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Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.1

Question 1.
Evaluate \(\int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x\)
Solution:

Question 2.
Evaluate \(\int\left(1+x+\frac{x^{2}}{2 !}\right) d x\)
Solution:

Question 3.
Evaluate \(\int \frac{3 x^{3}-2 \sqrt{x}}{x} d x\)
Solution:

Question 4.
Evaluate ∫(3x² – 5)² dx
Solution:
∫(3x² – 5)² dx
= ∫(9x⁴ – 30x² + 25) dx
= 9∫x⁴ dx – 30∫x² dx + 25∫1 dx
= 9(\(\frac{x^{5}}{5}\)) – 30(\(\frac{x^{3}}{3}\)) + 25x + c
= \(\frac{9x^{5}}{5}\) – 10x³ + 25x + c.

Question 5.
Evaluate \(\int \frac{1}{x(x-1)} d x\)
Solution:

Question 6.
If f'(x) = x² + 5 and f(0) = -1, then find the value of f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(x² + 5) dx
= ∫x² dx + 5∫1 dx
= \(\frac{x^{3}}{3}\) + 5x + c
Now, f(0) = -1 gives
f(0) = 0 + 0 + c = -1
∴ c = -1
∴ from (1), f(x) = \(\frac{x^{3}}{3}\) + 5x – 1.

Question 7.
If f(x) = 4x³ – 3x² + 2x + k, f(0) = -1 and f(1) = 4, find f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(4x³ – 3x² + 2x + k) dx
= 4∫x³ dx – 3∫x² dx + 2∫x dx + k∫1 dx
= 4(\(\frac{x^{4}}{4}\)) – 3(\(\frac{x^{3}}{3}\)) + 2(\(\frac{x^{2}}{2}\)) + kx + c
∴ f(x) = x⁴ – x³ + x² + kx + c
Now, f(0) = 1 gives
f(0) = 0 – 0 + 0 + 0 + c = 1
∴ c = 1
∴ from (1), f(x) = x⁴ – x³ + x² + kx + 1
Further f(1) = 4 gives
f(1) = 1 – 1 + 1 + k + 1 = 4
∴ k = 2
∴ from (2), f(x) = x⁴ – x³ + x² + 2x + 1.

Question 8.
If f(x) = \(\frac{x^{2}}{2}\) – kx + 1, f(0) = 2 and f(3) = 5, find f(x).
Solution:
By the definition of integral

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 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 Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae^3x + Be^-3x
Solution:
y = Ae^3x + Be^-3x ……(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c₂x + c₁
Differentiating w.r.t. x, we get

(iii) y = (c₁ + c₂x) e^x
Solution:
y = (c₁ + c₂x) e^x


This is the required D.E.

(iv) y = c₁ e^3x+ c₂ e^2x
Solution:



This is the required D.E.

(v) y² = (x + c)³
Solution:
y² = (x + c)³
Differentiating w.r.t. x, we get

This is the required D.E.

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x² + y² = 2ax.
Solution:
x² + y² = 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x² + y² = [2x + 2y \(\frac{d y}{d x}\)]x = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y² – x² is the required D.E.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.

Question 4.
Find the differential equation whose general solution is x³ + y³ = 35ax.
Solution:

Question 5.
Form the differential equation from the relation x² + 4y² = 4b².
Sol ution:
x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Question 1.
Determine the order and degree of each of the following differential equations:
(i) \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
Solution:
The given D.E. is \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
This D.E. has highest order derivative \(\frac{d^{2} x}{d t^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

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

(iii) \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\) with power 1.
∴ the given D.E. is of order 4 and degree 1.

(iv) (y'”)² + 2(y”)² + 6y’ + 7y = 0
Solution:
The given D.E. is (y”‘)² + 2(y”)² + 6y’ + 7y = 0
This can be written as \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+6 \frac{d y}{d x}+7 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ the given D.E. is of order 3 and degree 2.

(v) \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
Solution:
The given D.E. is \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
On squaring both sides, we get
\(1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d y}{d x}\right)^{3}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d y}{d x}\right)^{5}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 5.
∴ the given D.E. is of order 1 and degree 5.

(vi) \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(vii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
i.e., \(\frac{d^{3} y}{d x^{3}}=9^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. is of order 3 and degree 1.

Question 2.
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:

Solution:
(i) xy = log y + k
Differentiating w.r.t. x, we get

Hence, xy = log y + k is a solution of the D.E. y'(1 – xy) = y².

(ii) y = xⁿ
Differentiating twice w.r.t. x, we get

This shows that y = xⁿ is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-n x \frac{d y}{d x}+n y=0\)

(iii) y = e^x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x = y
Hence, y = ex is a solution of the D.E. \(\frac{d y}{d x}\) = y.

(iv) y = 1 – log x
Differentiating w.r.t. x, we get

Hence, y = 1 – log x is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}=1\)

(v) y = ae^x + be^-x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = a(e^x) + b(-e^-x) = ae^x – be^-x
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = a(e^x) – b(-e^-x)
= ae^x + be^-x
= y
Hence, y = ae^x + be^-x is a solution of the D.E. \(\frac{d^{2} y}{d x^{2}}\) = y.

(vi) ax² + by² = 5
Differentiating w.r.t. x, we get

Hence, ax² + by² = 5 is a solution of the D.E.
\(x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}=y\left(\frac{d y}{d x}\right)\)

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Miscellaneous Exercise 3 Questions and Answers.

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

(I) Choose the correct alternative.

Question 1.
Regression analysis is the theory of
(a) Estimation
(b) Prediction
(c) Both a and b
(d) Calculation
Answer:
(c) Both a and b

Question 2.
We can estimate the value of one variable with the help of other known variable only if they are
(a) Correlated
(b) Positively correlated
(c) Negatively correlated
(d) Uncorrelated
Answer:
(a) Correlated

Question 3.
There are ________ types of regression equation
(a) 4
(b) 2
(c) 3
(d) 1
Answer:
(b) 2

Question 4.
In the regression equation of Y on X
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent.
Answer:
(a) X is independent and Y is dependent

Question 5.
In the regression equation of X on Y
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent
Answer:
(b) Y is independent and X is dependent

Question 6.
b_xy is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(b) Regression coefficient of X on Y

Question 7.
b_yx is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(a) Regression coefficient of Y on X

Question 8.
‘r’ is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(d) Correlation coefficient between X and Y

Question 9.
b_xy . b_yx = _________
(a) v
(b) y_x
(c) r²
(d) (y_y)²
Answer:
(c) r²

Question 10.
If b_yx > 1 then b_xy is ______
(a) > 1
(b) < 1
(c) > 0
(d) < 0
Answer:
(b) < 1

Question 11.
|b_xy + b_yx| > ______
(a) |r|
(b) 2|r|
(c) r
(d) 2r
Answer:
(b) 2|r|

Question 12.
b_xy and b_yx are ________
(a) Independent of change of origin and scale
(b) Independent of change of origin but not of the scale
(c) Independent of change of scale but not of origin
(d) Affected by change of origin and scale
Answer:
(b) Independent of change of origin but not of the scale

Question 13.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_yx = ________
(a) \(\frac{d}{c} b_{v u}\)
(b) \(\frac{c}{d} b_{v u}\)
(c) \(\frac{a}{b} b_{v u}\)
(d) \(\frac{b}{a} b_{v u}\)
Answer:
(a) \(\frac{d}{c} b_{v u}\)

Question 14.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_xy = ________
(a) \(\frac{d}{c} b_{u v}\)
(b) \(\frac{c}{d} b_{u v}\)
(c) \(\frac{a}{b} b_{u v}\)
(d) \(\frac{b}{a} b_{u v}\)
Answer:
(b) \(\frac{c}{d} b_{u v}\)

Question 15.
Corr(x, x) = ________
(a) 0
(b) 1
(c) -1
(d) can’t be found
Answer:
(b) 1

Question 16.
Corr (x, y) = ________
(a) corr(x, x)
(b) corr(y, y)
(c) corr(y, x)
(d) cov(y, x)
Answer:
(c) corr(y, x)

Question 17.
Corr\(\left(\frac{x-a}{c}, \frac{y-b}{d}\right)\) = -corr(x, y) if,
(a) c and d are opposite in sign
(b) c and d are same in sign
(c) a and b are opposite in sign
(d) a and b are same in sign
Answer:
(a) c and d are opposite in sign

Question 18.
Regression equation of X and Y is
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = b_xy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = b_yx (y – \(\bar{y}\))
Answer:
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))

Question 19.
Regression equation of Y and X is
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = b_xy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = b_yx (y – \(\bar{y}\))
Solution:
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))

Question 20.
b_yx = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)

Question 21.
b_xy = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)

Question 22.
Cov (x, y) = ________
(a) Σ(x – \(\bar{x}\))(y – \(\bar{y}\))
(b) \(\frac{\sum(x-\bar{x})(y-\bar{y})}{n}\)
(c) \(\frac{\sum x y}{n}-\bar{x} \bar{y}\)
(d) b and c both
Answer:
(d) b and c both

Question 23.
If b_xy < 0 and b_yx < 0 then ‘r’ is ________
(a) > 0
(b) < 0
(c) > 1
(d) not found
Answer:
(b) < 0

Question 24.
If equation of regression lines are 3x + 2y – 26 = 0 and 6x + y – 31 = 0 then means of x and y are ________
(a) (7, 4)
(b) (4, 7)
(c) (2, 9)
(d) (-4, 7)
Answer:
(b) (4, 7)

(II) Fill in the blanks:

Question 1.
If b_xy < 0 and b_yx < 0 then ‘r’ is ________
Answer:
negative

Question 2.
Regression equation of Y on X is ________
Answer:
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))

Question 3.
Regression equation of X on Y is ________
Answer:
(x – \(\bar{x}\)) = b_xy (y – \(\bar{y}\))

Question 4.
There are ______ types of regression equations.
Answer:
2

Question 5.
Corr (x₁ – x) = ______
Answer:
-1

Question 6.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_xy = ______
Answer:
\(\frac{c}{d} b_{u v}\)

Question 7.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_yx = ______
Answer:
\(\frac{d}{c} b_{v u}\)

Question 8.
|b_xy + b_yx| ≥ ______
Answer:
2|r|

Question 9.
If b_yx > 1 then b_xy is ______
Answer:
< 1

Question 10.
b_xy . b_yx = ______
Answer:
r²

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

Question 1.
Corr (x, x) = 1.
Answer:
True

Question 2.
Regression equation of X on Y is y – \(\bar{y}\) = b_xy (x – \(\bar{x}\)).
Answer:
False

Question 3.
Regression equation of Y on X is y – \(\bar{y}\) = b_yx (x – \(\bar{x}\)).
Answer:
True

Question 4.
Corr (x, y) = Corr (y, x).
Answer:
True

Question 5.
b_xy and b_yx are independent of change of origin and scale.
Answer:
False

Question 6.
‘r’ is the regression coefficient of Y on X.
Answer:
False

Question 7.
b_yx is the correlation coefficient between X and Y.
Answer:
False

Question 8.
If u = x – a and v = y – b then b_xy = b_uv.
Answer:
True

Question 9.
If u = x – a and v = y – b then r_xy = r_uv.
Answer:
True

Question 10.
In the regression equation of Y on X, b_yx represents the slope of the line.
Answer:
True

(IV) Solve the following problems.

Question 1.
The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y the percentage increase in weekly sales over the period just prior to the beginning of the campaign.

Find the equation of regression line to predict percentage increase in sales if the company has been in progress for 1.5 weeks.
Solution:
Let u = x – 3, v = y – 15


∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 14.67) = 2.6(x – 2.5)
y – 14.67 = 2.6x – 6.5
y = 2.6x + 8.17
When x = 1.5
y = (2.6)(1.5) + 8.17
= 3.9 + 8.17
= 12.07

Question 2.
The regression equation of y on x is given by 3x + 2y – 26 = 0. Find b_yx.
Solution:
Given, regression equation of Y on X is
3x + 2y – 26 = 0
∴ 2y = -3x + 26
∴ y = \(\frac{-3}{2}\)x + 13
∴ b_yx = \(\frac{-3}{2}\)

Question 3.
If for a bivariate data \(\bar{x}\) = 10, \(\bar{y}\) = 12, v(x) = 9, σ_y = 4 and r = 0.6. Estimate y when x = 5.
Solution:
Given, V(x) = 9
∴ σ_x = 3
b_yx = \(\frac{r \cdot \sigma_{y}}{\sigma_{x}}\)
= 0.6 × \(\frac{4}{3}\)
= 0.8
∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 12) = 0.8(5 – 10)
y – 12 = 0.8(-5)
y – 12 = -4
y = 8

Question 4.
The equation of the line of regression of y on x is v = \(\frac{2}{9} x\) and x on y is x = \(\frac{y}{2}+\frac{7}{6}\). Find (i) r (ii) \(\sigma_{y}^{2} \text { if } \sigma_{x}^{2}=4\).
Solution:

Question 5.
Identify the regression equations of x on y and y on x from the following equations.
2x + 3y = 6 and 5x + 7y – 12 = 0
Solution:

∴ Our assumption is correct
∴ Regression equation of Y on X is 2x + 3y = 6
∴ Regression equation of X on Y is 5x + 7y – 12 = 0

Question 6.
(i) If for a bivariate data b_yx = -1.2 and b_xy = -0.3 then find r.
(ii) From the two regression equations y = 4x – 5 and 3x = 2y + 5, find \(\bar{x}\) and \(\bar{y}\).
Solution:
r² = b_yx . b_xy
r² = (-1.2) × (-0.3)
r² = 0.36
r = ±0.6
Since, b_yx . b_xy are negative, r = -0.6
Also,(\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
y = 4x – 5, 3x = 2y + 5
8x – 2y = 10
3x – 2y = 5
on subtracting,
5x = 5
x = 1
Substituting x = 1 in y = 4x – 5
y = 4(1) – 5
y = -1
∴ \(\bar{x}\) = 1, \(\bar{y}\) = -1

Question 7.
The equation of the two lines of regression are 3x + 2y – 26 = 0 and 6x + y – 31 = 0. Find
(i) Means of X and Y
(ii) Correlation coefficient between X on Y
(iii) Estimate of Y for X = 2
(iv) var (X) if var (Y) = 36
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression lines
3x + 2y = 26
6x + y = 31
3x + 2y = 26 …….(i)
12x + 2y = 62 ……..(ii)
on subtracting,
-9x = -36
x = 4
Substituting x = 4 in equation (i)
3(4) + 2y = 26
2y = 14
y = 7
∴ \(\bar{x}\) = 4, \(\bar{y}\) = 7

Question 8.
Find the line of regression of X on Y for the following data:
n = 8, Σ(x_i – x)² = 36, Σ(y_i – y)² = 44, Σ(x_i – x)(y_i – y) = 24
Solution:

Question 9.
Find the equation of line regression of Y on X for the following data:
n = 8, Σ(x_i – \(\bar{x}\))(y_i – \(\bar{y}\)) = 120, \(\bar{x}\) = 20, \(\bar{y}\) = 36, σ_x = 2, σ_y = 3.
Solution:

Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 36) = 3.75(x – 20)
(y – 36) = 3.75x – 75
y = 3.75x – 39

Question 10.
The following result was obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.

and Σ(x_i – \(\bar{x}\))(y_i – \(\bar{x}\)) = 1120. Find the Prediction of blood pressure of a man of age 40 years.
Solution:

Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 140) = 0.7(40 – 50)
y – 140 = 0.7(-10)
y – 140 = -7
∴ y = 133

Question 11.
The equations of two regression lines are 10x – 4y = 80 and 10y – 9x = -40 Find:
(i) \(\bar{x}\) and \(\bar{y}\)
(ii) b_yx and b_xy
(iii) If var(Y) = 36, obtain var(X)
(iv) r
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression
10x – 4y = 80 ……(i)
-9x + 10y = -40 ……..(ii)
50x – 20y = 400
-18x + 20y = -80
32x = 320
x = 10
x = 10 in equation (i)
10(10) – 4y = 80
4y = 20
y = 5
∴ \(\bar{x}\) = 10, \(\bar{y}\) = 5


(iv) r² = b_yx . b_xy = 0.36
r = ±0.6
Since b_yx and b_xy are positive
∴ r = 0.6

Question 12.
If b_yx = -0.6 and b_xy = -0.216 then find correlation coefficient between X and Y comment on it.
Solution:
r² = b_yx . b_xy
r² = -0.6 × -0.216
r² = 0.1296
r = ±√0.1296
r = ± 0.36
Since b_yx and b_xy are negative
r = -0.36

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

(I) Choose the correct alternatives:

Question 1.
Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) \(\frac{3}{26}\) sq units
(c) 26 sq units
(d) 3 sq units
Answer:
(a) \(\frac{26}{3}\) sq units

Question 2.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) \(\frac{3}{28}\) sq units
Answer:
(c) \(\frac{28}{3}\) sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{3}{56}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer:
(c) \(\frac{56}{3}\) sq units

Question 4.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer:
(b) \(\frac{3124}{5}\) sq units

Question 5.
Using definite integration area of circle x² + y² = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer:
(c) 25π sq units

(II) Fill in the blanks:

Question 1.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is _________
Answer:
\(\frac{3124}{5}\) sq units

Question 2.
Using definite integration area of the circle x² + y² = 49 is ___________
Answer:
49π sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is _________
Answer:
\(\frac{56}{3}\) sq units

Question 4.
The area of the region bounded by the curve x² = y, the X-axis and the lines x = 3 and x = 9 is _________
Answer:
234 sq units

Question 5.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer:
\(\frac{28}{3}\) sq units

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

Question 1.
The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x d y=\int_{y=c}^{y=d} g(y) d y\)
Answer:
True

Question 2.
The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left|\int_{a}^{b} f(x) d x-\int_{b}^{a} g(x) d x\right|\)
Answer:
False

Question 3.
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
True

Question 4.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer:
False

Question 5.
The area of the portion lying above the X-axis is positive.
Answer:
True

(IV) Solve the following:

Question 1.
Find the area of the region bounded by the curve xy = c², the X-axis, and the lines x = c, x = 2c.
Solution:

= c² log(\(\frac{2 c}{c}\))
= c² . log 2 sq units.

Question 2.
Find the area between the parabolas y² = 7x and x² = 7y.
Solution:

For finding the points of intersection of the two parabolas,
we equate the values of y² from their equations.
From the equation x² = 7y, y² = \(\frac{x^{4}}{49}\)
∴ \(\frac{x^{4}}{49}\) = 7x
∴ x⁴ = 343x
∴ x⁴ – 343x = 0
∴ x(x³ – 343) = 0
∴ x = 0 or x³ = 343, i.e. x = 7
When x = 0, y = 0
When x = 7, 7y = 49
∴ y = 7
∴ the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) – (area of the region ODABO)
Now, area of the region ODACO = area under the parabola y² = 7x
i.e. y = √7 √x

Area of the region ODABO = Area under the parabola
x² = 7y
i.e. y = \(\frac{x^{2}}{7}\)

∴ required area = \(\frac{98}{3}-\frac{49}{3}=\frac{49}{3}\) sq units.

Question 3.
Find the area of the region bounded by the curve y = x² and the line y = 10.
Solution:

By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution:
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.

Question 5.
Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y d x\), where y = x²
= \(\int_{1}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{1}^{4}=\frac{4^{3}}{3}-\frac{1}{3}=\frac{64-1}{3}\)
= 21 sq units.

Question 6.
Find the area of the region bounded by the curve x² = 25y, y = 1, y = 4, and the Y-axis.
Solution:

Question 7.
Find the area of the region bounded by the parabola y² = 25x and the line x = 5.
Solution:

Given the equation of the parabola is y² = 25x
∴ y = 5√x …… [∵ IIn first quadrant, y > 0]
Required area = area of the region OQRPO
= 2(area of the region ORPO)