Apply to Aakash iACST Scholarship Test 2024
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
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 ‘r’
Assume S as the surface area of the spherical ball
Then, ……….(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 ‘r’
Let the area of the circle be A
Then ……..(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 = 2πr
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 ΔABE is a right-angled triangle
Applying of the Pythagoras theorem results into,
Let’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 45° 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 θ such that it increases twice as fast as its sine.
Explanation: Let x = sin θ
Let’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 man’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 ΔABE and ΔDCE
∠DEC = ∠AEB (same angle)
∠DCE = ∠ABD = 90°
Hence by AA similarity,
ΔABE≅ΔDCE
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
Let’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 ‘a’ the length of the side of the cube.
Take the volume of the cube ‘V’
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),
∴
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 ΔABC 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,
⇒ x+y = k, where k is any constant value
⇒ y = k-x………..(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 R’’(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 × 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
∴ 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 ‘a’ be the side of the cube
Then surface area of the cube = ….(i)
Take ‘r’ as the radius of the sphere
Then the surface area of the sphere = …(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 Δ 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 90°.
Hence ∠ACB = 90°
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 × 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
∴ 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 let’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 β, 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. – 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, ∞)
B. [-2, -1]
C. (-∞, -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 = π
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
f’(x)=2cos 2x
Put f’(x)=0, and get
2cos 2x =0
⇒ cos 2x=0
It is possible when
0≤x≤2π
Thus, sin 2x does not decrease or increase on
(ii) Let f(x)=tan x
Apply first derivative and get
f’(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
f’(x)=-sin x
But, sin x>0 for
And -sin x<0 for
Hence f’(x)<0 for
⇒ cos x is strictly decreasing on
(iv) Let f(x)=cos 3x
Apply first derivative and get
f’(x)=-3sin 3x
Put f’(x)=0, we get
-3sin 3x=0
⇒ sin 3x=0
Because sin θ=0 if θ=0, π, 2π, 3π
⇒ 3x=0,π, 2π, 3π
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 f’(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
⇒ x-1=0
⇒ 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 ………(i)
Take logarithm on both side
⇒ 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.
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
Apply to Aakash iACST Scholarship Test 2024
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.
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.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | Application of Derivatives |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
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.
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.
Yes, you can click the NCERT exemplar Class 10 Maths solutions chapter 1 pdf download in the given link on the page.
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.
Here are some options you can explore to get admission in a good school even though admissions might be closed for many:
Waitlist: Many schools maintain waitlists after their initial application rounds close. If a student who secured a seat decides not to join, the school might reach out to students on the waitlist. So, even if the application deadline has passed, it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.
Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush. Reach out to the schools directly to see if they have any open seats in 10th grade.
Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.
Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.
Best CBSE schools in Delhi: Click Here
In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:
Choosing Your Stream:
Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.
Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.
Engineering Subjects (for a Degree):
Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.
Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.
Here's why 2025 is more likely:
Looking Ahead (2025 Admissions):
Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore, with your academic background. Here's why:
High CAT Score: A score exceeding 99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes CAT scores heavily in the shortlisting process.
Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.
However, the shortlisting process is multifaceted:
Here's what you can do to strengthen your application:
Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.
Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.
Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
Get up to 90% scholarship on NEET, JEE & Foundation courses
Register FREE for ALLEN Digital Scholarship Admission Test (ADSAT)
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
Enrol in PW Vidyapeeth center for JEE coaching
Enrol in PW Vidyapeeth center for NEET coaching
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Applications for CBSE Class 12 supplementary exams 2023 have started. Read details here.
CBSE Class 12th result 2023 declared. Check now.