NCERT Solutions class 12 Maths Exercise 6.5

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NCERT Solutions class 12 Maths Exercise 6.5 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

Download NCERT solutions for Applications of Derivatives as PDF.

NCERT Solutions for Class 12 Maths Application of Derivatives

1. Find the maximum and minimum values, if any, of the following functions given by:

(i)

(ii)

(iii)

(iv)

Ans. (i) Given:

Since for all R

Adding 3 both sides,

Therefore, the minimum value of is 3 when , i.e.,

This function does not have a maximum value.

(ii) Given:

……….(i)

Since for all R

Subtracting 2 from both sides,

Therefore, minimum value of is and is obtained when , i.e.,

And this function does not have a maximum value.

(iii) Given: ……….(i)

Since for all R

Multiplying both sides by and adding 10 both sides,

[Using eq. (i)]

Therefore, maximum value of is 10 which is obtained when i.e.,

And therefore, minimum value of does not exist.

(iv) Given:

Therefore, maximum value and minimum value of do not exist.

NCERT Solutions class 12 Maths Exercise 6.5

2. Find the maximum and minimum values, if any, of the following functions given by:

(i)

(ii)

(iii)

(iv)

(v)

Ans. (i) Given: ……….(i)

Since for all R

Subtracting 1 from both sides,

Therefore, minimum value of is which is obtained when i.e.,

From eq. (i), maximum value of hence it does not exist.

(ii) Given:

Since for all R

Multiplying by both sides and adding 3 both sides,

Therefore, maximum value of is which is obtained when i.e.,

From eq. (i), minimum value of hence it does not exist.

(iii) Given: ……….(i)

Since for all R

Adding 5 to all sides,

Therefore, minimum value of is 4 and maximum value is 6.

(iv) Given:

Since for all R

Adding 3 to all sides,

Therefore, minimum value of is 2 and maximum value is 4.

(v) Given: ……….(i)

Since

Adding 1 to both sides,

Therefore, neither minimum value not maximum value of exists.

NCERT Solutions class 12 Maths Exercise 6.5

3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Ans. (i) Given:

and

Now

[Turning point]

Again, when , [Positive]

Therefore, is a point of local minima and local minimum value =

(ii) Given:

and

Now

or [Turning points]

Again, when ,

[Negative]

is a point of local maxima and local maximum value

And when [Positive]

is a point of local minima and local minimum value

(iii) Given: ……….(i)

and

Now

[Positive]

can have values in both I and III quadrant.

But, therefore, is only in I quadrant.

= [Turning point]

= = = [Negative]

is a point of local maxima and local maximum value

(iv) Given: ……….(i)

and

Now

[Negative]

can have values in both II and IV quadrant.

= or

and [Turning point]

At =

= = [Negative]

is a point of local maxima and local maximum value =

= =

= = [Positive]

is a point of local maxima and local maximum value =

= =

(v) Given:

and

Now

or [Turning points]

At [Negative]

is a point of local maxima and local maximum value is

At [Positive]

is a point of local minima and local minimum value is

(vi) Given:

= and

Now

= 0

But therefore is only the turning point.

is a point of local minima and local minimum value is

(vii) Given:

and

Now

[Turning point]

At [Negative]

is a point of local maxima and local maximum value is

(viii) Given:

= =

And

Now

= 0

is a point of local maxima and local maximum value is

Again

has local maximum value at

NCERT Solutions class 12 Maths Exercise 6.5

4. Prove that the following functions do not have maxima or minima:

(i)

(ii)

(iii)

Ans. (i) Given:

Now

But this gives no real value of Therefore, there is no turning point.

does not have maxima or minima.

(ii) Given:

Now

1 = 0

But this is not possible. Therefore, there is no turning point.

does not have maxima or minima.

(iii) Given:

Now

Here, values of are imaginary.

does not have maxima or minima.

5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i)

(ii)

(iii)

(iv)

Ans. (i) Given:

Now

Therefore, absolute minimum value of is and absolute maximum value is 8.

(ii) Given:

Now

[Positive]

is in I quadrant. [ ]

Therefore, absolute minimum value is and absolute maximum value is 1.

(iii) Given:

Now

Therefore, absolute minimum value is and absolute maximum value is 8.

(iv) Given:

Now

Therefore, absolute minimum value is and absolute maximum value is 19.

6. Find the maximum profit that a company can make, if the profit function is given by

Ans. Given: Profit function

and

Now

At , [Negative]

has a local maximum value at .

At , Maximum profit

= 41 + 16 – 8 = 49

7. Find both the maximum value and minimum value of on the interval [0, 3].

Ans. Let on [0, 3]

Now

Since is imaginary, therefore it is rejected.

is turning point.

At =

Therefore, absolute minimum value is and absolute maximum value is 25.

8. At what points on the interval does the function attain its maximum value?

Ans. Let

Now

Putting

Now

= =

Putting

Also and

Since attains its maximum value 1 at and

Therefore, the required points are and

9. What is the maximum value of the function

Ans. Let

Now

[Turning point]

If is even, then

If is odd, then

Therefore, maximum value of is and minimum value of is

10. Find the maximum value of in the interval [1, 3]. Find the maximum value of the same function in

Ans. Let

Now

or [Turning points]

For Interval [1, 3], is turning point.

Therefore, maximum value of is 89.

For Interval is turning point.

Therefore, maximum value of is 139.

11. It is given that at the function attains its maximum value, on the interval [0, 20]. Find the value of

Ans. Let

Since, attains its maximum value at in the interval [0, 2], therefore

NCERT Solutions class 12 Maths Exercise 6.5

12. Find the maximum and minimum value of on

Ans. Let

Now

= =

where Z

For But , therefore

For = and

For

But , therefore

Therefore, it is clear that the only turning point of given by which belong to given closed interval are,

nearly

Therefore, Maximum value = and minimum value = 0

NCERT Solutions class 12 Maths Exercise 6.5

13. Find two numbers whose sum is 24 and whose product is as large as possible.

Ans. Let the two numbers be and

According to the question,

…….(i)

And let is the product of and

[From eq. (i)]

and

Now to find turning point,

At , [Negative]

is a point of local maxima and is maximum at .

From eq. (i),

Therefore, the two required numbers are 12 and 12.

NCERT Solutions class 12 Maths Exercise 6.5

14. Find two positive integers and such that and is maximum.

Ans. Given: ……….(i)

Let P = [To be maximized] ……….(ii)

Putting from eq. (i), in eq. (ii),

P =

…..(iii)

Now

It is clear that changes sign from positive to negative as increases through 45.

Therefore, P is maximum when

Hence, is maximum when and

NCERT Solutions class 12 Maths Exercise 6.5

15. Find two positive integers and such that their sum is 35 and the product is a maximum.

Ans. Given:

……….(i)

Let

[From eq. (i)]

……….(ii)

Now

or or

Now is rejected because according to question, is a positive number.

Also is rejected because from eq. (i), = 35 – 35 = 0, but is positive.

Therefore, is only the turning point.

At ,

By second derivative test, will be maximum at when .

Therefore, the required numbers are 10 and 25.

NCERT Solutions class 12 Maths Exercise 6.5

16. Find two positive integers whose sum is 16 and sum of whose cubes is minimum.

Ans. Let the two positive numbers are and

……….(i)

Let

[From eq. (i)]

and

Now = 0

At is positive.

is a point of local minima and is minimum when .

Therefore, the required numbers are 8 and 8.

NCERT Solutions class 12 Maths Exercise 6.5

17. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Ans. Given: Each side of square piece of tin is 18 cm.

Let cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are and cm.

Let denotes the volume of the open box.

and

Now = 0

is rejected because at length = which is impossible.

is the turning point.

At , [Negative]

· is minimum at i.e., side of each square to be cut off from each corner for maximum volume is 3 cm.

NCERT Solutions class 12 Maths Exercise 6.5

18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Ans. Given: Dimensions of rectangular sheet are 45 cm and 24 cm.

Let cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are and cm.

Let denotes the volume of the open box.

and

Now = 0

is rejected because at length = which is impossible.

is the turning point.

At , [Negative]

· is minimum at i.e., side of each square to be cut off from each corner for maximum volume is 5 cm.

NCERT Solutions class 12 Maths Exercise 6.5

19. Show that of all the rectangles inscribed in a given fixed circle, the square has maximum area.

Ans. Let PQRS be the rectangle inscribed in a given circle with centre O and radius

Let and be the length and breadth of the rectangle, i.e., and

In right angled triangle PQR, using Pythagoras theorem,

PQ² + QR² = PR²

…..(i)

Let A be the area of the rectangle, then A = =

And

Now

= 0

At , [Negative]

At , area of rectangle is maximum.

And from eq. (i), ,

i.e.,

Therefore, the area of inscribed rectangle is maximum when it is square.

NCERT Solutions class 12 Maths Exercise 6.5

20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Ans. Let be the radius of the circular base and be the height of closed right circular cylinder.

Total surface area (S) =

= (say)

…..(i)

Volume of cylinder

= [From eq. (i)]

and

Now

At [Negative]

is maximum at .

From eq. (i),

= =

Height = Diameter

Therefore, the volume of cylinder is maximum when its height is equal to the diameter of its base.

NCERT Solutions class 12 Maths Exercise 6.5

21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area.

Ans. Let be the radius of the circular base and be the height of closed right circular cylinder.

According to the question, Volume of the cylinder

…(i)

Total surface area (S) =

= [From eq. (i)]

S =

and

Now

= = [Positive]

S is minimum when radius cm

From eq. (i)

= =

NCERT Solutions class 12 Maths Exercise 6.5

22. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Ans. Let meters be the side of square and meters be the radius of the circle.

Length of the wire = Perimeter of square + Circumference of circle

= 28

……….(i)

Area of square = and Area of circle =

Combined area (A) = + = +

and

Now

And [Positive]

A is minimum when

Therefore, the wire should be cut at a distance from one end.

NCERT Solutions class 12 Maths Exercise 6.5

23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is of the volume of the sphere.

Ans. Let O be the centre and R be the radius of the given sphere, BM = and AM =

In right angled triangle OMB, using Pythagoras theorem,

OM² + BM² = OB²

……….(i)

Volume of a cone inscribed in the given sphere

……….(ii)

and

Now

= [Negative]

is maximum at

From eq. (i)

Maximum volume of the cone

= (Volume of the sphere)

NCERT Solutions class 12 Maths Exercise 6.5

24. Show that the right circular cone of least curve surface and given volume has an altitude equal to time the radius of the base.

Ans. Let be the radius and be the height of the cone.

Volume of the cone (V) =

(say) ……….(i)

And Surface area of the cone (S) =

(say) ….(ii)

and

Now

……….(iii)

At [Positive]

is minimum when

From eq. (i),

= [From eq. (iii)]

Therefore, Surface area is minimum when height = (radius of base)

NCERT Solutions class 12 Maths Exercise 6.5

25. Show that the semi-vertical angle of the cone of the maximum value and of given slant height is

Ans. Let be the radius, be the height, be the slant height of given cone and be the semi-vertical angle of cone.

……….(i)

Volume of the cone (V) = ……….(ii)

V =

and

Now

= [Negative]

V is maximum at

From eq. (i),

Semi-vertical angle,

NCERT Solutions class 12 Maths Exercise 6.5

26. Show that the semi-vertical angle of the right circular cone of given surface area and maximum volume is

Ans. Let be the radius and be the height of the cone and semi-vertical angle be

Total Surface area of cone (S) =

(say)

……….(i)

Volume of cone (V) =

= [Using quotient rule]

……….(ii)

Now

[height can’t be negative]

is the turning point.

Since, , therefore, Volume is maximum at

From eq. (i),

Now Semi-vertical angle of the cone =

Choose the correct answer in the Exercises 27 to 29.

NCERT Solutions class 12 Maths Exercise 6.5

27. The point on the curve which is nearest to the point (0, 5) is:

(A)

(B)

(C) (0, 0)

(D) (2, 2)

Ans. Equation of the curve is ……….(i)

Let P be any point on the curve (i), then according to question,

Distance between given point (0, 5) and P = (say)

= [From eq. (i)]

= Z (say)

and

Now

[Positive]

Z is minimum and is minimum at

From eq. (i)

and are two points on curve (i) which are nearest to (0, 5).

Therefore, option (A) is correct.

28. For all real values of the minimum value of is:

(A) 0

(B) 1

(C) 3

(D)

Ans. Given: ……….(i)

Now

= 0

and [Turning points]

At ,

from eq. (i),

At ,

from eq. (i),

[Minimum value]

Therefore, option (D) is correct.

29. The maximum value of is:

(A)

(B)

(C) 1

(D)

Ans. Let

= , ……….(i)

Now

= 0

[Turning point] and it belongs to the given enclosed interval i.e., [0, 1].

At , from eq. (i),

At from eq. (i),

At , from eq. (i),

Maximum value of is 1.

Therefore, option (C) is correct.

NCERT Solutions class 12 Maths Exercise 6.5

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