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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.6 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

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
\(\int \frac{d x}{x}\) = k∫dt + c
∴ 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 country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given log 2 = 0.6912, log 3 = 1.0986]
Solution:
Let P be the population at time t years.
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
\(\int \frac{d P}{P}\) = k∫dt + c
∴ log P = kt + c
Initially i.e. when t = 0, let P = P₀
∴ log P₀ = k x 0 + c
∴ c = log P₀
∴ log P = kt + log P₀
∴ log P – log P₀ = kt
∴ log(\(\frac{P}{P_{0}}\)) = kt ……(1)
Since, the population doubles in 60 years, i.e. when t = 60, P = 2P₀

∴ the population becomes triple in 95.4 years (approximately).

Question 3.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Solution:
Let θ°C be the temperature of the body at time t minutes.
The room temperature is given to be 25°C.
Then by Newton’s law of cooling, \(\frac{d \theta}{d t}\), the rate of change of temperature, is proportional to (θ – 25).


∴ the temperature of the body will be 36.36°C after 1 hour.

Question 4.
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 2½ hours. [Take √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
\(\int \frac{d x}{x}\) = k∫dt + c
∴ 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
∴ log(\(\frac{2000}{1000}\)) = k
∴ k = log 2

∴ the number of bacteria after 2½ hours = 5656.

Question 5.
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
Solution:
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is \(\frac{d m}{d t}\) which is proportional to m.


∴ log(3)^-1 = -kt
∴ -log 3 = -kt
∴ t = \(\frac{1}{k}\) log 3
∴ the original mass will disintegrate to 0.5 gm when t = \(\frac{1}{k}\) log 3

Question 6.
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution:
Let x gm be the amount of the substance left at time t.
Then the rate of decay is \(\frac{d x}{d t}\), which is proportional to x.


∴ \(\frac{x}{25}=\frac{27}{125}\)
∴ x = \(\frac{27}{5}\)
∴ the amount left after 3 hours \(\frac{27}{5}\) gm.

Question 7.
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
\(\int \frac{1}{P} d P\) = k∫dt + c
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k × 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt
∴ log(\(\frac{P}{30000}\)) = kt …….(1)
Now, when t = 40, P = 40000

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

Question 8.
A body cools according to Newton’s law from 100°C to 60°C in 20 minutes. The temperature of the surroundings is 20°C. How long will it take to cool down to 30°C?
Solution:
Let θ°C be the temperature of the body at time t.
The temperature of the surrounding is given to be 20°C.
According to Newton’s law of cooling


∴ the body will cool down to 30°C in 60 minutes, i.e. in 1 hour.

Question 9.
A right circular cone has a height of 9 cm and a radius of the base of 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec, where A is the area of the water surface
at that instant, show that the vessel will be full in 75 seconds.
Solution:

Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr² sq cm.
Since height of the right circular cone is 9 cm and radius of the base is 5 cm.
\(\frac{r}{h}=\frac{5}{9}\)
∴ r = \(\frac{5}{9} h\)
∴ area of water surface, i.e. A = \(\pi\left(\frac{5}{9} h\right)^{2}\)
∴ A = \(\frac{25 \pi h^{2}}{81}\) ……..(1)
The water level, i.e. the rate of change of h is \(\frac{d h}{d t}\) rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec.

∴ t = \(\frac{81 \times 9 \times 25}{3 \times 81}\) = 75
Hence, the vessel will be full in 75 seconds.

Question 10.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution:
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = \(\frac{4}{3}\)πr³ and S = 4πr²
The rate at which the raindrop evaporates is \(\frac{d V}{d t}\) which is proportional to the surface area.
∴ \(\frac{d V}{d t}\) ∝ S
∴ \(\frac{d V}{d t}\) = -kS, where k > 0 ………(1)

On integrating, we get
∫dr = -k∫dt + c
∴ r = -kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = -k × 0 + c
∴ c = 3
∴ r = -kt + 3
When t = 1, r = 2
∴ 2 = -k × 1 + 3
∴ k = 1
∴ r = -t + 3
∴ r = 3 – t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.

Question 11.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Solution:
Let P be the population of the city at time t.
Then the rate of growth of 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 = 0.04
∴ \(\frac{d P}{d t}\) = (0.04)P
∴ \(\frac{1}{P}\) dP = (0.04)dt
On integrating, we get
\(\int \frac{1}{P} d P\) = (0.04) ∫dt + c
∴ log P = (0.04)t + c
Initially, i.e., when t = 0, P = 10000
∴ log 10000 = (0.04) × 0 + c
∴ c = log 10000
∴ log P = (0.04)t + log 10000
∴ log P – log 10000 = (0.04)t
∴ log(\(\frac{P}{10000}\)) = (0.04)t
When t = 25, then
∴ log(\(\frac{P}{10000}\)) = 0.04 × 25 = 1
∴ log(\(\frac{P}{10000}\)) = log e ……[∵ log e = 1]
∴ \(\frac{P}{10000}\) = e = 2.7182
∴ P = 2.7182 × 10000 = 27182
∴ the population of the city after 25 years will be 27,182.

Question 12.
Radium decomposes at a rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution:
Let x be the amount of the radium at time t.
Then the rate of decomposition is \(\frac{d x}{d t}\) which is proportional to x.


Hence, \(\left(10-\frac{p}{10}\right)^{2} \%\) of the amount will be left after 2 years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\)
Solution:
\(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\) …….(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = \(\frac{1}{x}\) and Q = x³ – 3

This is the general solution.

(ii) cos²x . \(\frac{d y}{d x}\) + y = tan x
Solution:

(iii) (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:

(iv) \(\frac{d y}{d x}\) + y . sec x = tan x
Solution:
\(\frac{d y}{d x}\) + y sec x = tan x
∴ \(\frac{d y}{d x}\) + (sec x) . y = tan x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = sec x and Q = tan x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \sec x d x}\)
= \(e^{\log (\sec x+\tan x)}\)
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q . (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + tan²x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + sec²x – 1) dx + c
∴ (sec x + tan x) . y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.

(v) x \(\frac{d y}{d x}\) + 2y = x² . log x
Solution:

(vi) (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.

(vii) (x + a) \(\frac{d y}{d x}\) – 3y = (x + a)⁵
Solution:

(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ \(\frac{d r}{d \theta}\) + (2r cot θ + sin 2θ) = 0
∴ \(\frac{d r}{d \theta}\) + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
\(\frac{d r}{d \theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ

This is the general solution.

(ix) 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.

(x) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)
Solution:

(xi) \(\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)
Solution:

Question 2.
Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition,
\(\frac{d y}{d x}\) = x + 3y – 1
∴ \(\frac{d y}{d x}\) – 3y = x – 1 ………(1)
This is the linear differential equation of the form

This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y = 0 in (2), we get
0 = 2 + c
∴ c = -2
∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e^3x i.e. 3(x + 3y) = 2 (1 – e^3x).

Question 3.
Find the equation of the curve passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\) having slope of the tangent to the curve at any point (x, y) is \(-\frac{4 x}{9 y}\).
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}=-\frac{4 x}{9 y}\)
∴ y dy = \(-\frac{4}{9}\) x dx
Integrating both sides, we get
∫y dy= \(-\frac{4}{9}\) ∫x dx
∴ \(\frac{y^{2}}{2}=-\frac{4}{9} \cdot \frac{x^{2}}{2}+c_{1}\)
∴ 9y² = -4x² + 18c₁
∴ 4x² + 9y² = c where c = 18c₁
This is the general equation of the curve.
But the required curve is passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\).
∴ by putting x = \(\frac{3}{\sqrt{2}}\) and y = √2 in (1), we get
\(4\left(\frac{3}{\sqrt{2}}\right)^{2}+9(\sqrt{2})^{2}=c\)
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x² + 9y² = 36.

Question 4.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Solution:
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
x + y = \(\frac{d y}{d x}\) + 5
∴ \(\frac{d y}{d x}\) – y = x – 5 ………(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = -1 and Q = x – 5
∴ I.F. = \(e^{\int P d x}=e^{\int-1 d x}=e^{-x}\)
∴ the solution of (1) is given by

This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 – 0 + c
∴ c = -2
∴ from (2), the equation of the required curve is y = 4 – x – 2e^x.

Question 5.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) – xy = x ……….. (1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = -x and Q = x
∴ I.F. = \(e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}\)
∴ the solution of (1) is given by
y . (I.F.) = ∫Q . (I.F.) dx + c

This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is 1 + y = \(2 e^{\frac{x^{2}}{2}}\).

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.5

Question 1.
Find the second order derivatives of the following:
(i) 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9
Solution:
Let y = 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9

(ii) e^2x . tan x
Solution:
Let y = e^2x . tan x

(iii) e^4x . cos 5x
Solution:
Let y = e^4x . cos 5x

(iv) x³ . log x
Solution:
Let y = x³ . log x

(v) log(log x)
Solution:
Let y = log(log x)

(vi) x^x
Solution:
y = x^x
log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 2.
Find \(\frac{d^{2} y}{d x^{2}}\) of the following:
(i) x = a(θ – sin θ), y = a (1 – cos θ)
Solution:
x = a(θ – sin θ), y = a (1 – cos θ)
Differentiating x and y w.r.t. θ, we get
\(\frac{d x}{d \theta}=a \frac{d}{d \theta}(\theta-\sin \theta)\) = a(1 – cos θ) …….(1)

(ii) x = 2at², y = 4at
Solution:
x = 2at², y = 4at
Differentiating x and y w.r.t. t, we get

(iii) x = sin θ, y = sin³θ at θ = \(\frac{\pi}{2}\)
Solution:
x = sin θ, y = sin³θ
Differentiating x and y w.r.t. θ, we get,

(iv) x = a cos θ, y = b sin θ at θ = \(\frac{\pi}{4}\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

Question 3.
(i) If x = at² and y = 2at, then show that \(x y \frac{d^{2} y}{d x^{2}}+a=0\)
Solution:
x = at², y = 2at ………(1)
Differentiating x and y w.r.t. t, we get

(ii) If y = \(e^{m \tan ^{-1} x}\), show that \(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-m) \frac{d y}{d x}=0\)
Solution:
y = \(e^{m \tan ^{-1} x}\) ……..(1)

(iii) If x = cos t, y = e^mt, show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0\)
Solution:
x = cos t, y = e^mt

(iv) If y = x + tan x, show that \(\cos ^{2} x \cdot \frac{d^{2} y}{d x^{2}}-2 y+2 x=0\)
Solution:
y = x + tan x

(v) If y = e^ax . sin (bx), show that y₂ – 2ay₁ + (a² + b²)y = 0.
Solution:
y = e^ax . sin (bx) ………(1)

(vi) If \(\sec ^{-1}\left(\frac{7 x^{3}-5 y^{3}}{7 x^{3}+5 y^{3}}\right)=m\), show that \(\frac{d^{2} y}{d x^{2}}=0\)
Solution:

(vii) If 2y = \(\sqrt{x+1}+\sqrt{x-1}\), show that 4(x² – 1)y₂ + 4xy₁ – y = 0.
Solution:
2y = \(\sqrt{x+1}+\sqrt{x-1}\) …… (1)
Differentiating both sides w.r.t. x, we get

(viii) If y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\), show that \(\left(x^{2}+a^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=0\)
Solution:
y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\) = \(m \log \left(x+\sqrt{x^{2}+a^{2}}\right)\)

(ix) If y = sin(m cos^-1x), then show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+m^{2} y=0\)
Solution:
y = sin(m cos^-1x)
sin^-1y = m cos^-1x
Differentiating both sides w.r.t. x, we get

(x) If y = log(log 2x), show that xy₂ + y₁(1 + xy₁) = 0.
Solution:
y = log(log 2x)

(xi) If x² + 6xy + y² = 10, 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. x, we get

(xii) If x = a sin t – b cos t, y = a cos t + b sin t, Show that \(\frac{d^{2} y}{d x^{2}}=-\frac{x^{2}+y^{2}}{y^{3}}\)
Solution:
x = a sin t – b cos t, y = a cos t + b sin t
Differentiating x and y w.r.t. t, we get

Question 4.
Find the nth derivative of the following:
(i) (ax + b)^m
Solution:
Let y = (ax + b)^m

(ii) \(\frac{1}{x}\)
Solution:
Let y = \(\frac{1}{x}\)

(iii) e^ax+b
Solution:
Let y = e^ax+b

(iv) a^px+q
Solution:
Let y = a^px+q

(v) log(ax + b)
Solution:
Let y = log(ax + b)
Then \(\frac{d y}{d x}=\frac{d}{d x}[\log (a x+b)]\)

(vi) cos x
Solution:
Let y = cos x

(vii) sin(ax + b)
Solution:
Let y = sin(ax + b)

(viii) cos(3 – 2x)
Solution:

(ix) log(2x + 3)
Solution:

(x) \(\frac{1}{3 x-5}\)
Solution:
Let y = \(\frac{1}{3 x-5}\)

(xi) y = e^ax . cos (bx + c)
Solution:
y = e^ax . cos (bx + c)

(xii) y = e^8x . cos (6x + 7)
Solution:

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Question 1.
Find \(\frac{d y}{d x}\) if
(i) x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Solution:
x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Differentiating x and y w.r.t. m, we get
\(\frac{d x}{d m}=\frac{d}{d m}\left(\sqrt{a^{2}+m^{2}}\right)\)

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\), where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\) ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Solution:
x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Put t = tan θ Then θ = tan^-1t

(viii) x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Solution:
x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Put t = cos θ. Then θ = cos^-1t
x = cos^-1(4cos³θ – 3cos θ)

Question 2.
Find \(\frac{d y}{d x}\), if
(i) x = cosec²θ, y = cot³θ at θ = \(\frac{\pi}{6}\)
Solution:
x = cosec²θ, y = cot³θ
Differentiating x and y w.r.t. θ, we get

(ii) x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{3}\)
Solution:
x = a cos³θ, y = a sin³θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\)) at t = 1
Solution:
x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\))
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = \(\frac{\pi}{4}\)
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t = \(\frac{1}{2}\)
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

Question 3.
(i) If x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\), then show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution:
x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\)

(ii) If x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\), then show that \(\frac{d y}{d x}=-\frac{y \log x}{x \log y}\)
Solution:
x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\)
log x = log \(e^{\sin 3 t}\), log y = log \(e^{\cos 3 t}\)
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\), then show that y2 – \(\frac{d y}{d x}\) = 0.
Solution:
x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\)

(iv) If x = a cos³t, y = a sin³t, then show that \(\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}\)
Solution:
x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3), show that \(\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}\)
Solution:
x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3)

(vi) If x = log (1 + t²), y = t – tan^-1t, show that \(\frac{d y}{d x}=\frac{\sqrt{e^{x}-1}}{2}\)
Solution:
x = log (1 + t²), y = t – tan^-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\), show that sin x + \(\frac{d y}{d x}\) = 0
Solution:
x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\)
Differentiating x and y w.r.t. t, we get

(viii) If x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\), show that \(\frac{d x}{d y}=-\frac{b^{2} y}{a^{2} x}\)
Solution:
x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\)

Question 4.
(i) Differentiate x sin x w.r.t tan x.
Solution:
Let u = x sinx and v = tan x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

(ii) Differentiate \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) w.r.t \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)

(iii) Differentiate \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\) w.r.t \(\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\)
Solution:

(iv) Differentiate \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) w.r.t. tan^-1x
Solution:
Let u = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) and v = tan^-1x
Then we want to find \(\frac{d u}{d v}\)
Put x = tan θ. Then θ = tan^-1x.

(v) Differentiate 3x w.r.t. log_x3.
Solution:
Let u = 3x and v = log_x3.
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(3^{x}\right)=3^{x} \cdot \log 3\)

(vi) Differentiate \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) w.r.t. sec^-1x.
Solution:
Let u = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) and v = sec^-1x
Then we want to find \(\frac{d u}{d v}\).
Differentiating u and v w.r.t. x, we get

(vii) Differentiate x^x w.r.t. x^{sin x}.
Solution:
Let u = x^x and v = x^{sin x}
Then we want to find \(\frac{d u}{d x}\).
Take, u = x^x
log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get

(viii) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Solution:
Let u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and v = \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)
u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Put x = tan θ. Then θ = tan^-1x and

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 4.2 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 Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3

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

Question 1.
y = \(x^{x^{2 x}}\)
Solution:

Question 2.
y = \(x^{e^{x}}\)
Solution:

Question 3.
y = \(e^{x^{x}}\)
Solution:

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

Question 1.
y = \(\left(1+\frac{1}{x}\right)^{x}\)
Solution:

Question 2.
y = (2x + 5)^x
Solution:

Question 3.
y = \(\sqrt[3]{\frac{(3 x-1)}{(2 x+3)(5-x)^{2}}}\)
Solution:

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

Question 1.
y = (log x)^x + x^{log x}
Solution:

Question 2.
y = x^x + a^x
Solution:

Question 3.
y = \(10^{x^{x}}+10^{x^{10}}+10^{10^{x}}\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2

1. Find the rate of change of demand (x) of a commodity with respect to price (y) if:

Question 1.
y = 12 + 10x + 25x²
Solution:
Given y = 12 + 10x + 25x²

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{1}{10+50 x}\)

Question 2.
y = 18x + log(x – 4)
Solution:
Given y = 18x + log (x – 4)

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{x-4}{18 x-71}\)

Question 3.
y = 25x + log(1 + x²)
Solution:
Given y = 25x + log(1 + x²)

Hence, the rate of change of demand (x) with respect to price (y) \(\frac{d x}{d y}=\frac{1+x^{2}}{25 x^{2}+2 x+25}\)

2. Find the marginal demand of a commodity where demand is x and price is y.

Question 1.
y = xe^-x + 7
Solution:

Question 2.
y = \(\frac{x+2}{x^{2}+1}\)
Solution:

Question 3.
y = \(\frac{5 x+9}{2 x-10}\)
Solution: