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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Edited By Ramraj Saini | Updated on Sep 14, 2023 08:10 PM IST | #CBSE Class 12th
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NCERT Application-Of-Derivatives Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives are comprehensively discussed here. These NCERT solutions are created by expert team at Careers360 keeping in mind of latest syllabus of CBSE 2023-24. In the previous chapter, you have already learnt the differentiation of inverse trigonometric functions, exponential functions, logarithmic functions, composite functions, implicit functions, etc. In this article you will get NCERT Class 12 maths solutions chapter 6 with in depth explanation that will help you in understanding application of derivatives class 12.

This Story also Contains
  1. NCERT Application-Of-Derivatives Class 12 Questions And Answers
  2. NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives - Important Formulae
  3. NCERT Application-Of-Derivatives Class 12 Questions And Answers (Intext Questions and Exercise)
  4. Application-Of-Derivatives Class 12 NCERT solutions - Topics
  5. NCERT solutions for class 12 maths - Chapter wise
  6. NCERT solutions for class 12 subject wise
  7. NCERT solutions class wise
  8. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

In class 12 chapter 6 questions are based on the topics like finding the rate of change of quantities, equations of tangent, and normal on a curve at a point are covered in the application of derivatives class 12 NCERT solutions. Also, check NCERT solutions for class 12 other subjects.

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NCERT Application-Of-Derivatives Class 12 Questions And Answers PDF Free Download

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives - Important Formulae

>> Definition of Derivatives: Derivatives measure the rate of change of quantities.

Rate of Change of a Quantity:

The derivative is used to find the rate of change of one quantity concerning another. For a function y = f(x), the average rate of change in the interval [a, a+h] is:

(f(a + h) - f(a)) / h

Approximation:

Derivatives help in finding approximate values of functions. The linear approximation method, proposed by Newton, involves finding the equation of the tangent line.

Linear approximation equation: L(x) = f(a) + f'(a)(x - a)

Tangents and Normals:

A tangent to a curve touches it at a single point and has a slope equal to the derivative at that point.

Slope of tangent (m) = f'(x)

The equation of the tangent line is found using: m = (y2 - y1) / (x2 - x1)

The normal to a curve is perpendicular to the tangent.

The slope of normal (n) = -1 / f'(x)

The equation of the normal line is found using: -1 / m = (y2 - y1) / (x2 - x1)

Maxima, Minima, and Point of Inflection:

Maxima and minima are peaks and valleys of a curve. The point of inflection marks a change in the curve's nature (convex to concave or vice versa).

To find maxima, minima, and points of inflection, use the first derivative test:

  • Find f'(c) = 0.

  • Check the sign change of f'(x) on the interval.

  • Maxima when f'(x) changes from +ve to -ve, f(c) is the maximum.

  • Minima when f'(x) changes from -ve to +ve, f(c) is the minimum.

  • Point of inflection when the sign of f'(x) doesn't change.

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Increasing and Decreasing Functions:

An increasing function tends to reach the upper corner of the x-y plane, while a decreasing function tends to reach the lower corner.

For a differentiable function f(x) in the interval (a, b):

  • If f(x1) ≤ f(x2) when x1 < x2, it's increasing.

  • If f(x1) < f(x2) when x1 < x2, it's strictly increasing.

  • If f(x1) ≥ f(x2) when x1 < x2, it's decreasing.

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If f(x1) > f(x2) when x1 < x2, it's strictly decreasing.

Free download Class 12 Maths Chapter 6 Question Answer for CBSE Exam.

NCERT Application-Of-Derivatives Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT class 12 maths chapter 6 question answer: Exercise - 6.1

Question:1 a) Find the rate of change of the area of a circle with respect to its radius r when
r = 3 cm

Answer:

Area of the circle (A) = πr2
Rate of change of the area of a circle with respect to its radius r = dAdr = d(πr2)dr = 2πr
So, when r = 3, Rate of change of the area of a circle = 2π(3) = 6π
Hence, Rate of change of the area of a circle with respect to its radius r when r = 3 is 6π

Question:1 b) Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm

Answer:

Area of the circle (A) = πr2
Rate of change of the area of a circle with respect to its radius r = dAdr = d(πr2)dr = 2πr
So, when r = 4, Rate of change of the area of a circle = 2π(4) = 8π
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is 8π

Question:2 . The volume of a cube is increasing at the rate of 8cm3/s . How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

The volume of the cube(V) = x3 where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of 8cm3/s

we can write dVdt=dVdx.dxdt ( By chain rule)

dVdt=8=dVdx.dxdt

dx3dx.dxdt=8 3x2.dxdt=8

dxdt=83x2 - (i)
Now, we know that the surface area of the cube(A) is 6x2

dAdt=dAdx.dxdt=d6x2dx.dxdt=12x.dxdt - (ii)

from equation (i) we know that dxdt=83x2

put this value in equation (i)
We get,
dAdt=12x.83x2=32x
It is given in the question that the value of edge length(x) = 12cm
So,
dAdt=3212=83cm2/s

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

Answer:

Radius of a circle is increasing uniformly at the rate (drdt) = 3 cm/s
Area of circle(A) = πr2
dAdt=dAdr.drdt (by chain rule)
dAdt=dπr2dr.drdt=2πr×3=6πr
It is given that the value of r = 10 cm
So,
dAdt=6π×10=60π cm2/s
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s

Question:4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

It is given that the rate at which edge of cube increase (dxdt) = 3 cm/s
The volume of cube = x3
dVdt=dVdx.dxdt (By chain rule)
dVdt=dx3dx.dxdt=3x2.dxdt=3x2×3=9x2cm3/s
It is given that the value of x is 10 cm
So,
dVdt=9(10)2=9×100=900 cm3/s
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is 900 cm3/s

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

Answer:

Given = drdt=5 cm/s

To find = dAdt at r = 8 cm

Area of the circle (A) = πr2
dAdt=dAdr.drdt (by chain rule)
dAdt=dπr2dr.drdt=2πr×5=10πr=10π×8=80π cm2/s
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is 80π cm2/s

Question:6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer:

Given = drdt=0.7 cm/s
To find = dCdt , where C is circumference
Solution :-

we know that the circumference of the circle (C) = 2πr
dCdt=dCdr.drdt (by chain rule)
dCdt=d2πrdr.drdt=2π×0.7=1.4π cm/s
Hence, the rate of increase of its circumference is 1.4π cm/s

Question:7(a) . The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rate of change of

the perimeter of rectangle

Answer:

Given = Length x of a rectangle is decreasing at the rate (dxdt) = -5 cm/minute (-ve sign indicates decrease in rate)
the width y is increasing at the rate (dydt) = 4 cm/minute
To find = dPdt and at x = 8 cm and y = 6 cm , where P is perimeter
Solution:-

Perimeter of rectangle(P) = 2(x+y)
dPdt=d(2(x+y))dt=2(dxdt+dydt)=2(5+4)=2 cm/minute
Hence, Perimeter decreases at the rate of 2 cm/minute

Question:7(b) The length x of a rectangle is decreasing at the rate of 5 cm/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 the area of the rectangle.

Answer:

Given same as previous question
Solution:-
Area of rectangle = xy
dAdt=d(xy)dt=(xdydt+ydxdt)=(8×4+6×(5))=(3230)=2 cm2/minute
Hence, the rate of change of area is 2 cm2/minute

Question:8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer:

Given = dVdt=900 cm3/s
To find = drdt at r = 15 cm
Solution:-

Volume of sphere(V) = 43πr3
dVdt=dVdr.drdt=d(43πr3)dr.drdt=43π×3r2×drdt

dVdt=4πr2×drdt
drdt=dVdt4πr2=9004π×(15)2=900900π=1π cm/s
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1π cm/s

Question:9 . A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answer:

We need to find the value of dVdr at r =10 cm
The volume of the sphere (V) = 43πr3
dVdr=d(43πr3)dr=43π×3r2=4πr2=4π(10)2=4π×100=400π cm3/s
Hence, the rate at which its volume is increasing with the radius when the later is 10 cm is 400π cm3/s

Question:10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer:

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given that dxdt=2 cm/s
We need to find the rate at which the height of the ladder decreases (dhdt)
length of ladder(L) = 5m and x = 4m (given)
By Pythagoras theorem, we can say that
h2+x2=L2
h2=L2x2
h =L2x2
Differentiate on both sides w.r.t. t
dhdt=d(L2x2)dx.dxdt=122x52x2.dxdt=x25x2dxdt
at x = 4

dhdt=42516×2=43×2=83 cm/s
Hence, the rate at which the height of ladder decreases is 83 cm/s

Question:11. 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.

Answer:

We need to find the point at which dydt=8dxdt
Given the equation of curve = 6y=x3+2
Differentiate both sides w.r.t. t
6dydt=d(x3)dx.dxdt+0
=3x2.dxdt
dydt=8dxdt (required condition)
6×8dxdt=3x2.dxdt
3x2.dxdt=48dxdt x2=483=16
x=±4
when x = 4 , y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11 and
when x = -4 , y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3} So , the coordinates are
(4,11) and (4,313)

Question:12 The radius of an air bubble is increasing at the rate of 1 /2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that drdt=12 cm/s
We know that the shape of the air bubble is spherical
So, volume(V) = 43πr3
dVdt=dVdr.drdt=d(43πr3)dr.drdt=4πr2×12=2πr2=2π×(1)2=2π cm3/s
Hence, the rate of change in volume is 2π cm3/s

Question:13 A balloon, which always remains spherical, has a variable diameter 32(2x+1) Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = 43πr3
Diameter = 32(2x+1)
So, radius(r) = 34(2x+1)
dVdx=d(43πr3)dx=d(43π(34(2x+1))3)dx=43π×3×2764(2x+1)2×2
=278π(2x+1)2

Question:14 Sand is pouring from a pipe at the rate of 12 cm 3 /s. The falling sand forms a cone on the ground in such a way that 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?

Answer:

Given = dVdt=12 cm3/s and h=16r
To find = dhdt at h = 4 cm
Solution:-

Volume of cone(V) = 13πr2h
dVdt=dVdh.dhdt=d(13π(6h)2h)dh.dhdt=13π×36×3h2.dhdt=36π×(4)2.dhdt
dVdt=576π.dhdt

Question:15 The total cost C(x) in Rupees associated with the production of x units of an
item is given by C(x)=0.007x30.003x2+15x+4000

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = dCdx
C(x)=0.007x30.003x2+15x+4000
dCdx=d(.007x3.003x2+15x+400)dx=3×.007x22×.003x+15
=.021x2.006x+15
Now, at x = 17
MC =.021(17)2.006(17)+15
=6.069.102+15
=20.967
Hence, marginal cost when 17 units are produced is 20.967

Question:16 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

Answer:

Marginal revenue = dRdx
R(x)=13x2+26x+15
dRdx=d(13x2+26x+15)dx=13×2x+26=26(x+1)
at x = 7
dRdx=26(7+1)=26×8=208
Hence, marginal revenue when x = 7 is 208

Question:17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π

Answer:

Area of circle(A) = πr2
dAdr=d(πr2)dr=2πr
Now, at r = 6cm
dAdr=2π×6=12πcm2/s
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is 12πcm2/s
Hence, the correct answer is B

Question:18 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

Answer:

Marginal revenue = dRdx
R(x)=3x2+36x+5
dRdx=d(3x2+36x+5)dx=3×2x+36=6(x+6)
at x = 15
dRdx=6(15+6)=6×21=126
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is D


NCERT class 12 maths chapter 6 question answer: Exercise: 6.2

Question:1 . Show that the function given by f (x) = 3x + 17 is increasing on R.

Answer:

Let x1andx2 are two numbers in R
x1<x23x1<3x23x1+17<3x2+17f(x1)<f(x2)
Hence, f is strictly increasing on R

Question:2. Show that the function given by f(x)=e2x is increasing on R.

Answer:

Let x1 and x2 are two numbers in R
x1 < x22x1<2x2e2x1<e2x2f(x1)<f(x2)
Hence, the function f(x)=e2x is strictly increasing in R

Question:3 a) Show that the function given by f (x) = sinx is increasing in (0,π/2)

Answer:

Given f(x) = sinx
f(x)=cosx
Since, cosx>0 for each x ϵ(0,π2)
f(x)>0
Hence, f(x) = sinx is strictly increasing in (0,π2)

Question:3 b) Show that the function given by f (x) = sinx is

decreasing in (π2,π)

Answer:

f(x) = sin x
f(x)=cosx
Since, cosx<0 for each x ϵ(π2,π)
So, we have f(x)<0
Hence, f(x) = sin x is strictly decreasing in (π2,π)

Question:3 c) Show that the function given by f (x) = sinx is neither increasing nor decreasing in (0,π)

Answer:

We know that sin x is strictly increasing in (0,π2) and strictly decreasing in (π2,π)
So, by this, we can say that f(x) = sinx is neither increasing or decreasing in range (0,π)

Question:4(a). Find the intervals in which the function f given by f(x)=2x23x is increasing

Answer:

f(x)=2x23x
f(x)=4x3
Now,
f(x)=0
4x - 3 = 0
x=34
1628071298489 So, the range is (,34) and (34,)
So,
f(x)<0 when x ϵ(,34) Hence, f(x) is strictly decreasing in this range
and
f(x)>0 when xϵ(34,) Hence, f(x) is strictly increasing in this range
Hence, f(x)=2x23x is strictly increasing in xϵ(34,)

Question:4(b) Find the intervals in which the function f given by f(x)=2x23x is
decreasing

Answer:

f(x)=2x23x
f(x)=4x3
Now,
f(x)=0
4x - 3 = 0
x=34
1651257732514 So, the range is (,34) and (34,)
So,
f(x)<0 when x ϵ(,34) Hence, f(x) is strictly decreasing in this range
and
f(x)>0 when xϵ(34,) Hence, f(x) is strictly increasing in this range
Hence, f(x)=2x23x is strictly decreasing in xϵ(,34)

Question:5(a) Find the intervals in which the function f given by f(x)=2x33x236x+7 is
increasing

Answer:

It is given that
f(x)=2x33x236x+7
So,
f(x)=6x26x36
f(x)=0
6x26x36=06(x2x6)
x2x6=0
x23x+2x6=0
x(x3)+2(x3)=0
(x+2)(x3)=0
x = -2 , x = 3

So, three ranges are there (,2),(2,3) and (3,)
Function f(x)=6x26x36 is positive in interval (,2),(3,) and negative in the interval (-2,3)
Hence, f(x)=2x33x236x+7 is strictly increasing in (,2)(3,)
and strictly decreasing in the interval (-2,3)

Question:5(b) Find the intervals in which the function f given by f(x)=2x33x236x+7 is
decreasing

Answer:

We have f(x)=2x33x236x+7

Differentiating the function with respect to x, we get :

f(x)=6x26x36

or =6(x3)(x+2)

When f(x) = 0 , we have :

0 =6(x3)(x+2)

or (x3)(x+2) = 0

17567
So, three ranges are there (,2),(2,3) and (3,)
Function f(x)=6x26x36 is positive in the interval (,2),(3,) and negative in the interval (-2,3)

So, f(x) is decreasing in (-2, 3)

Question:6(a) Find the intervals in which the following functions are strictly increasing or
decreasing:
x2+2x5

Answer:

f(x) = x2+2x5
f(x)=2x+2=2(x+1)
Now,
f(x)=02(x+1)=0x=1

The range is from (,1) and (1,)
In interval (,1) f(x)=2(x+1) is -ve
Hence, function f(x) = x2+2x5 is strictly decreasing in interval (,1)
In interval (1,) f(x)=2(x+1) is +ve
Hence, function f(x) = x2+2x5 is strictly increasing in interval (1,)

Question:6(b) Find the intervals in which the following functions are strictly increasing or
decreasing

106x2x2

Answer:

Given function is,
f(x)=106x2x2
f(x)=64x
Now,
f(x)=0
6+4x=0
x=32

So, the range is (,32) and (32,)
In interval (,32) , f(x)=64x is +ve
Hence, f(x)=106x2x2 is strictly increasing in the interval (,32)
In interval (32,) , f(x)=64x is -ve
Hence, f(x)=106x2x2 is strictly decreasing in interval (32,)

Question:6(c) Find the intervals in which the following functions are strictly increasing or
decreasing:

2x39x212x+1

Answer:

Given function is,
f(x)=2x39x212x+1
f(x)=6x218x12
Now,
f(x)=06x218x12=06(x2+3x+2)=0x2+3x+2=0x2+x+2x+2=0x(x+1)+2(x+1)=0(x+2)(x+1)=0x=2 and x=1

So, the range is (,2) ,(2,1) and (1,)
In interval (,2) (1,) , f(x)=6x218x12 is -ve
Hence, f(x)=2x39x212x+1 is strictly decreasing in interval (,2) (1,)
In interval (-2,-1) , f(x)=6x218x12 is +ve
Hence, f(x)=2x39x212x+1 is strictly increasing in the interval (-2,-1)

Question:6(d) Find the intervals in which the following functions are strictly increasing or
decreasing:

69xx2

Answer:

Given function is,
f(x)=69xx2
f(x)=92x
Now,
f(x)=092x=02x=9x=92

So, the range is (,92) and (92,)
In interval (,92) , f(x)=92x is +ve
Hence, f(x)=69xx2 is strictly increasing in interval (,92)
In interval (92,) , f(x)=92x is -ve
Hence, f(x)=69xx2 is strictly decreasing in interval (92,)

Question:6(e) Find the intervals in which the following functions are strictly increasing or
decreasing:

(x+1)3(x3)3

Answer:

Given function is,
f(x)=(x+1)3(x3)3
f(x)=3(x+1)2(x3)3+3(x3)2(x+1)3
Now,
f(x)=03(x+1)2(x3)3+3(x3)2(x+1)33(x+1)2(x3)2((x3)+(x+1))=0(x+1)(x3)=0           or          (2x2)=0x=1 and x=3              or            x=1
So, the intervals are (,1),(1,1),(1,3) and (3,)

Our function f(x)=3(x+1)2(x3)3+3(x3)2(x+1)3 is +ve in the interval (1,3) and (3,)
Hence, f(x)=(x+1)3(x3)3 is strictly increasing in the interval (1,3) and (3,)
Our function f(x)=3(x+1)2(x3)3+3(x3)2(x+1)3 is -ve in the interval (,1) and (1,1)
Hence, f(x)=(x+1)3(x3)3 is strictly decreasing in interval (,1) and (1,1)

Question:7 Show that y=log(1+x)2x2+x,x>1 is an increasing function of x throughout its domain.

Answer:

Given function is,
f(x)y=log(1+x)2x2+x
f(x)dydx=11+x2(2+x)(1)(2x)(2+x)2=11+x4+2x2x(2+x)2
=11+x4(2+x)2=(2+x)24(x+1)(x+1)(2+x)2
=4+x2+4x4x4(x+1)(2+x)2=x2(x+1)(2+x)2
f(x)=x2(x+1)(x+2)2
Now, for x>1 , is is clear that f(x)=x2(x+1)(x+2)2>0
Hence, f(x)y=log(1+x)2x2+x strictly increasing when x>1

Question:8 Find the values of x for which y=[x(x2)]2 is an increasing function.

Answer:

Given function is,
f(x)y=[x(x2)]2
f(x)dydx=2[x(x2)][(x2)+x]
=2(x22x)(2x2)
=4x(x2)(x1)
Now,
f(x)=04x(x2)(x1)=0x=0,x=2 and x=1
So, the intervals are (,0),(0,1),(1,2) and (2,)
In interval (0,1)and (2,) , f(x)>0
Hence, f(x)y=[x(x2)]2 is an increasing function in the interval (0,1)(2,)

Question:9 Prove that y=4sinθ(2+cosθ)θ is an increasing function of θin[0,π/2]

Answer:

Given function is,
f(x)=y=4sinθ(2+cosθ)θ

f(x)=dydθ=4cosθ(2+cosθ)(sinθ)4sinθ)(2+cosθ)21
=8cosθ+4cos2θ+4sin2θ(2+cosθ)2(2+cosθ)2
=8cosθ+4(cos2θ+sin2θ)4cos2θ4cosθ(2+cosθ)2
=8cosθ+44cos2θ4cosθ(2+cosθ)2
=4cosθcos2θ(2+cosθ)2
Now, for θ ϵ [0,π2]
4cosθcos2θ4cosθcos20and (2+cosθ)2>0
So, f(x)>0 for θ in [0,π2]
Hence, f(x)=y=4sinθ(2+cosθ)θ is increasing function in θ ϵ [0,π2]

Question:10 Prove that the logarithmic function is increasing on (0,)

Answer:

Let logarithmic function is log x
f(x)=logx
f(x)=1x
Now, for all values of x in (0,) , f(x)>0
Hence, the logarithmic function f(x)=logx is increasing in the interval (0,)

Question:11 Prove that the function f given by f(x)=x2x+1 is neither strictly increasing nor decreasing on (– 1, 1).

Answer:

Given function is,
f(x)=x2x+1
f(x)=2x1
Now, for interval (1,12) , f(x)<0 and for interval (12,1),f(x)>0
Hence, by this, we can say that f(x)=x2x+1 is neither strictly increasing nor decreasing in the interval (-1,1)

Question:12 Which of the following functions are decreasing on 0,π/2 (A)cosx(B)cos2x(C)cos3x(D)tanx

Answer:

(A)
f(x)=cosxf(x)=sinx
f(x)<0 for x in (0,π2)
Hence, f(x)=cosx is decreasing function in (0,π2)

(B)
f(x)=cos2xf(x)=2sin2x
Now, as
0<x<π20<2x<π
f(x)<0 for 2x in (0,π)
Hence, f(x)=cos2x is decreasing function in (0,π2)

(C)
f(x)=cos3xf(x)=3sin3x
Now, as
0<x<π20<3x<3π2
f(x)<0 for x ϵ (0,π3) and f(x)>0 x ϵ (π3,π2)
Hence, it is clear that f(x)=cos3x is neither increasing nor decreasing in (0,π2)

(D)
f(x)=tanxf(x)=sec2x
f(x)>0 for x in (0,π2)
Hence, f(x)=tanx is strictly increasing function in the interval (0,π2)

So, only (A) and (B) are decreasing functions in (0,π2)

Question:13 On which of the following intervals is the function f given by f(x)=x100+sinx1 decreasing ?
(A) (0,1) (B) π2,π (C) 0,π2 (D) None of these

Answer:

(A) Given function is,
f(x)=x100+sinx1
f(x)=100x99+cosx
Now, in interval (0,1)
f(x)>0
Hence, f(x)=x100+sinx1 is increasing function in interval (0,1)

(B) Now, in interval (π2,π)
100x99>0 but cosx<0
100x99>cosx100x99cosx>0 , f(x)>0
Hence, f(x)=x100+sinx1 is increasing function in interval (π2,π)

(C) Now, in interval (0,π2)
100x99>0 and cosx>0
100x99>cosx100x99cosx>0 , f(x)>0
Hence, f(x)=x100+sinx1 is increasing function in interval (0,π2)

So, f(x)=x100+sinx1 is increasing for all cases
Hence, correct answer is (D) None of these

Question:14 For what values of a the function f given by f(x)=x2+ax+1 is increasing on
[1, 2]?

Answer:

Given function is,
f(x)=x2+ax+1
f(x)=2x+a
Now, we can clearly see that for every value of a>2
f(x)=2x+a >0
Hence, f(x)=x2+ax+1 is increasing for every value of a>2 in the interval [1,2]

Question:15 Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f(x)=x+1/x is increasing on I.

Answer:

Given function is,
f(x)=x+1/x
f(x)=11x2
Now,
f(x)=011x2=0x2=1x=±1

So, intervals are from (,1),(1,1) and (1,)
In interval (,1),(1,) , 1x2<111x2>0
f(x)>0
Hence, f(x)=x+1/x is increasing in interval (,1)(1,)
In interval (-1,1) , 1x2>111x2<0
f(x)<0
Hence, f(x)=x+1/x is decreasing in interval (-1,1)
Hence, the function f given by f(x)=x+1/x is increasing on I disjoint from [–1, 1]

Question:16 Prove that the function f given by f(x)=logsinx is increasing on

(0,π/2)anddecreasingon(π/2,π)
Answer:

Given function is,
f(x)=logsinx
f(x)=1sinxcosx=cotx
Now, we know that cot x is+ve in the interval (0,π/2) and -ve in the interval (π/2,π)
f(x)>0 in (0,π2) and f(x)<0 in (π2,π)
Hence, f(x)=logsinx is increasing in the interval (0,π/2) and decreasing in interval (π/2,π)

Question:17 Prove that the function f given by f (x) = log |cos x| is decreasing on (0,π/2)
and increasing on (3π/2,2π)

Answer:

Given function is,
f(x) = log|cos x|
value of cos x is always +ve in both these cases
So, we can write log|cos x| = log(cos x)
Now,
f(x)=1cosx(sinx)=tanx
We know that in interval (0,π2) , tanx>0tanx<0
f(x)<0
Hence, f(x) = log|cos x| is decreasing in interval (0,π2)

We know that in interval (3π2,2π) , tanx<0tanx>0
f(x)>0
Hence, f(x) = log|cos x| is increasing in interval (3π2,2π)

Question:18 Prove that the function given by f(x)=x33x2+3x100 is increasing in R.

Answer:

Given function is,
f(x)=x33x2+3x100
f(x)=3x26x+3
=3(x22x+1)=3(x1)2
f(x)=3(x1)2
We can clearly see that for any value of x in R f(x)>0
Hence, f(x)=x33x2+3x100 is an increasing function in R

Question:19 The interval in which y=x2ex is increasing is

(A) (,) (B) (2,0) (C) (2,) (D) (0,2)

Answer:

Given function is,
f(x)y=x2ex
f(x)dydx=2xex+ex(x2)
xex(2x)
f(x)=xex(2x)
Now, it is clear that f(x)>0 only in the interval (0,2)
So, f(x)y=x2ex is an increasing function for the interval (0,2)
Hence, (D) is the answer


NCERT application-of-derivatives class 12 solutions: Exercise: 6.3

Question:1 . Find the slope of the tangent to the curve y=3x44xatx=4

Answer:

Given curve is,
y=3x44x
Now, the slope of the tangent at point x =4 is given by
(dydx)x=4=12x34
=12(4)34
=12(64)4=7684=764

Question:2 . Find the slope of the tangent to the curve x1x2,x2atx=10

Answer:

Given curve is,

y=x1x2
The slope of the tangent at x = 10 is given by
(dydx)x=10=(1)(x2)(1)(x1)(x2)2=x2x+1(x2)2=1(x2)2
at x = 10
=1(102)2=182=164
hence, slope of tangent at x = 10 is 164

Question:3 Find the slope of the tangent to curve y=x3x+1 at the point whose x-coordinate is 2.

Answer:

Given curve is,
y=x3x+1
The slope of the tangent at x = 2 is given by
(dydx)x=2=3x21=3(2)21=3×41=121=11
Hence, the slope of the tangent at point x = 2 is 11

Question:4 Find the slope of the tangent to the curve y=x33x+2 at the point whose x-coordinate is 3.

Answer:

Given curve is,
y=x33x+2
The slope of the tangent at x = 3 is given by
(dydx)x=3=3x23=3(3)23=3×93=273=24
Hence, the slope of tangent at point x = 3 is 24

Question:5 Find the slope of the normal to the curve x=acos3θ,y=asin3θatθ=π/4

Answer:

The slope of the tangent at a point on a given curve is given by
(dydx)
Now,
(dxdθ)θ=π4=3acos2θ(sinθ)=3a(12)2(12)=32a4
Similarly,
(dydθ)θ=π4=3asin2θ(cosθ)=3a(12)2(12)=32a4
(dydx)=(dydθ)(dxdθ)=32a432a4=1
Hence, the slope of the tangent at θ=π4 is -1
Now,
Slope of normal = 1slope of tangent = 11=1
Hence, the slope of normal at θ=π4 is 1

Question:6 Find the slope of the normal to the curve x=1asinθ,y=bcos2θatθ=π/2

Answer:

The slope of the tangent at a point on given curves is given by
(dydx)
Now,
(dxdθ)θ=π2=a(cosθ)
Similarly,
(dydθ)θ=π2=2bcosθ(sinθ)
(dydx)x=π2=(dydθ)(dxdθ)=2bcosθsinθacosθ=2bsinθa=2b×1a=2ba
Hence, the slope of the tangent at θ=π2 is 2ba
Now,
Slope of normal = 1slope of tangent = 12ba=a2b
Hence, the slope of normal at θ=π2 is a2b

Question:7 Find points at which the tangent to the curve y=x33x29x+7 is parallel to the x-axis.

Answer:

We are given :

y=x33x29x+7

Differentiating the equation with respect to x, we get :

dydx = 3x2  6x  9 + 0

or = 3(x2  2x  3)

or dydx = 3(x+1)(x3)

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

dydx = 0

or 0 = 3(x+1)(x3)

Thus, Either x = -1 or x = 3

When x = -1 we get y = 12 and if x =3 we get y = -20

So the required points are (-1, 12) and (3, -20).

Question:8 Find a point on the curve y=(x2)2 at which the tangent is parallel to the chord joining the points (2, 0) and

(4, 4).

Answer:

Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
m=y2y1x2x1=4042=42=2
As it is given that the tangent is parallel to the chord, so their slopes are equal
i.e. slope of the tangent = slope of the chord
Given the equation of the curve is y=(x2)2
dydx=2(x2)=2
(x2)=1x=1+2x=3
Now, when x=3 y=(32)2=(1)2=1
Hence, the coordinates are (3, 1)

Question:9 Find the point on the curve y=x311x+5 at which the tangent is y=x11

Answer:

We know that the equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that slope of the tangent at a point on the given curve is given by dydx
Given the equation of curve is
y=x311x+5
dydx=3x211
3x211=13x2=12x2=4x=±2
When x = 2 , y=2311(2)+5=822+5=9
and
When x = -2 , y=(2)311(22)+5=8+22+5=19
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is y=x11

Question:10 Find the equation of all lines having slope –1 that are tangents to the curve y=1x1,x1

Answer:

We know that the slope of the tangent of at the point of the given curve is given by dydx

Given the equation of curve is
y=1x1
dydx=1(1x)2
It is given thta slope is -1
So,
1(1x)2=1(1x)2=1=1x=±1x=0 and x=2
Now, when x = 0 , y=1x1=101=1
and
when x = 2 , y=1x1=1(21)=1
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx + c
1 = -1 X 2 + c
c = 3
Now equation of line is
y = -x + 3
y + x - 3 = 0

Question:11 Find the equation of all lines having slope 2 which are tangents to the curve y=1x3,x3

Answer:

We know that the slope of the tangent of at the point of the given curve is given by dydx

Given the equation of curve is
y=1x3
dydx=1(x3)2
It is given that slope is 2
So,
1(x3)2=2(x3)2=12=x3=±12
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2 to the curve y=1x3

Question:12 Find the equations of all lines having slope 0 which are tangent to the curve
y=1x22x+3

Answer:

We know that the slope of the tangent at a point on the given curve is given by dydx

Given the equation of the curve as
y=1x22x+3
dydx=(2x2)(x22x+3)2
It is given thta slope is 0
So,
(2x2)(x22x+3)2=02x2=0=x=1
Now, when x = 1 , y=1x22x+3=1122(1)+3=112+3=12

Hence, the coordinates are (1,12)
Equation of line passing through (1,12) and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is
y=12

Question:13(i) Find points on the curve x29+y216=1 at which the tangents are parallel to x-axis

Answer:

Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by dydx
Given the equation of the curve is
x29+y216=19y2=144(116x2)
18ydydx=32x
dydx=(32x)18y=0x=0
From this, we can say that x=0
Now. when x=0 , 029+y216=1y216=1y=±4
Hence, the coordinates are (0,4) and (0,-4)

Question:13(ii) Find points on the curve x29+y216=1 at which the tangents are parallel to y-axis

Answer:

Parallel to y-axis means the slope of the tangent is , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by dydx
Given the equation of the curve is
x29+y216=19y2=144(116x2)
18ydydx=144(132x)
dydx=32x18y=
Slope of normal = dxdy=18y32x=0
From this we can say that y = 0
Now. when y = 0, x29+02161=x=±3
Hence, the coordinates are (3,0) and (-3,0)

Question:14(i) Find the equations of the tangent and normal to the given curves at the indicated
points:
y=x46x3+13x210x+5at(0,5)

Answer:

We know that Slope of the tangent at a point on the given curve is given by dydx
Given the equation of the curve
y=x46x3+13x210x+5
dydx=4x318x2+26x10
at point (0,5)
dydx=4(0)318(0)2+26(0)10=10
Hence slope of tangent is -10
Now we know that,
slope of normal=1slope of tangent=110=110
Now, equation of tangent at point (0,5) with slope = -10 is
y=mx+c5=0+cc=5
equation of tangent is
y=10x+5y+10x=5
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
y=mx+c5=0+cc=5
equation of normal is
y=110x+510yx=50

Question:14(ii) Find the equations of the tangent and normal to the given curves at the indicated
points:
y=x46x3+13x210x+5at(1,3)

Answer:

We know that Slope of tangent at a point on given curve is given by dydx
Given equation of curve
y=x46x3+13x210x+5
dydx=4x318x2+26x10
at point (1,3)
dydx=4(1)318(1)2+26(1)10=2
Hence slope of tangent is 2
Now we know that,
slope of normal=1slope of tangent=12
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
3=12×1+c
c=72
equation of normal is
y=12x+722y+x=7

Question:14(iii) Find the equations of the tangent and normal to the given curves at the indicated
points:

y=x3at(1,1)

Answer:

We know that Slope of the tangent at a point on the given curve is given by dydx
Given the equation of the curve
y=x3
dydx=3x2
at point (1,1)
dydx=3(1)2=3
Hence slope of tangent is 3
Now we know that,
slope of normal=1slope of tangent=13
Now, equation of tangent at point (1,1) with slope = 3 is
y=mx+c1=1×3+cc=13=2
equation of tangent is
y3x+2=0
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
1=13×1+c
c=43
equation of normal is
y=13x+433y+x=4

Question:14(iv) Find the equations of the tangent and normal to the given curves at the indicated points

y=x2at(0,0)

Answer:

We know that Slope of the tangent at a point on the given curve is given by dydx
Given the equation of the curve
y=x2
dydx=2x
at point (0,0)
dydx=2(0)2=0
Hence slope of tangent is 0
Now we know that,
slope of normal=1slope of tangent=10=
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = is

y=x×+0x=yx=0

Question:14(v) Find the equations of the tangent and normal to the given curves at the indicated points:

x=cost,y=sintatt=π/4

Answer:

We know that Slope of the tangent at a point on the given curve is given by dydx
Given the equation of the curve
x=cost,y=sint
Now,
dxdt=sint and dydt=cost
Now,
(dydx)t=π4=dydtdxdt=costsint=cott==cotπ4=1
Hence slope of the tangent is -1
Now we know that,
slope of normal=1slope of tangent=11=1
Now, the equation of the tangent at the point t=π4 with slope = -1 is
x=cosπ4=12 and

y=sinπ4=12
equation of the tangent at

t=π4 i.e. (12,12) is


yy1=m(xx1)y12=1(x12)2y+2x=2y+x=2
Similarly, the equation of normal at t=π4 with slope = 1 is
x=cosπ4=12 and

y=sinπ4=12
equation of the tangent at

t=π4 i.e. (12,12) is
yy1=m(xx1)y12=1(x12)2y2x=0yx=0x=y

Question:15(a) Find the equation of the tangent line to the curve y=x22x+7 which is parallel to the line 2xy+9=0

Answer:

Parellel to line 2xy+9=0 means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the slope of tangent at a given point to given curve is given by dydx
Given equation of curve is
y=x22x+7
dydx=2x2=2x=2
Now, when x = 2 , y=(2)22(2)+7=44+7=7
Hence, the coordinates are (2,7)
Now, equation of tangent paasing through (2,7) and with slope m = 2 is
y = mx + c
7 = 2 X 2 + c
c = 7 - 4 = 3
So,
y = 2 X x+ 3
y = 2x + 3
So, the equation of tangent is y - 2x = 3

Question:15(b) Find the equation of the tangent line to the curve y=x22x+7 which is perpendicular to the line 5y15x=13.

Answer:

Perpendicular to line 5y15x=13.y=3x+135 means slope of tangent=1slope of line
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
slope of tangent=1slope of line=13
Now, we know that the slope of the tangent at a given point to given curve is given by dydx
Given the equation of curve is
y=x22x+7
dydx=2x2=13x=56
Now, when x=56 , y=(56)22(56)+7=2536106+7=21736
Hence, the coordinates are (56,21736)
Now, the equation of tangent passing through (2,7) and with slope m=13 is
y=mx+c21736=13×56+cc=22736
So,
y=13x+2273636y+12x=227
Hence, equation of tangent is 36y + 12x = 227

Question:16 Show that the tangents to the curve y=7x3+11 at the points where x = 2 and x = – 2 are parallel .

Answer:

Slope of tangent = dydx=21x2
When x = 2
dydx=21x2=21(2)2=21×4=84
When x = -2
dydx=21x2=21(2)2=21×4=84
Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve y=7x3+11 is parallel

Question:17 Find the points on the curve y=x3 at which the slope of the tangent is equal to the y-coordinate of the point.

Answer:

Given equation of curve is y=x3
Slope of tangent = dydx=3x2
it is given that the slope of the tangent is equal to the y-coordinate of the point
3x2=y
We have y=x3
3x2=x33x2x3=0x2(3x)=0x=0        and          x=3
So, when x = 0 , y = 0
and when x = 3 , y=x3=33=27

Hence, the coordinates are (3,27) and (0,0)

Question:18 For the curve y=4x32x5 , find all the points at which the tangent passes
through the origin.

Answer:

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is y=4x32x5
Slope of tangent =

dydx=12x210x4
Now, equation of tangent is
Yy=m(Xx)
at (0,0) Y = 0 and X = 0
y=(12x310x4)(x)
y=12x310x5
and we have y=4x32x5
4x32x5=12x310x5
8x58x3=08x3(x21)=0x=0      and       x=±1
Now, when x = 0,

y=4(0)32(0)5=0
when x = 1 ,

y=4(1)32(1)5=42=2
when x= -1 ,

y=4(1)32(1)5=4(2)=4+2=2
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

Question:19 Find the points on the curve x2+y22x3=0 at which the tangents are parallel
to the x-axis.

Answer:

parellel to x-axis means slope is 0
Given equation of curve is
x2+y22x3=0
Slope of tangent =
2ydydx=2x2dydx=1xy=0x=1
When x = 1 ,

y2=x22x3=(1)22(1)3=15=4
y=±2
Hence, the coordinates are (1,2) and (1,-2)

Question:20 Find the equation of the normal at the point (am2,am3) for the curve ay2=x3.

Answer:

Given equation of curve is
ay2=x3y2=x3a
Slope of tangent

2ydydx=3x2adydx=3x22ya
at point (am2,am3)
dydx=3(am2)22(am3)a=3a2m42a2m3=3m2
Now, we know that
Slope of normal=1Slope of tangent=23m
equation of normal at point (am2,am3) and with slope 23m
yy1=m(xx1)yam3=23m(xam2)3ym3am4=2(xam2)3ym+2x=3am4+2am2
Hence, the equation of normal is 3ym+2x=3am4+2am2

Question:21 Find the equation of the normals to the curve y=x3+2x+6 which are parallel
to the line x+14y+4=0.

Answer:

Equation of given curve is
y=x3+2x+6
Parellel to line x+14y+4=0y=x14414 means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with our given equation. we get,
m=114
Slope of tangent = dydx=3x2+2
We know that
Slope of normal=1Slope of tangent=13x2+2
13x2+2=114
3x2+2=143x2=12x2=4x=±2
Now, when x = 2, y=(2)3+2(2)+6=8+4+6=18
and
When x = -2 , y=(2)3+2(2)+6=84+6=6
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope 114
yy1=m(xx1)y18=114(x2)14y252=x+2x+14y=254
Similarly, the equation of at point (-2,-6) with slope 114

yy1=m(xx1)y(6)=114(x(2))14y+84=x2x+14y+86=0
Hence, the equation of the normals to the curve y=x3+2x+6 which are parallel
to the line x+14y+4=0.

are x +14y - 254 = 0 and x + 14y +86 = 0

Question:22 Find the equations of the tangent and normal to the parabola y2=4ax at the point (at2,2at).

Answer:

Equation of the given curve is
y2=4ax

Slope of tangent = 2ydydx=4adydx=4a2y
at point (at2,2at).
dydx=4a2(2at)=4a4at=1t
Now, the equation of tangent with point (at2,2at). and slope 1t is
yy1=m(xx1)y2at=1t(xat2)yt2at2=xat2xyt+at2=0

We know that
Slope of normal=1Slope of tangent=t
Now, the equation of at point (at2,2at). with slope -t
yy1=m(xx1)y2at=(t)(xat2)y2at=xt+at3xt+y2atat3=0
Hence, the equations of the tangent and normal to the parabola

y2=4ax at the point (at2,2at). are
xyt+at2=0    and    xt+y2atat3=0  respectively

Question:23 Prove that the curves x=y2 and xy = k cut at right angles* if8k2=1.

Answer:

Let suppose, Curve x=y2 and xy = k cut at the right angle
then the slope of their tangent also cut at the right angle
means,
(dydx)a×(dydx)b=1 -(i)
2y(dydx)a=1(dydx)a=12y
(dydx)b=kx2
Now these values in equation (i)
12y×kx2=1k=2yx2k=2(xy)(x)k=2k(k23)    (x=y2y2y=ky=k13 and x=k23)2(k23)=1(2(k23))3=138k2=1
Hence proved

Question:24 Find the equations of the tangent and normal to the hyperbola
x2a2y2b2=1 at the point (x0,y0)

Answer:

Given equation is
x2a2y2b2=1y2a2=x2b2a2b2
Now ,we know that
slope of tangent = 2ya2dydx=2xb2dydx=xb2ya2
at point (x0,y0)
dydx=x0b2y0a2
equation of tangent at point (x0,y0) with slope xb2ya2
yy1=m(xx1)yy0=x0b2y0a2(xx0)yy0a2y02a2=xx0b2x02b2xx0b2yy0a2=x02b2y02a2
Now, divide both sides by a2b2
xx0a2yy0b2=(x02a2y02b2)
=1                    (x02a2y02b2=1)
xx0a2yy0b2=1
Hence, the equation of tangent is

xx0a2yy0b2=1
We know that
Slope of normal=1slope of tangent=y0a2x0b2
equation of normal at the point (x0,y0) with slope y0a2x0b2
yy1=m(xx1)yy0=y0a2x0b2(xx0)yy0y0a2+xx0x0b2=0

Question:25 Find the equation of the tangent to the curve y=3x2 which is parallel to the line 4x2y+5=0.

Answer:

Parellel to line 4x2y+5=0y=2x+52 means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the slope of the tangent at a given point to given curve is given by dydx
Given the equation of curve is
y=3x2
dydx=12.33x2=323x2
323x2=232=(43x2)29=16(3x2)3x2=9163x=916+23x=4116x=4148
Now, when

x=4148 , y=3x2y=3×41482=41162=916=±34

but y cannot be -ve so we take only positive value
Hence, the coordinates are

(4148,34)
Now, equation of tangent paasing through

(4148,34) and with slope m = 2 is
yy1=m(xx1)y34=2(x4148)48y36=2(48x41)48x24y=411848x24y=23
Hence, equation of tangent paasing through (4148,34) and with slope m = 2 is 48x - 24y = 23

Question:26 The slope of the normal to the curve y=2x2+3sinxatx=0 is
(A) 3 (B) 1/3 (C) –3 (D) -1/3

Answer:

Equation of the given curve is
y=2x2+3sinx
Slope of tangent = dydx=4x+3cosx
at x = 0
dydx=4(0)+3cos0=0+3
dydx=3
Now, we know that
Slope of normal=1 Slope of tangent=13
Hence, (D) is the correct option

Question:27 The line y=x+1 is a tangent to the curve y2=4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)

Answer:

The slope of the given line y=x+1 is 1
given curve equation is
y2=4x
If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent = 2ydydx=4dydx=2y
dydx=2y=1y=2
Now, when y = 2, x=y24=224=44=1
Hence, the coordinates are (1,2)

Hence, (A) is the correct answer


NCERT application-of-derivatives class 12 solutions: Exercise 6.4

Question:1(i) Using differentials, find the approximate value of each of the following up to 3
places of decimal. 25.3

Answer:

Lets suppose y=x and let x = 25 and Δx=0.3
Then,
Δy=x+Δxx
Δy=25+0.325
Δy=25.35
25.3=Δy+5
Now, we can say that Δy is approximate equals to dy
dy=dydxΔxdy=12x.(0.3)              (y=x and Δx=0.3)dy=1225.(0.3)dy=110.(0.3)dy=0.03
Now,
25.3=Δy+525.3=0.03+525.3=5.03
Hence, 25.3 is approximately equals to 5.03

Question:1(ii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

49.5

Answer:

Lets suppose y=x and let x = 49 and Δx=0.5
Then,
Δy=x+Δxx
Δy=49+0.549
Δy=49.57
49.5=Δy+7
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=12x.(0.5)              (y=x and Δx=0.5)dy=1249.(0.5)dy=114.(0.5)dy=0.035
Now,
49.5=Δy+749.5=0.035+749.5=7.035
Hence, 49.5 is approximately equal to 7.035

Question:1(iii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

0.6

Answer:

Lets suppose y=x and let x = 1 and Δx=0.4
Then,
Δy=x+Δxx
Δy=1+(0.4)1
Δy=0.61
0.6=Δy+1
Now, we cam say that Δy is approximately equals to dy
dy=dydxΔxdy=12x.(0.4)              (y=x and Δx=0.4)dy=121.(0.4)dy=12.(0.4)dy=0.2
Now,
0.6=Δy+10.6=(0.2)+10.6=0.8
Hence, 0.6 is approximately equal to 0.8

Question:1(iv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(0.009)1/3

Answer:

Lets suppose y=(x)13 and let x = 0.008 and Δx=0.001
Then,
Δy=(x+Δx)13(x)13
Δy=(0.008+0.001)13(0.008)13
Δy=(0.009)130.2
(0.009)13=Δy+0.2
Now, we cam say that Δy is approximately equals to dy
dy=dydxΔxdy=13(x)23.(0.001)              (y=(x)13 and Δx=0.001)dy=13(0.008)23.(0.001)dy=10.12.(0.001)dy=0.008
Now,
(0.009)13=Δy+0.2(0.009)13=(0.008)+0.2(0.009)13=0.208
Hence, (0.009)13 is approximately equal to 0.208

Question:1(v) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

(0.999)1/10

Answer:

Lets suppose y=(x)110 and let x = 1 and Δx=0.001
Then,
Δy=(x+Δx)110(x)110
Δy=(10.001)110(1)110
Δy=(0.999)1101
(0.999)110=Δy+1
Now, we cam say that Δy is approximately equals to dy
dy=dydxΔxdy=110(x)910.(0.001)              (y=(x)110 and Δx=0.001)dy=110(1)910.(0.001)dy=110.(0.001)dy=0.0001
Now,
(0.999)110=Δy+1(0.999)110=(0.0001)+1(0.999)110=0.9999=0.999 upto three decimal place
Hence, (0.999)110 is approximately equal to 0.999 (because we need to answer up to three decimal place)

Question:1(vi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(15)1/4

Answer:

Let's suppose y=(x)14 and let x = 16 and Δx=1
Then,
Δy=(x+Δx)14(x)14
Δy=(161)14(16)14
Δy=(15)142
(15)14=Δy+2
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=14(x)34.(1)              (y=(x)14 and Δx=1)dy=14(16)34.(1)dy=14×8.(1)dy=132.(1)dy=0.031
Now,
(15)14=Δy+2(15)14=(0.031)+2(15)14=1.969
Hence, (15)14 is approximately equal to 1.969

Question:1(vii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(26)1/3

Answer:

Lets suppose y=(x)13 and let x = 27 and Δx=1
Then,
Δy=(x+Δx)13(x)13
Δy=(271)13(27)13
Δy=(26)133
(26)13=Δy+3
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=13(x)23.(1)              (y=(x)13 and Δx=1)dy=13(27)23.(1)dy=13×9.(1)dy=127.(1)dy=0.037
Now,
(27)13=Δy+3(27)13=(0.037)+3(27)13=2.963
Hence, (27)13 is approximately equal to 2.963

Question:1(viii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(255)1/4

Answer:

Let's suppose y=(x)14 and let x = 256 and Δx=1
Then,
Δy=(x+Δx)14(x)14
Δy=(2561)14(256)14
Δy=(255)144
(255)14=Δy+4
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=14(x)34.(1)              (y=(x)14 and Δx=1)dy=14(256)34.(1)dy=14×64.(1)dy=1256.(1)dy=0.003
Now,
(255)14=Δy+4(255)14=(0.003)+4(255)14=3.997
Hence, (255)14 is approximately equal to 3.997

Question:1(ix) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(82)1/4

Answer:

Let's suppose y=(x)14 and let x = 81 and Δx=1
Then,
Δy=(x+Δx)14(x)14
Δy=(81+1)14(81)14
Δy=(82)143
(82)14=Δy+3
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=14(x)34.(1)              (y=(x)14 and Δx=1)dy=14(81)34.(1)dy=14×27.(1)dy=1108.(1)dy=.009
Now,
(82)14=Δy+3(8214=(0.009)+3(82)14=3.009
Hence, (82)14 is approximately equal to 3.009

Question:1(x) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(401)1/2

Answer:

Let's suppose y=(x)12 and let x = 400 and Δx=1
Then,
Δy=(x+Δx)12(x)12
Δy=(400+1)12(400)12
Δy=(401)1220
(401)12=Δy+20
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=12(x)12.(1)              (y=(x)12 and Δx=1)dy=12(400)12.(1)dy=12×20.(1)dy=140.(1)dy=0.025
Now,
(401)12=Δy+20(401)12=(0.025)+20(401)12=20.025
Hence, (401)12 is approximately equal to 20.025

Question:1(xi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(0.0037)1/2

Answer:

Lets suppose y=(x)12 and let x = 0.0036 and Δx=0.0001
Then,
Δy=(x+Δx)12(x)12
Δy=(0.0036+0.0001)12(0.0036)12
Δy=(0.0037)120.06
(0.0037)12=Δy+0.06
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=12(x)12.(0.0001)              (y=(x)12 and Δx=0.0001)dy=12(0.0036)12.(0.0001)dy=12×0..06.(0.0001)dy=10.12.(0.0001)dy=0.0008
Now,
(0.0037)12=Δy+0.06(0.0037)12=(0.0008)+0.06(0.0037)12=0.0608
Hence, (0.0037)12 is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(26.57)1/3

Answer:

Lets suppose y=(x)13 and let x = 27 and Δx=0.43
Then,
Δy=(x+Δx)13(x)13
Δy=(270.43)13(27)13
Δy=(26.57)133
(26.57)13=Δy+3
Now, we cam say that Δy is approximately equals to dy
dy=dydxΔxdy=13(x)23.(0.43)              (y=(x)13 and Δx=0.43)dy=13(27)23.(0.43)dy=13×9.(0.43)dy=127.(0.43)dy=0.0159=0.016(approx.)
Now,
(26.57)13=Δy+3(26.57)13=(0.016)+3(26.57)13=2.984
Hence, (0.0037)12 is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xiii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(81.5)1/4

Answer:

Lets suppose y=(x)14 and let x = 81 and 0.5
Then,
Δy=(x+Δx)14(x)14
Δy=(81+0.5)14(81)14
Δy=(81.5)143
(81.5)14=Δy+3
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=14(x)34.(0.5)              (y=(x)14 and Δx=0.5)dy=14(81)34.(0.5)dy=14×27.(0.5)dy=1108.(0.5)dy=.004
Now,
(81.5)14=Δy+3(8214=(0.004)+3(82)14=3.004
Hence, (81.5)14 is approximately equal to 3.004

Question:1(xiv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

(3.968)3/2

Answer:

Let's suppose y=(x)32 and let x = 4 and Δx=0.032
Then,
Δy=(x+Δx)32(x)32
Δy=(40.032)32(4)32
Δy=(3.968)328
(3.968)32=Δy+8
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=3(x)122.(0.032)              (y=(x)32 and Δx=0.032)dy=3(4)122.(0.032)dy=3×22.(0.032)dy=0.096
Now,
(3.968)32=Δy+8(3.968)32=(0.096)+8(3.968)32=7.904
Hence, (3.968)32 is approximately equal to 7.904

Question:1(xv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(32.15)1/5

Answer:

Lets suppose y=(x)15 and let x = 32 and Δx=0.15
Then,
Δy=(x+Δx)15(x)15
Δy=(32+0.15)15(32)15
Δy=(32.15)152
(32.15)15=Δy+2
Now, we can say that Δy is approximately equal to dy
dy=dydxΔxdy=15(x)45.(0.15)              (y=(x)15 and Δx=0.15)dy=15(32)45.(0.15)dy=15×16.(0.15)dy=0.1580dy=0.001
Now,
(32.15)15=Δy+2(32.15)15=(0.001)+2(32.15)15=2.001
Hence, (32.15)15 is approximately equal to 2.001

Question:2 Find the approximate value of f (2.01), where f(x)=4x2+5x+2.

Answer:

Let x = 2 and Δx=0.01
f(x+Δx)=4(x+Δx)2+5(x+Δx)+2
Δy=f(x+Δx)f(x)f(x+Δx)=Δy+f(x)
We know that Δy is approximately equal to dy
dy=dydx.Δxdy=(8x+5).(0.01)         (y=f(x)=4x2+5x+2 and Δx=0.01)dy=0.08x+0.05
f(x+Δx)=Δy+f(x)f(x+Δx)=0.08x+0.05+4x2+5x+2f(x+Δx)=0.08(2)+0.05+4(2)2+5(2)+2f(x+Δx)=0.16+0.05+16+10+2f(x+Δx)=28.21
Hence, the approximate value of f (2.01), where f(x)=4x2+5x+2. is 28.21

Question:3 Find the approximate value of f (5.001), where f(x)=x37x2+15.

Answer:

Let x = 5 and Δx=0.001
f(x+Δx)=(x+Δx)37(x+Δx)2+15
Δy=f(x+Δx)f(x)f(x+Δx)=Δy+f(x)
We know that Δy is approximately equal to dy
dy=dydx.Δxdy=(3x214x).(0.001)         (y=f(x)=x37x2+15 and Δx=0.001)dy=0.003x20.014x
f(x+Δx)=Δy+f(x)f(x+Δx)=0.003x20.014x+x37x2+15f(x+Δx)=0.003(5)20.014(5)+(5)37(5)2+15f(x+Δx)=0.0750.07+125175+15f(x+Δx)=34.995
Hence, the approximate value of f (5.001), where f(x)=x37x2+15 is 34.995

Question:4 Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

Answer:

Side of cube increased by 1% = 0.01x m
Volume of cube = x3 m3
we know that Δy is approximately equal to dy
So,
dy=dydx.Δxdy=3x2(0.01x)                (y=x3 and Δx=0.01x)dy=0.03x3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is 0.03x3 m3

Question:5 Find the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%.

Answer:

Side of cube decreased by 1% (Δx) = -0.01x m
The surface area of cube = 6a2 m2
We know that, (Δy) is approximately equal to dy

dy=dydx.Δxdy=12a(0.01x)                (y=6a2 and Δx=0.01x)dy=12x(0.01x)dy=0.12x2 m2
Hence, the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is 0.12x2 m2

Question:6 If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Answer:

Error in radius of sphere (Δr) = 0.02 m
Volume of sphere = 43πr3
Error in volume (ΔV)
dV=dVdr.ΔrdV=4πr2.Δr             (V=43πr3,r=7 and Δr=0.02)dV=4π(7)2(0.02)dV=4π(49)(0.02)dV=3.92π
Hence, the approximate error in its volume is 3.92π m3

Question:7 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Answer:

Error in radius of sphere (Δr) = 0.03 m
The surface area of sphere = 4πr2
Error in surface area (ΔA)
dA=dAdr.ΔrdA=8πr.Δr             (A=4πr2,r=9 and Δr=0.03)dA=8π(9)(0.03)dA=2.16π
Hence, the approximate error in its surface area is 2.16π m2

Question:8 If f(x)=3x2+15x+5 , then the approximate value of f (3.02) is
(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Answer:

Let x = 3 and Δx=0.02
f(x+Δx)=3(x+Δx)2+15(x+Δx)+5
Δy=f(x+Δx)f(x)f(x+Δx)=Δy+f(x)
We know that Δy is approximately equal to dy
dy=dydx.Δxdy=(6x+15).(0.02)         (y=f(x)=3x2+15x+5 and Δx=0.02)dy=0.12x+0.3
f(x+Δx)=Δy+f(x)f(x+Δx)=0.12x+0.3+3x2+15x+5f(x+Δx)=0.12(3)+0.3+3(3)2+15(3)+5f(x+Δx)=0.36+0.3+27+45+5f(x+Δx)=77.66
Hence, the approximate value of f (3.02) is 77.66
Hence, (D) is the correct answer

Question:9 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(A) 0.06 x3 m3 (B) 0.6 x3 m3 (C) 0.09 x3 m3 (D) 0.9 x3 m3

Answer:

Side of cube increased by 3% = 0.03x m
The volume of cube = x3 m3
we know that Δy is approximately equal to dy
So,
dy=dydx.Δxdy=3x2(0.03x)                (y=x3 and Δx=0.03x)dy=0.09x3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is 0.09x3 m3
Hence, (C) is the correct answer


NCERT application-of-derivatives class 12 solutions: Exercise: 6.5

Question:1(i) Find the maximum and minimum values, if any, of the following functions
given by
( f(x)=(2x1)2+3

Answer:

Given function is,
f(x)=(2x1)2+3
(2x1)20(2x1)2+33
Hence, minimum value occurs when
(2x1)=0x=12
Hence, the minimum value of function f(x)=(2x1)2+3 occurs at x=12
and the minimum value is
f(12)=(2.121)2+3
=(11)2+30+3=3
and it is clear that there is no maximum value of f(x)=(2x1)2+3

Question:1(ii) Find the maximum and minimum values, if any, of the following functions
given by

f(x)=9x2+12x+2

Answer:

Given function is,
f(x)=9x2+12x+2
add and subtract 2 in given equation
f(x)=9x2+12x+2+22f(x)=9x2+12x+42f(x)=(3x+2)22
Now,
(3x+2)20(3x+2)222 for every x ϵ R
Hence, minimum value occurs when
(3x+2)=0x=23
Hence, the minimum value of function f(x)=9x2+12x+2 occurs at x=23
and the minimum value is
f(23)=9(23)2+12(23)+2=48+2=2

and it is clear that there is no maximum value of f(x)=9x2+12x+2

Question:1(iii) Find the maximum and minimum values, if any, of the following functions
given by

f(x)=(x1)2+10

Answer:

Given function is,
f(x)=(x1)2+10
(x1)20(x1)2+1010 for every x ϵ R
Hence, maximum value occurs when
(x1)=0x=1
Hence, maximum value of function f(x)=(x1)2+10 occurs at x = 1
and the maximum value is
f(1)=(11)2+10=10

and it is clear that there is no minimum value of f(x)=9x2+12x+2

Question:1(iv) Find the maximum and minimum values, if any, of the following functions
given by
g(x)=x3+1

Answer:

Given function is,
g(x)=x3+1
value of x3 varies from <x3<
Hence, function g(x)=x3+1 neither has a maximum or minimum value

Question:2(i) Find the maximum and minimum values, if any, of the following functions
given by
f(x)=|x+2|1

Answer:

Given function is
f(x)=|x+2|1
|x+2|0|x+2|11 x ϵ R
Hence, minimum value occurs when |x + 2| = 0
x = -2
Hence, minimum value occurs at x = -2
and minimum value is
f(2)=|2+2|1=1
It is clear that there is no maximum value of the given function x ϵ R

Question:2(ii) Find the maximum and minimum values, if any, of the following functions
given by
g(x)=|x+1|+3

Answer:

Given function is
g(x)=|x+1|+3
|x+1|0|x+1|+33 x ϵ R
Hence, maximum value occurs when -|x + 1| = 0
x = -1
Hence, maximum value occurs at x = -1
and maximum value is
g(1)=|1+1|+3=3
It is clear that there is no minimum value of the given function x ϵ R

Question:2(iii) Find the maximum and minimum values, if any, of the following functions
given by
h(x)=sin(2x)+5

Answer:

Given function is
h(x)=sin(2x)+5
We know that value of sin 2x varies from
1sin2x1
1+5sin2x+51+54sin2x+56
Hence, the maximum value of our function h(x)=sin(2x)+5 is 6 and the minimum value is 4

Question:2(iv) Find the maximum and minimum values, if any, of the following functions
given by
f(x)=|sin4x+3|

Answer:

Given function is
f(x)=|sin4x+3|
We know that value of sin 4x varies from
1sin4x1
1+3sin4x+31+32sin4x+342|sin4x+3|4
Hence, the maximum value of our function f(x)=|sin4x+3| is 4 and the minimum value is 2

Question:2(v) Find the maximum and minimum values, if any, of the following functions
given by
h(x)=x+1,xϵ(1,1)

Answer:

Given function is
h(x)=x+1
It is given that the value of x ϵ(1,1)
So, we can not comment about either maximum or minimum value
Hence, function h(x)=x+1 has neither has a maximum or minimum value

Question:3(i) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
f(x)=x2

Answer:

Given function is
f(x)=x2f(x)=2xf(x)=02x=0x=0
So, x = 0 is the only critical point of the given function
f(0)=0 So we find it through the 2nd derivative test
f(x)=2f(0)=2f(0)>0
Hence, by this, we can say that 0 is a point of minima
and the minimum value is
f(0)=(0)2=0

Question:3(ii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
g(x)=x33x

Answer:

Given function is
g(x)=x33xg(x)=3x23g(x)=03x23=0x=±1
Hence, the critical points are 1 and - 1
Now, by second derivative test
g(x)=6x
g(1)=6>0
Hence, 1 is the point of minima and the minimum value is
g(1)=(1)33(1)=13=2
g(1)=6<0
Hence, -1 is the point of maxima and the maximum value is
g(1)=(1)33(1)=1+3=2

Question:3(iii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
h(x)=sinx+cosx, 0<x<π2

Answer:

Given function is
h(x)=sinx+cosxh(x)=cosxsinxh(x)=0cosxsinx=0cosx=sinxx=π4      as x ϵ (0,π2)
Now, we use the second derivative test
h(x)=sinxcosxh(π4)=sinπ4cosπ4h(π4)=1212h(π4)=22=2<0
Hence, π4 is the point of maxima and the maximum value is h(π4) which is 2

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

f(x)=sinxcosx

Answer:

Given function is
h(x)=sinxcosxh(x)=cosx+sinxh(x)=0cosx+sinx=0cosx=sinxx=3π4      as x ϵ (0,2π)
Now, we use second derivative test
h(x)=sinx+cosxh(3π4)=sin3π4+cos3π4h(3π4)=(12)12h(π4)=22=2<0
Hence, π4 is the point of maxima and maximum value is h(3π4) which is 2

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

f(x)=x36x2+9x+15

Answer:

Givrn function is
f(x)=x36x2+9x+15f(x)=3x212x+9f(x)=03x212x+9=03(x24x+3)=0x24x+3=0x2x3x+3=0x(x1)3(x1)=0(x1)(x3)=0x=1      and       x=3
Hence 1 and 3 are critical points
Now, we use the second derivative test
f(x)=6x12f(1)=612=6<0
Hence, x = 1 is a point of maxima and the maximum value is
f(1)=(1)36(1)2+9(1)+15=16+9+15=19
f(x)=6x12f(3)=1812=6>0
Hence, x = 1 is a point of minima and the minimum value is
f(3)=(3)36(3)2+9(3)+15=2754+27+15=15

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

g(x)=x2+2x,x>0

Answer:

Given function is
g(x)=x2+2xg(x)=122x2g(x)=0122x2=0x2=4x=±2 ( but as x>0 we only take the positive value of x i.e. x = 2)
Hence, 2 is the only critical point
Now, we use the second derivative test
g(x)=4x3g(2)=423=48=12>0
Hence, 2 is the point of minima and the minimum value is
g(x)=x2+2xg(2)=22+22=1+1=2

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

g(x)=1x2+2

Answer:

Gien function is
g(x)=1x2+2g(x)=2x(x2+2)2g(x)=02x(x2+2)2=0x=0
Hence., x = 0 is only critical point
Now, we use the second derivative test
g(x)=2(x2+2)2(2x)2(x2+2)(2x)((x2+2)2)2g(0)=2×4(2)4=816=12<0
Hence, 0 is the point of local maxima and the maximum value is
g(0)=102+2=12

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

f(x)=x1x,0<x<1

Answer:

Given function is
f(x)=x1x
f(x)=1x+x(1)21x
=1xx21x23x21xf(x)=023x21x=03x=2x=23
Hence, x=23 is the only critical point
Now, we use the second derivative test
f(x)=(1)(21x)(2x)(2.121x(1))(21x)2
=21x21x+x1x4(1x)
=3x4(1x)1x
f"(23)>0
Hence, it is the point of minima and the minimum value is
f(x)=x1xf(23)=23123f(23)=2313f(23)=233f(23)=239

Question:4(i) Prove that the following functions do not have maxima or minima:
f(x)=ex

Answer:

Given function is
f(x)=ex
f(x)=exf(x)=0ex=0
But exponential can never be 0
Hence, the function f(x)=ex does not have either maxima or minima

Question:4(ii) Prove that the following functions do not have maxima or minima:

g(x)=logx

Answer:

Given function is
g(x)=logx
g(x)=1xg(x)=01x=0
Since log x deifne for positive x i.e. x>0
Hence, by this, we can say that g(x)>0 for any value of x
Therefore, there is no c ϵ R such that g(c)=0
Hence, the function g(x)=logx does not have either maxima or minima

Question:4(iii) Prove that the following functions do not have maxima or minima:

h(x)=x3+x2+x+1

Answer:

Given function is
h(x)=x3+x2+x+1
h(x)=3x2+2x+1h(x)=03x2+2x+1=02x2+x2+2x+1=02x2+(x+1)2=0
But, it is clear that there is no c ϵ R such that f(c)=0
Hence, the function h(x)=x3+x2+x+1 does not have either maxima or minima

Question:5(i) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
f(x)=x3,xϵ[2,2]

Answer:

Given function is
f(x)=x3
f(x)=3x2f(x)=03x2=0x=0
Hence, 0 is the critical point of the function f(x)=x3
Now, we need to see the value of the function f(x)=x3 at x = 0 and as x ϵ [2,2] we also need to check the value at end points of given range i.e. x = 2 and x = -2
f(0)=(0)3=0f(2=(2)3=8f(2)=(2)3=8
Hence, maximum value of function f(x)=x3 occurs at x = 2 and value is 8
and minimum value of function f(x)=x3 occurs at x = -2 and value is -8

Question:5(ii) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:

f(x)=sinx+cosx,xϵ[0,π]

Answer:

Given function is
f(x)=sinx+cosx
f(x)=cosxsinxf(x)=0cosxsinx=0cos=sinxx=π4 as x ϵ [0,π]
Hence, x=π4 is the critical point of the function f(x)=sinx+cosx
Now, we need to check the value of function f(x)=sinx+cosx at x=π4 and at the end points of given range i.e. x=0 and x=π
f(π4)=sinπ4+cosπ4
=12+12=22=2
f(0)=sin0+cos0=0+1=1
f(π)=sinπ+cosπ=0+(1)=1
Hence, the absolute maximum value of function f(x)=sinx+cosx occurs at x=π4 and value is 2
and absolute minimum value of function f(x)=sinx+cosx occurs at x=π and value is -1

Question:5(iii) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
f(x)=4x12x2,xϵ[2,92]

Answer:

Given function is
f(x)=4x12x2
f(x)=4xf(x)=04x=0x=4
Hence, x = 4 is the critical point of function f(x)=4x12x2
Now, we need to check the value of function f(x)=4x12x2 at x = 4 and at the end points of given range i.e. at x = -2 and x = 9/2
f(4)=4(4)12(4)2
=1612.16=168=8
f(2)=4(2)12.(2)2=82=10
f(92)=4(92)12.(92)2=18818=638
Hence, absolute maximum value of function f(x)=4x12x2 occures at x = 4 and value is 8
and absolute minimum value of function f(x)=4x12x2 occures at x = -2 and value is -10

Question:5(iv) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

f(x)=(x1)2+3,xϵ[3,1]

Answer:

Given function is
f(x)=(x1)2+3
f(x)=2(x1)f(x)=02(x1)=0x=1
Hence, x = 1 is the critical point of function f(x)=(x1)2+3
Now, we need to check the value of function f(x)=(x1)2+3 at x = 1 and at the end points of given range i.e. at x = -3 and x = 1
f(1)=(11)2+3=02+3=3

f(3)=(31)2+3=(4)2+3=16+3=19
f(1)=(11)2+3=02+3=3
Hence, absolute maximum value of function f(x)=(x1)2+3 occurs at x = -3 and value is 19
and absolute minimum value of function f(x)=(x1)2+3 occurs at x = 1 and value is 3

Question:6 . Find the maximum profit that a company can make, if the profit function is
given by p(x)=4172x18x2

Answer:

Profit of the company is given by the function
p(x)=4172x18x2
p(x)=7236xp(x)=07236x=0x=2
x = -2 is the only critical point of the function p(x)=4172x18x2
Now, by second derivative test
p(x)=36<0
At x = -2 p(x)<0
Hence, maxima of function p(x)=4172x18x2 occurs at x = -2 and maximum value is
p(2)=4172(2)18(2)2=41+14472=113
Hence, the maximum profit the company can make is 113 units

Question:7 . Find both the maximum value and the minimum value of
3x48x3+12x248x+25 on the interval [0, 3].

Answer:

Given function is
f(x)=3x48x3+12x248x+25
f(x)=12x324x2+24x48f(x)=012(x32x2+2x4)=0x32x2+2x4=0
Now, by hit and trial let first assume x = 2
(2)32(2)2+2(2)488+44=0
Hence, x = 2 is one value
Now,
x32x2+2x4x2=(x2+2)(x2)(x2)=(x2+2)
x2=2 which is not possible
Hence, x = 2 is the only critical value of function f(x)=3x48x3+12x248x+25
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 0 and x = 3
1628071745575 =3×168×8+12×496+25=4864+4896+25=39

f(3)=3(3)48(3)3+12(3)248(3)+25=3×818×27+12×9144+25=243216+108144+25=16

f(0)=3(0)48(0)3+12(0)248(0)+25=25
Hence, maximum value of function f(x)=3x48x3+12x248x+25 occurs at x = 0 and vale is 25
and minimum value of function f(x)=3x48x3+12x248x+25 occurs at x = 2 and value is -39

Question:8 . At what points in the interval [0,2π] does the function sin2x attain its maximum value?

Answer:

Given function is
f(x)=sin2x
f(x)=2cos2xf(x)=02cos2x=0as x ϵ[0,2π]0<x<2π0<2x<4πcos2x=0 at 2x=π2,2x=3π2,2x=5π2and2x=7π2
So, values of x are
x=π4,x=3π4,x=5π4 and x=7π4 These are the critical points of the function f(x)=sin2x
Now, we need to find the value of the function f(x)=sin2x at x=π4,x=3π4,x=5π4 and x=7π4 and at the end points of given range i.e. at x = 0 and x=π

f(x)=sin2xf(π4)=sin2(π4)=sinπ2=1

f(x)=sin2xf(3π4)=sin2(3π4)=sin3π2=1

f(x)=sin2xf(5π4)=sin2(5π4)=sin5π2=1

f(x)=sin2xf(7π4)=sin2(7π4)=sin7π2=1

f(x)=sin2xf(π)=sin2(π)=sin2π=0

f(x)=sin2xf(0)=sin2(0)=sin0=0

Hence, at x=π4 and x=5π4 function f(x)=sin2x attains its maximum value i.e. in 1 in the given range of x ϵ [0,2π]

Question:9 What is the maximum value of the function sinx+cosx ?

Answer:

Given function is
f(x)=sinx+cosx
f(x)=cosxsinxf(x)=0cosxsinx=0cos=sinxx=2nπ+π4 where n ϵ I
Hence, x=2nπ+π4 is the critical point of the function f(x)=sinx+cosx
Now, we need to check the value of the function f(x)=sinx+cosx at x=2nπ+π4
Value is same for all cases so let assume that n = 0
Now
f(π4)=sinπ4+cosπ4
=12+12=22=2

Hence, the maximum value of the function f(x)=sinx+cosx is 2

Question:10. Find the maximum value of 2x324x+107 in the interval [1, 3]. Find the
the maximum value of the same function in [–3, –1].

Answer:

Given function is
f(x)=2x324x+107
f(x)=6x224f(x)=06(x24)=0x24=0x2=4x=±2 we neglect the value x =- 2 because x ϵ [1,3]
Hence, x = 2 is the only critical value of function f(x)=2x324x+107
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
f(2)=2(2)324(2)+107=2×848+107=1648+107=75

f(3)=2(3)324(3)+107=2×2772+107=5472+107=89

f(1)=2(1)324(1)+107=2×124+107=224+107=85
Hence, maximum value of function f(x)=2x324x+107 occurs at x = 3 and vale is 89 when x ϵ [1,3]
Now, when x ϵ [3,1]
we neglect the value x = 2
Hence, x = -2 is the only critical value of function f(x)=2x324x+107
Now, we need to check the value at x = -2 and at the end points of given range i.e. x = -1 and x = -3
f(1)=2(1)324(1)+107=2×(1)+24+107=2+24+107=129

f(2)=2(2)324(2)+107=2×(8)+48+107=16+48+107=139

f(3)=2(3)324(3)+107=2×(27)+72+107=54+72+107=125
Hence, the maximum value of function f(x)=2x324x+107 occurs at x = -2 and vale is 139 when x ϵ [3,1]

Question:11. It is given that at x = 1, the function x462x2+ax+9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Answer:

Given function is
f(x)=x462x2+ax+9
Function f(x)=x462x2+ax+9 attains maximum value at x = 1 then x must one of the critical point of the given function that means
f(1)=0
f(x)=4x3124x+af(1)=4(1)3124(1)+af(1)=4124+a=a120
Now,
f(1)=0a120=0a=120
Hence, the value of a is 120

Question:12 . Find the maximum and minimum values of x+sin2xon[0,2π]

Answer:

Given function is
f(x)=x+sin2x
f(x)=1+2cos2xf(x)=01+2cos2x=0as x ϵ [0,2π]0<x<2π0<2x<4πcos2x=12 at 2x=2nπ±2π3 where n ϵ Zx=nπ±π3x=π3,2π3,4π3,5π3 as x ϵ [0,2π]
So, values of x are
x=π3,2π3,4π3,5π3 These are the critical points of the function f(x)=x+sin2x
Now, we need to find the value of the function f(x)=x+sin2x at x=π3,2π3,4π3,5π3 and at the end points of given range i.e. at x = 0 and x=2π

f(x)=x+sin2xf(π3)=π3+sin2(π3)=π3+sin2π3=π3+32

f(x)=x+sin2xf(2π3)=2π3+sin2(2π3)=2π3+sin4π3=2π332

f(x)=x+sin2xf(4π3)=4π3+sin2(4π3)=4π3+sin8π3=4π3+32

f(x)=x+sin2xf(5π3)=5π3+sin2(5π3)=5π3+sin10π3=5π332

f(x)=x+sin2xf(2π)=2π+sin2(2π)=2π+sin4π=2π

f(x)=x+sin2xf(0)=0+sin2(0)=0+sin0=0

Hence, at x=2π function f(x)=x+sin2x attains its maximum value and value is 2π in the given range of x ϵ [0,2π]
and at x= 0 function f(x)=x+sin2x attains its minimum value and value is 0

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

Answer:

Let x and y are two numbers
It is given that
x + y = 24 , y = 24 - x
and product of xy is maximum
let f(x)=xy=x(24x)=24xx2f(x)=242xf(x)=0242x=0x=12
Hence, x = 12 is the only critical value
Now,
f(x)=2<0
at x= 12 f(x)<0
Hence, x = 12 is the point of maxima
Noe, y = 24 - x
= 24 - 12 = 12
Hence, the value of x and y are 12 and 12 respectively

Question:14 Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

Answer:

It is given that
x + y = 60 , x = 60 -y
and xy3 is maximum
let f(y)=(60y)y3=60y3y4
Now,
f(y)=180y24y3f(y)=0y2(1804y)=0y=0 and y=45

Now,
f(y)=360y12y2f(0)=0
hence, 0 is neither point of minima or maxima
f(y)=360y12y2f(45)=360(45)12(45)2=8100<0
Hence, y = 45 is point of maxima
x = 60 - y
= 60 - 45 = 15
Hence, values of x and y are 15 and 45 respectively

Question:15 Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.

Answer:

It is given that
x + y = 35 , x = 35 - y
and x2y5 is maximum
Therefore,
let f(y)=(35y)2y5=(122570y+y2)y5f(y)=1225y570y6+y7
Now,
f(y)=6125y4420y5+7y6f(y)=0y4(6125420y+7y2)=0y=0 and (y25)(y35)y=25,y=35
Now,
f(y)=24500y32100y4+42y5

f(35)=24500(35)32100(35)4+42(35)5=105043750>0
Hence, y = 35 is the point of minima

f(0)=0
Hence, y= 0 is neither point of maxima or minima

f(25)=24500(25)32100(25)4+42(25)5=27343750<0
Hence, y = 25 is the point of maxima
x = 35 - y
= 35 - 25 = 10
Hence, the value of x and y are 10 and 25 respectively

Question:16 . Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Answer:

let x an d y are positive two numbers
It is given that
x + y = 16 , y = 16 - x
and x3+y3 is minimum
f(x)=x3+(16x)3
Now,
f(x)=3x2+3(16x)2(1)
f(x)=03x23(16x)2=03x23(256+x232x)=03x23x2+96x768=096x=768x=8
Hence, x = 8 is the only critical point
Now,
f(x)=6x6(16x)(1)=6x+966x=96f(x)=96
f(8)=96>0
Hence, x = 8 is the point of minima
y = 16 - x
= 16 - 8 = 8
Hence, values of x and y are 8 and 8 respectively

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

Answer:

It is given that the side of the square is 18 cm
Let assume that the length of the side of the square to be cut off is x cm
So, by this, we can say that the breath of cube is (18-2x) cm and height is x cm
Then,
Volume of cube (V(x)) = x(182x)2
V(x)=(182x)2+(x)2(182x)(2)
V(x)=0(182x)24x(182x)=0324+4x272x72x+8x2=012x2144x+324=012(x212x+27)=0x29x3x+27=0(x3)(x9)=0x=3 and x=9 But the value of x can not be 9 because then the value of breath become 0 so we neglect value x = 9
Hence, x = 3 is the critical point
Now,
V(x)=24x144V(3)=24×3144.       =72144=72V(3)<0
Hence, x = 3 is the point of maxima
Hence, the length of the side of the square to be cut off is 3 cm so that the volume of the box is the maximum possible

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

Answer:

It is given that the sides of the rectangle are 45 cm and 24 cm
Let assume the side of the square to be cut off is x cm
Then,
Volume of cube V(x)=x(452x)(242x)
V(x)=(452x)(242x)+(2)(x)(242x)+(2)(x)(452x)
1080+4x2138x48x+4x290x+4x212x2276x+1080
V(x)=012(x223x+90)=0x223x+90=0x218x5x+23=0(x18)(x5)=0x=18 and x=5
But x cannot be equal to 18 because then side (24 - 2x) become negative which is not possible so we neglect value x= 18
Hence, x = 5 is the critical value
Now,
V(x)=24x276V(5)=24×5276V(5)=156<0
Hence, x = 5 is the point of maxima
Hence, the side of the square to be cut off is 5 cm so that the volume of the box is maximum

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

Answer:

Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively
and the radius of the circle is r
1628071829107 Now, by Pythagoras theorem
a=l2+b2
a = 2r
4r2=l2+b2l=4r2b2
Now, area of reactangle(A) = l × b
A(b)=b(4r2b2)
A(b)=4r2b2+b.(2b)24r2b2=4r2b2b24r2b2=4r22b24r2b2
A(b)=04r22b24r2b2=04r2=2b2b=2r
Now,
A(b)=4b(4r2b2)(4r22b2).(12(4r2b2)32.(2b))(4r2b2)2A(2r)=(4b)×2r(2r)2=22br<0
Hence, b=2r is the point of maxima
l=4r2b2=4r22r2=2r
Since, l = b we can say that the given rectangle is a square
Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area

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

Answer:

Let r be the radius of the base of cylinder and h be the height of the cylinder
we know that the surface area of the cylinder (A)=2πr(r+h)
h=A2πr22πr
Volume of cylinder
(V)=πr2h=πr2(A2πr22πr)=r(A2πr22)
V(r)=(A2πr22)+(r).(2πr)=A2πr24πr22=A6πr22
V(r)=0A6πr22=0r=A6π
Hence, r=A6π is the critical point
Now,
V(r)=6πrV(A6π)=6π.A6π=A6π<0
Hence, r=A6π is the point of maxima
h=A2πr22πr=22πA6π2πA6π=4πA6π2πA6π=2πA6π=2r
Hence, the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base

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

Answer:

Let r be the radius of base and h be the height of the cylinder
The volume of the cube (V) = πr2h
It is given that the volume of cylinder = 100 cm3
πr2h=100h=100πr2
Surface area of cube(A) = 2πr(r+h)
A(r)=2πr(r+100πr2)
=2πr(πr3+100πr2)=2πr3+200r=2πr2+200r
A(r)=4πr+(200)r2A(r)=04πr3=200r3=50πr=(50π)13
Hence, r=(50π)13 is the critical point
A(r)=4π+400rr3A((50π)13)=4π+400((50π)13)2>0
Hence, r=(50π)13 is the point of minima
h=100πr2=100π((50π)13)2=2.(50π)13
Hence, r=(50π)13 and h=2.(50π)13 are the dimensions of the can which has the minimum surface area

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

Answer:

Area of the square (A) = a2
Area of the circle(S) = πr2
Given the length of wire = 28 m
Let the length of one of the piece is x m
Then the length of the other piece is (28 - x) m
Now,
4a=xa=x4
and
2πr=(28x)r=28x2π
Area of the combined circle and square f(x) = A + S
=a2+πr2=(x4)2+π(28x2π)2
f(x)=2x16+(28x)(1)2πf(x)=xπ+4x1128πf(x)=0xπ+4x1128π=0x(π+4)=112x=112π+4
Now,
f(x)=18+12πf(112π+4)=18+12π>0
Hence, x=112π+4 is the point of minima
Other length is = 28 - x
= 28112π+4=28π+112112π+4=28ππ+4
Hence, two lengths are 28ππ+4 and 112π+4

Question:23 Prove that the volume of the largest cone that can be inscribed in a sphere of radius r is 8/27 of the volume of the sphere.

Answer:

1651257838832 Volume of cone (V) = 13πR2h
Volume of sphere with radius r = 43πr3
By pythagoras theorem in ΔADC we ca say that
OD2=r2R2OD=r2R2h=AD=r+OD=r+r2R2
V = 13πR2(r+r2+R2)=13πR2r+13πR2r2+R2
13πR2(r+r2R2)V(R)=23πRr+23πRr2R2+13πR2.2R2r2R2V(R)=013πR(2r+2r2R2R2r2R2)=013πR(2rr2R2+2r22R2R2r2R2)=0R0 So,2rr2R2=3R22r2Square both sides4r44r2R2=9R4+4r412R2r29R48R2r2=0R2(9R28r2)=0R0 So,9R2=8r2R=22r3
Now,
V(R)=23πr+23πr2R2+23πR.2R2r2R23πR2r2R2(1)(2R)(r2+R2)32V(22r3)<0
Hence, point R=22r3 is the point of maxima
h=r+r2R2=r+r28r29=r+r3=4r3
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r3
Volume = =13πR2h=13π8r29.4r3=827.43πr3=827× volume of sphere
Hence proved

Question:24 Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 time the radius of the base.

Answer:

Volume of cone(V)

13πr2hh=3Vπr2
curved surface area(A) = πrl
l2=r2+h2l=r2+9V2π2r4
A=πrr2+9V2π2r4=πr21+9V2π2r6

dAdr=2πr1+9V2π2r6+πr2.121+9V2π2r6.(6r5)9V2π2r7dAdr=02πr1+9V2π2r6+πr2.121+9V2π2r6.(6)9V2π2r7=02π2r6(1+9V2π2r6)=27V22π2r6(π2r6+9V2π2r6)=27V22π2r6+18V2=27V22π2r6=9V2r6=9V22π2
Now , we can clearly varify that
d2Adr2>0
when r6=9V22π2
Hence, r6=9V22π2 is the point of minima
V=2πr33
h=3Vπr2=3.2πr33πr2=2r
Hence proved that the right circular cone of least curved surface and given volume has an altitude equal to 2 time the radius of the base

Question:25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan12

Answer:

1628071922992 Let a be the semi-vertical angle of cone
Let r , h , l are the radius , height , slent height of cone
Now,
r=lsina and h=lcosa
we know that
Volume of cone (V) = 13πr2h=13π(lsina)2(lcosa)=πl3sin2acosa3
Now,
dVda=πl33(2sinacosa.cosa+sin2a.(sina))=πl33(2sinacos2asin3a)
dVda=0πl33(2sinacos2asin3a)=02sinacos2asin3a=02sinacos2a=sin3atan2a=2a=tan12
Now,
d2Vda2=πl33(2cosacos2a+2cosa(2cosasina+3sin2acosa))
Now, at a=tan12
d2Vdx2<0
Therefore, a=tan12 is the point of maxima
Hence proved

Question:26 Show that semi-vertical angle of the right circular cone of given surface area and maximum volume is sin1(1/3)

Answer:

1628071965473 Let r, l, and h are the radius, slant height and height of cone respectively
Now,
r=lsina and h=lcosa
Now,
we know that
The surface area of the cone (A) = πr(r+l)
A=πlsinal(sina+1)l2=Aπsina(sina+1)l=Aπsina(sina+1)
Now,
Volume of cone(V) =

13πr2h=13πl3sin2acosa=π3.(Aπsina(sina+1))32.sin2acosa
On differentiate it w.r.t to a and after that
dVda=0
we will get
a=sin113
Now, at a=sin113
d2Vda2<0
Hence, we can say that a=sin113 is the point if maxima
Hence proved

Question:27 The point on the curve x2=2y which is nearest to the point (0, 5) is

(A)(22,4)(B)(22,0)(C)(0,0)(D)(2,2)

Answer:

Given curve is
x2=2y
Let the points on curve be (x,x22)
Distance between two points is given by
f(x)=(x2x1)2+(y2y1)2
=(x0)2+(x225)2=x2+x445x2+25=x444x2+25
f(x)=x38x2x444x2+25f(x)=0x38x2x444x2+25=0x(x28)=0x=0 and x2=8x=22
f(x)=12((3x28)(x444x2+25(x38x).(x38x)2x444x2+25(x444x2+25)2))
f(0)=8<0
Hence, x = 0 is the point of maxima
f(22)>0
Hence, the point x=22 is the point of minima
x2=2yy=x22=82=4
Hence, the point (22,4) is the point on the curve x2=2y which is nearest to the point (0, 5)
Hence, the correct answer is (A)

Question:28 For all real values of x, the minimum value of 1x+x21+x+x2
is
(A) 0 (B) 1 (C) 3 (D) 1/3

Answer:

Given function is
f(x)=1x+x21+x+x2
f(x)=(1+2x)(1+x+x2)(1x+x2)(1+2x)(1+x+x2)2
=1xx2+2x+2x2+2x312x+x+2x2x22x3(1+x+x2)2=2+2x2(1+x+x2)2
f(x)=02+2x2(1+x+x2)2=0x2=1x=±1
Hence, x = 1 and x = -1 are the critical points
Now,
f(x)=4x(1+x+x2)2(2+2x2)2(1+x+x2)(2x+1)(1+x+x2)4f(1)=4×(3)234=49>0
Hence, x = 1 is the point of minima and the minimum value is
f(1)=11+121+1+12=13

f(1)=4<0
Hence, x = -1 is the point of maxima
Hence, the minimum value of
1x+x21+x+x2 is 13
Hence, (D) is the correct answer

Question:29 The maximum value of [x(x1)+1]1/3,0x1
(A)(13)1/3(B)1/2(C)1(D)0

Answer:

Given function is
f(x)=[x(x1)+1]1/3
f(x)=13.[(x1)+x].1[x(x1)+1]23=2x13[x(x1)+1]23
f(x)=02x13[x(x1)+1]23=0x=12
Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0
f(12)=[12(121)+1]1/3=(34)13
f(0)=[0(01)+1]1/3=(1)13=1
f(1)=[1(11)+1]1/3=(1)13=1
Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1

option c is correct


Application-of-derivatives class 12 NCERT solutions - Miscellaneous Exercise

Question:1(a) Using differentials, find the approximate value of each of the following:

(17/81)1/4

Answer:

Let y=x14 and x=1681 and Δx=181
Δy=(x+Δx)14x14
=(1681+181)14(1681)14
(1781)14=Δy+23
Now, we know that Δy is approximate equals to dy
So,
dy=dydx.Δx=14x34.181       (y=x14 and Δx=181)=14(1681)34.181=274×8.181=196
Now,
(1781)14=Δy+23=196+23=6596=0.677
Hence, (1781)14 is approximately equal to 0.677

Question:1(b) Using differentials, find the approximate value of each of the following:
(33)1/5

Answer:

Let y=x15 and x=32 and Δx=1
Δy=(x+Δx)15x15
=(32+1)15(32)15
(33)14=Δy+12
Now, we know that Δy is approximately equals to dy
So,
dy=dydx.Δx=15x65.1       (y=x15 and Δx=1)=15(32)65.1=15×64.1=1320
Now,
(33)15=Δy+12=1320+12=159320=0.497
Hence, (33)15 is approximately equals to 0.497

Question:2. Show that the function given by f(x)=logxx has maximum at x = e.

Answer:

Given function is
f(x)=logxx
f(x)=1x.1x+logx1x2=1x2(1logx)
f(x)=01x2(1logx)=01x20 So logx=1x=e
Hence, x =e is the critical point
Now,
f(x)=2xx3(1logx)+1x2(1x)=1x3(2x+2xlogx1)f(e)=1e3<0
Hence, x = e is the point of maxima

Question:3 . The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

Answer:

It is given that the base of the triangle is b
and let the side of the triangle be x cm , dxdt=3cm/s
We know that the area of the triangle(A) = 12bh
now, h=x2(b2)2
A=12bx2(b2)2
dAdt=dAdx.dxdt=12b2x2x2(b2)2.(3)
Now at x = b
dAdx=12b2b3b2.(3)=3b
Hence, the area decreasing when the two equal sides are equal to the base is 3b cm2/s

Question:4 Find the equation of the normal to curve x2=4y which passes through the point (1, 2).

Answer:

Given the equation of the curve
x2=4y
We know that the slope of the tangent at a point on the given curve is given by dydx
4dydx=2x dydx=x2
We know that
Slope of normal=1Slope of tangent=1x2=2x
At point (a,b)
Slope=2a
Now, the equation of normal with point (a,b) and Slope=2a

yy1=m(xx1)yb=2a(xa)
It is given that it also passes through the point (1,2)
Therefore,
2b=2a(1a)2aba=2a2ba=2b=2a -(i)
It also satisfies equation x2=4yb=a24 -(ii)
By comparing equation (i) and (ii)
2a=a24a3=8a=2
b=2a=22=1
Slope=2a=22=1

Now, equation of normal with point (2,1) and slope = -1

yy1=m(xx1)y1=1(x2)y+x=3
Hence, equation of normal is x + y - 3 = 0

Question:5 . Show that the normal at any point θ to the curve x=acosθ+aθsinθ,y=asinθaθcosθ is at a constant distance from the origin.

Answer:

We know that the slope of tangent at any point is given by dydx
Given equations are
x=acosθ+aθsinθ,y=asinθaθcosθ
dxdθ=asinθ+asinθaθcosθ=aθcosθ
dydθ=acosθacosθ+aθ(sinθ)=aθsinθ
dydx=dydθdxdθ=aθsinθaθcosθ=tanθ
We know that
Slope of normal=1Slope of tangent=1tanθ
equation of normal with given points and slope
y2y1=m(x2x1)yasinθ+aθcosθ=1tanθ(xacosθaθsinθ)ysinθasin2θ+aθcosθsinθ=xcosθ+acos2θ+aθsinθcosθysinθ+xcosθ=a
Hence, the equation of normal is ysinθ+xcosθ=a
Now perpendicular distance of normal from the origin (0,0) is
D=|(0)sinθ+(0)cosθa|sin2θ+cos2θ=|a|=a= constant                               (sin2x+cos2x=1)
Hence, by this, we can say that

the normal at any point θ to the curve x=acosθ+aθsinθ,y=asinθaθcosθ

is at a constant distance from the origin

Question:6(i) Find the intervals in which the function f given by f(x)=4sinx2xxcosx2+cosx is

increasing

Answer:

Given function is
f(x)=4sinx2xxcosx2+cosx
f(x)=(4cosx2cosx+xsinx)(2+cosx)(4sinx2xxcosx)(sinx)(2+cosx)2
=4cosxcos2x2+cosx
f(x)=04cosxcos2x2+cosx=0cosx(4cosx)=0cosx=0       and           cosx=4
But cosx4
So,
cosx=0x=π2 and 3π2
Now three ranges are there (0,π2),(π2,3π2) and (3π2,2π)
In interval (0,π2) and (3π2,2π) , f(x)>0

Hence, the given function f(x)=4sinx2xxcosx2+cosx is increasing in the interval (0,π2) and (3π2,2π)
in interval ,(π2,3π2),f(x)<0 so function is decreasing in this inter

Question:6(ii) Find the intervals in which the function f given by f x is equal to

f(x)=4sinx2xxcosx2+cosx is

decreasing

Answer:

Given function is
f(x)=4sinx2xxcosx2+cosx
f(x)=(4cosx2cosx+xsinx)(2+cosx)(4sinx2xxcosx)(sinx)(2+cosx)2
=4cosxcos2x2+cosx
f(x)=04cosxcos2x2+cosx=0cosx(4cosx)=0cosx=0       and           cosx=4
But cosx4
So,
cosx=0x=π2 and 3π2
Now three ranges are there (0,π2),(π2,3π2) and (3π2,2π)
In interval (0,π2) and (3π2,2π) , f(x)>0

Hence, given function f(x)=4sinx2xxcosx2+cosx is increasing in interval (0,π2) and (3π2,2π)
in interval ,(π2,3π2),f(x)<0
Hence, given function f(x)=4sinx2xxcosx2+cosx is decreasing in interval ,(π2,3π2)

Question:7(i) Find the intervals in which the function f given by f(x)=x3+1x3,x0

Increasing

Answer:

Given function is
f(x)=x3+1x3
f(x)=3x2+3x2x4f(x)=03x2+3x2x4=0x4=1x=±1
Hence, three intervals are their (,1),(1,1) and(1,)
In interval (,1) and (1,),f)x>0
Hence, given function f(x)=x3+1x3 is increasing in interval (,1) and (1,)
In interval (-1,1) , f(x)<0
Hence, given function f(x)=x3+1x3 is decreasing in interval (-1,1)

Question:7(ii) Find the intervals in which the function f given by f(x)=x3+1x3,x0

decreasing

Answer:

Given function is
f(x)=x3+1x3
f(x)=3x2+3x2x4f(x)=03x2+3x2x4=0x4=1x=±1
1651257893638 Hence, three intervals are their (,1),(1,1) and(1,)
In interval (,1) and (1,),f)x>0
Hence, given function f(x)=x3+1x3 is increasing in interval (,1) and (1,)
In interval (-1,1) , f(x)<0
Hence, given function f(x)=x3+1x3 is decreasing in interval (-1,1)

Question:8 Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1 with its vertex at one end of the major axis.

Answer:

1628072034896 Given the equation of the ellipse
x2a2+y2b2=1
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
m=±ba.a2n2
Therefore, Now
Coordinates of A = (n,ba.a2n2)
Coordinates of B = (n,ba.a2n2)
Now,
Length AB(base) = 2ba.a2n2
And height of triangle ABC = (a+n)
Now,
Area of triangle = 12bh
A=12.2ba.a2n2.(a+n)=aba2n2+bna2n2
Now,
dAdn=abna2n2+na2n2bn2a2n2
Now,
dAdn=0abna2n2+na2n2bn2a2n2=0abn+n(a2n2)bn2=0n=a,a2
but n cannot be zero
therefore, n=a2
Now, at n=a2
d2Adn2<0
Therefore, n=a2 is the point of maxima
Now,
b=2ba.a2(a2)2=3b
h=(a+n)=a+a2=3a2
Now,
Therefore, Area (A) =12bh=123b3a2=33ab4

Question:9 A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Answer:

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8 m3
h = 2m (given)
lb = 4 = l=4b
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
A(b)=4+2h(4b+b)
A(b)=2h(4b2+1)A(b)=02h(4b2+1)=0b2=4b=2
Now,
A(b)=2h(4×2bb3)A(2)=8>0
Hence, b = 2 is the point of minima
l=4b=42=2
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 = 4 m2
building of tank costs Rs 70 per sq metres for the base
Therefore, for 4 m2 Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) = 16 m2
building of tank costs Rs 45 per square metre for sides
Therefore, for 16 m2 Rs = 16 X 45 = 720 Rs
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs

Question:10 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer:

It is given that
the sum of the perimeter of a circle and square is k = 2πr+4a=ka=k2πr4
Let the sum of the area of a circle and square(A) = πr2+a2
A=πr2+(k2πr4)2
A(r)=2πr+2(k2πr16)(2π)A(r)=02π(8rk2πr8)=0r=k82π
Now,
A(r)=2π(82π8)=0A(k82π)>0
Hence, r=k82π is the point of minima
a=k2πr4=k2πk82π4=2k82π=2r
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Question:11 A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answer:

Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle (r=l2)
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle

= l+2b+πl2
1628072096595
l+2b+πl2=10l=2(102b)2+π
Area of window id given by (A) = lb+π2(l2)2
=2(102b)2+πb+π2(102b2+π)2
A(b)=208b2+π+π2.2(102b2+π).(2)2+π
=208b2+π2π(102b(2+π)2)A(b)=0208b2+π=2π(102b(2+π)2)40+20π16b8πb=20π4πb40=4b(π+4)b=10π+4
Now,
A(b)=82+π+4π(2+π)2=168π+4π(2+π)2=164π(2+π)2A(10π+4)<0
Hence, b = 5/2 is the point of maxima
l=2(102b)2+π=2(102.104+π)2+π=204+π
r=l2=202(4+π)=104+π
Hence, these are the dimensions of the window to admit maximum light through the whole opening

Question:12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a23+b23)32

Answer:

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle

1628072130108 Let the angle between AC and BC is θ
So, the angle between AD and ED is also θ
Now,
CD = b cosecθ
And
AD = asecθ
AC = H = AD + CD
= asecθ + b cosecθ
dHdθ=asecθtanθbcotθcosecθdHdθ=0asecθtanθbcotθcosecθ=0asecθtanθ=bcotθcosecθasin3θ=bcos3θtan3θ=batanθ=(ba)13
Now,
d2Hdθ2>0
When tanθ=(ba)13
Hence, tanθ=(ba)13 is the point of minima
secθ=aa23+b23a13 and cosecθ=ba23+b23b13

AC = aa23+b23a13+ ba23+b23b13 = (a23+b23)32
Hence proved

Question:13 Find the points at which the function f given by f(x)=(x2)4(x+1)3 has (i) local maxima (ii) local minima (iii) point of inflexion

Answer:

Given function is
f(x)=(x2)4(x+1)3
f(x)=4(x2)3(x+1)3+3(x+1)2(x2)4f(x)=04(x2)3(x+1)3+3(x+1)2(x2)4=0(x2)3(x+1)2(4(x+1)+3(x2))x=2,x=1 and x=27
Now, for value x close to 27 and to the left of 27 , f(x)>0 ,and for value close to 27 and to the right of 27 f(x)<0
Thus, point x = 27 is the point of maxima
Now, for value x close to 2 and to the Right of 2 , f(x)>0 ,and for value close to 2 and to the left of 2 f(x)<0
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point of inflexion

Question:14 Find the absolute maximum and minimum values of the function f given by
f(x)=cos2x+sinx,xϵ[0,π]

Answer: Given function is
f(x)=cos2x+sinx
f(x)=2cosx(sinx)+cosxf(x)=02cosxsinx+cosx=0cosx(12sinx)=0eithercosx=0      and       sinx=12x=π2          and          x=π6     as x ϵ[0,π]
Now,
f(x)=2(sinx)sinx2cosxcosx+(sinx)f(x)=2sin2x2cos2xsinxf(π6)=32<0
Hence, the point x=π6 is the point of maxima and the maximum value is
f(π6)=cos2π6+sinπ6=34+12=54
And
f(π2)=1>0
Hence, the point x=π2 is the point of minima and the minimum value is
f(π2)=cos2π2+sinπ2=0+1=1

Question:15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

Answer:

1628072169934 The volume of a cone (V) = 13πR2h
The volume of the sphere with radius r = 43πr3
By Pythagoras theorem in ΔADC we ca say that
OD2=r2R2OD=r2R2h=AD=r+OD=r+r2R2
V = 13πR2(r+r2+R2)=13πR2r+13πR2r2+R2
13πR2(r+r2R2)V(R)=23πRr+23πRr2R2+13πR2.2R2r2R2V(R)=013πR(2r+2r2R2R2r2R2)=013πR(2rr2R2+2r22R2R2r2R2)=0R0 So,2rr2R2=3R22r2Square both sides4r44r2R2=9R4+4r412R2r29R48R2r2=0R2(9R28r2)=0R0 So,9R2=8r2R=22r3
Now,
V(R)=23πr+23πr2R2+23πR.2R2r2R23πR2r2R2(1)(2R)(r2+R2)32V(22r3)<0
Hence, the point R=22r3 is the point of maxima
h=r+r2R2=r+r28r29=r+r3=4r3
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r3

Question:16 Let f be a function defined on [a, b] such that f(x)>0 , for all xϵ(a,b) . Then prove that f is an increasing function on (a, b).

Answer:

Let's do this question by taking an example
suppose
f(x)=x3>0,(a.b)
Now, also
f(x)=3x2>0,(a,b)
Hence by this, we can say that f is an increasing function on (a, b)

Question:17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R3 . Also, find the maximum volume.

Answer:

1628072213939 The volume of the cylinder (V) = πr2h
By Pythagoras theorem in ΔOAB
OA=R2r2
h = 2OA
h=2R2r2
V=2πr2R2r2
V(r)=4πrR2r2+2πr2.2r2R2r2V(r)=04πrR2r22πr3R2r2=04πr(R2r2)2πr3=06πr3=4πrR2r=6R3
Now,
V(r)=4πR2r2+4πr.2r2R2r26πr2R2r2.(1)2r2(R2r2)32V(6R3)<0
Hence, the point r=6R3 is the point of maxima
h=2R2r2==2R22R23=2R3
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R3
and maximum volume is
V=πr2h=π2R23.2R3=4πR333

Question:18 Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is

427πh3tan2α

Answer:

1628072251851 Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively

Volume of cylinder = πr2h
Volume of cone = 13πR2h
Now, we have
R=htana
Now, since ΔAOG\andΔCEG are similar
OAOG=CEEG
hR=hRr
h=h(Rr)R
h=h(htanar)htana=htanartana
Now,
V=πr2h=πr2.htanartana=πr2hπr3tana
Now,
dVdr=2πrh3πr2tanadVdr=02πrh3πr2tana=02πrh=3πr2tanar=2htana3
Now,
d2Vdr2=2πh6πrtana
at r=2htana3
d2Vdr2=2πh4πh<0
Hence, r=2htana3 is the point of maxima
h=htanartana=htana2htana3tana=13h
Hence proved
Now, Volume (V) at h=13h and r=2htana3 is
V=πr2h=π(2htana3)2.h3=427.πh3tan2a
hence proved

Question:19 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314
cubic metre per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

Answer:

It is given that
dVdt=314 m3/h
Volume of cylinder (V) = πr2h=100πh           (r=10m)
dVdt=100πdhdt314=100πdhdtdhdt=3.14π=1 m/h
Hence, (A) is correct answer

Question:20 The slope of the tangent to the curve x=t2+3t8,y=2t22t5 at the point
(2,– 1) is

A ) 22/7

B ) 6/7

C ) 7/6

D ) -6 /7

Answer:

Given curves are
x=t2+3t8 and y=2t22t5
At point (2,-1)
t2+3t8=2t2+3t10=0t2+5t2t10=0(t+5)(t2)=0t=2 and t=5
Similarly,
2t22t5=12t22t4=02t24t+2t4=0(2t+2)(t2)=0t=1 and t=2
The common value between two is t = 2
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a given point is given by dydx
dydt=4t2
dxdt=2t+3
(dydx)t=2=dydtdxdt=4t22t+3=824+3=67
Hence, (B) is the correct answer

Question:21 The line y is equal to mx+1 is a tangent to the curve y2=4x if the value of m is
(A) 1

(B) 2

(C) 3

(D)1/2

Answer:

Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by dydx
Given the equation of the curve is
y2=4x
2ydydx=4dydx=2y
Put this value of m in the given equation
y=2y.y24+1          (y2=4x and m=2y)y=y2+1y2=1y=2
m=2y=22=1
Hence, value of m is 1
Hence, (A) is correct answer

Question:22 T he normal at the point (1,1) on the curve 2y+x2=3 is
(A) x + y = 0

(B) x – y = 0

(C) x + y +1 = 0

(D) x – y = 1

Answer:

Given the equation of the curve
2y+x2=3
We know that the slope of the tangent at a point on the given curve is given by dydx
2dydx=2xdydx=x
We know that
Slope of normal=1Slope of tangent=1x=1x
At point (1,1)
Slope=11=1
Now, the equation of normal with point (1,1) and slope = 1

yy1=m(xx1)y1=1(x1)xy=0
Hence, the correct answer is (B)

Question:23 The normal to the curve x2=4y passing (1,2) is

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

Answer:

Given the equation of the curve
x2=4y
We know that the slope of the tangent at a point on the given curve is given by dydx
4dydx=2x dydx=x2
We know that
Slope of normal=1Slope of tangent=1x2=2x
At point (a,b)
Slope=2a
Now, the equation of normal with point (a,b) and Slope=2a

?
It is given that it also passes through the point (1,2)
Therefore,
2b=2a(1a)2aba=2a2ba=2b=2a -(i)
It also satisfies equation x2=4yb=a24 -(ii)
By comparing equation (i) and (ii)
2a=a24a3=8a=2
b=2a=22=1
Slope=2a=22=1

Now, equation of normal with point (2,1) and slope = -1

yy1=m(xx1)y1=1(x2)y+x=3
Hence, correct answer is (A)

Question:24 The points on the curve 9y2=x3 , where the normal to the curve makes equal intercepts with the axes are

A)(4,±83).B)(4,83).C)(4,±38).D)(±4,38)

Answer:

Given the equation of the curve
9y2=x3
We know that the slope of the tangent at a point on a given curve is given by dydx
18ydydx=3x2dydx=x26y
We know that
Slope of normal=1Slope of tangent=1x26y=6yx2
At point (a,b)
Slope=6ba2
Now, the equation of normal with point (a,b) and Slope=6ba2

yy1=m(xx1)yb=6ba2(xa)ya2ba2=6bx+6abya2+6bx=6ab+a2by6b+aba+x6a+a26=1
It is given that normal to the curve makes equal intercepts with the axes
Therefore,
6b+aba=6a+a266b(6+a)=a2(6+a)a2=6b
point(a,b) also satisfy the given equation of the curve
9b2=a39(a26)2=a39.a436=a3a=4
9b2=439b2=64b=±83
Hence, The points on the curve 9y2=x3 , where the normal to the curve makes equal intercepts with the axes are (4,±83)
Hence, the correct answer is (A)

If you are looking for application of derivatives class 12 ncert solution of exercises then they are listed below.

More about class 12 application-of-derivatives ncert solutions

If you are good at differentiation, NCERT Class 12 maths chapter 6 alone has 11% weightage in 12 board final examinations, which means you can score very easily with basic knowledge of maths and basic differentiation. After going through class 12 maths ch 6, you can build your concepts to score well in exams.

Class 12 maths chapter 6 seems to be very easy but there are chances of silly mistakes as it requires knowledge of other chapters also. So, practice all the NCERT questions on your own, you can take help of these NCERT solutions for class 12 maths chapter 6 application of derivatives. There are five exercises with 102 questions in chapter 6 class 12 maths. All these questions are explained in this Class 12 maths chapter 6 NCERT solutions article.

Also read,

What is the derivative?

The derivative dS/dt is the rate of change of distance(S) with respect to the time(t). In a similar manner, whenever one quantity (y) varies with another quantity (x), and also satisfy y=f(x) ,then \frac{dy}{dx} or f^{'}(x) represents the rate of change of y with respect to x and \dpi{100} \frac{dy}{dx} ]_{x=x_{o}} or f^{'}(x_{o}) represents the rate of change of y with respect to x at x=x_{o} . Let's take an example of a derivative

Example- Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm.
Solution- The area A of a circle with radius r is given by A=\pi r^{2} . Therefore, the rate of change of the area (A) with respect to its radius(r) is given by - \frac{dA}{dr}=\frac{d}{dr}(\pi r^{2})=2\pi r When r=5cm,\:\:\: \frac{dA}{dr}=10\pi Thus, the area of the circle is changing at the rate of 10\pi \:\:cm^2/s

Application-Of-Derivatives Class 12 NCERT solutions - Topics

6.1 Introduction

6.2 Rate of Change of Quantities

6.3 Increasing and Decreasing Functions

6.4 Tangents and Normals

6.5 Approximations

6.6 Maxima and Minima

6.6.1 Maximum and Minimum Values of a Function in a Closed Interval

NCERT solutions for class 12 maths - Chapter wise

NCERT solutions for class 12 subject wise

NCERT solutions class wise

NCERT Solutions for Class 12 maths chapter 6 PDFs are very helpful for the preparation of this chapter. Here are some tips to get command on this application of derivatives solutions.

NCERT Class 12 maths chapter 6 Tips

  • First cover the differentials and then go for its applications.

  • Solve the NCERT problems first with examples, NCERT Solutions for Class 12 maths chapter 6 PDF will help in this.

  • Try to make figures first and label it, if required. This will help in solving the problem easily.

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. What are the Important Topics Covered in NCERT Solutions Class 12 Maths Chapter 6?

NCERT maths chapter 6 class 12 solutions outlines the crucial uses of derivatives. The NCERT Solutions for Class 12 Maths Chapter 6 covers concepts such as utilizing derivatives to calculate the rate of change of quantities, determining ranges, and finding the equation of tangent and normal lines to a curve at a particular point. The ultimate goal of these solutions is to encourage students to practice and enhance their mathematical skills, aiding their overall academic progress.

2. Where can I find the complete solutions of NCERT Class 12 maths chapter 6?

you can directly download by clicking on the given link NCERT solutions for class 12 Maths. you can also get these solutions freely from careers360 official website. these solutions are make you comfortable with applications of derivative's problems and build your confidence that help you in exam to score well.

3. Can you provide a brief summary of class 12 maths chapter 6 solutions?

maths chapter 6 class 12 ncert solutions includes six main topics and a miscellaneous section with questions and answers at the end. The topics covered in this chapter are:

  • 6.1 - Introduction

  • 6.2 - Rate of Change of Quantities

  • 6.3 - Increasing and Decreasing Functions

  • 6.4 - Tangents and Normals

  • 6.5 - Approximations

  • 6.6 - Maxima and Minima

ch 6 maths class 12 ncert solutions are very important to get good hold in these topics.

4. What is the weightage of the chapter Application of Derivatives for CBSE board exam?

Application of derivatives has 11% weightage in the CBSE 12th board final exam. Having good weightage this chapter become more important for board as well as some premiere exams like JEE Main and JEE Advance. Therefor it is advise to students to make good hold on the concepts of this chapter.

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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