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

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of \(\left(\frac{d y}{d x}\right)^{3}-\frac{d^{3} y}{d x^{3}}+y e^{x}=0\) are respectively.
(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
Answer:
(a) 3, 1

Question 2.
The order and degree of \(\left[1+\left(\frac{d y}{d x}\right)^{3}\right]^{\frac{2}{3}}=8 \frac{d^{3} y}{d x^{3}}\) are respectively
(a) 3, 1
(c) 3, 3
(b) 1, 3
(d) 1, 1
Answer:
(c) 3, 3

Question 3.
The differential equation of y = k₁ + \(\frac{k_{2}}{x}\) is
(a) \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
(d) \(x \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
Answer:
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

Question 4.
The differential equation of y = k₁ e^x + k₂ e^-x is
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
(b) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=0\)
(d) \(\frac{d^{2} y}{d x^{2}}+y=0\)
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)

Question 5.
The solution of \(\frac{d y}{d x}\) = 1 is
(a) x + y = c
(b) xy = c
(c) x² + y² = c
(d) y – x = c
Answer:
(d) y – x = c

Question 6.
The solution of \(\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=0\) is
(a) x³ + y³ = 7
(b) x² + y² = c
(c) x³ + y³ = c
(d) x + y = c
Answer:
(c) x³ + y³ = c

Question 7.
The solution of x \(\frac{d y}{d x}\) = y log y is
(a) y = ae^x
(b) y = be^2x
(c) y = be^-2x
(d) y = e^ax
Answer:
(d) y = e^ax

Question 8.
Bacterial increases at a rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
(a) 4 hours
(b) 6 hours
(c) 8 hours
(d) 10 hours
Answer:
(b) 6 hours

Question 9.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is
(a) x
(b) -x
(c) e^x
(d) e^-x
Answer:
(c) e^x

Question 10.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is e^-x, then its solution is
(a) ye^-x = x + c
(b) ye^x = x + c
(c) ye^x = 2x + c
(d) ye^-x = 2x + c
Answer:
(a) ye^-x = x + c

(II) Fill in the blanks:

Question 1.
The order of highest derivative occurring in the differential equation is called ________ of the differential equation.
Answer:
order

Question 2.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called ________ of the differential equation.
Answer:
degree

Question 3.
A solution of differential equation that can be obtained from the general solution by giving particular values to the arbitrary constants is called _________ solution.
Answer:
particular

Question 4.
Order and degree of a differential equation are always _________ integers.
Answer:
positive

Question 5.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is _________
Answer:
e^-x

Question 6.
The differential equation by eliminating arbitrary constants from bx + ay = ab is _________
Answer:
\(\frac{d^{2} y}{d x^{2}}=0\)

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

Question 1.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is e^-x.
Answer:
True

Question 2.
The order and degree of a differential equation are always positive integers.
Answer:
True

Question 3.
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
Answer:
True

Question 4.
The order of highest derivative occurring in the differential equation is called the degree of the differential equation.
Answer:
False

Question 5.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called the order of the differential equation.
Answer:
False

Question 6.
The degree of the differential equation \(e^{\frac{d y}{d x}}=\frac{d y}{d x}+c\) is not defined.
Answer:
True

(IV) Solve the following:

Question 1.
Find the order and degree of the following differential equations:
(i) \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given differential equation is \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
∴ \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3
∴ order = 3 and degree = 3

(ii) \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
Solution:
The given differential equation is \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ order = 1, degree = 2.

Question 2.
Verify that y = log x + c is a solution of the differential equation \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\).
Solution:
y = log x + c
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{x}+0=\frac{1}{x}\)
∴ x\(\frac{d y}{d x}\) = 1
Differentiating again w.r.t. x, we get
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \times 1=0\)
∴ \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
This shows that y = log x + c is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)

Question 3.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = 1 + x + y + xy
Solution:
\(\frac{d y}{d x}\) = 1 + x + y + xy
∴ \(\frac{d y}{d x}\) = (1 + x) + y(1 + x) = (1 + x)(1 + y)
∴ \(\frac{1}{1+y}\) dy = (1 + x) dx
Integrating, we get
∫\(\frac{1}{1+y}\) dy = ∫(1 + x) dx
∴ log|1 + y| = x + \(\frac{x^{2}}{2}\) + c
This is the general solution.

(ii) \(e^{d y / d x}=x\)
Solution:

∴ from (1), the general solution is
y = x log x – x + c, i.e. y = x(log x – 1) + c.

(iii) dr = ar dθ – θ dr
Solution:
dr = ar dθ – θ dr
∴ dr + θ dr = ar dθ
∴ (1 + θ) dr = ar dθ
∴ \(\frac{d r}{r}=\frac{a d \theta}{1+\theta}\)
On integrating, we get
\(\int \frac{d r}{r}=a \int \frac{d \theta}{1+\theta}\)
∴ log |r| = a log |1 + θ| + c
This is the general solution.

(iv) Find the differential equation of the family of curves y = e^x (ax + bx²), where a and b are arbitrary constants.
Solution:
y = e^x (ax + bx²)
ax + bx² = ye^-x …….(1)
Differentiating (1) w.r.t. x twice and writing \(\frac{d y}{d x}\) as y₁ and \(\frac{d^{2} y}{d x^{2}}\) as y₂, we get


This is the required differential equation.

Question 4.
Solve \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) when x = \(\frac{2}{3}\) and y = \(\frac{1}{3}\).
Solution:

Question 5.
Solve y dx – x dy = -log x dx.
Solution:
y dx – x dy = -log x dx
∴ y dx – x dy + log x dx = 0
∴ x dy = (y + log x) dx
∴ \(\frac{d y}{d x}=\frac{y+\log x}{x}=\frac{y}{x}+\frac{\log x}{x}\)
∴ \(\frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x}\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 6.
Solve y log y \(\frac{d x}{d y}\) + x – log y = 0.
Solution:

Question 7.
Solve (x + y) dy = a² dx
Solution:

Question 8.
Solve \(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\)
Solution:
\(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\) ……..(1)
This is a linear differential equation of the form

This is the general solution.

Question 9.
The rate of growth of the population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lakh, when will the city have a population of 400000?
Solution:
Let P be the population at time t years.
Then the rate of growth of the population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\)= k dt
On integrating, we get
\(\int \frac{d P}{P}=k \int d t\)
∴ log P = kt + c
The population doubled in 25 years and present population is 1,00,000.
∴ initial population was 50,000
i.e. when t = 0, P = 50000
∴ log 50000 = k × 0 + c
∴ c = log 50000
∴ log P = kt + log 50000
When t = 25, P = 100000
∴ log 100000 = k × 25 + log 50000
∴ 25k = log 100000 – log 50000 = log(\(\frac{100000}{50000}\))
∴ k = \(\frac{1}{25}\) log 2
∴ log P = \(\frac{t}{25}\) log 2 + log 50000
If P = 400000, then
log 400000 = \(\frac{t}{25}\) log 2 + log 50000
∴ log 400000 – log 50000 = \(\frac{t}{25}\) log 2
∴ log(\(\frac{400000}{50000}\)) = \(\log (2)^{t / 25}\)
∴ log 8 = \(\log (2)^{t / 25}\)
∴ 8 = \((2)^{t / 25}\)
∴ \((2)^{t / 25}\) = (2)³
∴ \(\frac{t}{25}\) = 3
∴ t = 75
∴ the population will be 400000 in (75 – 25) = 50 years.

Question 10.
The resale value of a machine decreases over a 10 years period at a rate that depends on the age of the machine. When the machine is x years old, the rate at which its value is changing is ₹ 2200(x – 10) per year. Express the value of the machine as a function of its age and initial value. If the machine was originally worth ₹ 1,20,000 how much will it be worth when it is 10 years old?
Solution:
Let V be the value of the machine after x years.
Then rate of change of the value is \(\frac{d V}{d x}\) which is 2200(x – 10)
∴ \(\frac{d V}{d x}\) = 2200(x – 10)
∴ dV = 2200(x – 10) dx
On integrating, we get
∫dV = 2200∫(x – 10) dx
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + c
Initially, i.e. at x = 0, V = 120000
∴ 120000 = 2200 × 0 + c = c
∴ c = 120000
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + 120000 …….(1)
This gives value of the machine in terms of initial value and age x.
We have to find V when x = 10.
When x = 10, from (1)
V = 2200[\(\frac{100}{2}\) – 100] + 120000
= 2200 [-50] + 120000
= -110000 + 120000
= 10000
Hence, the value of the machine after 10 years will be ₹ 10000.

Question 11.
Solve y² dx + (xy + x²) dy = 0
Solution:

Question 12.
Solve x²y dx – (x³ + y³) dy = 0
Solution:

Question 13.
Solve yx \(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 14.
Solve (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:
(x + 2y³) \(\frac{d y}{d x}\) = y
∴ x + 2y³ = y \(\frac{d x}{d y}\)

This is the general solution.

Question 15.
Solve y dx – x dy + log x dx = 0
Solution:
y dx – x dy + log x dx = 0
∴ (y + log x) dx = x dy


This is the general solution.

Question 16.
Solve \(\frac{d y}{d x}\) = log x dx
Solution:
\(\frac{d y}{d x}\) = log x dx
∴ dy = log x dx
On integrating, we get
∫dy = ∫log x . 1 dx
∴ y = (log x) ∫1 dx – \(\int\left[\left\{\frac{d}{d x}(\log x)\right\} \cdot \int 1 d x\right] d x\)
∴ y = (log x) . x – \(\int \frac{1}{x} \cdot x d x\)
∴ y = x log x – ∫1 dx
∴ y = x log x – x + c
This is the general solution.

Question 17.
y log y \(\frac{d x}{d y}\) = log y – x
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.6

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
∴ log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ……(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? (Given: \(\sqrt{\frac{3}{2}}\) = 1.2247)
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, k is a constant
∴ \(\frac{d P}{P}\) = k dt
Integrating, we get
∫\(\frac{d P}{P}\) = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 40000
∴ log 40000 = 0 + c
∴ c = log 40000
∴ log P = kt + log 40000
∴ log P – log 40000 = kt
∴ log(\(\frac{P}{40000}\)) = kt ………(1)
When t = 40, P = 60000

∴ population after 60 years will be 73482.

Question 3.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after \(\frac{5}{2}\) hours. [Given: √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt …….(1)
Now, when t = 1, x = 2 × 1000 = 2000

∴ number of bacteria after \(\frac{5}{2}\) hours = 5656.

Question 4.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
∫\(\frac{1}{P}\)dP = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k x 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 5.
The rate of depreciation \(\frac{d V}{d t}\) of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹ 1,00,000 in the first year. Find the value after 6 years.
Solution:
Let V be the value of the machine at the end of t years.
Then \(\frac{d V}{d t}\), the rate of depreciation, is inversly proportional to (t + 1)².

Initially, i.e. when t = 0, V = 800000
∴ 800000 = \(\frac{k}{1}\) + c = k + c ………(1)
Now, when t = 1, V = 800000 – 100000 = 700000
∴ 700000 = \(\frac{k}{1+1}\) + c = \(\frac{k}{2}\) + c ……(2)
Subtracting (2) from (1), we get
100000 = \(\frac{1k}{2}\)
∴ k = 200000
∴ from (1), 800000 = 200000 + c
∴ c = 600000 200000
∴ V = \(\frac{200000}{t+1}\) + 600000
When t = 6,
V = \(\frac{200000}{7}\) + 600000
= 28571.43 + 600000
= 628571.43 ~ 628571
Hence, the value of the machine after 6 years will be ₹ 6,28,571.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.5

Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}+y=e^{-x}\)
Solution:
\(\frac{d y}{d x}+y=e^{-x}\) …….(1)
This is the linear differential equation of the form

This is the general solution.

Question 2.
\(\frac{d y}{d x}\) + y = 3
Solution:
\(\frac{d y}{d x}\) + y = 3
This is the linear differential equation of the form

This is the general solution.

Question 3.
x\(\frac{d y}{d x}\) + 2y = x² . log x.
Solution:
x\(\frac{d y}{d x}\) + 2y = x² . log x
∴ \(\frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 4.
(x + y)\(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 5.
y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 6.
\(\frac{d y}{d x}\) + 2xy = x
Solution:
\(\frac{d y}{d x}\) + 2xy = x ………(1)
This is the linear differential equation of the form

This is the general solution.

Question 7.
(x + a) \(\frac{d y}{d x}\) = -y + a
Solution:
(x + a) \(\frac{d y}{d x}\) + y = a
∴ \(\frac{d y}{d x}+\left(\frac{1}{x+a}\right) y=\frac{a}{x+a}\) ……..(1)
This is the linear differential equation of the form

This is the general solution.

Question 8.
dy + (2y) dx = 8 dx
Solution:
dy + (2y) dx = 8 dx
∴ \(\frac{d y}{d x}\) + 2y = 8 …….(1)
This is the linear differential equation of the form

This is the general solution.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.4

Solve the following differential equations:

Question 1.
x dx + 2y dy = 0
Solution:
x dx + 2y dy = 0
Integrating, we get
∫x dx + 2 ∫y dy = c₁
∴ \(\frac{x^{2}}{2}+2\left(\frac{y^{2}}{2}\right)=c_{1}\)
∴ x² + 2y² = c, where c = 2c₁
This is the general solution.

Question 2.
y² dx + (xy + x²) dy = 0
Solution:
y² dx + (xy + x²) dy = 0
∴ (xy + x²) dy = -y² dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get



This is the general solution.

Question 3.
x²y dx – (x³ + y³) dy = 0
Solution:
x²y dx – (x³ + y³) dy = 0
∴ (x³ + y³) dy = x²y dx
∴ \(\frac{d y}{d x}=\frac{x^{2} y}{x^{3}+y^{3}}\) ……(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)


This is the general solution.

Question 4.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:



This is the general solution.

Question 5.
(x² – y²) dx + 2xy dy = 0
Solution:
(x² – y²) dx + 2xy dy = 0
∴ 2xy dy = -(x² – y²) dx = (y² – x²) dx
∴ \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) ………(1)

Question 6.
xy\(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 7.
x²\(\frac{d y}{d x}\) = x² + xy – y²
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.3

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = x²y + y
Solution:
\(\frac{d y}{d x}\) = x²y + y
∴ \(\frac{d y}{d x}\) = y(x² + 1)
∴ \(\frac{1}{y}\) dy = (x² + 1) dx
Integrating, we get
∫\(\frac{1}{y}\) dy = ∫(x² + 1) dx
∴ log |y|= \(\frac{x^{3}}{3}\) + x + c
This is the general solution.

(ii) \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)
Solution:
\(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)

This is the general solution.

(iii) (x² – yx²) dy + (y² + xy²) dx = 0
Solution:
(x² – yx²) dy + (y² + xy²) dx = 0
∴ x²(1 – y) dy + y²(1 + x) dx = 0
∴ \(\frac{1-y}{y^{2}} d y+\frac{1+x}{x^{2}} d x=0\)
Integrating, we get


This is the general solution.

(iv) \(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)
Solution:
\(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)

∴ 2y² log |x + 1| = 2cy² – 1 is the required solution.

Question 2.
For each of the following differential equations find the particular solution:
(i) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0.
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

∴ the general solution is
log |1 + x²| + log |1 – y²| = log c, where c₁ = log c
∴ log |(1 + x²)(1 – y²) | = log c
∴ (1 + x²)(1 – y²) = c
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(ii) (x + 1) \(\frac{d y}{d x}\) -1 = 2e^-y, when y = 0, x = 1.
Solution:

∴ log |2 + e^y| = log |c(x + 1)|
∴ 2 + e^y = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰ = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is
2 + e^y = \(\frac{3}{2}\)(x + 1)
∴ 4 + 2e^y = 3x + 3
∴ 3x – 2e^y – 1 = 0

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, when x = e, y = e².
Solution:

∴ from (1), the general solution is
log |x log x| – log |y| = log c, where c₁ = log c
∴ log |\(\frac{x \log x}{y}\)| = log c
∴ \(\frac{x \log x}{y}\) = c
∴ x log x = cy
This is the general solution.
Now, y = e², when x = e
e log e = ce²
1 = ce ……[∵ log e = 1]
c = \(\frac{1}{e}\)
∴ the particular solution is x log x = (\(\frac{1}{e}\)) y
∴ y = e^x log x

(iv) \(\frac{d y}{d x}\) = 4x + y + 1, when y = 1, x = 0.
Solution:
\(\frac{d y}{d x}\) = 4x + y + 1
Put 4x + y + 1 = v

∴ log |v + 4| = x + c
∴ log |4x + y + 1 + 4| = x + c
i.e. log |4x + y + 5| = x + c
This is the general solution.
Now, y = 1 when x = 0
∴ log|0 + 1 + 5| = 0 + c,
i.e. c = log 6
∴ the particular solution is
log |4x + y + 5| = x + log 6
∴ \(\log \left|\frac{4 x+y+5}{6}\right|\) = x

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 Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.2

Evaluate the following.

Question 1.
\(\int x \sqrt{1+x^{2}} d x\)
Solution:

Question 2.
\(\int \frac{x^{3}}{\sqrt{1+x^{4}}} d x\)
Solution:

Question 3.
\(\int\left(e^{x}+e^{-x}\right)^{2}\left(e^{x}-e^{-x}\right) d x\)
Solution:

Question 4.
\(\int \frac{1+x}{x+e^{-x}} d x\)
Solution:

Question 5.
∫(x + 1)(x + 2)⁷(x + 3) dx
Solution:
Let I = ∫(x + 1)(x + 2)⁷(x + 3) dx
= ∫(x + 2)⁷ (x + 1)(x + 3) dx
= ∫(x + 2)⁷ [(x + 2) – 1][(x + 2) + 1] dx
= ∫(x + 2)⁷ [(x + 2)² – 1] dx
= ∫[(x + 2)⁹ – (x + 2 )⁷] dx
= ∫(x + 2 )⁹ dx – ∫(x + 2)⁷ dx
= \(\frac{(x+2)^{10}}{10}\) – \(\frac{(x+2)^{8}}{8}\) + c

Question 6.
\(\int \frac{1}{x \log x} d x\)
Solution:
Put log x = t
∴ \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{d x}{x \cdot \log x}=\int \frac{1}{\log x} \cdot \frac{1}{x} d x\)
= ∫\(\frac{1}{t}\) dt
= log |t| + c
= log|log x| + c.

Question 7.
\(\int \frac{x^{5}}{x^{2}+1} d x\)
Solution:

Question 8.
\(\int \frac{2 x+6}{\sqrt{x^{2}+6 x+3}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x}+x} d x\)
Solution:

Question 10.
\(\int \frac{1}{x\left(x^{6}+1\right)} d x\)
Solution:

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.