# NCERT Exemplar Class 12 Maths Solutions Chapter 6 Applications Of Derivatives

NCERT Exemplar Solutions For Class 12 Maths Chapter 6 Applications Of Derivatives- For those who want to learn more about Calculus Mathematics, the best way to do it is by solving the NCERT Exemplar Class 12 Maths solutions Chapter 6. Students, who want to learn the topic of applications of derivatives, should attempt all the questions, mentioned in the chapter. Solving questions will not only help in understanding the elements but will also help in solving questions in the exam to score well. However, many tend to find it difficult to solve the questions and thus, require timely guidance. Also, read NCERT Class 12 Maths solutions

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Question:1

Answer:

Given: When a spherical ball salt is dissolved, the rate of decrease of the volume at any instant is proportional to the surface

To prove: The rate of decrease of radius is constant at any given time

Explanation: Take the radius of the spherical ball at any time t be â€˜râ€™

Assume S as the surface area of the spherical ball

Then, â€¦â€¦â€¦.(i)

Take the volume of the spherical ball be V

Then,

According to the given criteria,

The rate of decrease of volume is indicated by the negative sign It can also be written as

Here K is the proportional constant

After substitution of values from equation (i) and (ii), we get

When the constant term is taken outside the LHS, we get

After the derivatives are applied with respect to t, we get

After cancelling of the like terms, we get

Hence the rate of decrease of the radius of the spherical ball is constant.

Hence Proved

Question:2

Answer:

Given: A circle with uniformly increasing area rate

To prove: relation between perimeter and radius is inversely proportional

Explanation: Take the radius of circle â€˜râ€™

Let the area of the circle be A

Then â€¦â€¦..(i)

After considering the give criteria of area increasing at a uniform rate,

Substitute the value of equation (i) into above equation,

Differentiating with respect to t results in

Let P as the perimeter of the circle,

P = 2Ï€r

Now differentiate the perimeter with respect to t,

Apply all the derivatives,

Substituting of equation (ii) in the above equation,

Cancelling out of the like terms results into

Now covert it into proportionality,

Hence in the given conditions, the relation between perimeter of circle and radius is inversely proportional.

Hence Proved

Question:3

Answer:

Given: a boy of n height 1.5 m is flying a kite at a height of 151.5 m. The kite is moving with a speed of 10m/s. And the kite is 250 m away from the boy.

To find: the letting out speed of the string

Explanation: the above situation is explained by the figure,

Referring to the above figure,

Height of the kite, H = AD = 151.5 m

Height of the boy, b = BC = 1.5 m

x = CD = BE

Distance between kite and boy = AB = y =250

So, we need to calculate the increasing rate of the string

From figure, h = AE

= AD-ED

= 151.5-1.5

= 150m

The figure implies that the Î”ABE is a right-angled triangle

Applying of the Pythagoras theorem results into,

Letâ€™s differentiate the equation (i) with respect to time,

After using the differentiation sum rule, we get

since the height is not increasing, it indicates that it is constant, thus

Let's apply the derivative with respect to t

since the speed of the kite is 10 m/s so

When y = 250

After substituting the corresponding values in equation (iii), we get

Therefore, the letting out speed of the string is 8 m/s

Question:4

Answer:

Given: two men A and B start with velocities v at the same time from the junction of the two roads inclined at 45Â° to each other

To find: The rate of separation of the two men

Explanation:

The distance x travelled by A and B on any given time t will be same as they have velocity.

Hence

Apply the derivatives with respect to t,

Now take out the constant terms, we get

After value substitution for equation (i) we get

Thus, the above given amount is the rate of separation of the two roads.

Question:5

Find an angle which increases twice as fast as its sine.

Answer:

Given: a condition

To find: the angle Î¸ such that it increases twice as fast as its sine.

Explanation: Let x = sin Î¸

Letâ€™s differentiate with respect to t,

Now applying the derivative results into,

As this is given in the question

Let's substitute this value in equation (i)

After cancelling the like terms, we get

But given this possibility occurs only when

Hence the angle is

Question:6

Find the approximate value of .

Answer:

Given:

And as the nearest integer to 1.999 is 2 ,

Hence,

Therefore, the function becomes,

After applying of first derivative, we get

Now let

Now we know,

From equations (i) and (ii), substituting of functions results in,

Substitution of values of a and h, we get

So, the approximate value of = 31.92.

Question:7

Answer:

Given: A hollow spherical shell having an internal radius of 3cm and external radii of 3.0005 cm

To find: The amount of metal used in the formation of the spherical shell.

Explanation: Let the r and R be the internal and external radii respectively.

So, it is given,

R = 3.0005 and r = 3

Let V be the volume of the hollow shell.

So according to question,

To get the approximate value of , differentiation is carried out

But the integer nearest to 3.0005 is 3,

So 3.0005 = 3+0.0005

So let a = 3 and h = 0.0005

Hence,

Let the function becomes,

Now applying first derivative, we get

Now let

Now we know,

Now substituting the function from (ii) and (iii), we get

Substituting the values of a and h, we get

So, substituting of the value in equation (i),

Hence, the approximate volume of the metal in the hollow spherical shell is .

Question:8

Answer:

Given: a 2m tall man walks at the rate of m/s towards a m tall street light.

To find: calculate the rate of movement of the tip of the shadow and also the rate of change in the length of the shadow when he is m from the base of the light.

Explanation:

Here the street light is AB =

And man is DC = 2m

Let BC = x m and CE = y m

The rate of the manâ€™s walk towards the streetlight is , and as the man is moving towards the street light, the entity carries a negative charge

Hence, ....(i)

Now consider Î”ABE and Î”DCE

âˆ DEC = âˆ AEB (same angle)

âˆ DCE = âˆ ABD = 90Â°

Hence by AA similarity,

Î”ABEâ‰…Î”DCE

Hence by CPCT,

Apply the first derivative with respect to t,

After substituting the value in the above equation from equation (i), we get

So, the rate of movement of tip of the shadow is -1m/s, i.e., the length of the shadow is decreasing at the rate of 1m/s.

Let BE = z

So from fig,

z = x+y

Letâ€™s apply the first derivative with respect to t on the above equation.

So, the rate of tip of the shadow moving towards the light source is of m/s.

Question:9

Answer:

Given: L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and

To find: the rate of water running out of the pool at the last 5s. and also the average rate of water flowing at during the first 5s.

Explanation: Take the rate of Water running out given by

Given

Differentiation of the above given equation results in,

Now, to find the speed of water running out at the end of 5s, we need to find the value of equation (i) with t=5

Therefore, 2000 L/s is the rate of water running out at the end of 5s

To calculate the initial rate we need to take t=o in equation (i)

Equation (ii) tells about the final rate of water flowing whereas the equation (iii) is the initial rate.

Thus, the average rate during 5s is

After calculation, the final rate of water flowing out of the pool in 5s is 3000L/s.

Question:10

Answer:

Given: a cube with volume increasing at a constant rate

To prove: the relation between the increase int eh surface area with the length of the side is inversely proportional.

Explanation: Let â€˜aâ€™ the length of the side of the cube.

Take the volume of the cube â€˜Vâ€™

Then

As mentioned in the question, the rate of volume increase is constant, then

After substituting in the above equation, the values from equation (i) we get

Differentiating the equation with respect to t,

Take S as the surface area of the cube, then

After differentiating the surface area with respect to t, we get

Applying the derivatives, we get

After substituting value from equation (ii) in the given equation we get

After taking out the like terms,

Now converting it to proportional, we get

Therefore, the relation between the length and the side of the cube is inversely proportional in the given condition.

Hence Proved.

Question:11

Answer:

Given: two squares of sides x and y, such that

To find: The rate of change of area of both the squares with respect to each other

Explanation: Take and as the area of first and 2nd square respectively

Thus, the area of the 1st square will be

Differentiating the equation with respect to time, we get

And the area of the second square is

But given,

Now substituting the known value in equation (ii),

After differentiating equation (iii) with respect to time it results into,

Apply the power rule of differentiation to get,

Applying the sum rule of differentiation, we get

Since we need to find the rate of change of area of both the squares with respect to each other, which is

Substituting the known values from equation (i) and (iv), we get

By cancelling the like terms, we get

Question:12

Find the condition that the curves and 2xy = k intersect orthogonally.

Answer:

Given: two curves and 2xy = k

To find: to track the condition where both the curves intersect orthogonally

Explanation: Given 2xy = k

Put in the value of y in another curve equation, i.e., we get

Putting both the sides under cube root, we get

Substituting equation (ii) in equation (i), we get

is the point of intersection of two curves

Now given

After differentiating the equation with respect to x, we get

After tracking the value of differentiation at the point of intersection i.e., at we

get

Also given 2xy = k

Differentiating this with respect to x, we get

Then again finding the about given differentiation value at the point of intersection i.e., at , we get

But the orthogonal interaction of two curves occurs if

m1.m2 = -1

Then Substituting the values from equation (iii) and equation (iv), we get

This condition proves to fulfill the orthogonal interaction point for the two curves.

Question:13

Prove that the curves and touch each other.

Answer:

Given: two curves and

To prove: two curves meet each other at a point

Explanation:

Now given

Differentiating this with respect to x, we get

Also given xy = 4

Differentiating this with respect to x, we get

The using the product rule of differentiation, we get

But the touch of 2 curves is possible if

Now substituting the values from equation (ii) and equation (ii), we get

Now substituting

When

when

Therefore, (2,2) and (-2, -2) is the intersection point of the two curve

Substituting these points of intersection equation (i) and equation (ii), we get

For (2,2),

Thus, the condition for both the curves to touch is possible if that they have same slope

Hence the two given curves touch each other.

Hence proved

Question:14

Answer:

Given: curve

To find: point coordinates on which tangent is equally inclined to the axis on the curve

Explanation: given

After differentiating with respect to,

Now using the sum rule of differentiation

Then differentiating the equation, we get

The given curve has this tangent

As mentioned in the question tangent is equally inclined to the axis,

Substituting values in the curve equation from equation (ii)

When y = 4, then x = 4 from equation (ii)

Show the points on the curve at which the tangent equally inclined to the axis has the coordinates (4,4).

Question:15

Find the angle of intersection of the curves and

Answer:

Given: the curves and

To find: the interaction angle between two curves

Explanation: acknowledging first curve

when the above curve is differentiated with respect to X

second curve differentiated with respect to X

Given Substituting the other curve equation with this

When , we get

When we get

Hence the intersection points are since angle of intersection can be found using the formula

i.e.,

Substituting the values from equation (i) and equation (ii), we get

For the equation gets converted into,

Hence, the angle at which the curve intersect at

Question:16

Prove that the curves and touch each other at the point (1, 2).

Answer:

Given: two curves and

To prove: Two curves have the possibilities of meeting at a point(1,2)

Explanation:

Now given

Differentiating the above equation with respect to x

Now using the sum rule of differentiation

Also given

Differentiating the value with respect to x, we get

Using a product rule of differentiation, we get

Finding the solution of the above equation at point (1,2), we get

From equation (i) and (ii),

âˆ´

Therefore it is possible for both the curves to touch each other at point(1,2).

Hence proved

Question:17

Find the equation of the normal lines to the curve which are parallel to the line .

Answer:

Given: equation of the curve equation of line

To find: the equation of the normal lines to the given curve which are parallel to the given line

Explanation:

Now given equation of curve as

Differentiating the equation with respect to X

Assume the slope of the normal curve to be m_{2} is given by

Substituting value from equation (i), we get

The known equation of the line is

the slope of this line will be

since, slope of normal to the curve should be equal to the slope of the line which is parallel to the curve,

Substituting values from equation (ii) and (iii), we get

After putting y=x in the equation of the curve, we get

But from equation (iv)

y=x

Therefore, the points at which normal to the given curve is parallel to the given line are (2, 2) and (-2, -2)

Thus, the equations of the normal can be calculated by

Question:18

At what points on the curve , the tangents are parallel to the y-axis?

Answer:

Given: equation of a curve

To find: the points on the curve , the tangents are parallel to the y-axis

Explanation: the given equation of curve as

Differentiating the equation with respect to X

Sense the tangents are parallel to the axis as mentioned in the question

Thus,

Putting y = 2 in curve equation, we get

After the splitting of the middle term,

Thus, the needed points are(-1, 2) and (3, 2).

Hence the points on the curve , the tangents are parallel to the y-axis are (-1, 2) and (3, 2).

Question:19

Show that the line touches the curve at the point where the curve intersects the axis of y.

Answer:

Given: equation of line the curve intersects the y-axis

To show: the line touches the curve at the point where the curve intersects the axis of y

Explanation: given the curve intersects the y-axis, i.e., at x = 0

Now differentiate the given curve equation with respect to x, i.e.,

Then considering the line equation,

Line touches the curve only if their slopes are equal

From equation (i) and (ii), we see that

Hence, the line touches the curve at the point where the curve intersects the axis of y.

Question:20

Answer:

Given:

To show: the mentioned function increases in R

Explanation: Given

Substituting the first derivative in respect to x

But the derivative of 2x is 2,so

But the derivative of ,so

When the sum rule is applied to the last part we get,

Then to calculate any real value x, the above value of f(x) is larger than or equal to zero

Hence

And we know, if , then f(x) is increasing function.

Hence, the given function is an increasing function in R.

Question:21

Show that for is decreasing in R.

Answer:

Given:

To show: the above function is decreasing in R.

Explanation: Given

The first derivative is applied with respect to x,

By using the sum rule of differentiation, we get

Removing all the constant terms, we get

But the derivative of sin X = cos x and that of cos x = -sin x, so

Multiplying and dividing RHS by 2,

And we know, if , then f(x) is decreasing function.

Therefore, the given function is decreasing function in R.

Question:22

Show that is an increasing function in

Answer:

Given:

To show: the given function is increasing in

Explanation: Given

First derivative is applied with respect to x,

Using the differentiation rule for , results into

Now use the sum rule of differentiation,

To make f(x) to be increasing function,

But this is possible only when

Hence, the given function is increasing function in

Question:23

At what point, the slope of the curve is maximum? Also find the maximum slope.

Answer:

Given:

To find: the point in curve where the slope is maximum and the maximum value of the slope.

Explanation: given

The slope of the curve can be found by calculating the first derivative of the known curve equation,

Thus, slope of the curve is

Then using the derivative,

To know the critical point, it is necessary to have the value of the slope,

Using the derivative leads to,

When the second derivative is equated to 0 it gives the critical point, i.e.,

Then finding the third derivative of the curve,

i.e.,

Putting the values of the derivative results in

As the third derivative is less than 0, so the maximum slope of the given curve is at x=1.

Equating the first derivative with x=1 leads to the maximum value of the slope, i.e.,

Therefore, the slope of the curve is maximum at x=1, and the maximum value of the slope is 12 .

Question:24

Prove that has maximum value at .

Answer:

Given:

To prove: the given function has maximum value at

Explanation: given

While calculating the first derivative we get,

Then putting the derivative, we get

Critical point ca be calculated by equating the derivative with 0,

This is possible only when

The second derivative if the c=function can be calculated by,

Putting the value of derivative, we get

Then, we will substitute in the above equation, we get

Putting the corresponding value, we get

Hence f(x) has a maximum value at .

Hence proved.

Question:25

Answer:

Given: a right-angled triangle with the sum of the lengths of its hypotenuse and side.

To show: at this angle the area of the triangle is maximum

Explanation:

Let Î”ABC be the right-angled triangle,

Let hypotenuse, AC = y,

side, BC = x, AB = h

then the calculation of sum of the side and hypotenuse is done using,

â‡’ x+y = k, where k is any constant value

â‡’ y = k-xâ€¦â€¦â€¦..(i)

Take A as the area of the triangle, as we know

Then using the Pythagoras theorem, we get

Putting the value from equation (i) in above equation, we get

Applying the values from equation (iii) into equation (ii), we get

The above equation is differentiated with respect to x,

Then the constant terms are taken out,

Power rule pf differentiation is applied on the second part of the above equation,

Once again, differentiating equation (iv) with respect to x, we get

Using the product rule of differentiation,

Then using the power rule of differentiation,

Putting , in above equation, we get

Hence the maximum value of A is at

We know,

Then from figure,

Applying the value of y=k -x from equation (i), we get

Putting the value of we get

This possibility is present when

Therefore, the area of the triangle is maximum only when the angle between them is

Question:26

Answer:

Given: function

To find: the points of local maxima, local minima and the points of inflection of f(x) and also to find the corresponding local maximum and local minimum values.

Explanation: given

Calculating the first derivative of f(x), i.e.,

Equating the first derivative with 0 to find out the critical point,

Then splitting the middle term, we get

Now we will find the corresponding y value by putting the numerous values of x in given function

Hence the point is (0,-1)

Hence the point is (1,0)

Hence the point is (3,-28)

Therefore, we see that

At x = 3, y has minimum value = -28. Hence x = 3 is point of local minima.

At x = 1, y has maximum value = 0. Hence x = 1 is point of local maxima.

And at x = 0, y has neither maximum nor minimum value, hence this is point of inflection.

Question:27

Answer:

Given: a telephone company in a town has 500 subscribers and collects fixed charges of Rs 300/- per subscriber per year, company increase the annual subscription and for every increase of Re 1/- one subscriber will discontinue the service

To find: the best increase amount for the company to earn maximum profit

Explanation: company has 500 subscribers, and collects 300 per subscriber per year.

Let x as the increase in annual subscription by the company

As per the question, the number of subscribers to discontinue the service will be x

The total revenue earned after the increment would be calculated by,

We need to calculate the first derivative of the above equation,

The critical point is calculated by equating the first derivative with 0,

Then we calculate the second derivative of the total revenue function, i.e., again differentiate equation (i), i.e.,

Hence Râ€™â€™(100) is also less than 0,

Therefore, R(x) is maximum at x = 0, i.e.,

Thus, the required increase on the subscription fee for the company to make profit is by Rs 100.

Question:28

If the straight-line touches the curve then prove that

Answer:

Given: equation of straight-line , equation of curve and the straight line touches the curve

To prove:

Explanation: the know line equation is,

We know that, if a line y = mx+c touches the eclipse, then required condition is

Then putting the corresponding values, we get

Removing the like terms we get,

Hence, proved.

Question:29

Answer:

Given: a cardboard box that is open and square in shape has area

To show: cubic units is the maximum volume of the box.

Explanation:

Take the side of the square be x cm and

Take the height the box be y cm.

So, the total area of the cardboard used is

A = area of square base + 4x area of rectangle

But it is given this is equal to , hence

According to the given condition the area of the square base will be

V = base Ã— height

Since the base is square, the volume is

Then putting the values of equation (i) in equation (ii), we get

Calculation of the first derivative of the equation,

Removing all the constant terms

Using the sum rule of differentiation, we get

Removing all the constant terms, we get

After differentiating the equation, we get

We need to calculate the second derivative to find out the maximum value of x , so for that let equating above equation with 0, we get

Differentiating equation (iii) again with respect to x, we get

Removing all the constant terms results into

Using the differentiation rule of sum, we get

, the above equation becomes,

Thus, the volume (V) is maximum at

âˆ´ The box has a maximum value of

So, the box has a maximum value of is cubic units.

Hence, proved.

Question:30

Answer:

Given: rectangle of perimeter 36cm

To find: to estimate the dimensions of a rectangle in a way that it can sweep out maximum amount of volume when resolved to about one of its sides. Also, to find the maximum volume

Explanation: x and y can be the length and the breadth of the rectangle

The known perimeter of the rectangle is 36cm

Now when the rectangle revolve about side y it will form a cylinder with y as the height and x as the radius, then if the volume of the cylinder is V, then we know

Applying the value from equation (i) in above equation we get

Then find out the first derivative of the given equation,

Taking out the constant terms from equation followed by using the sum rule of differentiation,

To calculate critical point, we will equate the first derivative to 0, i.e.,

By differentiating the second equation, second derivative of the volume equations can be easily calculated,

Taking out the constant terms from equation followed by using the sum rule of differentiation,

Now substituting x = 12 (from equation (iii)), we get

Hence at x = 12, V will have maximum value.

The maximum value of V can be found by substituting x = 12 in

i. e

Therefore, the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides equal to 12cm.

And the maximum volume is .

Question:31

Answer:

Given: The combined surface area of a cube and sphere are constant

To find: the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum

Explanation: Let â€˜aâ€™ be the side of the cube

Then surface area of the cube = â€¦.(i)

Take â€˜râ€™ as the radius of the sphere

Then the surface area of the sphere = â€¦(ii)

According to the question, the surface area of both the figures is added, thus adding the equation (i) and (ii), we get

As the formula of volume of cube is

Plus the volume of a sphere is

Hence adding both the volumes will result into,

Then putting the values from equation (iii) in above equation,

After finding the first derivative of the volume, we get

After taking out the constant terms along with using the sum rule of differentiating,

Using the power rule of differentiation,

Now we know,

Hence

To find the second derivative of this volume equation, we cam simply differentiate the equation (ii),

After removing the constant terms, we apply the sum rule of differentiation,

Using the product rule of differentiation,

Again, the power rule of differentiation is used,

Differentiating the equation, we get

Hence for

The substituting, in equation (iii), we get

Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,

a:2r

Hence the required ratio is

a:2r = 1:1

Question:32

Answer:

Given: a circle with AB as diameter and C is any point on the circle

To show: area of Î” ABC is maximum, when it is isosceles

Explanation: If we take r as the radius of the circle, the diameter will become 2r= AB

This indicates that any angle in a semicircle is 90Â°.

Hence âˆ ACB = 90Â°

Now let AC = x and BC = y

Using the Pythagoras theorem in this right-angled triangle ABC,

Then putting the values from the equation (i), we get

By finding the first derivative of the area,

Simultaneously using the product rule of differentiation and also taking out the constant terms,

Using the power rule of differentiation,

Critical point can be calculated by putting the first derivative equal to 0

Hence

Differentiating the equation (ii) will give us the second derivative of the equation

Removing all the constant terms and then using the product rule of differentiation,

After using the power rule of differentiation,

For in above equation, we get

Thus, for , the area of is maximum

The maximum value can be calculated by substituting in equation (i),

i.e., the two sides of the are equal

Hence, the area of is maximum, when it is isosceles

Hence, proved.

Question:33

Answer:

Given: A metal box with a square base and vertical sides is to contain . The material for the top and bottom costs Rs and the material for the sides costs Rs

To find: the minimum cost of the box

.

Take x cm as the side of the square

Take y cm as the vertical side of the metal box

According to the given information in the question, the formula used volume for square base is

V=base Ã— height

Due to its square base, the formula of the volume is

This is equal to . So, volume becomes

Then we need to calculate the total area of the metal box.

Area of top and bottom

The mentioned material for the top and bottom costs Rs , thus, the material cost for top and bottom becomes

Cost of top and bottom =Rs. 5()

Area of one side of the metal box = xy cm

The total sides present in the metal box are 4, so

Thus, the total area of all the sides of the metal box = 4xy

The cost of the material for sides is Rs

âˆ´ Cost of all the sides of the metal box =Rs. 2.50(4xy)

The overall area of the metal box will be

This will make the cost of the box to be

Putting the value of y from equation (i) in the above equation,

Both the sides are differentiated with respect to x

Using differentiation rule of sum, we get

Then using the derivative, we get

Take c=0 to find the minimum value of x by apply second derivative test, so the above equation is equated with 0

Solving this we get

Again, differentiating equation (ii) with respect to x,

Using the differentiation rule of sum,

At x=8, the above equation becomes,

Now at x=8, , so as per the second derivative test, x is a point of local minima and will be minimum value of C.

Hence least cost becomes

Hence the least cost of the metal box is Rs. 1920

Question:34

Given: x, 2x and are the sides of a rectangular parallelepiped. The sum of the surface areas of a sphere and a rectangular parallelepiped is given to be constant.

To prove: if x is equal to three times the radius of the sphere, the sum of their volumes is minimum and find the minimum value of the sum of their volumes

Surface area of rectangular parallelepiped:

Let radius of sphere be r cm, then surface area is

Now sum of the surface areas is,

Now given that the sum of the surface areas is constant, so

Now, differentiate (i) with respect to r and get

Apply differentiation rule of sum and get

Take the constant terms out and get

Apply derivative and get

Let V denote the sum of volumes of both the shapes, so

The first derivative of volume must be equal to 0 for minima or maxima

Differentiate (iii) with respect to r and get

Apply differentiation rule of sum and get

Take constant terms out and get

Apply derivative and get

Substitute value of from (ii) and get

i.e., the radius of the sphere is 1/3 of x.

Hence proved

Now letâ€™s find the second derivative value at x=3r.

Now, apply derivative with respect to r to (iv) and get

Apply differentiation rule of sum and get

Take constant terms out and get

The first part is applied the differentiation rule of product, so

Substitute value of from (ii) and get

Substitutex=3r and get

It is positive;so V is minimum when , and the minimum value of Volume

can be obtained by substituting in equation (iii), we get

Therefore, it Is the minimum value of the sum of their volumes.

Question:35

Answer:

Let x cm be the side of the equilateral triangle, then the area of the triangle is

Also, the rate of side increasing at instant of time t is

Differentiate area with respect to time t and get

Take the constants out and get,

Apply the derivative and get

Substitute given value of and get

Now, put side

Hence, the rate at which the area increases is

So the correct answer is option C.

Question:36

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is:

A. radian/sec

B. radian/sec

C. 20 radian/sec

D. 10 radian/sec

Answer:

Let 5m/500cm be the length of the ladder, which is the hypotenuse of the right triangle formed in the above figure.

Now let the angle between the ladder and the floor be Î², so

Differentiate both sides with respect to time t and get

Apply derivatives and get

Now, the top of the ladder slides downwards at the rate of

So the equation is

Now,

Hence the rate at which the angle between the floor and the ladder is decreasing is radian/sec

So the correct answer is option B.

Question:37

The curve has at (0, 0)

A. a vertical tangent (parallel to y-axis)

B. a horizontal tangent (parallel to x-axis)

C. an oblique tangent

D. no tangent

Answer:

Given

Differentiate both sides with x and get

Apply power rule and get

Now at (0,0)

So the curve at (0,0) has vertical tangent parallel to Y-axis.

Hence the correct answer is option A.

Question:38

The equation of normal to the curve which is parallel to the line is

Answer:

Given the equation of the line is

Differentiate both sides with x and get

Apply sum rule and 0 is the differentiation of constant, so

Take the constants out and get

Apply power rule and get

Hence, the slope of the given curve is provided.

Also, the slope of the normal to the curve is

Now,

After differentiating with respect to x

Therefore, the slope is

Now, because the normal to the curve is parallel to this line, that means the slope of the line must be equal to slope of the normal to the given curve,

Substitute the value of the given equation

When x=2, the equation is

When x=-2, the equation is

So, the points are at which normal is parallel to the given line.

And required equation at is

Hence the equation of normal to the curve is

So the correct answer is option C

Question:39

If the curve and , cut orthogonally at (1, 1), then the value of a is:

A. 1

B. 0

C. â€“ 6

D. 6

Answer:

Given the fact that curve and , cut orthogonally at (1, 1)

Differentiate on both sides with x and get

Apply sum rule and also 0 is the derivative of the constant, so

Apply power rule and get

Putting (1,1)

Differentiate on both sides with x and get

Apply power rule and get

Putting (1,1)

Both curves cut orthogonally at (1,1), 50

So from (i) and (ii), we get

Hence when the curves cut orthogonally at (1, 1), then the value of a is 6.

So the correct answer is option D.

Question:40

If and if x changes from 2 to 1.99, what is the change in y

A. 0.32

B. 0.032

C. 5.68

D. 5.968

Answer:

A)

Given

Differentiate on both sides with x and get

Apply power rule and get

Now, value of x changes from 2 to 1.99, so the change in x is

So the change in y is,

Substitute corresponding values and get

Now at x=2, the change in y becomes

Therefore, change in y is 0.32.

Question:41

The equation of tangent to the curve where it crosses x-axis is:

Answer:

A)

Given the equation of the curve is

Both the sides are differentiated with respect to x,

Using the power rule

As the derivative of a constant is always 0 we get

Again, using the power rule

The mentioned curve passes through the x -axis, i.e., y=0

Thus, the curve equation becomes

As the point of passing for the given curve is (2,0)

So the equation (i) at point (2,0) is,

So, the slope of tangent to the curve is

Therefore, the equation of tangent of the curve passing through (2,0) is given by

Thus, the equation of tangent to the curve , where it crosses x-axis is.

Hence, the correct option is option A

Question:42

The points at which the tangents to the curve are parallel to x-axis are:

A. (2, -2), (-2, -34)

B. (2, 34), (-2, 0)

C. (0, 34), (-2, 0)

D. (2, 2), (-2, 34)

Answer:

D)

Given the equation of the curve is

Differentiating on both sides with respect to x, we get

Applying the sum rule of differentiation, we get

We know derivative of a constant is 0,so above equation becomes

Applying the power rule we get

Thus, the slope of line parallel to the x -axis is given by

So equating equation (i) to 0 we get

When x=2, the given equation of curve becomes,

When x=-2, the given equation of curve becomes,

Hence, the points at which the tangents to the curve are parallel to x-axis are (2, 2) and (-2, 34).

So, the correct option is option D.

Question:43

The tangent to the curve at the point (0, 1) meets x-axis at:

A. (0, 1)

B.

C. (2, 0)

D. (0, 2)

Answer:

Given the equation of the curve is

Differentiating on both sides with respect to x, we get

Applying the exponential rule of differentiation, we get

As it is given the curve has tangent at (0,1), so the curve passes through the point (0,1), so above equation at (0,1), becomes

So, the slope of the tangent to the curve at point (0,1) is 2

Hence the equation of the tangent is given by

It is given that the tangent to the curve at the point (0,1) meet x-axis i.e., y=0

So the equation on tangent becomes,

Hence, the required point is

Therefore, the tangent to the curve at the point (0,1) meets x -axis at

So, the correct option is option B.

Question:44

The slope of tangent to the curve at the point (2, -1) is:

A.

B.

C.

D.

Answer:

Curve of the given equation is

With respect to t, while differentiating on both sides, we get

After application of the sum rule of differentiation, we get

Constant's derivative is 0, so above equation becomes

Power Rule application leads to

With respect to t, we differentiate on both side and get

Sum Rule application leads to

The Constant's derivative is 0, so the equation becomes

Applying power rule

We know,

Substitute values from equation (i) and (ii)

The point through which the curve passes is (2,-1), now, substitute the same and get

Split the middle term

Take 2 as common

Split the middle term again

In equation (iii) and (iv), 2 is common

So, t=2

So, the slope of the tangent at t=2 is as follows

Therefore, the slope of tangent at the point (2,-1) is

So, the correct answer is option B.

Question:45

The two curves and intersect at an angle of

A.

B.

C.

D.

Answer:

Given the curve and

Differentiate on both the sides with respect to x

Apply the sum rule and also 0 is the the derivative of the constant, so it becomes

Apply power rule and get

Apply product rule and get

Differentiate on both the sides with respect to x and get

Apply the sum rule and also 0 is the derivative of the constant, so

Apply power rule and get

Apply product rule and get

Since the product of the slopes is -1, it means that both the curves are intersecting at right angle i.., they are making angle with each other.

So, the correct answer is option C

Question:46

The interval on which the function is decreasing is:

A. [-1, âˆž)

B. [-2, -1]

C. (-âˆž, -2]

D. [-1, 1]

Answer:

Given

Apply first derivative and get

Apply power rule and get

Split the middle term and get

Now f'(x)=0 gives

x=-1, -2

Three intervals are made when these points divide the real number line

So, the interval on which the function decreases is [-2, -1].

So, the correct answer is option B.

Question:47

Let the be defined by , then f(x) :

A. has a minimum at x = Ï€

B. has a maximum, at x = 0

C. is a decreasing function

D. is an increasing function

Answer:

Given if

Apply the first derivative and get

Apply power rule and get

Now, 1 is the maximum value of sin x.

So, function f is an increasing function.

So the correct answer is option D.

Question:48

y = decreases for the values of x given by :

A.

B.

C.

D.

Answer:

Given

Apply first derivative and get

Apply sum rule of differentiation and get

Apply power rule and get

Now, split middle term and get

Now, gives us

x=1, 3

The points divide this real number line into three intervals

So, the interval on which the function decreases is (1,3) i.e., 1<x<3

So the correct answer is option A.

Question:

The function is strictly

A. increasing in

B. decreasing in

C. decreasing in

D. decreasing in

Answer:

Given

Apply the first derivative and get

Apply sum rule and get

Then apply power rule and get

Now apply the derivative,

Now, and

Hence

Therefore,

Hence f(x) is increasing when

and , when

Hence f(x) is decreasing when

Now

Hence, f(x) is decreasing in

So the correct answer is option B

Question:50

Which of the following functions is decreasing on .

A. sin2x

B. tan x

C. cos x

D. cos 3x

Answer:

(i) Let f(x)=sin 2x

Apply first derivative and get

fâ€™(x)=2cos 2x

Put fâ€™(x)=0, and get

2cos 2x =0

â‡’ cos 2x=0

It is possible when

0â‰¤xâ‰¤2Ï€

Thus, sin 2x does not decrease or increase on

(ii) Let f(x)=tan x

Apply first derivative and get

fâ€™(x)=

Now. square of every number is always positive,

So, tan x is increasing function in

(iii) Let f(x)=cos x

Apply first derivative and get

fâ€™(x)=-sin x

But, sin x>0 for

And -sin x<0 for

Hence fâ€™(x)<0 for

â‡’ cos x is strictly decreasing on

(iv) Let f(x)=cos 3x

Apply first derivative and get

fâ€™(x)=-3sin 3x

Put fâ€™(x)=0, we get

-3sin 3x=0

â‡’ sin 3x=0

Because sin Î¸=0 if Î¸=0, Ï€, 2Ï€, 3Ï€

â‡’ 3x=0,Ï€, 2Ï€, 3Ï€

so we write it on number line as

Now, this point into 2 disjoint intervals.

i.e.

case 1 : for

So wher

Also,

From equation (a), we get

sin 3x <0 for

case 2: for

Now

Also,

Equation (b) gives

Hence, cos 3 x does not decrease or increase on

So, the correct answer is option C i.e., is decreasing in

Question:51

The function

A. always increases

B. always decreases

C. never increases

D. sometimes increases and sometimes decreases.

Answer:

Given

Apply first derivative and get

Apply sum rule and get

Apply derivative,

Square of every number is always positive,

So

So always increases.

So the correct answer is option A

Question:52

If x is real, the minimum value of is

A. -1

B. 0

C. 1

D. 2

Answer:

Let

Apply first derivative and get

Apply derivative,

Hence the minimum value of f(x) at x=4 is given by

So, if x is real, 1 is the minimum value of

So the correct answer is option C.

Question:53

The smallest value of the polynomial in [0, 9] is

A. 126

B. 0

C. 135

D. 160

Answer:

Let

Apply first derivative and get

Apply derivative,

Split middle term and get

Now we find the values of f(x) at x=0, 4, 8, 9

Hence we find that 0 is the absolute minimum value of f(x) in [0,9] at x=0.

So the correct answer is option B.

Question:54

The function , has

A. two points of local maximum

B. two points of local minimum

C. one maxima and one minima

D. no maxima or minima

Answer:

Let

Apply first derivative and get

Apply derivative,

Put f'(x)=0, and get

Split middle term and get

Now we find the values of f(x) at x=-1, 2

Hence from above we find that the point of local maxima is x=-1 and 11 is the maximum value of f(x).

Whereas the point of local minima is x=2 and -16 is the minimum value of f(x).

So, the correct answer is option C.

Hence, the given function has 1 minima and 1 maxima.

Question:55

The maximum value of sin x cos x is

A.

B.

C.

D.

Answer:

Let f(x)= sin x cos x

sin2x=2sin x cos x

Apply first derivative and get

Apply derivative,

Put and get

Equate the angles and get

Now we find second derivative by deriving equation (i) and get

Apply derivative,

Now we find the value of we get

But so above equation becomes

Hencee at, is maximum and is the point of maxima.

Now we will find the maximum value of by substituting in we get

So, maximum value of is

So, the correct answer is option B

Question:56

At , is:

A. maximum

B. minimum

C. zero

D. neither maximum nor minimum

Answer:

Given

Apply first derivative and get

Apply sum rule and take the constant terms out and get

Apply derivative,

And we found fâ€™(x) at not equal to 0.

So cannot be point of minima or maxima.

Hence, at is not minima nor maxima.

So, the correct answer is option D.

Question:57

Maximum slope of the curve is:

A. 0

B. 12

C. 16

D. 32

Answer:

Given equation of curve is

Apply first derivative and get

Apply sum rule and $\varrho$ is the differentiation of the constant term, so

Apply power rule and get

Hence, it is the slope of the curve.

Now to find out the second derivative of the given curve, we will differentiate equation (i) once again

Apply sum rule and 0 is the differentiation of the constant term so

Apply power rule and get

Now we will find the critical point by equating the second derivative to 0, we get

-6(x-1) =0

â‡’ x-1=0

â‡’ x=1

Now, to find out the third derivative of the given curve, we will differentiate equation (ii) once again

Apply sum rule and 0 is the differentiation of the constant term, so

Apply power rule and get

Hence, maximum slope is at

Now, substitute in (i), and get

Therefore, 12 is the maximum slope of the curve .

So, the correct answer is option B.

Question:58

has a stationary point at

A. x = e

B.

C. x = 1

D.

Answer:

Given equation is

Let â€¦â€¦â€¦(i)

Take logarithm on both side

â‡’ log y=x log x

Apply first derivative and get

Apply product rule and get

Apply first derivative and get

Substitute value of y from (i) and get

Now we find the critical point by equating (i) to 0 and get

Equate the terms and get

Therefore f(x) has a stationary point at

So, the correct answer is option B.

Question:59

The maximum value of is:

A.

B.

C.

D.

Answer:

Let

Take logarithm on both side

Applying first derivative and get

Apply product rule and get

Applying first derivative and get

Now we find critical point by equating (i) to 0

Therefore f(x) has a stationary point at .

i.e the maximum value of

So, the correct answer is option C.

Question:60

Fill in the blanks in each of the following

The curves and touch each other at the point_____.

Answer:

Given the first curve is

Applying first derivative and get

Apply sum rule and 0 is the differentiation of the constant term is 0,so

Apply power rule and get

This is the slope of the first curve; let be equal to this.

The second curve is

Applying first derivative and get

Apply sum rule and 0 is the differentiation of the constant term, so

This is the slope of the second curve; let be equal to this.

Now the slopes must be equal, because they touch each other, i.e.,

Split middle term and get

Substitute in both the equations and get

For first curve,

For second curve,

Substitute x=3 in both equations

For first curve,

For second curve,

Hence at x=3 both curves don't touch

So, the curves and do not touch each other.

Question:61

Answer:

Given curve is

Apply first derivative and get

It is the slope of tangent

Substitute (0,0) in slope and get

Hence, -1 is the slope of the normal to the curve at (0,0)

Hence the equation is

Question:62

Answer:

Given

Apply first derivative and get

Apply sum rule and 0 is the differentiation of the constant term, so

Apply first derivative and get

Also f(x) increases on R

This is possible when

Hence

The values of a increases on

Question:63

Fill in the blanks in each of the following

The function decreases in the interval _______.

Answer:

Given

After applying derivative, we get

Apply quotient rule and 0 is the differentiation of the constant term, so

Equate this with 0 and get

and are the intervals formed by these two critical numbers

(i) in the interval

(ii) in the interval (-1,0),

is decreasing in (-1,0)

(iii) in the interval (0,1),

is increasing in

(iii) in the interval

is decreasing in

Therefore, the function decreases in the interval .

Question:64

Fill in the blanks in each of the following

The least value of the function is ______.

Answer:

Given

After applying the derivative

Apply sum rule and get

Apply quotient rule on second part and get

Equate it with 0 and get

Now given by second derivative,

Apply derivative and get

Apply sum rule and get

Apply quotient rule on the second part and get

Now, equate it with

The least value of f(x) is

Multiply and divide by and get

Therefore, the least value of function is

With the NCERT Exemplar Class 12 Maths solutions Chapter 6, you will have a better grasp of the topics and how to solve the questions. Our team of experts will help in expanding the questions in a way that will make it easier for the students to make sense of. Also, the solutions provided are following CBSE standards and guidelines.

You can easily download the pdf copy of NCERT Exemplar Class 12 Maths solutions Chapter 6 by using the Maths NCERT exemplar Class 12 solutions chapters 6 pdf download feature which includes the solutions for the in-chapter questions and final exercise questions. Our guidance team and teachers have solved the questions in a way that is easy to understand and simple to grasp.

The language is simple and the process is shown in a step-by-step manner while being thorough enough for the students to understand every question in detail. Solving the questions before exams during the preparatory phase can help in getting an idea of what will be asked and how to manage the score well in mathematics.

## Subtopics in NCERT Exemplar Class 12 Maths solutions Chapter 6

The sub-topics that are covered in this chapter are:

Introduction

Rate of change of quantities

Increasing and decreasing functions

Normals and Tangents

Approximations

Maxima and minima

6.6.1 Minimum and maximum value of the function is a closed interval

## What you will learn from NCERT Exemplar Class 12 Maths solutions Chapter 6?

In NCERT Exemplar Class 12 Mathematics Chapter 6 Applications of Derivatives, the students will come to know in detail about derivatives of functions and numbers. Also, one will know about the applications and uses of these derivatives. One will learn about the most crucial part of the derivatives topic, and that is the rate of change. One will learn about how one or two quantities change due to change in some other quantity.

Students will get a clear picture of how derivatives affect functions after referring to NCERT Exemplar Class 12 Maths solutions Chapter 6. It will teach whether the function is increasing, decreasing or strictly increasing/decreasing. This chapter will also cover Rolle's Theorem, LaGrange's Theorem of Mean Value, finding tangent line equations, and all its cases.

They will learn in-depth about approximations and how to find approximate value to some of the quantities by reading NCERT Exemplar Class 12 Maths solutions Chapter 6. One of the most crucial things that the students will cover is details about finding maxima and minima of functions, points of local etc. They will also learn about the closed function's absolute maxima and minima.

## NCERT Exemplar Class 12 Maths Solutions

## Important Topics to cover from NCERT Exemplar Class 12 Maths solutions Chapter 6

In NCERT Exemplar Class 12 Mathematics Solutions Chapter 6, the students will learn in detail about the derivatives and their fundamentals and applications. Several topics are covered in this, which has significance in higher calculus and even other subjects as it is a very common topic in exams.

In NCERT Exemplar Class 12 Maths solutions Chapter 6, the students will come face to face with several topics like decreasing and increasing functions, minima and maxima, approximations, normal, tangents, change of quantities and its rate which are very commonly asked in the exams.

### NCERT Exemplar Class 12 Solutions

### Also, check NCERT Solutions for questions given in book:

## Frequently Asked Question (FAQs) - NCERT Exemplar Class 12 Maths Solutions Chapter 6 Applications Of Derivatives

**Question: **What is the best way to prepare for an exam from NCERT books?

**Answer: **

First of all, you should read the chapters well for the exam. You must also go through the questions after the end of each chapter.

**Question: **Who has prepared the solutions?

**Answer: **

The Class 12 Maths NCERT exemplar solutions chapters 6 are prepared by us who are experts in mathematics. The solutions are prepared after understanding and reference from professional books.

**Question: **Can I download the solutions for this chapter?

**Answer: **

Yes, you can click the NCERT exemplar class 10 maths solutions chapter 1 pdf download in the given link on the page.

**Question: **Are these solutions helpful for board examinations?

**Answer: **

Yes, the NCERT Exemplar Class 12 Solutions for Mathematics Chapter 6 will provide you with the ability to ace the board exams if you utilise it properly.

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