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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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3

Question 1.
Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.
(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]
(ii) f(x) = tan x, x ∈ [0, π]
(iii) f(x) = x – 2 log x, x ∈ [2, 7]
Solution:
(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]
f(-1) = 1
f(1) = 1
⇒ f(-1) = f(1) = 1
But f(x) is not differentiable at x = 0
∴ Rolle’s theorem is not applicable.

(ii) f (x) = tan x
f(x) is not continuous at x = \(\frac{\pi}{2}\).
So Rolle’s Theorem is not applicable.

(iii) f(x) = x – 2 log x
f(x) = x – 2 log x
f(2) = 2 – 2 log 2 = 2 – log 4
f(7) = 7 – 2 log 7 = 7 – log 49
f(2) ≠ f(7)
So Rolle’s theorem is not applicable.

Question 2.
Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:
(i) f(x) = x² – x, x ∈ [0, 1]
(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]
(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]
Solution:
(i) f(x) = x² – x, x ∈ [0, 1]
f(0) = 0, f(1) = 0
⇒ f(0) = f(1) = 0
f(x) is continuous on [0, 1]
f(x) is differentiable on (0, 1)
Now, f'(x) = 2x – 1
Since, the tangent is parallel to x-axis then
f'(x) = 0 ⇒ 2x – 1 = 0
x = \(\frac { x }{ 3 }\) ∈ (0, 1)

(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]
f(-1) = \(\frac { 1+2 }{ -1+2 }\) = 3
f(6) = \(\frac { 36-12 }{ 8 }\) = \(\frac { 24 }{ 8 }\) = 3
⇒ f (-1) = 3 = f(6)
f(x) is continuous on [- 1, 6]
f(x) is differentiable on (- 1, 6)
Now, f'(x)

Since the tangent is parallel to the x-axis.
f'(x) = 0
⇒ x² + 4x – 4 = 0
⇒ x = –\(\frac { 4±\sqrt{16+16} }{ 2 }\)
x = –\(\frac { 4±4√2 }{ 2 }\) = -2 ± 2√2
x = -2 ± 2√2 ∈ (-1, 6)

(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]
f(0) = 0, f(9) = √9 – \(\frac { 9 }{ 3 }\) = 3 – 3 = 0
⇒ f(0) = 0 = f(9)
f(x) is continuous on [0, 9]
f(x) is differentiable on (0, 9)
Now f'(x) = \(\frac { 1 }{ 2√x }\) – \(\frac { 1 }{ 3 }\)
Since, the tangent is parallel to x-axis.
f'(x) = 0

Question 3.
Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals:
(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]
(ii) f(x) = |3x + 1|, x ∈ [-1, 3]
Solution:
(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]
f(0) = undefined
∴ f(x) is not continuous at x = 0
Hence, Lagrange’s mean value theorem is not applicable.

(ii) f(x) =|3x + 1|, x ∈ [-1, 3]
The function is not differentiable at x = \(\frac{-1}{3}\).
So Lagrange’s mean value theorem is not applicable in the given interval.

Question 4.
Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
(i) f(x) = x³ – 3x + 2, x ∈ [-2, 2]
(ii) f(x) = (x – 2) (x – 7), x ∈ [3, 11]
Solution:
f(x) = x³ – 3x + 2, x ∈ [-2, 2]
f(x) is continuous in [- 2, 2]
f(x) is differentiable in (- 2, 2)
f(-2) = (-2)³ – 3 (-2) + 2 = – 8 + 6 + 2 = 0
f(2) = (2)³ -3(2) + 2 = 8 – 6 + 2 = 4
∴ f(x) is defined in the given interval.
Given that tangent is parallel to the secant line of the curve between x = -2 and x = 2.

(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]
f(x) is continuous in [3, 11]
f(x) is differentiable in (3, 11)
f(3) = (3 – 2) (3 – 7) = (1) (-4) = -4
f(11) = (11 – 2) (11 – 7) = (9) (4) = 36
∴ f(x) is defined in the given interval.
Given that the tangent is parallel to the secant line ofthe curve between x = 3 and x = 11.
∴ f'(c) = \(\frac { f(b)-f(a) }{ b-a }\)
2c – 9 = \(\frac { 36+4 }{ 11-3 }\) where f'(x) = 2x – 9
2x – 9 = \(\frac { 40 }{ 8 }\) = 5
2c = 14 ⇒ c = 7 ∈ (3, 11)
∴ x = 7.

Question 5.
Show that the value in the conclusion of the mean value theorem for
(i) f(x) = \(\frac { 1 }{ x }\) on a closed interval of positive numbers [a, b] is \(\sqrt { ab }\)
(ii) f(x) = Ax² + Bx + C on any interval [a, b] is \(\frac { a+b }{ 2 }\)
Solution:

(ii) f(x) = Ax² + Bx + C, x ∈ [a, b]
f'(x) = 2Ax + B
By Mean Value Theorem,

Question 6.
A race car driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?
Solution:
Let a = 0, b = 2 and the interval is [0, 2] and f(0) = 20 (given)
We need to find f(2)
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c) [b – a]
f(b) – 20 ≤ 150(2 – 0)
f(b) ≤ 300 + 20
f(b) ≤ 320
∴ Maximum distance f(2) = 320 km.

Question 7.
Suppose that for a function f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) – f(1) ≤ 3.
Solution:
Given: For f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4
∴ a = 1, b = 4.
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c) (b – a)
f(4) – f(1) ≤ 1(4 – 1)
f(4) – f(1) ≤ 3
Hence Proved.

Question 8.
Does there exist a differentiable function f(x) such that f(0) = -1, f(2) = 4 and f (x) ≤ 2 for all x. Justify your answer.
Solution:
Given:For f(x), f'(x) ≤ 2, f(0) = -1, f(2) = 4
∴ a = 0, b = 2
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c)(b – a)
f(2) – f(0) ≤ f'(c) (2 – 0)
\(\frac { 4+1 }{ 2 }\) ≤ f'(c) ⇒ \(\frac { 5 }{ 2 }\) ≤ f'(c) ≤ 2 (given)
f(x) cannot be a differentiable function in (0, 2) as f'(x) cannot be 2.5.

Question 9.
Show that there lies a point on the curve f(x) = x(x + 3)e^-π/2, -3 ≤ x ≤ 0 where tangent drawn is parallel to the x-axis.
Solution:
f(x) = x(x + 3)e^-π/2, -3 ≤ x ≤ 0
f(x) is continuous in [-3, 0]
f(x) is differentiable in (- 3, 0)
f(-3) = -3 (-3 + 3)e^-π/2 = 0
f(0) = 0
⇒ f(-3) = f(0) = 0
Since the tangent is parallel to x-axis.
f'(c) = 0
e^-π/2 (2c + 3) = 0 where f'(x) = e^-π/2 (2x + 3)
2c + 3 = 0
c = –\(\frac { 3 }{ 2 }\) ∈ (-3, 0)
∴ The point lies on the curve.

Question 10.
Using Mean Value Theorem prove that for, a > 0, b > 0, |e^-a – e^-b| < |a – b|
Solution:
Let f(x) = e^-x
f'(x) = e^-x
By Lagrange’s Mean Value Theorem,

Hence Proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.
Find the slope of the tangent to the following curves at the respective given points.
(i) y = x⁴ + 2x² – x at x = 1
(ii) x = a cos³ t, y = b sin³ t at t = \(\frac { π }{ 2 }\)
Solution:
(i) y = x⁴ + 2x² – x
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 4x – 1
Slope of the tangent (\(\frac { dy }{ dx }\))_(x=1)
= 4(1)³ + 4(1) – 1
= 4 + 4 – 1 = 7
(ii) x = a cos³ t, y = b sin³ t
Differenriating w.r.t. ‘t’
\(\frac { dx }{ dt }\) = – 3a cos² t sin t
\(\frac { dy }{ dt }\) = 3b sin² t sin t

Question 2.
Find the point on the curve y = x² – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
Solution:
y = x² – 5x + 4
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 2x – 5
Given line 3x + y = 7
Slope of the line = –\(\frac { 3 }{ 1 }\) = -3
Since the tangent is parallel to the line, their slopes are equal.
∴ \(\frac { dy }{ dx }\) = -3
⇒ 2x – 5 = -3
2x = 2
x = 1
When x = 1, y = (1)² – 5 (1) + 4 = 0
∴ Point on the curve is (1, 0).

Question 3.
Find the points on curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y = 1729.
Solution:
y = x³ – 6x² + x+ 3
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 3x² – 12x + 1
Slope of the normal = \(\frac { 1 }{ 3x^2 – 12x + 1 }\)
Given line is x + y = 1729
Slope of the line is – 1
Since the normal is parallel to the line, their slopes are equal.
\(\frac { 1 }{ 3x^2 – 12x + 1 }\) = -1
3x² – 12x + 1 = 1
3x² – 12x =0
3x(x – 4) = 0
x = 0, 4
When x = 0, y = (0)³ – 6(0)² + 0 + 3 = 3
When x = 4, y = (4)³ – 6(4)² + 4 + 3
= 64 – 96 + 4 + 3 = -25
∴ The points on the curve are (0, 3) and (4, -25).

Question 4.
Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal.
Solution:
y² – 4xy = x² + 5 ………… (1)
Differentiating w.r.t. ‘x’

When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.
\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac { x+2y }{ y-2x }\) = 0
⇒ x + 2y = 0
x = -2y
Substituting in (1)
y² – 4 (-2y) y = (-2y)² + 5
y² + 8y² = 4y² + 5
5y² = 5 ⇒ y² = 1
y = ±1
When y = 1, x = -2
When y = – 1, x = 2
∴ The points on the curve are (- 2, 1) and (2, -1).

Question 5.
Find the tangent and normal to the following curves at the given points on the curve.
(i) y = x² – x⁴ at (1, 0)
(ii) y = x⁴ + 2e^x at (0, 2)
(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 3 }\)
Solution:
(i) y = x² – x⁴ at (1, 0)
Differentiating w.r.t. ‘x’
\(\frac { dx }{ dy }\) = 2x – 4x³
Slope of the tangent ‘m’ = (\(\frac { dx }{ dy }\))_{(1, 0)}
= 2 (1) – 4 (1)³ = -2
Slope of the normal –\(\frac { 1 }{ m }\) = \(\frac { -1 }{ -2 }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m (x – x₁)
y – 0 = – 2 (x – 1)
y = -2x + 2
2x + y – 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 0 = \(\frac { 1 }{ 2 }\)(x – 1)
2y = x- 1
x – 2y – 1 = 0

(ii) y = x⁴ + 2e^x at (0, 2)
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 2e^x
Slope of the tangent ‘m’
(\(\frac { dy }{ dx }\))_{(0, 2)} = 4(0)³ + 2e⁰ = 2
Slope of the Normal –\(\frac { 1 }{ m }\) =-\(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – 2 = 2(x – 0)
⇒ y – 2 = 2x
⇒ 2x – y + 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\) (x – x₁)
y – 2 = –\(\frac { 1 }{ 2 }\)(x – 0)
2y – 4 = -x
x + 2y – 4 = 0

(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = x cos x + sin x
Slope of the tangent ‘m’ = (\(\frac { dy }{ dx }\))_{(π/2, π/2)}
= \(\frac { π }{ 2 }\) cos \(\frac { π }{ 2 }\) + sin \(\frac { π }{ 2 }\) = 1
Slope of the Normal –\(\frac { 1 }{ m }\) = -1
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = 1 (x – \(\frac { π }{ 2 }\))
⇒ x – y = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = -1(x – \(\frac { π }{ 2 }\))
⇒ y – \(\frac { π }{ 2 }\) = -x + \(\frac { π }{ 2 }\)
⇒ x + y – π = 0

(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 2 }\)
at t = \(\frac { π }{ 3 }\), x = cos \(\frac { π }{ 3 }\) = \(\frac { 1 }{ 2 }\)
at t = \(\frac { π }{ 3 }\), y = 2 sin² \(\frac { π }{ 3 }\) = 2(\(\frac { 3 }{ 4}\)) = \(\frac { 3 }{ 2 }\)
Point is (\(\frac { 1 }{ 2 }\), \(\frac { 3 }{ 2 }\))
Now x = cos t y = 2 sin² t
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -sin t; \(\frac { dy }{ dt }\) = 4 sin t cos t
Slope of the tangent

Slope of the Normal –\(\frac { 1 }{ m }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = -2(x – \(\frac { 1 }{ 2 }\))
⇒ 2y – 3 = – 4x + 2
⇒ 4x + 2y – 5 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)(x – \(\frac { 1 }{ 2 }\))
⇒ 2 (2y – 3) = 2x – 1
⇒ 4y – 6 = 2x – 1
⇒ 2x – 4y + 5 = 0

Question 6.
Find the equations of the tangents to the curve y = 1 + x³ for which the tangent is orthogonal with the line x + 12y = 12.
Solution:
Curve is y = 1 + x³
Differentiating w.r.t ‘x’,
Slope of the tangent ‘m’ = \(\frac { dy }{ dx }\) = 3x²
Given line is x + 12y = 12
Slope of the line is –\(\frac { 1 }{ 12 }\)
Since the tangent is orthogonal with the line, the slope of the tangent is 12.
∴ \(\frac { dy }{ dx }\) = 12
i.e 3x² = 12
x² = 4
x = ±2
When x = 2, y = 1 + 8 = 9 ⇒ point is (2, 9)
When x = -2, y = 1 – 8 = -7 ⇒ point is (-2, -7)
Equation of tangent with slope 12 and at the j point (2, 9) is
y – 9 = 12 (x – 2)
y – 9 = 12x – 24
12x – y – 15 = 0
Equation of tangent with slope 12 and at the point (-2, -7) is
y + 7 = 12 (x + 2)
y + 7 = 12x + 24
12x – y + 17 = 0

Question 7.
Find the equations of the tangents to the curve y = –\(\frac { x+1 }{ x-1 }\) which are parallel to the line x + 2y = 6.
Solution:

Given line is x + 2y = 6
Slope of the line = –\(\frac { 1 }{ 2 }\)
Since the tangent is parallel to the line, then the slope of the tangent is –\(\frac { 1 }{ 2 }\)
∴ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ (x-1)^2 }\) = –\(\frac { 1 }{ 2 }\)
(x – 1)² = 4
x – 1 = ±2
x = -1, 3
When x = – 1, y = 0 ⇒ point is (-1, 0)
When x = 3, y = 2 ⇒ point is (3, 2)
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (-1, 0) is
y – o = –\(\frac { 1 }{ 2 }\)(x + 1)
2y = -x – 1 ⇒ x + 2y + 1 = 0
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (3, 2) is 2
y – 2 = –\(\frac { 1 }{ 2 }\) (x – 3)
2y – 4 = -x + 3
x + 2y – 7 = 0.

Question 8.
Find the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t ∈ R at any point on the curve.
Solution:
x = 7 cos t and y = 2 sin t, t ∈ R
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -7 sin t and \(\frac { dy }{ dt }\) = 2 cos t
Slope of the tangent ‘m’
\(\frac { dy }{ dx }\) = \(\frac{\frac { dy }{ dt }}{\frac{ dx }{ dt }}\) = \(\frac { 2 cot t }{ -7 sin t }\)
Any point on the curve is (7 Cos t, 2 sin t)
Equation of tangent is y – y₁ = m (x – x₁)
y – 2 sint = –\(\frac { 2 cot t }{ 7 sin t }\) (x – 7 cos t)
7y sin t – 14 sin² t = -2x cos t + 14 cos² t
2x cos t + 7 y sin t – 14 (sin² t + cos² t) = 0
2x cos t + 7y sin t – 14 = 0
Now slope of normal is –\(\frac { 1 }{ 3 }\) = \(\frac { 7 sin t }{ 2 cos t }\)
Equation of normal is y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 2 sin t = \(\frac { 7 sin t }{ 2 cos t }\) (x – 7 cos t)
2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0

Question 9.
Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0
Solution:
Given curves are xy = 2 ……… (1)
x² + 4y = 0 ………. (2)
Now solving (1) and (2)
Substituting (1) in (2)
⇒ x² + 4(2/x) = 0
x³ + 8 = 0
x³ = -8
x = -2
Substituting in (1) ⇒ y = \(\frac { 2 }{ -2 }\) = -1
∴ Point of intersection of (1) and (2) is (-2, -1)
xy = 2 ⇒ y = \(\frac { 2 }{ x }\) ……….. (1)
Differentiating w.r.t. ‘x’

The angle between the curves

Question 10.
Show that the two curves x² – y² = r² and xy = c² where c, r are constants, cut orthogonally.
Solution:
Given curves are x² – y² = r² ……….. (1)
xy = c² …….. (2)
Let (x₁, y₁) be the point of intersection of the given curves.
(1) ⇒ x² – y² = r²
Differentiating w.r.t ‘x’,
2x – 2y \(\frac { dx }{ dy }\) = 0
\(\frac { dx }{ dy }\) = \(\frac { x }{ y }\)
now (\(\frac { dx }{ dy }\))(_x1,y1) = m₁ = \(\frac { x_1 }{ y_1 }\)
(2) ⇒ xy = c²
Differentiating w.r.t ‘x’,

Hence, the given curves cut orthogonally.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.1

Question 1.
A particle moves along a straight line in such a way that after t seconds its distance from the origin is s = 2t² + 3t metres.
(i) Find the average velocity between t = 3 and t = 6 seconds.
(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.
Solution:
s = 2t² + 3t
(i) Average velocity between t = 3 and t = 6 seconds
Now s(t) = 2t² + 3t
Average velocity

(ii) f(t) = 2t² + 3t
f'(t) = 4t + 3
f'(3) = 4(3) + 3 = 15
f'(6) = 4(3) + 3 = 15

Question 2.
A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of s = 16t² in t seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?
Solution:
(i) The camera falls a distance of s = 16t² in t sec.
s = 400 ft
∴ 16t² =400
t² = \(\frac { 400 }{ 16 }\) = 25
t = 5 sec
∴ Camera falls for 5 sec before it hits the ground.

(ii) In 5 sec camera falls 400 ft (given)
∴ Average velocity in 2 sec
= \(\frac { s(5)-s(3) }{ 5-3 }\)
= \(\frac { 16(5^2)-16(3^2) }{ 2 }\)
= \(\frac { 400-144 }{ 2 }\)
= \(\frac { 256 }{ 2 }\)
= 128 ft/sec

(iii) f(t) = 16t²
f'(t) = 32t
f'(t) at t = 5 = 32(5)
= 160 ft/sec

Question 3.
A particle moves along a line according to the law s(t) = 2t³ – 9t² + 12t – 4, where t ≥ 0.
(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle’s acceleration each time the velocity is zero.
Solution:
s (t) = 2t³ – 9t² + 12t – 4, t ≥ 0
velocity v = \(\frac { ds }{ dt }\) = 6t² – 18t + 12
When the particle changes its direction, v = 0
6t² – 18t + 12 = 0 (÷6)
t² – 3t + 2 = 0
(t – 2) (t – 1) = 0
t = 1, 2
∴ When time t = 1 sec and t = 2 sec, the particle changes its direction.

(ii) The distance travelled in the first 4 seconds is
|s(0) – s(1)| + |s(1) – s(2)| + |s(2) – s(3)| + |s(3) – s(4)|
Here, s(t) = 2t³ – 9t² + 12t – 4
s(0) = -4
s(1) = 1
s(2) = 0
s(3) = 5
and s(4) = 28
∴ Distance travelled in the first 4 seconds
= |-4 – 1| + |1 – 0| + |0 – 5| + |5 – 28|
= 5 + 1 + 5 + 23 = 34 m

(iii) s (t) = 2t³ – 9t² + 12t – 4
velocity v = \(\frac { ds }{ dt }\) = 6t² – 18t + 12
v = 0 ⇒ 6(t² – 3t + 2) = 0 ⇒ t = 1, 2
Acceleration = \(\frac { d^2s }{ dt^2 }\) = 12t – 18
at t = 1, Acceleration = 12(1) – 18 = -6m/sec²
at t = 2, Acceleration = 12 (2) – 18 = 6 m/sec²

Question 4.
If the volume of a cube of side length x is v = x³. Find the rate of change of the volume with respect to x when x = 5 units.
Solution:
volume of a cube v = x³
Rate of change \(\frac { dv }{ dx }\) = 3x²
When x = 5 units, \(\frac { dv }{ dx }\) = 3(5)² = 3(25) = 75 units.

Question 5.
If the mass m(x) (in kilograms) of a thin rod of length x (in metres) is given by, m(x) = \(\sqrt { 3x }\) then what is the rate of change of mass with respect to the length when it is x = 3 and x = 27 metres.
Solution:

Question 6.
A stone is dropped into a pond causing ripples in the from of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?
Solution:
radius = r, Rate of changes of radius \(\frac { dr }{ dt }\) = 2 and
given r = 5 cm
Area of circle A = πr²
Differentiating w.r.t ‘t’,
\(\frac { dA }{ dt }\)= 2πr\(\frac { dr }{ dt }\)
= 2π (5) (2)
= 20 π
∴ Area of circle (ripple) is increasing at the rate of 20 π cm²/sec.

Question 7.
A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shore line. How fast is the beam moving along the shoreline when it makes an angle of 45° with the shore?
Solution:

Time for one revolution = 10 sec
Now, angular velocity \(\frac { dv }{ dt }\) = \(\frac { 2π }{ 10 }\) = \(\frac { π }{ 5 }\)
From the figure, tan 45° = \(\frac { AB }{ OA }\)
1 = \(\frac { x }{ 5 }\) ⇒ x = 5
Again, tan θ = \(\frac { x }{ 5 }\)
x = 5 tan θ
Differentiating w.r.t. ‘t’
\(\frac { dx }{ dt }\) = 5 sec² θ \(\frac { dθ }{ dt }\)
= 5 sec² (45°) (\(\frac { π }{ 5 }\))
= (√2)² π = 2π
∴ The beam is moving at the rate of 2π km/sec.

Question 8.
A conical water tank with a vertex down of 12 metres height has a radius of 5 metres at the top. If water flows into the tank at a rate of 10 cubic m/min, how fast is the depth of the water increases when the water is 8 metres deep?
Solution:

From the figure \(\frac { r}{ h }\) = \(\frac { 5 }{ 12 }\)
r = \(\frac { 5h }{ 12 }\)
given rate of change of volume \(\frac { dV }{ dt }\) = 10
When h = 8 to find \(\frac { dh }{ dt }\)
Volume of cone V = \(\frac { 1 }{ 3 }\) πr² h

The depth of the water increasing at the rate of \(\frac { 9 }{ 10π }\) m/min

Question 9.
A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall.
(i) How fast is the top of the ladder moving down the wall?
(ii) At what rate, the area of the triangle formed by the ladder, wall, and floor is changing?
Solution:

Let the height of the wall where the ladder touches are ‘y’ m.
The bottom of the ladder is at a distance of ‘x’ m from the wall.
Given x = 8, \(\frac { dx }{ dt }\) = 5
x² + y² = 17²
(Pythagoras Theorem)
y² = 17² – x² = 289 – 64 = 225
∴ y = 15
Differentiating w.r.t. ‘t’

(i) The top of the ladder is moving down the wall at the rate of \(\frac { 8 }{ 3 }\) m/sec
(ii) Area of triangle formed by the ladder, wall and the floor is A = \(\frac { 1 }{ 2 }\) xy
differentiating w.r.t. ‘t’

∴ Area of the triangle is increasing at the rate of 26.83 m²/sec.

Question 10.
A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
Solution:

given x = 0.8, y = 0.6, \(\frac { dy }{ dt }\) = -60
and \(\frac { ds }{ dt }\) = 20
from the figure
S² = x² + y²,
S² = (0.8)² + (0.6)² = 0.64 + 0.36 = 1
S² = 1 ⇒ S = 1
S² = x² + y²,
Differentiating w.r.t. ‘t’

∴ Speed of the car is 70 km/hr.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Choose the most suitable answer from the given four alternatives

Question 1.
If \(\overline { a }\) and \(\overline { b }\) are parallel vectors, then [\(\overline { a }\), \(\overline { c }\), \(\overline { b }\)] is equal to
(a) 2
(b) -1
(c) 1
(d) 0
Solution:
(d) 0
Hint:
Since \(\overline { a }\) and \(\overline { b }\) are parallel ⇒ \(\overline { a }\) = λ\(\overline { b }\)
[ \(\overline { a }, \overline { c }, \overline { b }\)] = [λ\(\overline { b }, \overline { c }, \overline { b }\) ]
= λ[ \(\overline { b }, \overline { c }, \overline { b }\) ]
= λ(0) = 0

Question 2.
If a vector \(\overline { α }\) lies in the plane of \(\overline { ß }\) and \(\overline { γ }\), then
(a) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 1
(b) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = -1
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
(d) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 2
Solution:
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
Hint:
If \(\overline { α }\) lies in \(\overline { ß }\) & \(\overline { γ }\) plane
we have [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0

Question 3.
If \(\overline { a }\).\(\overline { b }\) = \(\overline { b }\).\(\overline { c }\) = \(\overline { c }\).\(\overline { a }\) = 0, then the value of [ \(\overline { a }, \overline { b }, \overline { c }\) ] is
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(b) \(\frac { 1 }{ 3 }\)|\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(c) 1
(d) -1
Solution:
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
Hint:

Question 4.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three unit vectors such that \(\overline { a }\) is perpendicular to \(\overline { b }\) and is parallel to \(\overline { c }\) then \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) is equal to
(a) \(\overline { a }\)
(b) \(\overline { b }\)
(c) \(\overline { c }\)
(d) \(\overline { 0 }\)
Solution:
(b) \(\overline { b }\)
Hint:

Question 5.
If [ \(\overline { a }, \overline { b }, \overline { c }\) ] = 1 then the value of

(a) 1
(b) -1
(c) 2
(d) 3
Solution:
(a) 1
Hint:

Question 6.
The volume of the parallelepiped with its edges represented by the vectors \(\hat { i }\) + \(\hat { j }\), \(\hat { i }\) + 2\(\hat { j }\), \(\hat { i }\) + \(\hat { j }\) + π\(\hat { k }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { π }{ 3 }\)
(c) π
(d) \(\frac { π }{ 4 }\)
Solution:
(c) π
Hint:
\(\left|\begin{array}{lll}
1 & 1 & 0 \\
1 & 2 & 0 \\
1 & 1 & \pi
\end{array}\right|\) = π\(\left|\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right|\)
= π (2 – 1) = π

Question 7.
If \(\overline { a }\) and \(\overline { b }\) are unit vectors such that [\(\overline { a }\), \(\overline { b }\), \(\overline { a }\) × \(\overline { b }\)] = \(\frac { 1 }{ 4 }\), then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(a) \(\frac { π }{ 6 }\)
Hint:

Question 8.
If \(\overline { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + \(\hat { j }\), \(\overline { c }\) = \(\hat { i }\) and (\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) – λ\(\overline { a }\) + µ\(\overline { b }\) then the value of λ + µ is
(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:
\(\overline { a }\).\(\overline { c }\) = 1 and \(\overline { b }\).\(\overline { c }\) = 1
(\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) = (\(\overline { c }\) × \(\overline { a }\))\(\overline { b }\) – (\(\overline { c }\) × \(\overline { b }\))\(\overline { a }\) = λ\(\overline { a }\) + µ\(\overline { b }\)
⇒ µ = c; a = 1λ = -(\(\overline { c }\).\(\overline { b }\)) = -1
µ + λ = 1 – 1 = 0

Question 9.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are non-coplanar, non-zero vectors
such that [\(\overline { a }\), \(\overline { b }\), \(\overline { c }\)] = 3, then {[\(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\)]²} is equal to
is equal to
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:

= 3⁴ = 81

Question 10.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three non-coplanar vectors such that \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = \(\frac { \overline{b}+\overline{c} }{ √2 }\) then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { 3π }{ 4 }\)
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:
(b) \(\frac { 3π }{ 4 }\)
Hint:

Question 11.
If the volume of the parallelepiped with \(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { b }\) × \(\overline { c }\)), (\(\overline { b }\) × \(\overline { c }\)) × (\(\overline { c }\) × \(\overline { a }\)) and (\(\overline { c }\) × \(\overline { a }\)) × (\(\overline { a }\) × \(\overline { b }\)) as coterminous edges is
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:
Given volume of the parallelepiped with

Question 12.
Consider the vectors \(\overline { a }\), \(\overline { b }\), \(\overline { c }\), \(\overline { d }\) such that (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = \(\overline { 0 }\) Let P₁ and P₂ be the planes determined by the pairs of vectors \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\), \(\overline { d }\) respectively. Then the angle between P₁ and P₂ is
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Solution:
(a) 0°
Hint:
A vector perpendicular to the plane P₁ of a, b is \(\overline { a }\) × \(\overline { b }\),
A vector perpendicular to the plane P₂ of c and d is \(\overline { c }\) × \(\overline { d }\)
∴ (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = 0
⇒ (\(\overline { a }\) × \(\overline { b }\)) || \(\overline { c }\) × \(\overline { d }\)
⇒ The angle between the planes is \(\overline { 0 }\)

Question 13.
If \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = (\(\overline { a }\) × \(\overline { b }\)) × \(\overline { c }\) where \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are any three vectors such that \(\overline { b }\).\(\overline { c }\) ≠ 0 and \(\overline { a }\).\(\overline { b }\) ≠ 0, then \(\overline { a }\) and \(\overline { c }\) are
(a) perpendicular
(b) parallel
(c) inclined at angle \(\frac { π }{ 3 }\)
(d) inclined at an angle \(\frac { π }{ 6 }\)
Solution:
(b) parallel
Hint:

Question 14.
If \(\overline { a }\) = 2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + 2\(\hat { j }\) – 5\(\hat { k }\), \(\overline { c }\) = 3\(\hat { i }\) + 5\(\hat { j }\) – \(\hat { k }\) then \(\overline { a }\) vector perpendicular to a and lies in the plane containing \(\overline { b }\) and \(\overline { c }\) is
(a) -17\(\hat { i }\) + 21\(\hat { j }\) – 97\(\hat { k }\)
(b) 17\(\hat { i }\) + 21\(\hat { j }\) – 123\(\hat { k }\)
(c) -17\(\hat { i }\) – 21\(\hat { j }\) + 97\(\hat { k }\)
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Solution:
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Hint:
A vector ⊥r to \(\overline { a }\) and lies in the plane containing \(\overline { b }\) and \(\overline { c }\)

Question 15.
The angle between the lines \(\frac { x-2 }{ 3 }\) = \(\frac { y+1 }{ -2 }\), z = 2 and \(\frac { x-1 }{ 1 }\) = \(\frac { 2y+3 }{ 3 }\) = \(\frac { z+5 }{ 2 }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(d) \(\frac { π }{ 2 }\)
Hint:

Question 16.
If the line \(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ -5 }\) = \(\frac { z+2 }{ 2 }\) lies in the plane x + 3y – αz + ß = 0 then (α + ß) is
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Solution:
(b) (-6, 7)
Hint:
\(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ 5 }\) = \(\frac { z+2 }{ 2 }\) = λ ⇒ (3λ + 2, -5λ + 1, 2λ – 2)
which lie in x + 3y – αz + ß = 0
(3λ + 2) + 3(-5λ + 1) – α(2λ – 2) + ß = 0
3λ + 2 – 15λ + 3 – 2αλ + 2α + ß = 0.
(-12λ – 2αλ) + 2α + ß + 5 = 0.
-12λ – 2αλ = 0
2αλ = -12λ
α = -6
2α+ ß +5 = 0
-12 + ß + 5 = 0
ß – 7 = 0
ß = 7
(α, ß) = (-6, 7)

Question 17.
The angle between the line \(\overline { r }\) = (\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)) + t(2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\)) and the plane \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\)) + 4 = 0 is
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Solution:
(c) 45°
Hint:

Question 18.
The co-ordinates of the point where the line \(\overline { r }\) = (6(\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\)) + t(-\(\hat { i }\) + \(\hat { k }\) meets the plane \(\overline { r }\) ((\(\hat { i }\) + (\(\hat { j }\) – (\(\hat { k }\)) = 3 are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Solution:
(d) (5, -1, 1)
Hint:
Given \(\overline { r }\) = (6(\(\hat { i }\) – (\(\hat { j }\) – 3(\(\hat { k }\)) + t(-(\(\hat { i }\) + (\(\hat { k }\))
\(\frac { x-6 }{ -1 }\) = \(\frac { y+1 }{ 0 }\) = \(\frac { z+3 }{ 4 }\) = t ⇒ (-t + 6, -1, 4t – 3)
which meets x + y – z = 3
-t + 6 – 1 – 4t + 3 = 3
-5t + 5 = 0
5t = 5
t = 1
∴ Co-ordinate is (5, -1, 1)

Question 19.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
(x₁, y₁, z₁) = (o, 0, o)
(a, b, c) = (3, -6, 2); d = 7.
d = \(\frac { ax_1+by_1+cz_1+d }{ \sqrt{a^2+b^2+c^2} }\) = \(\frac { 7 }{ \sqrt{9+36+4} }\) = \(\frac { 7 }{ 7 }\) = 1

Question 20.
The distance between the planes
x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is
(a) \(\frac { √7 }{ 2√2 }\)
(b) \(\frac { 7 }{ 2 }\)
(c) \(\frac { √7 }{ 2 }\)
(d) \(\frac { 7 }{ 2√2 }\)
Solution:
(a) \(\frac { √7 }{ 2√2 }\)
Hint:
x + 2y + 3z+7 = 0
2x + 4y + 6z + 7 = 0
(÷ 2) x + 2y + 3z + \(\frac { 7 }{ 2 }\) = 0
(1) and (2) are parallel planes

Question 21.
If the direction cosines of a line are \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\)
(a) c = ±3
(b) c = ±√3
(c) c > 0
(d) 0 < c < 1
Solution:
(b) c = ±√3
Hint:
cos²α + cos²ß + cos²γ = 1
\(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) = 1
\(\frac { 3 }{ c ^2}\) = 1
c² = 3
c = ±√3

Question 22.
The vector equation \(\overline { r }\) = (\(\hat { i }\) – 2\(\hat { j }\) – \(\hat { k }\)) + t(6\(\hat { i }\) – \(\hat { k }\)) represents a straight line passing through the points
(a) (0, 6, -1) and (1, -2, -1)
(b) (0, 6, -1) and (-1, -4, -2)
(c) (1, -2, -1) and (1, 4, -2)
(d) (1, -2, -1) and (0, -6, 1)
Solution:
(c) (1, -2, -1) and (1, 4, -2)
Hint:
Given vector equation is

Question 23.
If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k = Q, then the values of k are
(a) ±3
(b) ±6
(c) -3, 9
(d) 3, -9
Solution:
(d) 3, -9
Hint:

Question 24.
If the planes \(\overline { r }\) (2\(\hat { i }\) – λ\(\hat { j }\) + \(\hat { k }\)) = 3 and \(\overline { r }\) (4\(\hat { i }\) + \(\hat { j }\) – µ\(\hat { k }\)) = 5 are parallel, then the value of λ and µ are
(a) \(\frac { 1 }{ 2 }\), -2
(b) –\(\frac { 1 }{ 2 }\), 2
(c) –\(\frac { 1 }{ 2 }\), -2
(d) \(\frac { 1 }{ 2 }\), 2
Solution:
(c) –\(\frac { 1 }{ 2 }\), -2
Hint:

Question 25.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(\frac { 1 }{ 5 }\), then the value of λ is
(a) 2√3
(b) 3√2
(c) 0
(d) 1
Solution:
(a) 2√3
Hint:

5 = \(\sqrt { 4+9+λ^2 }\)
25 = 4 + 9 + λ²
25 = 13 + λ²
λ² = 12
λ = 2√3