy + by² = 0, then show that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution:
ax² + 2hxy + by² = 0 ……..(1)
∴ ax² + hxy + hxy + by² = 0
∴ x(ax + hy) + y(hx + by) = 0
∴ x(ax + hy) = -y(hx + by)
∴ \(\frac{a x+h y}{h x+b y}=-\frac{y}{x}\) …….(2)
Differentiating (1) w.r.t. x, we get

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6

1. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = √x
Solution:

Question 2.
y = x⁵
Solution:

Question 3.
y = x^-7
Solution:

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

Question 1.
y = e^x
Solution:

Question 2.
y = e^(2x+1)
Solution:

Question 3.
y = e^{log x}
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5

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

Question 1.
x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 2.
x = 2at², y = at⁴
Solution:

Question 3.
x = e^3t, y = e^(4t+5)
Solution:
x = e^3t, y = e^(4t+5)
Differentiating x and y w.r.t. t, we get

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

Question 1.
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\)
Solution:
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 2.
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²)
Solution:
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 3.
Differentiate 5^x with respect to log x.
Solution:
Let u = 5^x and v = log x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(5^{x}\right)=5^{x} \cdot \log 5\)

3. Solve the following:

Question 1.
If x = \(a\left(1-\frac{1}{t}\right)\), y = \(a\left(1+\frac{1}{t}\right)\), then show that \(\frac{d y}{d x}\) = -1
Solution:

Question 2.
If x = \(\frac{4 t}{1+t^{2}}\), y = \(3\left(\frac{1-t^{2}}{1+t^{2}}\right)\), then show that \(\frac{d y}{d x}=-\frac{9 x}{4 y}\)
Solution:

Question 3.
If x = t . log t, y = t^t, then show that \(\frac{d y}{d x}\) – y = 0.
Solution:
x = t log t
Differentiating w.r.t. t, we get

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.1

Question 1.
Construct a matrix A = [a_ij]_3×2 whose elements aij isgiven by
(i) a_ij = \(\frac{(i-j)^{2}}{5-i}\)
Solution:

(ii) a_ij = i – 3j
Solution:
a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2
a₁₂ = 1 – 3(2) = 1 – 6 = -5
a₂₁ = 2 – 3(1) = 2 – 3 = -1
a₂₂ = 2 – 3(2) = 2 – 6 = -4
a₃₁ = 3 – 3(1) = 3 – 3 = 0
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

(iii) a_ij = \(\frac{(i+j)^{3}}{5}\)
Solution:

Question 2.
Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular matrix:

Solution:
(i) Since, all the elements below the diagonal are zero, it is an upper triangular matrix.
(ii) This matrix has only one column, it is a column matrix.
(iii) This matrix has only one row, it is a row matrix.
(iv) Since, diagonal elements are equal and non-diagonal elements are zero, it is a scalar matrix.
(v) Since, all the elements above the diagonal are zero, it is a lower triangular matrix.
(vi) Since, all the non-diagonal elements are zero, it is a diagonal matrix.
(vii) Since, diagonal elements are 1 and non-diagonal elements are 0, it is an identity (or unit) matrix.

Question 3.
Which of the following matrices are singular or non-singular:
(i) \(\left[\begin{array}{ccc}
a & b & c \\
p & q & r \\
2 a-p & 2 b-q & 2 c-r
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
Solution:
Let C = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |C| = \(\left|\begin{array}{rrr}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right|\)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49
= 52 ≠ 0
∴ C is a non-singular matrix.

(iv) \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:
Let D = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |D| = \(\left|\begin{array}{rr}
7 & 5 \\
-4 & 7
\end{array}\right|\)
= 49 – (-20)
= 69 ≠ 0
∴ D is a non-singular matrix.

Question 4.
Find k, if the following matrices are singular:
(i) \(\left[\begin{array}{cc}
7 & 3 \\
-2 & K
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & K
\end{array}\right]\)
Since, A is a singular matrix, |A| = 0
∴ \(\left|\begin{array}{rr}
7 & 3 \\
-2 & k
\end{array}\right|\) = 0
∴ 7k – (-6) = 0
∴ 7k = -6
∴ k = \(-\frac{6}{7}\)

(ii) \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{~K} & 1 \\
10 & 9 & 1
\end{array}\right]\)
Solution:
Let B = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{~K} & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since, B is a singular matrix, |B| = 0
∴ \(\left|\begin{array}{rrr}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6.

(iii) \(\left[\begin{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let C = \(\left[\begin{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since, C is a singular matrix, |C| = 0
∴ \(\left|\begin{array}{crr}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\) = 0
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3(-6 – 1) = 0
∴ 8k – 8 – 20 – 21 = 0
∴ 8k = 49
∴ k = \(\frac{49}{8}\)

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

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

(I) Choose the correct alternative:

Question 1.
Which of the following is not a statement?
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) 2 is only even prime number
(d) Come here
Answer:
(d) Come here

Question 2.
Which of the following is an open statement?
(a) x is a natural number
(b) Give me a glass of water
(c) Wish you best of luck
(d) Good morning to all
Answer:
(a) x is a natural number

Question 3.
Let p ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r). Then this law is known as
(a) Commutative law
(b) Associative law.
(c) De Morgan’s law
(d) Distributive law
Answer:
(d) Distributive law

Question 4.
The false statement in the following is:
(a) p ∧ (~p) is a contradiction
(b) (p → q) ↔ (~q → ~p) is a contradiction
(c) ~(~p) ↔ p is a tautology
(d) p ∨ (~p) ↔ p is a tautology.
Answer:
(b) (p → q) ↔ (~q → ~p) is a contradiction

Question 5.
Consider the following three statements
p : 2 is an even number.
q : 2 is a prime number.
r : Sum of two prime numbers is always even.
Then, the symbolic statement (p ∧ q) → ~r means:
(a) 2 is an even and prime number and the sum of two prime numbers is always even.
(b) 2 is an even and prime number and the sum of two prime numbers is not always even.
(c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.
(d) If 2 is an even and prime number, then the sum of two prime numbers is also even.
Answer:
(c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.

Question 6.
If p : He is intelligent.
q : He is strong.
Then, symbolic form of statement: ‘It is wrong that, he is intelligent or strong’ is
(a) ~p ∨ ~p
(b) ~(p ∧ q)
(c) ~(p ∨ q)
(d) p ∨ ~q
Answer:
(c) ~(p ∨ q)

Question 7.
The negation of the proposition ‘If 2 is prime, then 3 is odd’, is
(a) If 2 is not prime, then 3 is not odd
(b) 2 is prime and 3 is not odd
(c) 2 is not prime and 3 is odd
(d) If 2 is not prime, then 3 is odd
Answer:
(b) 2 is prime and 3 is not odd

Question 8.
The statement (~p ∧ q) ∨ ~q is
(a) p ∨ q
(b) p ∧ q
(c) ~(p ∨ q)
(d) ~(p ∧ q)
Answer:
(d) ~(p ∧ q)
Hint:
(~p ∧ q) ∨ ~q = (~p ∨ ~q) ∧ (q ∨ ~q)
= (~p ∨ ~q) ∧ t
= ~p ∨ ~q
= ~(p ∧ q)

Question 9.
Which of the following is always true?
(a) ~(p → q) ≡ ~q → ~p
(b) ~(p ∨ q) ≡ ~p ∨ ~q
(c) ~(p → q) ≡ p ∧ ~q
(d) ~(p ∧ q) ≡ ~p ∧ ~q
Answer:
(c) ~(p → q) ≡ p ∧ ~q

Question 10.
~(p ∨ q) ∨ (~p ∧ q) is logically equivalent to
(a) ~p
(b) p
(c) q
(d) ~q
Answer:
(a) ~p
Hint:
~(p ∨ q) ∨ (~p ∧ q) ≡ (~p ∧ ~q) ∨ (~p ∧ q)
≡ ~p ∧ (~q ∨ q)
≡ ~p ∧ t
≡ ~p

Question 11.
If p and q are two statements, then (p → q) ↔ (~q → ~p) is
(a) contradiction
(b) tautology
(c) neither (a) nor (b)
(d) none of these
Answer:
(b) tautology

Question 12.
If p is the sentence ‘This statement is false’, then
(a) truth value of p is T
(b) truth value of p is F
(c) p is both true and false
(d) p is neither true nor false
Answer:
(d) p is neither true nor false

Question 13.
Conditional p → q is equivalent to
(a) p → ~q
(b) ~p ∨ q
(c) ~p → ~q
(d) p ∨ ~q
Answer:
(b) ~p ∨ q

Question 14.
Negation of the statement ‘This is false or That is true’ is
(a) That is true or This is false
(b) That is true and This is false
(c) This is true and That is false
(d) That is false and That is true
Answer:
(c) This is true and That is false

Question 15.
If p is any statement, then (p ∨ ~p) is a
(a) contingency
(b) contradiction
(c) tautology
(d) none of them
Answer:
(c) tautology

(II) Fill in the blanks:

Question 1.
The statement q → p is called as the ___________ of the statement p → q.
Answer:
Converse

Question 2.
Conjunction of two statements p and q is symbolically written as
Answer:
p ∧ q

Question 3.
If p ∨ q is true, then truth value of ~p ∨ ~q is ___________
Answer:
False

Question 4.
Negation of ‘some men are animal’ is ___________
Answer:
All men are not animal.
OR
No men are animals.

Question 5.
Truth value of if x = 2, then x² = -4 is ___________
Answer:
False

Question 6.
Inverse of statement pattern p → q is given by ___________
Answer:
~p → ~q

Question 7.
p ↔ q is false when p and q have ___________ truth values.
Answer:
Different

Question 8.
Let p : The problem is easy. r : It is not challenging. Then verbal form of ~p → r is ___________
Answer:
If the problem is not easy, then it is not challenging.

Question 9.
Truth value of 2 + 3 = 5 if and only if -3 > -9 is ___________
Answer:
T [Hint: T ↔ T = T]

(III) State whether each of the following is True or False:

Question 1.
Truth value of 2 + 3 < 6 is F.
Answer:
False

Question 2.
There are 24 months in a year is a statement.
Answer:
True

Question 3.
p ∧ q has truth value F if both p and q have truth value F.
Answer:
False

Question 4.
The negation of 10 + 20 = 30 is, it is false that 10 + 20 ≠ 30.
Answer:
False

Question 5.
Dual of (p ∧ ~q) ∨ t is (p ∨ ~q) ∨ c.
Answer:
False

Question 6.
Dual of ‘John and Ayub went to the forest’ is ‘John or Ayub went to the forest.’
Answer:
True

Question 7.
‘His birthday is on 29th February’ is not a statement.
Answer:
True

Question 8.
x² = 25 is true statement.
Answer:
False

Question 9.
The truth value of ‘√5 is not an irrational number’ is T.
Answer:
False

Question 10.
p ∧ t = p.
Answer:
True

(IV) Solve the following:

Question 1.
State which of the following sentences are statements in logic:
(i) Ice cream Sundaes are my favourite.
Solution:
It is a statement.

(ii) x + 3 = 8, x is variable.
Solution:
It is a statement.

(iii) Read a lot to improve your writing skill.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) z is a positive number.
Solution:
It is an open sentence, hence it is not a statement.

(v) (a + b)² = a² + 2ab + b² for all a, b ∈ R.
Solution:
It is a statement.

(vi) (2 + 1)² = 9.
Solution:
It is a statement.

(vii) Why are you sad?
Solution:
It is an interrogative sentence, hence it is not a statement.

(viii) How beautiful the flower is!
Solution:
It is an exclamatory sentence, hence it is not a statement.

(ix) The square of any odd number is even.
Solution:
It is a statement.

(x) All integers are natural numbers.
Solution:
It is a statement.

(xi) If x is a real number, then x² ≥ 0.
Solution:
It is a statement.

(xii) Do not come inside the room.
Solution:
It is an imperative sentence, hence it is not a statement.

(xiii) What a horrible sight it was!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question 2.
Which of the following sentences are statements? In case of a statement, write down the truth value:
(i) What is a happy ending?
Solution:
It is an interrogative sentence, hence it is not a statement.

(ii) The square of every real number is positive.
Solution:
It is a statement that is false, hence its truth value is F.

(iii) Every parallelogram is a rhombus.
Solution:
It is a statement that is true, hence its truth value is T.

(iv) a² – b² = (a + b)(a – b) for all a, b ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is T.

(v) Please carry out my instruction.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) The Himalayas is the highest mountain range.
Solution:
It is a statement that is true, hence its truth value is T.

(vii) (x – 2)(x – 3) = x² – 5x + 6 for all x ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is T.

(viii) What are the causes of rural unemployment?
Solution:
It is an interrogative sentence, hence it is not a statement.

(ix) 0! = 1.
Solution:
It is a statement that is true, hence its truth value is T.

(x) The quadratic equation ax² + bx + c = 0 (a ≠ 0) always has two real roots.
Solution:
It is a statement that is false, hence its truth value is F.

Question 3.
Assuming the first statement as p and second as q, write the following statements in symbolic form:
(i) The Sun has set and Moon has risen.
Solution:
Let p : The Sun has set.
q : Moon has risen.
Then the symbolic form of the given statement is p ∧ q.

(ii) Mona likes Mathematics and Physics.
Solution:
Let p : Mona likes Mathematics.
q : Mona likes Physics.
Then the symbolic form of the given statement is p ∧ q.

(iii) 3 is a prime number if 3 is a perfect square number.
Solution:
Let p : 3 be a prime number.
q : 3 is a perfect square number.
Then the symbolic form of the given statement is p ↔ q.

(iv) Kavita is brilliant and brave.
Solution:
Let p : Kavita is brilliant.
q : Kavita is brave.
Then the symbolic form of the given statement is p ∧ q.

(v) If Kiran drives a car, then Sameer will walk.
Solution:
Let p : Kiran drives a car.
q : Sameet will walk.
Then the symbolic form of the given statement is p → q.

(vi) The necessary condition for the existence of a tangent to the curve of the function is continuity.
Solution:
The given statement can be written as:
‘If the function is continuous, then the tangent to the curve exists.’
Let p : The function is continuous.
q : Tangent to the curve exists.
Then the symbolic form of the given statement is p → q.

(vii) To be brave is necessary and sufficient condition to climb Mount Everest.
Solution:
Let p : To be brave.
q : Climb Mount Everest.
Then the symbolic form of the given statement is p ↔ q.

(viii) x³ + y³ = (x + y)³, iff xy = 0.
Solution:
Let p : x³ + y³ = (x + y)³.
q : xy = 0.
Then the symbolic form of the given statement is p ↔ q.

(ix) The drug is effective though it has side effects.
Solution:
Let p : The drug is effective.
q : It has side effects.
Then the symbolic form of the given statement is p ∧ q.

(x) If a real number is not rational, then it must be irrational.
Solution:
Let p : A real number is not rational.
q : It must be irrational.
Then the symbolic form of the given statement is p → q.

(xi) It is not true that Ram is tall and handsome.
Solution:
Let p : Ram is tall.
q : Ram is handsome.
Then the symbolic form of the given statement is ~(p ∧ q).

(xii) Even though it is not cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is not cloudy and it is still raining,
Let p : It is not cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

(xiii) It is not true that intelligent persons are neither polite nor helpful.
Solution:
Let p : Intelligent persons are neither polite nor helpful.
Then the symbolic form of the given statement is ~p.

(xiv) If the question paper is not easy, then we shall not pass.
Solution:
Let p : The question paper is not easy.
q : We shall not pass.
Then the symbolic form of the given statement is p → q.

Question 4.
If p : Proof is lengthy.
q : It is interesting.
Express the following statements in symbolic form:
(i) Proof is lengthy and it is not interesting.
(ii) If the proof is lengthy, then it is interesting.
(iii) It is not true that the proof is lengthy but it is interesting.
(iv) It is interesting iff the proof is lengthy.
Solution:
The symbolic form of the given statements are:
(i) p ∧ ~q
(ii) p → q
(iii) ~(p ∧ q)
(iv) q ↔ p

Question 5.
Let p : Sachin win the match.
q : Sachin is a member of the Rajya Sabha.
r : Sachin is happy.
Write the verbal statement for each of the following:
(i) (p ∧ q) ∨ r
Solution:
Sachin wins the match and he is a member of the Rajya Sabha or Sachin is happy.

(ii) p → r
Solution:
If Sachin wins the match, then he is happy.

(iii) ~p ∨ q
Solution:
Sachin does not win the match or he is a member of the Rajya Sabha.

(iv) p → (q ∨ r)
Solution:
If Sachin wins the match, then he is a member of the Rajya Sabha or he is happy.

(v) p → q
Solution:
If Sachin wins the match, then he is a member of the Rajya Sabha.

(vi) (p ∧ q) ∧ ~r
Solution:
Sachin wins the match and he is a member of the Rajya Sabha but he is not happy.

(vii) ~(p ∨ q) ∧ r
Solution:
It is false that Sachin wins the match or he is a member of the Rajya Sabha but he is happy.

Question 6.
Determine the truth values of the following statements:
(i) 4 + 5 = 7 or 9 – 2 = 5.
Solution:
Let p : 4 + 5 = 7.
q : 9 – 2 = 5.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …….[F ∨ F ≡ F]

(ii) If 9 > 1, then x² – 2x + 1 = 0 for x = 1.
Solution:
Let p : 9 > 1.
q : x² – 2x + 1 = 0 for x = 1.
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]

(iii) x + y = 0 is the equation of a straight line if and only if y² = 4x is the equation of the parabola.
Solution:
Let p : x + y = 0 is the equation of a straight line.
q : y² = 4x is the equation of the parabola.
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]

(iv) It is not true that 2 + 3 = 6 or 12 + 3 = 5.
Solution:
Let p : 2 + 3 = 6.
q : 12 + 3 = 5.
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) ≡ ~F ≡ T]

Question 7.
Assuming the following statements
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth values of the following:

(i) Stock prices are not high or stocks are rising.
Solution:
p and q are true, i.e. T.
∴ ~p and ~q are false, i.e. F.
The given statement in symbolic form is ~p ∨ q.
Since, ~T ∨ T ≡ F ∨ T ≡ T, the given statement is true.
Hence, its truth value is ‘T’.

(ii) Stock prices are high and stocks are rising if and only if stock prices are high.
Solution:
The given statement in symbolic form is (p ∧ q) ↔ p.
Since (T ∧ T) ↔ T ≡ T ↔ T ≡ T, the given statement is true.
Hence, its truth value is ‘T’.

(iii) If stock prices are high, then stocks are not rising.
Solution:
The given statement in symbolic form is p → ~q.
Since, T → ~T ≡ T → F ≡ F, the given statement is false.
Hence, its truth value is ‘F’.

(iv) It is false that stocks are rising and stock prices are high.
Solution:
The given statement in symbolic form is ~(q ∧ p).
Since, ~(T ∧ T) ≡ ~T ≡ F, the given statement is false.
Hence, its truth value is ‘F’.

(v) Stock prices are high or stocks are not rising iff stocks are rising.
Solution:
The given statement in symbolic form is (p ∨ ~q) ↔ q.
Since (T ∨ ~T) ↔ T ≡ (T ∨ F) ↔ T
≡ T ↔ T
≡ T, the given statement is true.
Hence, its truth value is ‘T’.

Question 8.
Rewrite the following statements without using conditional:
[Hint: P → q ≡ ~p ∨ q]
(i) If price increases, then demand falls.
(ii) If demand falls, then the price does not increase.
Solution:
Since, p → q ≡ ~p ∨ q, the given statements can be written as:
(i) Price does not increase or demand falls.
(ii) Demand does not fall or price does not increase.

Question 9.
If p, q, r are statements with truth values T, T, F respectively, determine the truth values of the following:
(i) (p ∧ q) → ~p
Solution:
Truth values of p, q, r are T, T, F respectively.
(p ∧ q) → ~p ≡ (T ∧ T) → ~T
≡ T → F
≡ F
Hence, the truth value of the given statement is false, i.e. F.

(ii) p ↔ (q → ~p)
Solution:
p ↔ (q → ~p) ≡ T ↔ (T → ~T)
≡ T ↔ (T → F)
≡ T ↔ F
≡ F
Hence, the truth value of the given statement is false, i.e. F.

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

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

(v) ~[(p → q) ↔ (p ∧ ~q)]
Solution:
~[(p → q) ↔ (p ∧ ~q)]
≡ ~[(T → T) ↔ (T ∧ ~T)]
≡ ~[T ↔ (T ∧ F)]
≡ ~[T ↔ F]
≡ ~F
≡ T.
Hence, the truth value of the given statement is true, i.e. T.

Question 10.
Write the negations of the following:
(i) If ΔABC is not equilateral, then it is not equiangular.
Solution:
Let p : ΔABC is not equilateral.
q : It is not equiangular.
Then the symbolic form of the given statement is p → q.
Since, ~(p → q) ≡ p ∧ ~q, the negation of the given statement is:
‘ΔABC is not equilateral and it is equiangular.’

(ii) Ramesh is intelligent and he is hard working.
Solution:
Let p : Ramesh is intelligent.
q : He is hard working.
Then the symbolic form of the given statement is p ∧ q.
Since, ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is:
‘Ramesh is not intelligent or he is not hard-working.’

(iii) A angle is a right angle if and only if it is of measure 90°.
Solution:
Let p : An angle is a right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p ↔ q.
Since, ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of the given statement is:
‘An angle is a right angle and it is not of measure 90° or an angle is of measure 90° and it is not a right angle.’

(iv) Kanchanjunga is in India and Everest is in Nepal.
Solution:
Let p : Kanchenjunga is in India.
q : Everest is in Nepal.
Then the symbolic form of the given statement is p ∧ q.
Since, ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is:
‘Kanchenjunga is not in India or Everest is not in Nepal.’

(v) If x ∈ A ∩ B, then x ∈ A and x ∈ B.
Solution:
Let p : x ∈ A ∩ B, q : x ∈ A, r : x ∈ B.
Then the symbolic form of the given statement is P → (q ∧ r)
Since, ~(p → q) ≡ p ∧ ~q and ~(p ∧ q)= ~p ∨ ~q,
the negation of the given statement is:
‘x ∈ A ∩ B and x ∉ A or x ∉ B.

Question 11.
Construct the truth table for each of the following statement patterns:
(i) (p ∧ ~q) ↔ (q → p)
Solution:
(p ∧ ~q) ↔ (q → p)

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

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
(p ∧ r) → (p ∨ ~q)

(iv) (p ∨ r) → ~(q ∧ r)
Solution:
(p ∨ r) → ~(q ∧ r)

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

Question 12.
What is a tautology? What is a contradiction? Show that the negation of a tautology is a contradiction and the negation of a contradiction is a tautology.
Solution:
Tautology: A statement pattern that has all the entries in the last column of its truth table as T is called a tautology.
For example:

In the above truth table for the statement p ∨ ~p,
we observe that all the entries in the last column are T.
Hence, the statement p ∨ ~p is a tautology.

Contradiction: A statement pattern that has all the entries in the last column of its truth table as F is called a contradiction.
For example:

In the above truth table for the statement p ∧ ~p,
we observe that all the entries in the last column are F.
Hence, the statement p ∧ ~p is a contradiction.

To show that the negation of a tautology is a contradiction and vice versa:
A tautology is true on every row of its truth table.
Since, ~T = F and ~F = T, when we negate a tautology, the resulting statement is false on every row of its table.
i.e. the negation of tautology is a contradiction.
Similarly, the negation of a contradiction is a tautology.

Question 13.
Determine whether the following statement patterns is a tautology or a contradiction or a contingency:
(i) [(p ∧ q) ∨ (~p)] ∨ [p ∧ (~q)]
Solution:
[(p ∧ q) ∨ (~p)] ∨ [p ∧ (~q)]

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

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

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

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

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

(iv) [~(p ∧ q) → p] ↔ [(~p) ∧ (~q)]
Solution:
[~(p ∧ q) → p] ↔ [(~p) ∧ (~q)]

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

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

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

Question 14.
Using the truth table, prove the following logical equivalences:
(i) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

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

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

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

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

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

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

The entries in columns 3 and 10 are identical.
∴ p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)

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

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

Question 15.
Write the converse, inverse, contrapositive of the following statements:
(i) If 2 + 5 = 10, then 4 + 10 = 20.
Solution:
Let p : 2 + 5 = 10.
q : 4 + 10 = 20.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If 4 + 10 = 20, then 2 + 5 = 10.
Inverse: ~p → ~q is the inverse of p → q
i.e. If 2 + 5 ≠ 10, then 4 + 10 ≠ 20.
Cotrapositive: ~q → ~p is the contrapositive of p → q,
i.e. If 4 +10 ≠ 20, then 2 + 5 ≠ 10.

(ii) If a man is a bachelor, then he is happy.
Solution:
Let p : A man is a bachelor.
q : He is happy.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If a man is happy, then he is a bachelor.
Inverse: ~p → ~q is the inverse of p → q
i.e. If a man is not a bachelor, then he is not happy.
Contrapositive: ~q → ~p is the contrapositive of p → q
i.e., If a man is not happy, then he is not a bachelor.

(iii) If I do not work hard, then I do not prosper.
Solution:
Let p : I do not work hard.
q : I do not prosper.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If I do not prosper, then I do not work hard.
Inverse: ~p → ~q is the inverse of p → q
i.e. If I work hard, then I prosper.
Contrapositive: ~q → ~p is the contrapositive of p → q
i.e. If I prosper, then I work hard.

Question 16.
State the dual of each of the following statements by applying the principle of duality:
(i) (p ∧ ~q) ∨ (~p ∧ q) ≡ (p ∨ q) ∧ ~(p ∧ q)
(ii) p ∨ (q ∨ r) ≡ ~[(p ∧ q) ∨ (r ∨ s)]
(iii) 2 is an even number or 9 is a perfect square.
Solution:
The duals are given by:
(i) (p ∨ ~q) ∧ (~p ∨ q) ≡ (p ∧ q) ∨ ~(p ∨ q)
(ii) p ∧ (q ∧ r) ≡ ~[(p ∨ q) ∧ (r ∧ s)]
(iii) 2 is an even number and 9 is a perfect square.

Question 17.
Rewrite the following statements without using the connective ‘If … then’:
(i) If a quadrilateral is a rhombus, then it is not a square.
(ii) If 10 – 3 = 7, then 10 × 3 ≠ 30.
(iii) If it rains, then the principal declares a holiday.
Solution:
Since, p → q ≡ ~p ∨ q the given statements can be written as:
(i) A quadrilateral is not a rhombus or it is not a square.
(ii) 10 – 3 ≠ 7 or 10 × 3 ≠ 30.
(iii) It does not rain or the principal declares a holiday.

Question 18.
Write the dual of each of the following:
(i) (~p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q)
(ii) (p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
(iii) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
(iv) ~(p ∨ q) ≡ ~p ∧ ~q.
Solution:
The duals are given by:
(i) (~p ∨ q) ∧ (p ∨ ~q) ∧ (~p ∨ ~q)
(ii) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
(iii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
(iv) ~(p ∧ q) ≡ ~p ∧ ~q

Question 19.
Consider the following statements:
(i) If D is a dog, then D is very good.
(ii) If D is very good, then D is a dog.
(iii) If D is not very good, then D is not a dog.
(iv) If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.
Solution:
Let p : D is a dog. and q : D is very good.
Then the given statements in the symbolic form are:
(i) p → q
(ii) q → p
(iii) ~q → ~p
(iv) ~p → ~q

The entries in columns (i) and (iii) are identical. Hence, these statements are equivalent.
∴ the statements (i) and (iii) have the same meaning.
Similarly, the entries in columns (ii) and (iv) are identical. Hence, these statements are equivalent.
∴ the statements (ii) and (iv) have the same meaning.

Question 20.
Express the truth of each of the following statements by Venn diagrams:
(i) All men are mortal.
Solution:
Let U : a set of all human being
A : set of all men
B : set of all mortals.
Then the Venn diagram represents the truth of the given statement is as below:

(ii) Some persons are not politicians.
Solution:
Let U : set of all human being
A : set of all persons
B : set of all politicians.
Then the Venn diagram represents the truth of the given statement is as follows:

(iii) Some members of the present Indian cricket are not committed.
Solution:
Let U : set of all human being
X : set of all members of present Indian cricket
Y : set of all committed members of the present Indian cricket.
Then the Venn diagram represents the truth of the given statement is as below:

(iv) No child is an adult.
Solution:
Let U : set of all human beings
C : set of all children
A : set of all adults.
Then the Venn diagram represents the truth of the given statement is as below:

Question 21.
If A = {2, 3, 4, 5, 6, 7, 8}, determine the truth value of each of the following statements:
(i) ∃ x ∈ A, such that 3x + 2 > 9.
Solution:
Clearly x = 3, 4, 5, 6, 7, 8 ∈ A satisfy 3x + 2 > 9.
So, the given statement is true, hence its truth value is T.

(ii) ∀x ∈ A, x² < 18.
Solution:
x = 5, 6, 7, 8 ∈ A do not satisfy x² < 18.
So the given statement is false, hence its truth value is F.

(iii) ∃x ∈ A, such that x + 3 < 11.
Solution:
Clearly x = 2, 3, 4, 5, 6, 7 ∈ A which satisfy x + 3 < 11.
So, the given statement is True, hence its truth value is T.

(iv) ∀x ∈ A, x² + 2 ≥ 5.
Solution:
x² + 2 ≥ 5 for all x ∈ A.
So, the given statement is true, hence its truth value is T.

Question 22.
Write the negations of the following statements:
(i) 7 is a prime number and the Taj Mahal is in Agra.
Solution:
Let p : 7 be a prime number.
q : Taj Mahal is in Agra.
Then the symbolic form of the given statement is p ∧ q.
Since, (p ∧ q) ≡ ~p ∨ ~q,
the negation of the given statement is:
‘7 is not a prime number or Taj Mahal is not in Agra.’

(ii) 10 > 5 and 3 < 8.
Solution:
Let p : 10 > 5.
q : 3 < 8.
Then the symbolic form of the given statement is P ∧ q.
Since, ~(p ∧ q) = ~p ∨ ~q, the negation of the given statement is:
’10 ≤ 5 or 3 ≥ 8′
OR
’10 ≯ 5 or 3 ≮ 8′

(iii) I will have tea or coffee.
Solution:
The negation of the given statement is:
‘I will not have tea and coffee.’

(iv) ∀n ∈ N, n + 3 > 9.
Solution:
The negation of the given statement is:
‘∃n ∈ N, such that n + 3 ≯ 9.’
OR
‘∃n ∈ N, such that n + 3 ≤ 9.’

(v) ∃x ∈ A, such that x + 5 < 11.
Solution:
The negation of the given statement is:
‘∀x ∈ A, x + 5 ≮ 1.’
OR
‘∀x ∈ A, x + 5 ≥ 11.’