NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

NCERT Solutions for class 12 maths exercise 6.1 of chapter 6-Application of derivatives are created by an expert CBSE teacher by step by step method. NCERT Solutions for class 12 maths exercise 6.1-Application of derivatives is very easy to understand here because all NCERT solutions are explained here beautifully.Class 12 maths NCERT solutions of exercise 6.1 of chapter 6 are the solutions of unsolved questions of NCERT maths text book prescribed by India’s most popular board CBSE.In chapter 6 you will study about the application of derivatives based on the different kinds of geometrical figures.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

You can also study

Exercise 6.2 – Application of Derivatives

Exercise 6.3 -Application of Derivatives

Exercise 6.4- Application of Derivatives

Exercise 6.5 – Application of Derivatives

NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1. Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm (b) r = 4 cm

Ans. The area of circle is given as following

A = π r²

The rate of change of area of the circle is with respect to radius = Differentiation of area with respect to radius

Differentiating area(A) with respect to radius(r)

$\boldsymbol{\frac{dA}{dr}=\pi \frac{d}{dr}\left ( r^{2} \right )}$

$\boldsymbol{\frac{dA}{dr}=\pi .2r=2\pi r}$

(a) When r = 3 cm, the rate of change of area is

$\boldsymbol{\left ( \frac{dA}{dr} \right )_{r=3}=2\pi \times 3=6\pi\: cm^{2}}$

(b) When r= 4, the rate of change of area is

$\boldsymbol{\left ( \frac{dA}{dr} \right )_{r=4}=2\pi \times 4=8\pi\: cm^{2}}$

Q2.The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm ?

Ans. Let the side of the cube is = x and its volume is = V

The volume of cube is given as follows

V = x³

$\boldsymbol{\frac{dV}{dt}=\frac{d}{dt}\left ( x^{3} \right )}$

$\boldsymbol{\frac{d}{dt}\left ( x^{3} \right )=\frac{d}{dt}\left ( x^{3} \right ).\frac{dx}{dx}}$

$\boldsymbol{\frac{dV}{dt}=\frac{d}{dx}\left ( x^{3} \right ).\frac{dx}{dt}}$

$\boldsymbol{=3x^{2}.\frac{dx}{dt}}$

$\boldsymbol{\frac{dV}{dt}=3x^{2}.\frac{dx}{dt}}$

Since we are given the rate of change of volume with respect to time

$\boldsymbol{\frac{dV}{dt}=8}$

$\boldsymbol{\frac{dx}{dt}=\frac{8}{3x^{2}}}$

Now, we have to evaluate the increasing rate of surface area

Surface area of the cube is given by following formula

S = 6x²

Diffrentiating both sides with respect to t

$\boldsymbol{\frac{dS}{dt}=6\frac{d}{dt}\left ( x^{2} \right )}$

$\boldsymbol{\frac{dS}{dt}=6\frac{d}{dx}\left ( x^{2} \right ).\frac{dx}{dt}}$

$\boldsymbol{\frac{dS}{dt}=12x.\frac{dx}{dt}}$

Putting the value of dx/dt = 8/3x²

$\boldsymbol{\frac{dS}{dt}=12x.\frac{8}{3x^{2}}=\frac{32}{x}}$

When the age of cube is = 12 cm, then rate of increasing surface area

$\boldsymbol{\left ( \frac{ds}{dt} \right )_{x=12}=\frac{32}{12}=\frac{8}{3}}$

Hence,the increasing rate of surface area is= 8/3 cm²/s

Q3.The radius of the circle is increasing uniformly at the rate of 3 cm/s .Find the rate at which the area of circle is increasing when the radius is 10 cm.

Ans. We are given the increasing rate of radius = 3 cm/s

Let the radius of circle is = r

The rate of increasing radius

$\boldsymbol{\frac{dr}{dt}= 3\: cm/s}$

The area of the circle is given by

A = π r²

$\boldsymbol{\frac{dA}{dt}=\frac{d}{dt}\left ( \pi r^{2} \right ).\frac{dr}{dr}}$

$\boldsymbol{\frac{dA}{dt}=\frac{d}{dr}\left ( \pi r^{2} \right ).\frac{dr}{dt}}$

$\boldsymbol{\frac{dA}{dt}=2\pi r.\frac{dr}{dt}}$

Putting the value of dr/dt = 3

$\boldsymbol{\frac{dA}{dt}=2\pi r\times 3=6\pi r}$

When the radius of circle is = 10 cm, the rate of increasing area of the circle is

$\boldsymbol{\left ( \frac{dA}{dt} \right )_{r=10}=6\pi \times 10=60\pi }$

Therefore the increasing rate of area is 60π when radius of circle is 10 cm

Q4.An edge of the variable cube is increasing at the rate of 3 cm/s .How fast is the volume of cube increasing when the edge if 10 cm long.

Ans.Let the edge of the variable cube at time t is = x

We are given the increasing rate of the edge is

$\boldsymbol{\frac{dx}{dt}=3\, cm/s}$

Volume of the cube is given by

V = x³

The rate of the increasing volume of the cube

$\boldsymbol{\frac{dV}{dt}=\frac{d}{dt}\left ( x^{3} \right )}$

$\boldsymbol{\frac{dV}{dt}=\frac{d}{dt}\left ( x^{3} \right ).\frac{dx}{dx}}$

$\boldsymbol{\frac{dV}{dt}=\frac{d}{dx}\left ( x^{3} \right ).\frac{dx}{dt}}$

$\boldsymbol{\frac{dV}{dt}=3x^{2}.\frac{dx}{dt}}$

Putting the value of dx/dt = 3 cm/s

$\boldsymbol{\frac{dV}{dt}=3x^{2}.\times 3=9x^{2}}$

The rate of increasing volume of cube when side, x= 10 cm

$\boldsymbol{\left ( \frac{dV}{dt} \right )_{x=10cm}=9\times 10^{2}=900}$

Hence, the rate of increasing volume of the cube is 900 cm³/s when side =10 cm

Q5. A stone is dropped into a quite lake and waves move in circles at the rate of 5 cm/s. At the instant when the radius of circular wave is 8 cm, how fast is the enclosed area increasing?

Ans.Let the radius of circular wave is = r at time t

We are given the increasing rate of change of radius of circular wave = 5 cm/s

The area of circle is given as following

A = π r²

The rate of change of enclosed area of the circle is with respect to radius = Differentiation of area with respect to time

Differentiating area(A) with respect to radius(t)

$\boldsymbol{\frac{dA}{dt}=\frac{d}{dt}\left ( \pi r^{2} \right )}$

$\boldsymbol{\frac{dA}{dt}=\frac{d}{dt}\left ( \pi r^{2} \right )\frac{dr}{dr}}$

$\boldsymbol{\frac{dA}{dt}=\frac{d}{dr}\left ( \pi r^{2} \right )\frac{dr}{dt}}$

Putting the value of dr/dt = 5 cm/s

$\boldsymbol{\frac{dA}{dt}=2\pi r\times 5=10\pi r}$

At the instant when radius is = 8 cm, the rate of increasing enclosed area of circle is

$\boldsymbol{\left ( \frac{dA}{dt} \right )_{r=8\: cm}=10\pi \times 8=80\pi }$

Hence, when the radius is 8 cm the rate of change of the enclosed area of the circular wave is = 80π cm²

NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q6.The radius of a circle is increasing at the rate of 0.7cm/s. what is the rate of its circumference?

Ans. Consider r cm be the radius of the circle at time t.

Rate of increase of radius of circle = 0.7cm/sec

$\boldsymbol{\frac{dr}{dt}=0.7\: cm/s}$

Rate of the circumference of the circle means, how fast circumference is increasing if the radius is increasing at the rate of 0.7 cm/s

The circumference of a circle is given by

l = 2πr

$\boldsymbol{\frac{dl}{dt}=\frac{d}{dt}\left ( 2\pi r \right )}$

$\boldsymbol{\frac{dl}{dt}=\frac{d}{dt}\left ( 2\pi r \right ).\frac{dr}{dr}}$

$\boldsymbol{\frac{dl}{dt}=\frac{d}{dr}\left ( 2\pi r \right ).\frac{dr}{dt}}$

Putting the value of dr/dt = 0.7 cm/s

$\boldsymbol{\frac{dl}{dt}=2\pi \times 0.7=1.4\pi }$

Therefore the rate of the circumference is = 1.4 cm/s, if the radius is increasing at the rate of 0.7 cm/s.

Q7.The length x of a rectangle is decreasing at the rate of 5cm/minute and the width,y is increasing at the rate of 4 cm/minute. When x = 8cm and y =6cm, find the rates of change of (a) the perimeter and (b)the area of the rectangle

Ans. The length of the rectangle is =x

We are given the decreasing rate of the length = -5 cm/minute

$\boldsymbol{\frac{dx}{dt}= -5\: cm/m}$

We are given increasing rate of width of the rectangle

$\boldsymbol{\frac{dy}{dt}=4 cm/m}$

The parameter of the rectangle, z is given by

z = 2x + 2y

$\boldsymbol{\frac{dz}{dt}=2\frac{dx}{dt}+2\frac{dy}{dt}}$

Putting the value of dx/dt and dy/dt

$\boldsymbol{\frac{dz}{dt}=2\times -5+2\times 4=-2}$

Therefore the rate of change of parameter is -2 cm/minute or parameter of rectangle is decreased at the rate of 2 cm/minute.

The area of rectangle, A is given by

A = xy

$\boldsymbol{\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}}$

Putting the value of dx/dt , dy/dt, x= 8 and y= 6

$\boldsymbol{\frac{dA}{dt}=8\times 4+6\times -5}$

$\boldsymbol{\frac{dA}{dt}=2}$

Since rate of change of area is positive,so the area of rectangle is increased at the rate of 2 cm²/minute.

Q8.A balloon, which always remains spherical on inflammation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15cm

Ans. Since the balloon always remains spherical on inflammation,so the volume of balloon, V is given as follows

$\boldsymbol{V=\frac{4}{3}\pi r^{3}}$

The rate of change of volume of balloon is given 900 cubic centimeter/second

$\boldsymbol{\frac{dV}{dt}=900\: cm^{3}/s}$

Differentiating the volume equation of baloon

$\boldsymbol{\frac{dV}{dt}= \frac{4\pi }{3}\frac{d}{dt}\left ( r^{3} \right ).\frac{dr}{dr}}$

$\boldsymbol{\frac{dV}{dt}= \frac{4\pi }{3}\frac{d}{dr}\left ( r^{3} \right ).\frac{dr}{dt}}$

Putting the value of dV/dt

$\boldsymbol{900=4\pi r^{2}.\frac{dr}{dt}}$

$\boldsymbol{\frac{dr}{dt}=\frac{900}{4\pi r^{2}}=\frac{225}{\pi r^{2}}}$

The rate of change of area = 225/πr²

Hence the rate of change of radius (dr/dt), when radius is 15 cm

$\boldsymbol{\frac{dr}{dt}=\frac{225}{\pi \left ( 15 \right )^{2}}=\frac{225}{225\pi }}$

$\boldsymbol{\left ( \frac{dr}{dt} \right )_{r=15}=\frac{1}{\pi }}$

Q9.A balloon, which always remains spherical has a variables radius. Find the rate at which its volume is increasing with the radius when the later is 10cm.

Ans.

The volume of balloon(V) with variable radius is given as follows

$\boldsymbol{V =\frac{4}{3}\pi r^{3}}$

The rate of increasing its volume with respect to radius is

$\boldsymbol{\frac{dV}{dr}=\frac{4\pi }{3}\frac{d}{dr}\left ( r^{3} \right )}$

$\boldsymbol{\frac{dV}{dr}=4\pi r^{2} }$

The rate of increasing the volume of the balloon when the radius is 10 cm

$\boldsymbol{\left ( \frac{dV}{dr} \right )_{r=10 cm}=4\pi \times 10^{2}=400\pi }$

Hence, the rate of the increasing volume of the radius is 400 cm³/ cm.

Q10. A ladder 5m long is leaning against a wall. The bottom of the ladder is pulled among the ground, away from the wall at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans. Let the height of the wall at which the ladder touches = y m and the distance of foot of the ladder on the ground = x m

We are given that the distance of foot of the ladder is increasing at the rate of 2 cm/s.

$\boldsymbol{\frac{dx}{dt}=2\: cm/s}$

Now,let’s find out the rate of the increasing height of the ladder i.e dy/dt

Applying the Pythagoras theorem

x² + y² = 5²

x² + y² =25

$\boldsymbol{y=\sqrt{25-x^{2}}}$

$\boldsymbol{\frac{dy}{dt}=\frac{d}{dt}.\left ( \sqrt{25-x^{2}} \right ).\frac{dx}{dx}}$

$\boldsymbol{\frac{dy}{dt}=\frac{d}{dx}.\left ( \sqrt{25-x^{2}} \right ).\frac{dx}{dt}}$

$\boldsymbol{\frac{dy}{dt}=\frac{-x}{\sqrt{25-x^{2}}}.\frac{dx}{dt}}$

$\boldsymbol{\frac{dy}{dt}=\frac{-2x}{\sqrt{25-x^{2}}}}$

The rate of increasing height when x = 4 m

$\boldsymbol{\left ( \frac{dy}{dt} \right )_{x=4m}=\frac{-2\times 4}{\sqrt{25-4^{2}}}=\frac{-8}{3}}$

$\boldsymbol{\left ( \frac{dy}{dt} \right )_{x=4m}=\frac{-2\times 4}{\sqrt{25-4^{2}}}=\frac{-8}{3}\: cm/s}$

Q11.A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Ans.We are given the equation of the curve

6y = x³ + 2

$\boldsymbol{y=\frac{x^{3}+2}{6}}$

Let’s find the rate of change of the position of the particle on x and y-axis

$\boldsymbol{\frac{dy}{dt}=\frac{1}{6}\frac{d}{dt}\left ( x^{3}+2 \right )}$

$\boldsymbol{\frac{dy}{dt}=\frac{1}{6}\frac{d}{dt}\left ( x^{3} \right ).\frac{dx}{dx}}$

$\boldsymbol{\frac{dy}{dt}=\frac{1}{6}\frac{d}{dx}\left ( x^{3} \right ).\frac{dx}{dt}}$

$\boldsymbol{\frac{dy}{dt}=\frac{3x^{2}}{6}.\frac{dx}{dt}}$

$\boldsymbol{\frac{dy}{dt}=\frac{x^{2}}{2}.\frac{dx}{dt}}$

We are given the condition that the y-coordinate is changing 8 times as fast as the x-coordinate.

$\boldsymbol{\frac{dy}{dt}=8\frac{dx}{dt}}$

Substituting this value in the differential equation we got

$\boldsymbol{8\frac{dx}{dt}=\frac{x^{2}}{2}.\frac{dx}{dt}}$

x² = 16 ⇒ x = ±4

Putting value of x = 4 and x = -4 in the equation of the curve

y =(4³ +2)/6 = 66/6 = 11 and y=(-4³ +2)/6 = -62/6=-31/3

Hence the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate are (4,11) and (-4, -31/3)

NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q12.The radius of an air bubble is increasing at the rate of the 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1cm?

Ans.We are given the radius of an air bubble is increasing at the rate of 1/2 cm/s.

$\boldsymbol{\frac{dr}{dt}=\frac{1}{2}\: cm/s}$

Volume of the bubble (V) is given as follows

$\boldsymbol{V=\frac{4}{3}\pi r^{3}}$

$\boldsymbol{\frac{dV}{dt}=\frac{4\pi }{3}\frac{d}{dt}\left ( r^{3} \right )}$

$\boldsymbol{\frac{dV}{dt}=\frac{4\pi }{3}\frac{d}{dt}\left ( r^{3} \right ).\frac{dr}{dr}}$

$\boldsymbol{\frac{dV}{dt}=\frac{4\pi }{3}\frac{d}{dr}\left ( r^{3} \right ).\frac{dr}{dt}}$

$\boldsymbol{\frac{dV}{dt}=4\pi r^{2}.\frac{dr}{dt}}$

Putting the value of dr/dt = 1/2

$\boldsymbol{\frac{dV}{dt}=4\pi r^{2}\times \frac{1}{2}=2\pi r^{2}}$

Rate of change of volume is =2πr²

At r = 1 cm the rate of change of volume is = 2π(1)²=2π cm³/s

Q13.A balloon, which always remains spherical , has a variable diameter 3/2(2x+1) .Find the rate of change of its volume with respect to x.

Ans. We are given variable diameter(2r)

$\boldsymbol{2r=\frac{3}{2}\left ( 2x+1 \right )}$

$\boldsymbol{r=\frac{3}{4}\left ( 2x+1 \right )}$

$\boldsymbol{\frac{dr}{dx}=\frac{3 }{4}\frac{d}{dx}\left ( 2x+1 \right )}$

$\boldsymbol{\frac{dr}{dx}=\frac{3 }{4}\frac{d}{dx}\left ( 2x+1 \right )=6}$

$\boldsymbol{\frac{dr}{dx}=\frac{3}{4}\frac{d}{dx}\left ( 2x+1 \right )=\frac{3}{2}}$

Volume (V) of the balloon is given as follows

$\boldsymbol{V=\frac{4}{3}\pi r^{3}}$

The rate of change of volume is

$\boldsymbol{\frac{dV}{dx}=\frac{4\pi }{3}\frac{d}{dx}\left ( r^{3} \right )}$

$\boldsymbol{\frac{dV}{dx}=\frac{4\pi }{3}\frac{d}{dx}\left ( r^{3} \right ).\frac{dr}{dr}}$

$\boldsymbol{\frac{dV}{dx}=\frac{4\pi }{3}\frac{d}{dr}\left ( r^{3} \right ).\frac{dr}{dx}}$

$\boldsymbol{\frac{dV}{dx}=4\pi r^{2}.\frac{dr}{dx}}$

Putting the value of dr/dx =3/2 and r= 3/4(2x+1)

$\boldsymbol{\frac{dV}{dx}=4\pi \frac{9}{16}\left ( 2x+1 \right )^{2}\times \frac{3}{2}}$

$\boldsymbol{\frac{dV}{dx}=\frac{27\pi }{8}\left ( 2x+1 \right )^{2}}$

Q14.Sand is pouring from a pipe at the rate of 12 cm³ /s.The falling sand forms a cone on the ground in such a way the height of the cone is always one-sixth of the radius of the base.How fast is the height of the sand cone increasing when the height is 4 cm.

Ans.Let the height of cone is = h and radius of base is = r at time t

The given relation between height and radius of base of the cone is

h = r/6

r = 6h

We are given the rate of increasing the volume of cone =12 cm³/s

$\boldsymbol{\frac{dV}{dt}= 12\: cm^{3}/s}$

The volume of cone is given as follows

$\boldsymbol{V=\frac{1}{3}\pi r^{2}h}$

Putting r = 6h

$\boldsymbol{V=\frac{1}{3}\pi \left ( 6h \right )^{2}h}$

V = 12πh³

Differentiating Volume with respect time

$\boldsymbol{\frac{dV}{dt}=12\pi \frac{d}{dt}\left ( h^{3} \right )}$

$\boldsymbol{\frac{dV}{dt}=12\pi \frac{d}{dt}\left ( h^{3} \right ).\frac{dh}{dh}}$

$\boldsymbol{\frac{dV}{dt}=12\pi \frac{d}{dh}\left ( h^{3} \right ).\frac{dh}{dt}}$

$\boldsymbol{\frac{dV}{dt}=36\pi h^{2}.\frac{dh}{dt}}$

Putting the value of dV/dt and h = 4 cm

$\boldsymbol{36\pi 4^{2}.\frac{dh}{dt}=12}$

$\boldsymbol{\frac{dh}{dt}=\frac{12}{36\pi 4^{2}}=\frac{1}{48\pi }}$

Hence the rate of increasing height of the cone is (1/48π) cm/s when height is 4 cm.

Q15.The total cost C(x) in rupees associated with the production of x units of an item given by C(x) = 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.

Ans. We are given the relationship between total cost C(x) and the number of units (x) produced

C = 0.007x³ – 0.003x² + 15x + 4000

We know the marginal cost is the rate of change in cost with respect to items produced,so differentiating C with respect to items produced(x)

$\boldsymbol{\frac{dC}{dx}=0.007\frac{d}{dx}\left ( x^{3} \right )-0.003\frac{d}{dx}\left ( x^{2} \right )+15\frac{d}{dx}.x}$

$\boldsymbol{\frac{dC}{dx}=0.007\times 3x^{2}-0.003\times 2x+15}$

$\boldsymbol{\frac{dC}{dx}=0.021x^{2}-0.006x+15}$

Putting x = 17, we have

$\boldsymbol{\frac{dC}{dx}=0.021\times 17^{2}-0.006\times 17+15}$

$\boldsymbol{\frac{dC}{dx}=6.069-0.102+15}$

$\boldsymbol{\frac{dC}{dx}=20.967}$

Hence the marginal cost of the production is Rs 20.987 when 17 units of the items are produced.

Q16.The total revenue in rupees received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Ans. We are given total revenue (R) from the sale of (x) units.

R = 13x² + 26x + 15

Marginal revenue is the rate of change in revenue with respect to sold units of the product, so differentiating R with respect to x.

$\boldsymbol{\frac{dR}{dx}=13\frac{d}{dx}\left ( x^{2} \right )+26\frac{d}{dx}.x}$

$\boldsymbol{\frac{dR}{dx}=26x+26}$

Marginal revenue ,when x = 7

$\boldsymbol{\frac{dR}{dx}=26\times 7+26=182+26=208}$

Hence,the marginal revenue is Rs 208.

NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q17.The rate of change the area of a circle with respect to its radius r at r = 6 cm is:

(A) 10 π

(B) 12π

(C)8π

(D) 11π

Ans. The area (A) of circle is given as follows

A = π r²

$\boldsymbol{\frac{dA}{dr}=\pi \frac{d}{dr}\left ( r^{2} \right )=2\pi r}$

Rate of change of area with respect to radius,when r =6 cm is

$\boldsymbol{\left ( \frac{dA}{dr} \right )_{r=6cm}=2\pi \times 6=12\pi }$

Hence,the correct answer is (B) 12π cm²

Q18.The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue when x= 15 is:

(A) 116 (B) 96 (C) 90 (D) 126

Ans. We are given total revenue R(x)

R= 3x² + 36x + 5.

$\boldsymbol{\frac{dR}{dx}=3\frac{d}{dx}\left ( x^{2} \right )+36\frac{d}{dx}.x}$

$\boldsymbol{\frac{dR}{dx}=6x+36}$

Marginal revenue when x= 15 is

$\boldsymbol{\left ( \frac{dR}{dx} \right )_{x=15}=6\times 15+36=126}$

Hence, the correct answer is (D) Rs 126.

You can compensate us

Paytm number 9891436286

The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals