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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 12 Discrete Mathematics Ex 12.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2

Question 1.
Let p : Jupiter is a planet and q: India is an island be any two simple statements. Give verbal sentence describing each of the following statements.
(i) ¬P
(ii) P ∧ ¬q
(iii) ¬p v q
(iv) p → ¬q
(v) p ↔ q
Solution:
p: Jupiter is a planet
q: India is an island
(i) ¬P : Jupiter is not a planet
(ii) P ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p v q : Jupiter is not a planet or India is an island
(iv) p → ¬q : Jupiter is a planet then India is not an island
(v) p ↔ q : Jupiter is a planet if and only if India is an island

Question 2.
Write each of the following sentences in symbolic form using statement variables p and q.
(i) 19 is not a prime number and all the angles of a triangle are equal.
(ii) 19 is a prime number or all the angles of a triangle are not equal.
(iii) 19 is a prime number and all the angles of a triangle are equal.
(iv) 19 is not a prime number.
Solution:
Let p : 19 is a prime number
q : All the angles of a triangle are equal
(i) 19 is not a prime number and all the angles of a triangle are equal ⇒ ¬p ∧ q
(ii) 19 is a prime number or all the angles of a triangle are not equal ⇒ p v ¬q
(iii) 19 is a prime number and all the angles of a triangle are equal ⇒ p ∧ q
(iv) 19 is not a prime number ⇒ ¬p

Question 3.
Determine the truth value of each of the following statements
(i) If 6 + 2 = 5, then the milk is white.
(ii) China is in Europe dr √3 is art integer
(iii) It is not true that 5 + 5 = 9 or Earth is a planet.
(iv) 11 is a prime number and all the sides of a rectangle are equal.
Solution:
(i) If 6 + 2 = 5, then milk is white,
p: 6 + 2 = 5(F)
q: Milk is white (T)
∴ p → q is having the truth value T

(ii) China is in Europe or √3 is an integer
p: China is in Europe (F)
q: √3 is an integer (F)
∴ p v q is having the truth value F

(iii) It is not true that 5 + 5 = 9 or Earth is a planet
p: 5 + 5 = 9 (F)
q: Earth is a planet (T)
∴ ¬p v q is having the truth value T

(iv) 11 is a prime number and all the sides of a rectangle are equal.
p : 11 is a prime number (T)
q: All the sides of a rectangle are equal (F)
∴ p ∧ q is having the truth value F

Question 4.
Which one of the following sentences is a proposition?

1. 4 + 7 = 12
2. What are you doing?
3. 3ⁿ ≤ 81, n ∈ N
4. Peacock is our national bird.
5. How tall this mountain is!

Solution:

1. is a proposition
2. not a proposition
3. is a proposition
4. is a proposition
5. not a proposition

Question 5.
Write the converse, inverse, and contrapositive of each of the following implication.
(i) If x and y are numbers such that x = y, then x² = y²
(ii) If a quadrilateral is a square then it is a rectangle.
Solution:
(i) Converse: If x and y are numbers such that x² = y² then x = y.
Inverse: If x and y are numbers such that x ≠ y then x² ≠ y².
Contrapositive: If x and v are numbers such that x² ≠ y² then x ≠ y.

(ii) Converse: If a quadrilateral is a rectangle then it is a square.
Inverse: If a quadrilateral is not a square then it is not a rectangle.
Contrapositive: If a quadrilateral is not a rectangle then it is not a square.

Question 6.
Construct the truth table for the following statements.
(i) ¬P ∧ ¬q
(ii) ¬(P ∧ ¬q)
(iii) (p v q) v ¬q
(iv) (¬p → r) ∧ (p ↔ q)
Solution:
(i) ¬P ∧ ¬q

(ii) ¬(P ∧ ¬q)

(iii) (p v q) v ¬q

(iv) (¬p → r) ∧ (p ↔ q)

Question 7.
Verify whether the following compound propositions are tautologies or contradictions or contingency.
(i) (P ∧ q) ∧¬ (p v q)
(ii) ((P v q) ∧¬p) → q
(iii) (p → q) ↔ (¬p → q)
(iv) ((p → q) ∧ (q → r)) → (p → r)
Solution:
(i) (P ∧ q) ∧¬ (p v q)

The entries in the last column are only F.
∴ The given statement is a contradiction

(ii) ((P v q) ∧¬p) → q

The entries in the last column are only T.
∴ The given statement is a Tautology

(iii) (p → q) ↔ (¬p → q)

The entries in the last column are a combination of T and F.
∴ The given statement is a contingency.

(iv) ((p → q) ∧ (q → r)) → (p → r)

All the entries in the last column are only T.
∴ The given statement is a tautology.

Question 8.
Show that (i) ¬(p ∧ q) ≡ ¬P v ¬q
(ii) ¬(p → q) ≡ p ∧¬q
Solution:
(i) ¬(p ∧ q) ≡ ¬P v ¬q

The entries in the columns corresponding to ¬(p ∧ q) and ¬P v ¬q are identical and hence they are equivalent.

(ii) ¬(p → q) ≡ p ∧ ¬q

The entries in the columns corresponding to ¬(p → q) and p ∧ ¬q are identical and hence they are equivalent.

Question 9.
Prove that q → p ≡ ¬p → ¬q
Solution:
q → p ≡ ¬p → ¬q

The entries in the columns corresponding to q → p and ¬p → ¬q are identical and hence they are equivalent.
∴ q → q = ¬p → ¬q
Hence proved.

Question 10.
Show that p → q and q → p are not equivalent
Solution:

From the table, it is clear that
p → q ≠ q → P

Question 11.
Show that ¬(p ↔ q) ≡ p ↔ ¬q
Solution:

From the table, it is clear that
¬(p ↔ q) ≡ p ↔ ¬q

Question 12.
Check whether the statement p → (q → p) is a tautology or a contradiction without using the truth table.
Solution:
P → (q → p)
≡ P → (¬q v p) [∵ implication law]
≡ ¬p v (¬q v p) [∵ implication law]
≡ ¬p v (p v ¬q) [∵ Commutative law]
≡ (¬p V p) v (¬p v ¬q) [∵ Distribution law]
≡ T v ¬(p ∧ q) ≡ T [Tautology]
Hence p → (q → p) is a Tautology.

Question 13.
Using the truth table check whether the statements ¬(p v q) v (¬p ∧ q) and ¬P are logically equivalent.
Solution:

From the table, it is clear that ¬P and
¬(p v q) v (¬p ∧ q) are logically equivalent
i.e. ¬(p v q) v (¬p ∧ q) ≡ ¬p

Question 14.
Prove p → (q → r) ≡ (p ∧ q) → r without using the truth table.
Solution:
P → (q → r)
≡ P → (¬q v r) [∵ implication law]
≡ ¬p v (¬q v r) [∵ implication law]
≡ (¬p v ¬q) v r [∵ Associative law]
≡ ¬(p ∧ p) v r [∵ DeMorgan’s law]
≡ (p ∧ q) → r ≡ T [∵ implication law]
Hence Proved.

Question 15.
Prove that p → (¬q v r) ≡ ¬p v (¬q v r) using truth table.
Solution:

From the table, it is clear that the column of p → (¬q v r) and ¬p v (¬q v r) are identical.
∴ p → (¬q v r) ≡ ¬p v (¬q v r)
Hence proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.2

Question 1.
Find the differential dy for each of the following functions.
(i) y = \(\frac { (1-2x)^3 }{ 3-4x }\)
(ii) y = (3 + sin2x)^2/3
(iii) y = e^{x2 – 5x +7} cos(x² – 1)
Solution:

(ii) y = (3 + sin2x)^2/3

Question 2.
Find df for f(x) = x² + 3x and evaluate it for
(i) x = 2 and dx = 0.1
(ii) x = 3 and dx = 0.02
Solution:
y = f(x) = x² + 3x
dy = (2x + 3) dx
(i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1)
= 7(0.1) = 0.7

(ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2)
= 9(0.02) = 0.18

Question 3.
Find Δf and df for the function f for the indicated values of x, Δx and compare:
(i) f(x) = x³ – 2x², x = 2, Δx = dx = 0.5
(ii) f(x) = x² + 2x + 3, x = -0.5, Δx = dx = 0.1
Solution:
(i) y = f(x) = x³ – 2x²
dy = (3x² – 4x) dx
dy (when x = 2 and dx = 0.5) = [3(2²) – 4(2)] (0.5)
= (12 – 8)(0.5) = 4(0.5) = 2
(i.e.,) df = 2
Now ∆f = f(x + ∆x) – f(x)
Here x = 2 and ∆x = 0.5
f(x) = x³ – 2x²
So f(x + ∆x) = f(2 + 0.5) = f(2.5) = (2.5)³ – 2 (2.5)² = (2.5)² [2.5 – 2] = 6.25 (0.5) = 3.125
f(x) = f(2) = 2³ – 2(2²) = 8 – 8 = 0
So ∆f = 3.125 – 0 = 3.125

(ii) y = f(x) = x² + 2x + 3
dy = (2x + 2) dx
dy (when x = – 0.5 and dx = 0.1)
= [2(-0.5) + 2] (0.1)
= (-1 + 2) (0.1) = (1) (0.1) = 0.1
(i.e.,) df = 0.1
Now ∆f = f(x + ∆x) – f(x)
Here x = -0.5 and ∆x = 0.1
x² + 2x + 3
f(x + ∆x) = f(-0.5 + 0.1) = f(-0.4)
= (-0.4)² + 2(-0.4) + 3
= 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36
f(x) = f(-0.5) = (-0.5)² + 2(-0.5) + 3
= 0.25 – 1 + 3 = 3.25 – 1 = 2.25
So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11

Question 4.
Assuming log₁₀ e = 0.4343, find an approximate value of Iog₁₀ 1003.
Solution:
Let f(x) = log ₁₀ x then
f ‘(x) = \(\frac { 1 }{ x }\) log₁₀ e (log₁₀ x = log₁₀ e log_e x)
f(x + Δx) – f(x) = f ‘(x) Δx
f(1003) – f(1000) = \(\frac { 0.4344 }{ 1000 }\) × 3
log₁₀ 1003 – log₁₀ 1000 = 0.0013029
log₁₀ 1003 = log₁₀ 10³ + 0.0013029
= 3 + 0.0013029
= 3.0013029
Approximate value of log₁₀ 1003 = 3.0013029

Question 5.
The trunk of a tree has a diameter of 30 cm. During the following year, the circumference grew 6 cm.
(i) Approximately how much did the tree diameter grow?
(ii) What is the percentage increase in the area of the cross-section of the tree?
Solution:
(i) Diameter of the trunk of the tree
D = 30 cm
Rate of change of circumference
ds = 6 cm per year
Circumference S = πD
dS = πdD
6 = πdD
\(\frac { 6 }{ π }\) = dD
Rate of increasing diameter = \(\frac { 6 }{ π }\) cm

Question 6.
An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is 5 mm and the radius to the outside of the shell is 5.3 mm, find the volume of the shell approximately.
Solution:
Radius of the inside shell = 5 mm
Radius of the outside shell = 5.3 mm
Volume V = \(\frac { 4 }{ 3 }\) πr³
dV = \(\frac { 4 }{ 3 }\) π3r²dr
= 4 π 5 × 5 × 0.3
= 100 π × 0.3
= 30 π
Approximate volume of the shell = 30 mm³

Question 7.
Assume that the cross-section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from 2 mm to 2.1 mm, how much is the cross-sectional area increased approximately?
Solution:
The radius of an artery section = 2 mm
dr = 2.1 – 2
= 0.1
Area A = πr²
dA = 2πrdr
= 2 × π × 2 × 0.1
= 0.4 π
Increased area = 0.4 π mm²

Question 8.
In a newly developed city, it is estimated that the voting population (in thousands) will increase according to V(t) =30 + 12t² – t³, 0 ≤ t ≤ 8 where t is the time in years. Find the approximate change in voters for the time change from 4 to 4 1/6 years.
Solution:
V(t) = 30 + 12t² – t³; dt = 4 \(\frac { 1 }{ 6 }\) – 4 = \(\frac { 1 }{ 6 }\)
V’(t) = (24t – 3t²)dt
= (24(4)-3 (4)²) × \(\frac { 1 }{ 6 }\)
= (96 – 48) × \(\frac { 1 }{ 6 }\)
= 48 × \(\frac { 1 }{ 6 }\)
= 8
Voters in thousands
∴ Approximate change of voters = 8 × 1000 = 8000

Question 9.
The relation between the number of words y a person learns in x hours is given by y = 52√x, 0 ≤ x ≤ 9. What is the approximate number of words learned when x changes from
(i) 1 to 1.1 hours?
(ii) 4 to 4.1 hours?
Solution:
y = 52 √x
dy = 52 × \(\frac { 1 }{ 2 }\) × x^-1/2 dx
x = 1, dx = 0.1
\(\frac { 26 }{ √x }\) × 0.1 = 26 × 0.1
= 2.6
≅ 3 words

(ii) y = 52 √y
dy = 52 × \(\frac { 1 }{ 2 }\) × x^-1/2 dx
x = 4, dx = 0.1
\(\frac { 26 }{ √4 }\) × 0.1 = 13 × 0.1
= 1.3
≅ 1 word

Question 10.
A circular plate expands uniformly under the influence of heat. If its radius increases from 10.5 cm to 10.75 cm, then find an approximate change in the area and the approximate percentage change in the area.
Solution:
Area of the circular plate A = πr²
= π × 10.5 × 105
= 110.25 π
dA = 2πrdr
= 2π × 10.5 × 0.251
= 5.25 π
Approximate percentage change in the area
= \(\frac { dA }{ A }\) × 100
= \(\frac { 5.25π }{ 110.25π }\) × 100
= 0.04761 × 100
= 4.76%

Question 11.
A coat of paint of thickness 0.2 cm is applied to the faces of cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.
Solution:
v = a³
so dv = a² da
dv (when) a = 10 cm and da = 0.20 cm
= 3(10²) (0.2)
300 × 0.2 = 60 cm³

Actual paint used = v at x + ∆x = 10.2 and x = 10 cm
= a³ at x + ∆x = 10.2 and x = 10
= (10.2)³ – (10) = 61.2 cm³

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1

Question 1.
Let f(x) = \(\sqrt[3] { x }\). Find the linear approximation at x = 27. Use the linear approximation to approximate \(\sqrt[3] { 27.2 }\)
Solution:
x = 27
f(x) = \(\sqrt[3] { 27 }\) = 3
We need to find the value of \(\sqrt[3] { 27.2 }\)
We know that
f(x₀ + Δx) = f(x₀) + f'(x₀) Δx

∴ Approximate value of \(\sqrt[3] { 27.2 }\) = 3.0074

Question 2.
Use the linear approximation to find approximate value of
(i) (123)^2/3
(ii) \(\sqrt[4] { 15 }\)
(iii) \(\sqrt[4] { 26 }\)
Solution:
(i) (123)^2/3
Let x₀ = 125, Δx = -2
f(x) = x^2/3, f(x₀) = 25
We know that
f(x₀ + Δx) = f(x₀) + f'(x₀) Δx

= 25 – 0.2666
(123)^2/3 = 24.7334

(ii) \(\sqrt[4] { 15 }\) = (15)^1/4
f(x) = x^1/4, f(x₀) = (16)^1/4 = 2
We know that
f(x₀ + Δx) = f(x₀) + f’(x₀) Δx

= 2 – 0.03125
(15)^1/4 = 1.96875

(iii) (26)^1/3
f(x) = x^1/3, f(x₀) = (27)^1/3 = 2, Δx = -1
We know that
f(x₀ + Δx) = f(x₀) + f’(x₀) Δx

= 3 – 0.370
(26)^1/3 = 2.963

Question 3.
Find a linear approximation for the following functions at the indicated points
(i) f(x) = x³ – 5x + 12, x₀ = 2
(ii) g(x) = \(\sqrt { x^2+9 }\), x₀ = -4
(iii) h(x) = \(\frac { x }{ x+1 }\), x₀ = 1
Solution:
(i) We know that the linear approximation
L(x) = f(x₀) + f’(x₀)(x – x₀)
f(x) = x³ – 5x + 12
f'(x) = 3x² – 5
f'(x₀) = f'(2) = 12 – 5 = 7
f(x₀) = f(2) = 8 – 10 + 12 = 10
L(x) = 10 + 7 (x – 2)
= 10 + 7x – 14
= 7x – 4

(ii) g(x) = \(\sqrt { x^2+9 }\), x₀ = -4
g(x₀) = g(14) = \(\sqrt {16+9 }\) = 5

(iii) h(x) = \(\frac { x }{ x+1 }\), x₀ = 1

Question 4.
The radius of a circular plate is measured as 12.65. cm instead of the actual length 12.5 cm find the following in calculating the area of the circular plate:
(i) Absolute error
(ii) Relative error
(iii) Percentage error
Solution:
Actual radius of the circular plate = 12.5 cm
Measured radius of the circular plate = 12.65
dr = 12.65 – 12.5
= 0.15
A = π r²
dA = 2π r dv
Change in Area
A(12.65) – A(12.5) = dA
= 2π × 12.5 × 0.15
= 3.75 π
Exact calculation of the Area changes gives
A(12.65) – A(12.5) = π(12.65)² – π(12.5)²
= 160.0225 π – 156.25 π
= 3.7725 π
Absolute error = 3.7725 π – 3.75 π
= 0.0225 π cm²
Relative error
= \(\frac { 3.7725π-3.75π }{ 3.7725π }\)
= \(\frac { 0.0225π }{ 3.7725π }\)
= 0.00596
= 0.006 cm²
Percentage error = Relative error × 100
= 0.006 × 100
= 0.6% .

Question 5.
A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:
(i) Change in the volume.
(ii) Change in the surface area
Solution:
(i) Given r = 10
dr = 10 – 9.8 = 0.2
Volume v = \(\frac { 4 }{ 3 }\)πr³
dv = \(\frac { 4 }{ 3 }\).3πr²dv
Change in volume
v(10) – v(9.8) = 4π(10)²(0.2)
= 80π cm³

(ii) Surface area of the sphere
S(r) = 4πr²
S'(r) = 8πr
Change in surface area at r = 10 is
= S'(r) [10 – 9.8]
= 8π (10) (0.2) = 16π cm²
∴ Surface Area decreases by 16π cm²

Question 6.
The time T, taken for a complete oscillation of a single pendulum with length l, is given by the equation T = 2π\(\sqrt{\frac { l }{ g }}\) where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of l.
Solution:
Given T = 2π\(\sqrt{\frac { l }{ g }}\)
On taking log both sides, we get
log T = log 2 + log π + \(\frac { 1 }{ 2 }\) log l – \(\frac { 1 }{ 2 }\) log g
On differentiating both sides w. r. to l, we get

So, the percentage error in T is 1%

Question 7.
Show that the percentage error in the nth root of a number is approximately \(\frac { 1 }{ n }\) times the percentage error in the number.
Solution:
Let x be the number
Let y = x^1/n
log y = \(\frac { 1 }{ n }\) log x
Taking differentiate on both sides, we have

= \(\frac { 1 }{ n }\) times the percentage error in the number.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.10

Choose the most suitable answer from the given four alternatives:

Question 1.
The volume of a sphere is increasing in volume at the rate of 3π cm³/ sec. The rate of change of its radius when radius is \(\frac { 1 }{ 2 }\) cm
(a) 3 cm/s
(b) 2 cm/s
(c) 1 cm/s
(d) \(\frac { 1 }{ 2 }\) cm/s
Solution:
(a) 3 cm/s
Hint:
Volume V = \(\frac { 4 }{ 3 }\) πr³
Given \(\frac { dV }{ dt }\) = 3π cm³/sec
Differentiating w.r.t. ‘t’ r = \(\frac { 1 }{ 2 }\) cm

Rate of change of radius is 3 cm/sec.

Question 2.
A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. Find the rate of change of the balloon’s angle of elevation in radian per second when the balloon is 30 metres above the ground.
(a) \(\frac { 3 }{ 25 }\) radan/sec
(b) \(\frac { 4 }{ 25 }\) radian/sec
(c) \(\frac { 1 }{ 5 }\) radian/sec
(d) \(\frac { 1 }{ 3 }\) radian/sec
Solution:
(b) \(\frac { 4 }{ 25 }\) radian/sec
Hint:

Question 3.
The position of a particle moving along a horizontal line of any time t is given by s(t) = 3t² – 2t – 8. The time at which the particle is at rest is
(a) t = 0
(b) t = \(\frac { 1 }{ 3 }\)
(c) t = 1
(d) t = 3
Solution:
(b) t = \(\frac { 1 }{ 3 }\)
Hint:
s(t) = 3t² – 2t – 8
Velocity V = \(\frac { ds }{ dt }\) = 6t – 2
When the particle comes to rest,
velocity V = 0
6t – 2 = 0
t = \(\frac { 1 }{ 3 }\)

Question 4.
A stone is thrown, up vertically. The height reaches at time t seconds is given by x = 80t – 16t². The stone reaches the maximum! height in time t seconds is given by
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(b) 2.5
Hint:
x = 80t – 16t²
Velocity V = \(\frac { ds }{ dt }\) = 80 – 32t
When it reaches the maximum height
V = 0 ⇒ 80 – 32t = 0
t = \(\frac { 80 }{ 32 }\) = 2.5

Question 5.
Find the point on the curve 6y = x³ + 2 at which y-coordinate changes 8 times as fast as x-coordinate is
(a) (4, 11)
(b) (4, -11)
(c) (-4, 11)
(d) (-4, -11)
Solution:
(a) (4, 11)
Hint:

x = ±4
When x = 4, y = 11
∴ Point on the curve is (4, 11).

Question 6.
The abscissa of the point on the curve f(x) = \(\sqrt { 8-2x }\) at which the slope of the tangent is -0.25?
(a) -8
(b) -4
(c) -2
(d) 0
Solution:
(b) -4
Hint:
f(x) = \(\sqrt { 8-2x }\)
f'(x) = –\(\frac { 2 }{ 2\sqrt { 8-2x } }\) = –\(\frac { 1 }{ \sqrt { 8-2x } }\)
Slope of the tangent is – 0.25
ie„ f'(x) = -0.25
–\(\frac { 1 }{ \sqrt { 8-2x } }\) = -0.25 = \(\frac { -1 }{ 4 }\)
\(\sqrt { 8-2x }\) = 4
8 – 2x = 16
x = –\(\frac { 8 }{ 2 }\) = -4
Abscissa x = -4

Question 7.
The slope of the line normal to the curve f(x) = 2 cos 4x at x = \(\frac { π }{ 12 }\) is
(a) -4√3
(b) -4
(c) –\(\frac { √3 }{ 12 }\)
(d) 4√3
Solution:
(c) –\(\frac { √3 }{ 12 }\)
Hint:
f(x) = 2 cos 4x
f'(x) = -8 sin 4x
Slope of the normal at x = \(\frac { π }{ 12 }\) is

Question 8.
The tangent to the curve y² – xy + 9 = 0 is vertical when
(a) y = 0
(b) y = ±√3
(c) –\(\frac { 1 }{ 2 }\)
(d) y = ±3
Solution:
(d) y = ±3
Hint:
y² – xy + 9 = 0 ……… (1)
2y\(\frac { dy }{ dx }\) – (x\(\frac { dy }{ dx }\) + y) = 0
\(\frac { dy }{ dx }\) (2y – x) = y
\(\frac { dy }{ dx }\) = \(\frac { y }{ 2y-x }\)
When the tangent is vertical \(\frac { dy }{ dx }\) = ∞
i.e., \(\frac { y }{ 2y-x }\) = \(\frac { 0 }{ 1 }\)
⇒ 2y – x = 0
2y = x
sub in (1)
y² – 2y² + 9 = 0
⇒ y² = 9
y = ±3

Question 9.
The angle between y² = x and x² = y at the origin is
(a) tan^-1 \(\frac { 3 }{ 4 }\)
(b) tan^-1 (\(\frac { 4 }{ 3 }\))
(c) \(\frac { π }{ 2 }\)
(d) \(\frac { π }{ 4 }\)
Solution:
(c) \(\frac { π }{ 2 }\)
Hint:
y² = x and x² = y are the standard forms of parabolas for which y-axis and x-axis are the two tangents respectively.
Angle between x-axis and y-axis is \(\frac { π }{ 2 }\)

Question 10.
The value of the limit \(\lim _{x \rightarrow 0}\) (cot x – \(\frac { 1 }{ x }\)) is
(a) 0
(b) 1
(c) 2
(d) ∞
Solution:
(d) ∞
Hint:

Question 11.
The function sin⁴ x + cos⁴ x is increasing in the interval
(a) [ \(\frac { 5π }{ 8 }\), \(\frac { 3π }{ 4 }\) ]
(b) [ \(\frac { π }{ 2 }\), \(\frac { 5π }{ 8 }\) ]
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
(d) [ 0, \(\frac { π }{ 4 }\) ]
Solution:
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
Hint:
f(x) = sin⁴x + cos⁴x
f'(x) = 4 sin³ x cos x – 4 cos³ x sin x
f'(x) = 0 ⇒ 4 sin x cos x (sin²x – cos²x) = 0
sin x = 0; cos x = 0; sin² – cos² x = 0
x = 0; x = \(\frac { π }{ 2 }\); sin² x = cos² x
x = \(\frac { π }{ 4 }\)

In [0, \(\frac { π }{ 2 }\) ], f'(x) = -ve ⇒ f(x) is decreasing
In [ \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\) ], f'(x) = +ve ⇒ f(x) is increasing

Question 12.
The number is given by Rolle’s theorem for the function x³ – 3x², x ∈ [0, 3] is
(a) 1
(b) √2
(c) \(\frac { 3 }{ 2 }\)
(d) 2
Solution:
(d) 2
Hint:
f(x) = x³ – 3x²
f'(x) = 3x² – 6x
f'(x) = 0
⇒ 3x (x – 2) = 0
x = 0, 2

Question 13.
The number given by the Mean value theorem for the function \(\frac { 1 }{ x }\), x ∈ [1, 9] is
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(c) 3
Hint:

Question 14.
The minimum value of the function |3 – x| + 9 is
(a) 0
(b) 3
(c) 6
(d) 9
Solution:
(d) 9
Hint:
f(x) = |3 – x | + 9
Minimum value of |3 – x | = 0
Minimum value of |3 – x| + 9 = 0 + 9 = 9 and No Maximum value.

Question 15.
The maximum slope of the tangent to the curve y = e^x sin x, x ∈ [0, 2π] is at
(a) x = \(\frac { π }{ 4 }\)
(b) x = \(\frac { π }{ 2 }\)
(c) x = π
(d) x = \(\frac { 3π }{ 2 }\)
Solution:
(b) x = \(\frac { π }{ 2 }\)
Hint:
y = e^x sin x, x ∈ [0, 2π] dy
Slope ‘S’ = \(\frac { dy }{ dx }\) = e^x cos x + e^x sin x
S = e^x (cos x + sin x)
\(\frac { dS }{ dx }\) = e^x (-sin x + cos x) + (cos x + sin x)e^x
= e^x (2 cos x)
For maximum or minimum,
\(\frac { dS }{ dx }\) = 0 ⇒ 2e^x cos x = 0
e^x = 0 is not possible
∴ cos x = 0
x = \(\frac { π }{ 2 }\)

Question 16.
The maximum value of the function x² e^-2x, x > 0 is
(a) \(\frac { 1 }{ e }\)
(b) \(\frac { 1 }{ 2e }\)
(c) \(\frac { 1 }{ e^2 }\)
(d) \(\frac { 4 }{ e^4 }\)
Solution:
(c) \(\frac { 1 }{ e^2 }\)
Hint:
Let f(x) = x²e^-2x, x > 0
f'(x) = -2x² e^-2x + e^-2x (2x)
f'(x) = 0 ⇒ -2xe^-2x(x – 1) = 0
x = 0 and x = 1
f(x) attains maximum at x = 1 as f”(x) < 0
when x = 1
∴ Maximum value f(1) = (1)² e^-2 = \(\frac { 1 }{ e^2 }\)

Question 17.
One of the closest points on the curve x² – y² = 4 to the point (6, 0) is
(a) (2, 0)
(b) (√5, 1)
(c) (3, √5)
(d) (\(\sqrt { 13 }\), -√3)
Solution:
(c) (3, √5)
Hint:
x² – y² = 4
y² = x² – 4
y = ±\(\sqrt { x^2-4 }\)
Any point on the curve is (x, ± \(\sqrt { x^2-4 }\))
Distance between (6, 0) and (x, ± \(\sqrt { x^2-4 }\)) is \(\sqrt { (x-6)^2+x^2-4 }\)
Substituting all’ the given options, we get minimum distance.
∴ Required point is (3, √5)

Question 18.
The maximum value of the product of two positive numbers’, when their sum of the squares is 200, is
(a) 100
(b) 25√7
(c) 28
(d) 24\(\sqrt { 14 }\)
Solution:
(a) 100
Hint:
Given x² + y² = 200
y² = 200 – x²
y = \(\sqrt { 200-x^2 }\)
Product P = xy = x\(\sqrt { 200-x^2 }\)
\(\frac { dP }{ dx }\) = \(\frac { x(-2x) }{ 2\sqrt{200-x^2} }\) + \(\sqrt{200-x^2}\) …. (1)
\(\frac { dP }{ dx }\) = 0 ⇒ -2x² + 200 = 0
x² = 100
x = 10
∴ y = \(\sqrt{200-100}\) = 10
∴ Maximum product is P = xy
= (10) (10)
= 100

Question 19.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Solution:
(d) has no points of inflection
Hint:
y = ax⁴ + bx²
\(\frac { dy }{ dx }\) = 4ax³ + 2bx
\(\frac { d^2y }{ dx^2 }\) = 12ax² + 2b
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 12ax² + 2b = 0
x² = –\(\frac { b }{ 6a }\)
x is unreal.
Hence no points of inflection.

Question 20.
The point of inflection of the curve y = (x – 1)³ is
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)
Solution:
(c) (1, 0)
Hint:
y = (x – 1)³
\(\frac { dy }{ dx }\) = 3(x – 1)²
\(\frac { d^2y }{ dx^2 }\) = 6(x – 1)
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 6(x – 1) = 0
x = 1
y = f(x)
\(\frac { dy }{ dx }\) = f'(x)
\(\frac { d^2y }{ dx^2 }\) = f”(x)
In (-∞, 1), f”(x) < 0, curve is concave down In (1, ∞), f”(x) > 0, curve is concave up
f”(x) changes its sign when passing through x = 1
when x = 1, y = 0
∴ (1, 0) is the point of inflection.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.9

Question 1.
Find the asymptotes of the following curves:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
(ii) f(x) = \(\frac { x^2 }{ x+1 }\)
(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)
(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)
Solution:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
The function becomes undefined when x = 1 and x = -1
∴ x = 1 and x = -1 are the vertical asymptotes

As ‘x’ gets larger (Positive or negative) the function, the function attaining the value 1.
∴ y = 1 is horizontal asymptote

(ii) f(x) = \(\frac { x^2 }{ x+1 }\)

The function becomes undefined when x = -1
∴ Vertical asymptote is x = -1 and there is no horizontal asymptote.
No horizontal asymptote exists for the curve. Oblique asymptote can be obtained by polynomial long division method.

Oblique (or) slant asymptote is y = x – 1

(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
No vertical asymptotes
Horizontal asymptotes RHL (Right Hand Limit)


y = 3 and y = -3 are the Horizontal asymptotes
Slant asymptotes’. No such slant asymptotes exist for the given curve.

(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)

When x = -3, the function becomes undefined.
∴ x = -3 is the vertical asymptote.
No Horizontal asymptote exist for the curve.
Oblique asymptote can be obtained by polynomial long division method

∴ y = x – 9 is the slant (or) oblique asymptote.

(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)

The function becomes undefined when x = 2.
∴ x = 2 is the vertical asymptote.
No Horizontal asymptote exist for the given curve.
Oblique asymptote can be obtained by polynomial long division method.
∴ y = \(\frac { x }{ 3 }\) + \(\frac { 8 }{ 3 }\) (or) 3y = x + 8 is the slant asymptote.

Question 2.
Sketch the graphs of the following functions
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)
(ii) y = x \(\sqrt { 4-x }\)
(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)
(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)
(v) y = \(\frac { x^3 }{ 24 }\) – log x
Solution:
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)

Factorizing we get
y = –\(\frac { 1 }{ 3 }\) (x – 1)² (x + 2) = f(x)
The domain and the range of the given function f(x) are the entire real line.
Putting y = 0, we get x = 1, 1, – 2. Hence the x-intercepts are (1, 0) and (- 2, 0) and by putting x = 0. We get y = –\(\frac { 2 }{ 3 }\). Therefore, the y-intercept is (0, –\(\frac { 2 }{ 3 }\))
f'(x) = \(\frac { (3x^2-3) }{ 3 }\) = -(x² – 1) = 1 – x²
f'(x) = 0 ⇒ 1 – x² = 0 ⇒ x = ±1
The critical points of the curve occur at x = ± 1 .
f”(x) = -2x
f”(1) = – 2 < 0, ∴ f(x) is maximum at x = 1 and the local maximum is f(1) = o
f”(-1) = 2 > 0, ∴ f(x) is minimum at x = -1 and the local minimum is
f(-1) = –\(\frac { 4 }{ 3 }\)
f”(x) = – 2x < 0 ∀ x > 0, ∴ The function is concave downward in the positive real line.
f”(x) = 2x > 0 ∀ x < 0, ∴ The function is concave upward in the negative real line.
Since f”(x) = 0 at x = 0 and f”(x) changes its sign when passing through x = 0.
Hence the point of inflection is (0, –\(\frac { 2 }{ 3 }\))
The curve has no asymptotes.

(ii) y = x\(\sqrt { 4-x }\)

y = x\(\sqrt { 4-x }\) = f(x)
where x > 4 the curve does not exist and it exists for x ≤ 4
∴ The domain is (-∞, 4] and the Range is (-∞, \(\frac { 16 }{ 3√3 }\) ]
The curve passes through the origin. The curve intersects x-axis at (4, 0).
f'(x) = –\(\frac { x }{ 2 \sqrt{4-x} }\) + \(\sqrt { 4-x }\) = \(\frac { 8-3x }{ 2 \sqrt{4-x} }\)
f'(x) = 0 ⇒ 8 – 3x = 0 ⇒ x = \(\frac { 8 }{ 3 }\)
∴ Critical point of the curve occur at x = \(\frac { 8 }{ 3 }\)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\)
f”(\(\frac { 8 }{ 3 }\)) = –\(\frac { 3√3 }{ 4 }\) < 0
∴ f(x) is maximum at x = \(\frac { 8 }{ 3 }\) and the local maximum f(\(\frac { 8 }{ 3 }\)) = \(\frac { 16 }{ 3√3 }\) and local minimum is 0 at x = 4 (from the graph)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\) < 0 ∀ x < 4
∴ The curve is concave downward in the negative real line.
No point of inflection exists.
As x → ∞, y → ±∞ , and hence the curve does not have any asymptotes.

(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)

The domain of the given function f(x) is (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)
ie. x < -2 (or) -2 < x < 2 (or) x > 2.
Range of f(x) is (-∞, –\(\frac { 1 }{ 4 }\)) ∪ (1, ∞)
i.,e. f(x) ≤ –\(\frac { 1 }{ 4 }\) (or) f(x) > 1.
Putting y = 0, x is unreal. Hence, there is no ‘x’ intercept.
By putting x = 0, we get y = –\(\frac { 1 }{ 4 }\).
∴ y intercept is (0, –\(\frac { 1 }{ 4 }\))
f'(x) = –\(\frac { 10x }{ (x^2-4)^2 }\)
f'(x) = 0 ⇒ x = 0,
∴ The critical point is at x = 0
f'(x) = \(\frac { 10(x^2-4)(3x^+4) }{ (x^2-4)^4 }\)
f'(0) = –\(\frac { 5 }{ 8 }\) < 0,
∴ f(x) is maximum at
x = 0. Hence the local maximum is f(0) = –\(\frac { 1 }{ 4 }\)
No points of inflection exist for the curve.
When x = ± 2, y = ∞
∴ Vertical asymptotes are x = 2 and x = -2 and Horizontal asymptote is y = 1.

(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)

The Domain of the function f(x) is the entire real line.
ie., (-∞, ∞) ⇒ -∞ < x < ∞ and the range is (0, 1) ie., 0 < f(x) < 1
No ‘x’ intercept for f(x) and when x = 0
y = \(\frac { 1 }{ 2 }\)
∴ The ‘y’ intercept is (0, \(\frac { 1 }{ 2 }\))
f'(x) = \(\frac { e^{-x} }{ (1+e^{-x})^2 }\)
f'(x) = 0 ⇒ which is absurd. Hence there is no extremum.
No vertical asymptote for the curve exist and the Horizontal asymptotes are y = 1 and y = 0.

(v) y = \(\frac { x^3 }{ 24 }\) – log x

The curve exists only for positive values of ‘x’ (x > 0) ie., domain is (0, ∞) and
The range is (\(\frac { 1 }{ 3 }\) – log e², ∞)
No intersection points are possible
f'(x) = \(\frac { x^2 }{ 8 }\) – \(\frac { 1 }{ x }\)
f'(x) = 0 ⇒ x³ – 8 = 0 ⇒ x = 2
∴ Critical point occur at x = 2
f'(x) = \(\frac { x }{ 4 }\) + \(\frac { 1 }{ x^2 }\)
f”(2) = \(\frac { 3 }{ 4 }\) > 0,
∴ f(x) is mimmum at x = 2 and the local minimum is f(2) = \(\frac { 1 }{ 3 }\) – log e²
No point of inflection exists.
No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis).

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.
Solution:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) = 12x – x²
To find the maximum product.
p'(x) = 12 – 2x
p”(x) = -2
p'(x) = 0 ⇒ 12 – 2x = 0 ⇒ 2x = 12
⇒ x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12 – x = 12 – 6 = 6
So the two numbers are 6, 6

Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.
Solution: Let the two positive numbers be ‘x’ and ‘y’
Given product is 20 ⇒ xy = 20 ⇒ y = \(\frac { 20 }{ x }\)
Sum S = x + y
S = x + \(\frac { 20 }{ x }\)
\(\frac { dS }{ dx }\) = 1 – \(\frac { 20 }{ x^2 }\)
For maximum or minimum, \(\frac { dS }{ dx }\) = 0
x² – 20 = 0 x² = 20
x = ±2√5
[x= -2√5 is not possible
\(\frac { d^2S }{ dx^2 }\) = \(\frac { 40 }{ x^3 }\)
at x = 2√5, \(\frac { d^2S }{ dx^2 }\) > 0
∴ Sum ‘S’ is minimum when x = 2√5
y = \(\frac { 20 }{ 2√5 }\) = 2√5
Minimum sum = 2√5 + 2√5 = 4√5

Question 3.
Find the smallest possible value of x² + y² given that x + y = 10.
Solution:
Given x + y = 10 ⇒ y = 10 – x
Let A = x² + y²
A = x² + (10 – x)²
\(\frac { dA }{ dx }\) = 2x + 2(10 – x)(-1)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ 2(2x – 10) = 0
x = 5
\(\frac { d^2A }{ dx^2 }\) = 4
at x = 5, \(\frac { d^2A }{ dx^2 }\) > 0
∴ A is minimum when x = 5
y = 10 – 5 = 5
∴ The smallest possible value of x² + y² is
(5)² + (5)² = 25 + 25 = 50

Question 4.
A garden is to be laid out in a rectangular area and protected by a wire fence. What is the largest possible area of the fenced garden with 40 meters of wire?
Solution:

Perimeter = 40 m
2(l + b) = 40 ⇒ l + b = 20
Let l = x m
b = (20 – x)m
Area = l × b = x(20 – x) = 20x – x²
To find the maximum area
A(x) = 20x – x²
A'(x) = 20 – 2x
A”(x) = -2
A'(x) = 0 ⇒ 20 – 2x = 0
⇒ x = 10
x = 10 is a maximum point
:. Maximum Area = x (20 – x)
= 10(20 – 10)
= 10 × 10 = 100 sq.m.

Question 5.
A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at the other sides of the page are 1 cm. What should be the dimensions’ of the page so that the area of the paper used is minimum?
Solution:
Let the width of the printed part be ‘x’ cm
Let the height Of the printed part be ‘y’ cm
Given, Area of the printed part = 24 cm²
i.e., xy = 24
y = \(\frac { 24 }{ x }\)

From the given data,
Width of the page
= x + 2(1) = x + 2 cm
Height of the page
= y + 2(1.5) = y + 3 cm
∴ Area of the paper
‘A’ = (x + 2) (y + 3)
= (x + 2) (\(\frac { 24 }{ x }\) + 3)
A = 24 + 3x + \(\frac { 48 }{ x }\) + 6
\(\frac { dA }{ dx }\) = 3 – \(\frac { 48 }{ x^2 }\)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
3x² – 48 = 0
x² = 16
x = ±4 [∵ x cannot be negative
∴ x = 4
Now, \(\frac { d^2A }{ dx^2 }\) = \(\frac { 96 }{ x^3 }\)
at x = 4, \(\frac { d^2A }{ dx^2 }\) > 0
∴ Area is minimum when x = 4
y = \(\frac { 24 }{ 4 }\) = 6
∴ Dimensions of the page:
Width of the page = x + 2 = 4 + 2 = 6 cm
Height of the page = y + 3 = 6 + 3 = 9 cm

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:

Let the length of the pasture be ‘x’ m
Let the breadth of the pasture be ‘y’ m
Given Area = 1,80,000
xy = 1,80,000
y = 1,80,000
For fencing, we need 2y + x
(one side is River)
Let P = 2y + x
P = 2(\(\frac { 180000 }{ 2 }\)) + x = \(\frac { 360000 }{ x }\) + x
\(\frac { dP }{ dx }\) = –\(\frac { 360000 }{ x^2 }\) + 1
For maximum or minimum,
\(\frac { dP }{ dx }\) = 0
⇒ – 360000 + x² = 0
x² = 360000
x = ±600
[x = -600 is not possible]
∴ x = 600
Now, \(\frac { d^2P }{ dx^2 }\) = \(\frac { 720000 }{ x^3 }\)
at x = 600, \(\frac { d^2P }{ dx^2 }\) > 0
∴ P is minimum when x = 600
y = \(\frac { 180000 }{ 600 }\) = 300
∴ Length of the minimum needed fencing material = 2y + x = 2(300) + 600 = 1200 m

Question 7.
Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of a radius of 10 cm.
Solution:
Let the length of the rectangle be ‘x’ cm
The breadth of the rectangle be ‘y’ cm

From the figure,
x² + y² = (20)² [Pythagoras Theorem
y² = 400 – x²
[∵ radius of the circle is 10 cm
y = \(\sqrt { 400-x^2 }\)
Now, Axea of the rectangle A = xy

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { -2x^2+400 }{ \sqrt{400-x^2} }\)
x² = 200
x = ±10√2
x = -10√2 is not possible
∴ x = 10√2

Area of the rectangle is maximum When x = 10√2
y = \(\sqrt { 400-200 }\) = \(\sqrt { 200 }\) = 10√2
∴ x = y = 10√2
Length of the rectangle = 10√2 cm
Breadth of the rectangle = 10√2 cm
(Note: A largest rectangle that can be inscribed in a circle is a square)

Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let x, y be the length and breadth of a rectangle and given perimeter is P (say)
ie. 2(x + y) = P
y = \(\frac { P }{ 2 }\) – x
Area of a rectangle ‘A’ = xy
A = x(\(\frac { P }{ 2 }\) – x) = \(\frac { P }{ 2 }\) x – x²
\(\frac { dA }{ dx }\) = \(\frac { P }{ 2 }\) – 2x
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { P }{ 2 }\) – 2x = 0
x = \(\frac { P }{ 4 }\)
Now, \(\frac { d^2A }{ dx^2 }\) = -2
at x = \(\frac { P }{ 4 }\), \(\frac { d^2A }{ dx^2 }\) < 0
∴ Area of the rectangle is maximum when x = \(\frac { P }{ 4 }\)
Now, y = \(\frac { P }{ 2 }\) – x = \(\frac { P }{ 2 }\) – \(\frac { P }{ 4 }\) = \(\frac { P }{ 4 }\)
∴ Length of a rectangle = \(\frac { P }{ 4 }\)
Breadth of a rectangle = \(\frac { P }{ 4 }\)
Since Length = Breadth, the rectangle is a square.
Hence Proved.

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm.
Solution:

Given radius of the semi-circle = ‘r’ cm
Let the length of the rectangle be ‘x’ cm
Let the breadth of the rectangle be ‘y’ cm
From the figure,

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
⇒ \(\frac { 1 }{ 2 }\) [ \(\frac { 4r^2-2x^2 }{ \sqrt{4x^2-x^2} }\) ] = 0
4r² – 2x² = 0
x² = 2r²
x = ± √2 r
x = -√2 r is not possible
∴ x = √2 r

∴ Length of the rectangle is √2 r cm
Breadth of the rectangle is \(\frac { r }{ √2 }\) cm

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq. cm. Determine the dimensions of the box for the maximum volume.
Solution:
Let ‘x’ be the length Of the box.
‘y’ be the height of the box.
Given, surface area = 108 sq.cm
i,e. 4(xy) + x² = 108
⇒ y = \(\frac { 108-x^2 }{ 4x }\)
Volume of the box V = x²y

maximum or minimum, \(\frac { dV }{ dx }\) = 0
⇒ 108 – 3x² = 0
x² = 36
x = ± 6 [x = – 6 is not possible
∴ x = 6
Now, \(\frac { d^2V }{ dx^2 }\) = –\(\frac { 6x }{ 4 }\) = –\(\frac { 3x }{ 2 }\)
at x = 6, \(\frac { d^2V }{ dx^2 }\) < 0
Volume of theboxis maximum when x = 6
y = \(\frac { 108-36 }{ 24 }\) = \(\frac { 72 }{ 24 }\) = 3
∴ Length of the box = 6 cm
Breadth of the box = 6 cm
Height of the box = 3 cm

Question 11.
The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
Solution:
Given r + h = 6
⇒ r = 6 – h
Volume V = πr²h
V = π(6 – h)²h
\(\frac { dV }{ dh }\) = π [(6 – h)² (1) + 2h(6 – h) (-1)] = π(6 – h)[6 – 3h]
For maximum or minimum,
\(\frac { dV }{ dh }\) = 0
⇒ π (6 – h) (6 – 3h) = 0
⇒ h = 6, h = 2
h = 6 is not possible as r + h = 6
∴ h = 2
\(\frac { d^2V }{ dh^2 }\) = π [(6 – h)(-3) + (6 – 3h)(-1)] = π [6h – 24]
at h = 2, \(\frac { d^2V }{ dh^2 }\) < 0
∴ Volume of the cylinder is maximum when h = 2 and r = 6 – 2 = 4
greatest value of V = π(4)² (2) = 32 π
least value of V = 0

Question 12.
A hollow cone with a base radius of a cm and’ height of b cm is placed on a table. Show that) the volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 }\) times the volume of the cone.
Solution:

Cone
heigthof the cone = b cm
base radius = a cm
Cylinder
Let the base radius be ‘r’ cm
height be ‘h’ cm
From the figure, \(\frac { h }{ a-r }\) = \(\frac { b }{ a }\)
(using similar triangles property
⇒ h = \(\frac { b }{ a }\) (a – r)
= b – \(\frac { b }{ a }\) r
Volume of cylinder V = πr²h

For maximum or minimum,
\(\frac { dV }{ dr }\) = 0
⇒ br(2a – 3r) = 0
r = 0 and r = \(\frac { 2a }{ 3 }\)
r = 0 is not possible
∴ r = \(\frac { 2a }{ 3 }\)

Hence proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 1.
Find intervals of concavity and points of inflection for the following functions:
(i) f(x) = x(x – 4)³
(ii) f(x) = sin x + cos x, 0 < x < 2π
(iii) f(x) = \(\frac { 1 }{ 2 }\)(e^x – e^-x)
Solution:
(i) f(x) = x(x – 4)³
f'(x) = 3x(x – 4)² + (x – 4)³(1)
= (x – 4)² (4x – 4) = 4 (x – 4)² (x – 1)
f'(x) = 4 [(x-4)² (1) + (x – 1) 2 (x – 4)]
= 4(x – 4)(x – 4 + 2x – 2)
= 4(x – 4)(3x – 6) = 12(x – 4)(x – 2)
f'(x) = 0 ⇒ 12 (x – 4) (x – 2) = 0
Critical points x = 2, 4

The intervals are (- ∞, 2), (2, 4) and (4, ∞)
In the interval (-∞, 2), f”(x) > 0 ⇒ Curve is Concave upward
In the interval (2, 4), f”(x) < 0 ⇒ Curve is Concave downward.
In the interval (4, ∞), f”(x) > 0 ⇒ Curve is Concave upward.
The curve is concave upward in
(-∞, 2), (4, ∞) it is concave downward in (2, 4).
f”(x) changes its sign when passing through x = 2 and x = 4
Now f(2) = 2 (2 – 4)³ = -16 and f(4) = 4 (4 – 4)³ = 0
∴ The points of inflection are (2, -16) and (4, 0).

(ii) f(x) = sin x + cos x, 0 < x < 2π
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f'(x) = 0 ⇒ sin x + cos x = 0
Critical points x = \(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)

The intervals are (0, \(\frac { 3π }{ 4 }\)), (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)) and (\(\frac { 7π }{ 4 }\), 2π)
In the interval (0, \(\frac { 3π }{ 4 }\)), f'(x) < 0 ⇒ curve is concave down.
In the interval (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)), f'(x) > 0 ⇒ curve is concave up.
In the interval (\(\frac { 7π }{ 4 }\), 2π), f'(x) < 0 ⇒ curve is concave down.
The curve is concave upward in (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)) and concave downward in (0, \(\frac { 3π }{ 4 }\)) and (\(\frac { 7π }{ 4 }\), 2π)
f'(x) changes its sign when passing through x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 7π }{ 4 }\)

(iii) f(x) = \(\frac { 1 }{ 2 }\) (e^x – e^-x)
f'(x) = \(\frac { 1 }{ 2 }\) (e^x + e^-x)
f”(x) = \(\frac { 1 }{ 2 }\) (e^x – e^-x)
f”(x) = 0 ⇒ \(\frac { 1 }{ 2 }\) (e^x – e^-x) = 0
Critical point x = 0
The intervals are (-∞, o) and (0, ∞)
In the interval (-∞, 0), f”(x) < 0 ⇒ curve is concave down.
In the interval (0, ∞), f(x) > 0 ⇒ curve is concave up.
∴ The curve is concave up in (0, ∞) and concave down in (-∞, 0).
f'(x) changes its sign when passing through x = 0
Now f(0) = – (e° – e°) = \(\frac { 1 }{ 2 }\) (1 – 1) = 0
∴ The point of inflection is (0, 0).

Question 2.
Find the local extrema for the following functions using second derivative test:
(i) f(x) = -3x⁵ + 5x³
(ii) f(x) = x log x
(iii) f(x) = x² e^-2x
Solution:
(i) f(x) = – 3x⁵ + 5x³
f'(x) = 0, f”(x) = -ve at x = a
⇒ x = a is a maximum point
f'(x) = 0, f”(x) = +ve at x = 6
⇒ x = b is a minimum point
f(x) = – 3x⁵ + 5x³
f’ (x) = -15x⁴ + 15x²
f”(x) = -60x³ + 30x
f'(x) = 0 ⇒ – 15x² (x² – 1) = 0
⇒ x = 0, +1, -1
at x = 0, f”(x) = 0
at x = 1, f”(x) = -60 + 30 = – ve
at x = -1, f”(x) = 60 – 30 = + ve
So at x = 1, f'(x) = 0 and f”(x) = -ve
⇒ x = 1 is a local maximum point.
and f(1) = 2
So the local maximum is (1, 2)
at x = -1, f'(x) = 0 and f”(x) = +ve
⇒ x = -1 is a local maximum point and f(-1) = -2.
So the local minimum point is (-1, -2)
∴ a local minimum is -2 and the local maximum is 2.

(ii) f(x) = x log x
f'(x) = x – \(\frac { 1 }{ 4 }\) + log x = 1 + log x
For maximum or minimum
f'(x) = 0 ⇒ 1 + log x = 0
⇒ log x = -1
x = e^-1 = \(\frac { 1 }{ e }\)
f”(x) = \(\frac { 1 }{ x }\)
at x = \(\frac { 1 }{ e }\), f”(x) > 0 ⇒ f(x) attains minimum.
∴ Local minimum f(\(\frac { 1 }{ e }\)) = \(\frac { 1 }{ e }\) log (\(\frac { 1 }{ e }\))
= \(\frac { 1 }{ e }\)(-1) = –\(\frac { 1 }{ e }\)

(iii) f(x) = x² e^-2x
f'(x) = x²[-2e^-2x] + e^-2x (2x)
= 2e^-2x (x – x²)
f”(x) = 2e^-2x(1 – 2x) + (x – ²) (-4e^-2x)
= 2e^-2x [(1 – 2x) + (x – x²) (- 2)]
= 2e^-2x [2x² – 4x + 1]
f'(x) = 0 ⇒ 2e^-2x(x – x²) = 0
⇒ x (1 – x) = 0
⇒ x = 0 or x = 1
at x = 0, f”(x) = 2 × 1 [0 – 0 + 1] = +ve
⇒ x = 0 is a local minimum point and the minimum value is f(0) = 0 at x = 1,
f”(x) = 2e^-2 [2 – 4 + 1] = -ve
⇒ x = 1 is a local maximum point and the maximum value is f(1) = \(\frac{1}{e^{2}}\)
Local maxima \(\frac{1}{e^{2}}\) and local minima = 0

Question 3.
For the function f(x) = 4x³ + 3x² – 6x + 1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.
Solution:
(x) = 4x³ + 3x² – 6x + 1
Monotonicity
f(x) = 4x³ + 3x² – 6x + 1
f'(x) = 12x² + 6x – 6
f'(x) = 0 ⇒ 6(2x² + x – 1) = 0
x = -1, \(\frac { 1 }{ 2 }\) (Stationary points)
∴ The intervals of monotonicity are (-∞, -1), (-1, \(\frac { 1 }{ 2 }\)) and (\(\frac { 1 }{ 2 }\), ∞)
In (-∞, -1), f'(x) > 0 ⇒ f(x) is strictly increasing
In (-1, \(\frac { 1 }{ 2 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing
In (\(\frac { 1 }{ 2 }\), ∞) f'(x) > 0 ⇒ f(x) is strictly increasing
f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -1
∴ Local maximum f(-1) = -4 + 3 + 6 + 1 = 6
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = \(\frac { 1 }{ 2 }\)
∴ Local minimum f(\(\frac { 1 }{ 2 }\))
= 4(\(\frac { 1 }{ 8 }\)) + 3(\(\frac { 1 }{ 4 }\)) – 6(\(\frac { 1 }{ 2 }\)) + 1
= \(\frac { 1 }{ 2 }\) + \(\frac { 3 }{ 4 }\) – + 1 = –\(\frac { 3 }{ 4 }\)
f(x) = 4x³ + 3x² – 6x + 1
f'(x) = 12x² + 6x – 6
f”(x) = 24x + 6
f’(x) = 0 ⇒ 24x + 6 = 0
x = –\(\frac { 6 }{ 24 }\) = –\(\frac { 1 }{ 4 }\) (critical points)
∴ The intervals are (∞, \(\frac { 1 }{ 4 }\)) and (\(\frac { 1 }{ 4 }\), ∞) f”(x) > 0
In the interval (-∞, –\(\frac { 1 }{ 4 }\)), f”(x) < 0 ⇒ curve is concave down.
In the interval (-\(\frac { 1 }{ 4 }\), ∞), f”(x) > 0 ⇒ curve is concave up.
The curve is concave upward in (-\(\frac { 1 }{ 4 }\), ∞) and concave downward in (-∞, –\(\frac { 1 }{ 4 }\))
f”(x) changes its sign when passing through x = –\(\frac { 1 }{ 4 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 1.
Find the absolute extrema of the following functions on the given closed interval.
(i) f(x) = x² – 12x + 10; [1, 2]
(ii) f(x) = 3x⁴ – 4x³; [-1, 2].
(iii) f(x)= 6x^{\(\frac { 4 }{ 3 }\)} – 3x^{\(\frac { 1 }{ 3 }\)}; [-1, 1]
(iv) f(x) = 2 cos x + sin 2x; [0, \(\frac { π }{ 2 }\) ]
Solution:
(i) f(x) = x² – 12x + 10;
f'(x) = 2x – 12
f'(x) = 0 ⇒ 2x – 12 = 0
x = 6 ∉ (1, 2)
Now, Evaluating f(x) at the end points x = 1, 2
f(1) = 1 – 12 + 10 = -1
f(2) = 4 – 24 + 10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

(ii) f(x) = 3x⁴ – 4x³
f'(x) = 12x³ – 12x²
f'(x) = 0 ⇒ 12x²(x – 1) = 0
⇒ x = 0 or x = 1
[Here x = 0, 1 ∈ [-1, 2]]
Now f (-1) = 4
f(0) = 0
f(1) = -1
f(2) = 16
so absolute maximum = 16 and absolute minimum = -1

Question 2.
Find the intervals of monotonicities and hence find the local extremum for the following functions:
(i) f(x) = 2x³ + 3x² – 12x
(ii) f(x) = \(\frac { x }{ x-5 }\)
(iii) f(x) = \(\frac { e^x }{ 1-e^x }\)
(iv) f(x) = \(\frac { x^3 }{ 3 }\) – log x
(v) f(x) = sin x cos x+ 5, x ∈ (0, 2π)
Solution:
(i) f(x) = 2x³ + 3x² – 12x
f'(x) = 6x² + 6x – 12
f'(x) = 0 ⇒ 6(x² + x – 2) = 0
(x + 2)(x – 1) = 0
Stationary points x = -2, 1

Now, the intervals of monotonicity are
(-∞, -2), (-2, 1) and (1, ∞)
In (-∞, -2), f'(x) > 0 ⇒ f(x) is strictly increasing.
In (-2, 1), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In (1, ∞), f'(x) > 0 ⇒ f(x) is strictly increasing.
f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -2.
Local maximum
f(-2) = 2 (-8) + 3 (4) – 12 (-2)
= -16 + 12 + 24 = 20
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1.
∴ Local minimum f(1) = 2 + 3 – 12 = -7

(ii) f(x) = \(\frac { x }{ x-5 }\)
f'(x) = \(\frac { (x-5)(1)-x(1) }{ (x-5)^2 }\) = –\(\frac { 5 }{ (x-5)^2 }\)
f'(x) = 0, which is absured
But in f(x) = \(\frac { x }{ x-5 }\)
The function is defined only when x < 5 or x > 5
∴ The intervals are (-∞, 5) and (5, ∞)
In the interval (-∞, 5), f'(x) < 0
In the interval (5, ∞), f'(x) < 0
∴ f(x) is strictly decreasing in (-∞, 5) and (5, ∞)

When x = 0, f(x) becomes undefined.
∴ x = 0 is an excluded value.
∴ The intervals are (-∞, 0) ∪ (0, ∞) in – (-∞, ∞), f'(x) > 0
∴ f(x) is strictly increasing in (- ∞, ∞) and there is no extremum.

(iv) f(x)= \(\frac { x^3 }{ 3 }\) – log x
f'(x) = x² – \(\frac { 1 }{ x }\)
f'(x) = 0 ⇒ x³ – 1 = 0 ⇒ x = 1
The intervals are (0, 1) and (1, ∞).
i.e., when x > 0, the function f(x) is defined in the interval (0, 1), f'(x) < 0
∴ f(x) is strictly decreasing in (0, 1) in the interval (1, ∞), f'(x) > 0
∴f(x) is strictly increasing in (1, ∞)
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1
∴ Local minimum
f(1) = \(\frac { 1 }{ 3 }\) – log 1 = \(\frac { 1 }{ 3 }\) – 0 = \(\frac { 1 }{ 3 }\)

(v) f(x) = sin x cos x + 5, x ∈ (0, 2π)
f'(x) = cos 2x
f'(x) = 0 ⇒ cos 2x = 0
Stationary points
x = \(\frac { π }{ 4 }\), \(\frac { 3π }{ 4 }\), \(\frac { 5π }{ 4 }\), \(\frac { π }{ 4 }\) ∈x = (0, 2π)

In the interval (0, \(\frac { π }{ 4 }\)), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval (\(\frac { π }{ 4 }\), \(\frac { 3π }{ 4 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval (\(\frac { 3π }{ 4 }\), \(\frac { 5π }{ 4 }\)), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval (\(\frac { 5π }{ 4 }\), \(\frac { 7π }{ 4 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval (\(\frac { 7π }{ 4 }\), 2π), f'(x) > 0 ⇒ f(x) is strictly increasing.
f'(x) changes its sign from positive to negative when passing through x = \(\frac { π }{ 4 }\) and x = \(\frac { 5π }{ 4 }\)
∴ f(x) attains local maximum at x = \(\frac { π }{ 4 }\) and \(\frac { 5π }{ 4 }\)

f'(x) changes its sign from negative to positive when passing through x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 7π }{ 4 }\)
∴ f(x) attains local maximum at x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 5π }{ 4 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 1.
Evaluate the following limits, if necessary use L’ Hôpital’s Rule:
\(\lim _{x \rightarrow 0}\) \(\frac { 1-cosx }{ x^2 }\)
Solution:
\(\lim _{x \rightarrow 0}\) \(\frac { 1-cosx }{ x^2 }\)

Question 2.
\(\lim _{x \rightarrow ∞}\) \(\frac { 2x^2-3 }{ x^2-5x+3 }\)
Solution:

Question 3.
\(\lim _{x \rightarrow ∞}\) \(\frac { x }{ log x }\)
Solution:
\(\lim _{x \rightarrow ∞}\) \(\frac { x }{ log x }\) [ \(\frac { ∞ }{ ∞ }\) indeterminate form
Applying L’ Hôpital’s Rule
\(\lim _{x \rightarrow ∞}\) \(\frac { 1 }{ \frac{1}{x} }\) = \(\lim _{x \rightarrow ∞}\) x = ∞

Question 4.
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { secx }{ tanx }\)
Solution:
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { secx }{ tanx }\) [ \(\frac { ∞ }{ ∞ }\) indeterminate form
Simplifying, we get
\(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { 1 }{ sinx }\) = \(\frac { 1 }{ sin \frac{π}{2} }\) = 1

Question 5.
\(\lim _{x \rightarrow ∞}\) e^-x√x
Solution:
\(\lim _{x \rightarrow ∞}\) e^-x√x [0 × ∞ indeterminate form
The other form is \(\lim _{x \rightarrow ∞}\) \(\frac { √x }{ e^x }\)
[0 × ∞ indeterminate form
Applying L’ Hôpital’s Rule
= \(\lim _{x \rightarrow ∞}\) \(\frac { 1 }{ 2 \sqrt{xe^x} }\)
= 0

Question 6.
\(\lim _{x \rightarrow ∞}\) (\(\frac { 1 }{ sinx }\) – \(\frac { 1 }{ x }\))
Solution:

Question 7.
\(\lim _{x \rightarrow 1}\) (\(\frac { 2 }{ x^2-1 }\) – \(\frac { x }{ x-1 }\))
Solution:

Question 8.
\(\lim _{x \rightarrow 0^+}\) x^x
Solution:
\(\lim _{x \rightarrow 0^+}\) x^x [0° indeterminate form
Let g(x) = x^x
Taking log on both sides
log g(x) = log x^x
log g(x) = x log x
\(\lim _{x \rightarrow 0^+}\) log g(x) = \(\lim _{x \rightarrow 0^+}\) x log x [0 × ∞ indeterminate form

Question 9.
\(\lim _{x \rightarrow ∞}\) (1 + \(\frac { 1 }{ x }\)) ^x
Solution:

Applying L’ Hôpital’s Rule

Exponentiating we get, \(\lim _{x \rightarrow ∞}\) g(x) = e¹ = e

Question 10.
\(\lim _{x \rightarrow \frac{π}{2}}\) (sin x) ^{tan x}
Solution:
\(\lim _{x \rightarrow \frac{π}{2}}\) (sin x)^{tan x} [1^∞ indeterminate form]
Let g(x) = (sin x) ^{tan x}
Taking log on both sides,
log g(x) = tan x log sin x
\(\lim _{x \rightarrow \frac{π}{2}}\) log g(x) = \(\lim _{x \rightarrow \frac{π}{2}}\) \(\frac { log sin x }{ cot x }\)
[ \(\frac { 0 }{ 0 }\) Indeterminate form
Applying L’ Hôpital’s Rule
= \(\lim _{x \rightarrow \frac{π}{2}}\) (\(\frac { cotx }{ -cosec^2x }\)) = -1
exponentiating, we get
\(\lim _{x \rightarrow \frac{π}{2}}\) g(x) = e^-1 = \(\frac { 1 }{ e }\)

Question 11.
\(\lim _{x \rightarrow 0^+}\) (cos x) ^{\(\frac { 1 }{ x^2 }\)}
Solution:
\(\lim _{x \rightarrow 0^+}\) (cos x) ^{\(\frac { 1 }{ x^2 }\)} [1^∞ indeterminate form
let g(x) = (cos x)^{\(\frac { 1 }{ x^2 }\)}
Taking log on both sides,

Question 12.
If an initial amount A₀ of money is invested at an interest rate r compounded n times a year, the value of the investment after t years is A = A₀(1 + \(\frac { r }{ n }\))^nt. If the interest is compounded continuously, (that is as n → ∞), show that the amount after t years is A = A₀e^rt.
Solution:

Applying L-Hospital’s Rule

Hence Proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 1.
Write the Maclaurin series expansion of thef following functions:
(i) e^x
(ii) sin x
(iii) cos x
(iv) log (1 – x); – 1 ≤ x ≤ 1
(v) tan^-1 (x); -1 ≤ x ≤ 1
(vi) cos² x
Solution:
(i) Let f(x) = e^x
f(x) = e^x f'(0) = e° = 1
f(x) = e^x f'(0) = e° = 1
f”(x) = e^x f”(0) = e° = 1
Maclaurin ‘s expansion is

(ii) Let f(x) = sin x
f(x) = sin x; f(0) = 0
f'(x) = cos x; f'(0) = 1
f”(x) = -sin x; f”(0) = 0
f”‘(x) = -cos x; f”'(0) = -1
f^IV(x) = sin x; f^IV(0) = 0
f^V(x) = cos x; f^V(0) = 1
f^VI(x) = -sin x; f^VI(0) = 0
f^VII(x) = -cos x; f^VII(0) = -1
Maclaurin ‘s expansion is

(iii) Let f(x) = cos x
f(x) = cos x ; f(0) = 1
f'(x) = -sin x ; f'(0) = 0
f”(x) = -cos x ; f”(0) = -1
f”'(x) = sin x ; f”'(0) = 0
f^IV(x) = cos x ; f^IV(0) = 1
f^V(x) = -sin x ; f^V(0) = 0
f^VI(x) = -cos x ; f^VI(0) = -1
Maclaurin ‘s expansion is

(iv) log (1 – x); – 1 ≤ x ≤ 1

Maclaurin ‘s expansion is

(v) tan^-1 (x); -1 ≤ x ≤ 1
f(x) = tan^-1 x ; f(0) = 0
f'(x) = \(\frac { 1 }{ 1+x^2 }\) f'(0) = 1
= 1 – x² + x⁴ – x⁶ + …..
f”(x) = -2x + 4x³ – 6x⁵ + ….. f”(0) = 0
f”'(x) = -2 + 12x² – 30x⁴ + ….. f”(0) = -2
f^IV(x) = 24x – 120x³ + …… f^IV(0) = 0
f^V(x) = 24 – 360x² + ….. f^V(0) = 24 .
f^VI(x) = -720x + ….. f^VI(0) = 0
f^VII(x) = -720 + … f^VII(0) = -720
Maclaurin ‘s expansion is

(vi) Let f(x) = cos² x
f(x) = cos² x ; f(0) = 1
f'(x) = -2 cos x sin x ; f'(0) = 0
= -sin 2 x
f”(x) = -2 cos 2x ; f”(0) = -2
f”‘(x) = 4 sin 2x ; f”‘(0) = 0
f^IV(x) = 8 cos 2x ; f^IV( 0) = 8
f^V(x) = -16 sin 2x ; f^V(0) = 0
f^VI(x) = -32 cos 2x ; f^VI(0) = -32
Maclaurin’s expansion is

Question 2.
Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0.
Solution:
Let f(x) = log x
Taylor series of f(x) is

Question 3.
Expand sin x ascending powers x – \(\frac { π }{ 4 }\) upto three non-zero terms.
Solution:
Let f(x) = sin x

Taylor series of f(x) is

Question 4.
Expand the polynomial f(x) = x² – 3x + 2 in power of x – 1.
Solution:
Let f(x) = x² – 3x + 2
f(x) = x² – 3x + 2 ; f(1) = 0
f'(x) = 2x – 3 ; f'(1) = -1
f”(x) = 2 ; f”(1) = 2
Taylor series of f(x) is
f(x) = \(\sum_{n=0}^{n=\infty}\) a_n (x – 1)ⁿ, where a_n = \(\frac { f^{(n)} (1)}{ n! }\)
∴ The required expansion is
x² – 3x + 2 = 0 – \(\frac { 1(x-1) }{ 1! }\) + \(\frac { 2(x-1)^2 }{ 2! }\)
= -(x – 1) + (x – 1)²