NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

 

NCERT solutions for class 12 maths chapter 6 Application of Derivatives: 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 solutions for class 12 maths chapter 6 application of derivatives. The derivative has many applications in various fields like social science, physics, optimization, science, engineering etc. CBSE NCERT solutions for class 12 maths chapter 6 applications of derivatives will cover questions on some specific applications like graph, functions and approximations. Questions based on the topics like finding the rate of change of quantities, equations of tangent and normal on a curve at a point, local minima and maxima of a function, intervals on which a function is increasing or decreasing, and approximate value of certain quantities are covered in the solutions of NCERT class 12 maths chapter 6 application of derivatives. Check all NCERT solutions at a single place which will help you to learn CBSE maths and science.

If you are good in differentiation, it won't take much effort to learn its applications. This chapter alone has 11% weightage in 12 board final examination, which means you can score very easily by your basic knowledge of maths and basic differentiation. This chapter 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 CBSE NCERT solutions for class 12 maths chapter 6 application of derivatives. There are five exercises with 102 questions. All these questions are explained in this solutions of NCERT class 12 maths chapter 6 application of derivatives article. 

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

 

Topics of NCERT Grade 12 Maths Chapter-6 Application of Derivatives

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 6 Application of Derivatives-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) =  \pi r^{2}
Rate of change of the area of a circle with respect to its radius r = \frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2 \pi r
So, when r = 3, Rate of change of the area of a circle = 2 \pi (3)  = 6 \pi
Hence, Rate of change of the area of a circle with respect to its radius r when r = 3 is  6 \pi

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) =  \pi r^{2}
Rate of change of the area of a circle with respect to its radius r = \frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2 \pi r
So, when r = 4, Rate of change of the area of a circle = 2 \pi (4)  = 8 \pi
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is  8 \pi

Question:2. The volume of a cube is increasing at the rate of 8 cm^3 /s. How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

The volume of the cube(V) =  x^{3}    where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of 8 cm^3 /s

we can write  \frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}       ( By chain rule)

\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}     

\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8      \Rightarrow 3x^{2}.\frac{dx}{dt} = 8

\frac{dx}{dt} = \frac{8}{3x^{2}}                                                        - (i)
Now, we know that the surface area of the cube(A) is   6x^{2}

\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt}                      - (ii)

from equation (i) we know that  \frac{dx}{dt} = \frac{8}{3x^{2}}

put this value in equation (i)
We get,
              \frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}
It is given in the question that the value of  edge length(x) = 12cm 
So,
    \frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} cm^2/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 \left ( \frac{dr}{dt} \right ) =  3 cm/s 
Area of circle(A) = \pi r^{2}
\frac{dA}{dt} =\frac{dA}{dr}.\frac{dr}{dt}                            (by chain rule)
\frac{dA}{dt} =\frac{d \pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 3 = 6\pi r
It is given that the value of r = 10 cm
So,
       \frac{dA}{dt} = 6\pi \times 10 = 60\pi \ cm^{2}/s
Hence,  the rate at which the area of the circle is increasing when the radius is 10 cm  is  60\pi \ cm^{2}/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  \left ( \frac{dx}{dt} \right )  = 3 cm/s
The volume of cube = x^{3} 
\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}             (By chain rule)
\frac{dV}{dt} = \frac{dx^{3}}{dx}.\frac{dx}{dt} = 3x^{2}.\frac{dx}{dt} = 3x^{2}\times 3 = 9x^{2} cm^{3}/s
It is given that the value of x is 10 cm 
So, 
      \frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900 \ cm^{3}/s
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is  900 \ cm^{3}/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 =   \frac{dr}{dt} = 5 \ cm/s

To find =  \frac{dA}{dt}    at  r = 8 cm
 
Area of the circle (A) = \pi r^{2}
\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt}                         (by chain rule)
\frac{dA}{dt} = \frac{d\pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ cm^{2}/s
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is   80\pi \ cm^{2}/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 =  \frac{dr}{dt} = 0.7 \ cm/s
To find =   \frac{dC}{dt}            , where C is circumference
Solution :-
 
we know that the circumference of the circle (C) = 2\pi r
\frac{dC}{dt} = \frac{dC}{dr}.\frac{dr}{dt}                              (by chain rule)
\frac{dC}{dt} = \frac{d2\pi r}{dr}.\frac{dr}{dt} = 2\pi \times 0.7 = 1.4\pi \ cm/s
Hence,  the rate of increase of its circumference is 1.4\pi \ 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 (\frac{dx}{dt})  =   -5  cm/minute                  (-ve sign indicates decrease in rate)
                the width y is increasing at the rate   (\frac{dy}{dt}) = 4 cm/minute
To find =  \frac{dP}{dt}    and      at  x = 8 cm and y = 6 cm                     , where P is perimeter 
Solution:-

Perimeter of rectangle(P) = 2(x+y)
\frac{dP}{dt} = \frac{d(2(x+y))}{dt} = 2\left ( \frac{dx}{dt} + \frac{dy}{dt} \right ) = 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
      \frac{dA}{dt} = \frac{d(xy)}{dt} = \left ( x\frac{dy}{dt} + y\frac{dx}{dt} \right ) = \left ( 8\times 4 + 6 \times (-5) \right ) = (32 -30) = 2 \ cm^{2}/minute
Hence, the rate of change of area is  2 \ cm^{2}/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 =  \frac{dV}{dt} = 900 \ cm^{3}/s
To find =   \frac{dr}{dt}   at r = 15 cm
Solution:-

Volume of sphere(V) =  \frac{4}{3}\pi r^{3}
\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})} {dr}.\frac{dr}{dt} = \frac{4}{3}\pi\times 3r^{2} \times \frac{dr}{dt}
                               
\frac{dV}{dt}= 4 \pi r^{2} \times \frac{dr}{dt}    
\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times(15)^{2}} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/s
Hence,  the rate at which the radius of the balloon increases when the radius is 15 cm is  \frac{1}{\pi} \ 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 \frac{dV}{dr}  at r =10 cm
The volume of the sphere (V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dr} = \frac{d(\frac{4}{3}\pi r^{3})}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi \ cm^{3}/s
Hence,  the rate at which its volume is increasing with the radius when the later is 10 cm is 400\pi \ cm^{3}/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 \frac{dx}{dt} = 2 \ cm/s
We need to find  the rate at  which the height of the ladder decreases (\frac{dh}{dt})
length of ladder(L) = 5m and x = 4m    (given)
By Pythagoras theorem, we can say that 
h^{2}+x^{2} = L^{2}
h^{2} = L^{2} - x^{2}
  h  = \sqrt{L^{2} - x^{2}}
Differentiate on both sides w.r.t.  t
\frac{dh}{dt} = \frac{d(\sqrt{L^{2} -x^{2}})}{dx}.\frac{dx}{dt} = \frac{1}{2}\frac{-2x}{\sqrt{5^{2}-x^{2}}}.\frac{dx}{dt} = \frac{-x}{\sqrt{25-x^{2}}}\frac{dx}{dt}
at x = 4

\frac{dh}{dt} = \frac{-4}{\sqrt{25-16}}\times 2 = \frac{-4}{3} \times 2 =\frac{-8}{3} \ cm/s
Hence, the rate at which the height of ladder decreases is \frac{8}{3} \ cm/s

Question:11. A particle moves along the curve 6y = x^3 + 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 \frac{dy}{dt} = 8\frac{dx}{dt}
Given the equation of curve =   6y = x^3 + 2
Differentiate both sides w.r.t. t
6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0
           = 3x^{2}.\frac{dx}{dt}
\frac{dy}{dt} = 8\frac{dx}{dt}   (required condition)
6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}
3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}\Rightarrow x^{2} = \frac{48}{3} = 16
x = \pm 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,\frac{-31}{3})     

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 \frac{dr}{dt} = \frac{1}{2} \ cm/s 
We know that the shape of the air bubble is spherical
So, volume(V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ cm^{3}/s
Hence, the rate of change in volume is   2\pi \ cm^{3}/s

Question:13 A balloon, which always remains spherical, has a variable diameter \frac{3}{2}( 2x +1) Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = \frac{4}{3}\pi r^{3}
Diameter = \frac{3}{2}(2x+1)
So, radius(r) = \frac{3}{4}(2x+1)
\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d(\frac{4}{3}\pi (\frac{3}{4}(2x+1))^{3})}{dx} = \frac{4}{3}\pi\times 3\times\frac{27}{64}(2x+1)^{2}\times 2
                                                                               = \frac{27}{8}\pi (2x+1)^{2}

Question:15 The total cost C(x) in Rupees associated with the production of x units of an
item is given by C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = \frac{dC}{dx} 
C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000
\frac{dC}{dx} = \frac{d(.007x^{3} - .003x^{2}+15x+400)}{dx} = 3\times .007x^{2} - 2\times.003x+15
                                                                                    = .021x^{2} - .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) = 13 x^2 + 26 x + 15

Find the marginal revenue when x = 7

Answer:

Marginal revenue =  \frac{dR}{dx}
R ( x) = 13 x^2 + 26 x + 15
\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)
at x = 7
\frac{dR}{dx} = 26(7+1) = 26\times8 = 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) = \pi r^{2}
\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r
Now, at r = 6cm
\frac{dA}{dr}= 2\pi \times 6 = 12\pi cm^{2}/s
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is 12\pi cm^{2}/s
Hence, the correct answer is B

Solutions of NCERT for class 12 maths chapter 6 Applications of Derivatives-Exercise: 6.2

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

Answer:

Let   x_1 and x_2  are two numbers in R
x_1 < x_2 \Rightarrow 3x_1 < 3 x_2 \Rightarrow 3x_1 + 17 < 3x_2+17 \Rightarrow f(x_1)< f(x_2)
Hence, f is strictly increasing on R

Question:2. Show that the function given by f(x) = e^{2x} is increasing on R.

Answer:

Let x_1 \ and \ x_2 are two numbers in R
x_1 \ < \ x_2 \Rightarrow 2x_1 < 2x_2 \Rightarrow e^{2x_1} < e^{2x_2} \Rightarrow f(x_1) < f(x_2)
Hence, the function f(x) = e^{2x} is  strictly increasing in R

Question:3 a) Show that the function given by f (x) = \sin x is  increasing in \left ( 0 , \pi /2 \right )

Answer:

Given f(x) = sinx
f^{'}(x) = \cos x
Since,     \cos x > 0 \ for \ each \ x\ \epsilon \left ( 0,\frac{\pi}{2} \right )
f^{'}(x) > 0
Hence, f(x) = sinx is strictly increasing in   \left ( 0,\frac{\pi}{2} \right )

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

decreasing in \left ( \frac{\pi}{2},\pi \right )

Answer:

f(x) = sin x
f^{'}(x) = \cos x
Since,  \cos x < 0  for each x \ \epsilon \left ( \frac{\pi}{2},\pi \right )
So, we have    f^{'}(x) < 0
Hence, f(x) = sin x is strictly decreasing in \left ( \frac{\pi}{2},\pi \right )

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

Answer:

We know that sin x is strictly increasing in \left ( 0,\frac{\pi}{2} \right )   and strictly decreasing in \left ( \frac{\pi}{2},\pi \right )
So, by this, we can say that f(x) = sinx is neither increasing or decreasing in range  \left ( 0,\pi \right )

Question:4(a). Find the intervals in which the function f given by f ( x) = 2x ^2 - 3 x is   increasing

Answer:

f ( x) = 2x ^2 - 3 x
f^{'}(x) = 4x - 3
Now,
   f^{'}(x) = 0
 4x - 3 = 0
     x = \frac{3}{4}

So, the range is  \left ( -\infty, \frac{3}{4} \right ) \ and \ \left ( \frac{3}{4}, \infty \right )
So,
        f(x)< 0   when  x \ \epsilon \left ( -\infty,\frac{3}{4} \right )        Hence, f(x) is strictly decreasing in this range
and
      f(x) > 0     when  x \epsilon \left ( \frac{3}{4},\infty \right )             Hence, f(x) is strictly increasing in this range
Hence, f ( x) = 2x ^2 - 3 x   is strictly increasing in   x \epsilon \left ( \frac{3}{4},\infty \right )

Question:4(b) Find the intervals in which the function f given by f ( x) = 2 x ^2 - 3 x is
decreasing
 

Answer:

f ( x) = 2x ^2 - 3 x
f^{'}(x) = 4x - 3
Now,
&nnbsp;  f^{'}(x) = 0
 4x - 3 = 0
     x = \frac{3}{4}

So, the range is  \left ( -\infty, \frac{3}{4} \right ) \ and \ \left ( \frac{3}{4}, \infty \right )
So,
        f(x)< 0   when  x \ \epsilon \left ( -\infty,\frac{3}{4} \right )        Hence, f(x) is strictly decreasing in this range
and
      f(x) > 0     when  x \epsilon \left ( \frac{3}{4},\infty \right )             Hence, f(x) is strictly increasing in this range
Hence, f ( x) = 2x ^2 - 3 x   is strictly decreasing  in   x \epsilon \left ( -\infty ,\frac{3}{4}\right )

Question:5(a) Find the intervals in which the function f given by f (x) = 2x^3 - 3x ^2 - 36 x + 7 is
increasing 

Answer:

It is given that
f (x) = 2x^3 - 3x ^2 - 36 x + 7
So,
f^{'}(x)= 6x^{2} - 6x - 36
f^{'}(x)= 0
6x^{2} - 6x -36 =0 \Rightarrow 6 (x^{2} - x-6)
x^{2} - x-6 = 0
x^{2} - 3x+2x-6 = 0
x(x-3) + 2(x-3) = 0\\
(x+2)(x-3) = 0
x = -2 ,  x = 3

So, three ranges are there   (-\infty,-2) , (-2,3) \ and \ (3,\infty)
Function    f^{'}(x)= 6x^{2} - 6x - 36  is positive in interval  (-\infty,-2) , (3,\infty)   and negative in the interval  (-2,3)
Hence, f (x) = 2x^3 - 3x ^2 - 36 x + 7  is strictly increasing in (-\infty,-2) \cup (3,\infty)
and  strictly decreasing in the interval  (-2,3) 

Question:5(b) Find the intervals in which the function f given by f ( x) = 2x ^3 - 3x ^2 - 36x + 7 is
  decreasing

Answer:

We have        f ( x) = 2x ^3 - 3x ^2 - 36x + 7

Differentiating the function with respect to x,  we get  :

                       f' ( x) = 6x ^2 - 6x - 36

or                                = 6\left ( x-3 \right )\left ( x+2 \right )

When          f'(x)\ =\ 0,   we have  :

                                              0\ = 6\left ( x-3 \right )\left ( x+2 \right )

or                                         \left ( x-3 \right )\left ( x+2 \right )\ =\ 0


So, three ranges are there   (-\infty,-2) , (-2,3) \ and \ (3,\infty)
Function    f^{'}(x)= 6x^{2} - 6x - 36  is positive in the interval  (-\infty,-2) , (3,\infty)   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:
 x ^2 + 2x -5

Answer:

f(x)  = x ^2 + 2x -5
f^{'}(x) = 2x + 2 = 2(x+1)
Now, 
f^{'}(x) = 0 \\ 2(x+1) = 0\\ x = -1

The range is from (-\infty,-1) \ and \ (-1,\infty)
In interval (-\infty,-1)   f^{'}(x)= 2(x+1) is -ve 
Hence, function f(x)  = x ^2 + 2x -5 is strictly decreasing in interval  (-\infty,-1)
In interval (-1,\infty)  f^{'}(x)= 2(x+1) is +ve
Hence, function f(x)  = x ^2 + 2x -5 is strictly increasing in interval  (-1,\infty)

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

 10 - 6x - 2x^2

Answer:

Given function is,
f(x) = 10 - 6x - 2x^2
f^{'}(x) = -6 - 4x
Now,
f^{'}(x) = 0
6+4x= 0
x= -\frac{3}{2}

So, the  range is (-\infty , -\frac{3}{2}) \ and \ (-\frac{3}{2},\infty)
In interval (-\infty , -\frac{3}{2})  ,  f^{'}(x) = -6 - 4x is +ve
Hence, f(x) = 10 - 6x - 2x^2 is strictly increasing in the interval  (-\infty , -\frac{3}{2})
In interval ( -\frac{3}{2},\infty)  , f^{'}(x) = -6 - 4x is -ve
Hence, f(x) = 10 - 6x - 2x^2 is strictly decreasing in interval  ( -\frac{3}{2},\infty)

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

- 2 x^3 - 9x ^2 - 12 x + 1

Answer:

Given function is,
f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}
f^{'}(x) = - 6 x^2 - 18x - 12
Now,
f^{'}(x) = 0\\ - 6 x^2 - 18x - 12 = 0\\ -6(x^{2}+3x+2) = 0 \\ x^{2}+3x+2 = 0 \\x^{2} + x + 2x + 2 = 0\\ x(x+1) + 2(x+1) = 0\\ (x+2)(x+1) = 0\\ x = -2 \ and \ x = -1

So, the range is (-\infty , -2) \ , (-2,-1) \ and \ (-1,\infty)
In interval  (-\infty , -2) \cup \ (-1,\infty)  , f^{'}(x) = - 6 x^2 - 18x - 12 is -ve
Hence, f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{} is strictly decreasing in interval  (-\infty , -2) \cup \ (-1,\infty)
In interval (-2,-1)  , f^{'}(x) = - 6 x^2 - 18x - 12 is +ve
Hence, f(x) = - 2 x^3 - 9x ^2 - 12 x + 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:

 6- 9x - x ^2

Answer:

Given function is,
f(x) = 6- 9x - x ^2
f^{'}(x) = - 9 - 2x
Now,
f^{'}(x) = 0\\ - 9 - 2x = 0 \\ 2x = -9\\ x = -\frac{9}{2}

So, the range is  (-\infty, - \frac{9}{2} ) \ and \ ( - \frac{9}{2}, \infty )
In interval (-\infty, - \frac{9}{2} ) ,  f^{'}(x) = - 9 - 2x  is +ve
Hence,  f(x) = 6- 9x - x ^2  is strictly increasing in interval  (-\infty, - \frac{9}{2} )
In interval ( - \frac{9}{2},\infty ) , f^{'}(x) = - 9 - 2x is -ve
Hence,  f(x) = 6- 9x - x ^2  is strictly decreasing in interval ( - \frac{9}{2},\infty )

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

 ( x+1) ^3 ( x-3) ^3

Answer:

Given function is,
f(x) = ( x+1) ^3 ( x-3) ^3
f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3
Now,
f^{'}(x) = 0 \\ 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^{3} \\ 3(x+1)^{2}(x-3)^{2}((x-3) + (x+1) ) = 0 \\ (x+1)(x-3) = 0 \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ (2x-2) = 0\\ x=-1 \ and \ x = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ x = 1
So, the intervals are  (-\infty,-1), (-1,1), (1,3) \ and \ (3,\infty)

Our function f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3   is +ve in the interval  (1,3) \ and \ (3,\infty)
Hence,  f(x) = ( x+1) ^3 ( x-3) ^3  is strictly increasing in the interval  (1,3) \ and \ (3,\infty)
Our function f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3   is -ve in the interval  (-\infty,-1) \ and \ (-1,1)
Hence,  f(x) = ( x+1) ^3 ( x-3) ^3  is strictly decreasing in interval  (-\infty,-1) \ and \ (-1,1)

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

Answer:

Given function is,
 f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }
f^{'}(x)\Rightarrow \frac{dy}{dx} = \frac{1}{1+x} - \frac{2 (2+x) - (1)(2x)} {(2+x)^{2} } = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^{2}}
                                                                                          = \frac{1}{1+x} - \frac{4}{(2+x)^2} = \frac{(2+x)^2 - 4(x+1)}{(x+1)(2+x)^{2}}
                                                                                          = \frac{4+x^{2} +4x -4x - 4}{(x+1)(2+x)^{2}} = \frac{x^{2} }{(x+1)(2+x)^{2}}
f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2}
Now, for  x > -1  , is is clear that   f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2} > 0  
Hence,   f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }  strictly increasing when  x > -1             

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

Answer:

Given function is,
f(x)\Rightarrow y = [x(x-2)]^{2}
f^{'}(x)\Rightarrow \frac{dy}{dx} = 2[x(x-2)][(x-2)+x]
                           = 2(x^2-2x)(2x-2)
                           = 4x(x-2)(x-1)
Now,
 f^{'}(x) = 0\\ 4x(x-2)(x-1) = 0\\ x=0 , x= 2 \ and \ x = 1
So, the intervals are (-\infty,0),(0,1),(1,2) \ and \ (2,\infty)
In interval  (0,1)and \ (2,\infty) ,  f^{'}(x)> 0
Hence, f(x)\Rightarrow y = [x(x-2)]^{2} is an increasing function in the interval  (0,1)\cup (2,\infty)

Question:9 Prove that   y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is an increasing function of    \theta\: \: in\: \: [ 0 , \pi /2 ]

Answer:

Given function is,
f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta

f^{'}(x) = \frac{dy}{d\theta} = \frac{4 \cos \theta(2+\cos \theta) - (-\sin \theta)4\sin \theta) }{(2+ \cos \theta )^2} - 1
                         = \frac{8 \cos \theta+4\cos^2 \theta + 4\sin^2 \theta - (2+ \cos \theta )^2 }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4(\cos^2 \theta + \sin^2 \theta) - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4 - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}                                                 
                         = \frac{4 \cos \theta-\cos^2 \theta }{(2+ \cos \theta )^2}
Now,  for  \theta \ \epsilon \ [0,\frac{\pi}{2}]
\\ 4 \cos \theta \geq \cos^2 \theta\\ 4 \cos \theta - \cos^2 \geq 0\\ and \ (2+\cos \theta)^2 > 0
So, f^{'}(x) > 0 \ for \ \theta \ in \ [0,\frac{\pi}{2}]
Hence, f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is increasing function in  \theta \ \epsilon \ [0,\frac{\pi}{2}]    

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

Answer:

Let logarithmic function is log x
f(x) = log x
f^{'}(x) = \frac{1}{x}
Now, for all values of x in ( 0 , \infty ) , f^{'}(x) > 0
Hence,  the logarithmic function f(x) = log x  is increasing in the interval  ( 0 , \infty )

Question:11 Prove that the function f given by f ( x) = x ^2 - x + 1  is neither strictly increasing nor decreasing on (– 1, 1).

Answer:

Given function is,
f ( x) = x ^2 - x + 1
f^{'}(x) = 2x - 1
Now, for interval (-1,\frac{1}{2}) ,  f^{'}(x) < 0        and for interval  (\frac{1}{2},1),f^{'}(x) > 0
Hence, by this, we can say that  f ( x) = x ^2 - x + 1  is neither strictly increasing nor decreasing in the interval (-1,1) 

Question:12 Which of the following functions are decreasing on 0 , \pi /2 (A) \cos x \\(B) \cos 2x \\ (C) \cos 3x \\ (D) \tan x

Answer:

(A)  
f(x) = \cos x \\ f^{'}(x) = -\sin x
f^{'}(x) < 0   for x in (0,\frac{\pi}{2})
Hence, f(x) = \cos x is decreasing function in  (0,\frac{\pi}{2})

(B)  
f(x) = \cos 2x \\ f^{'}(x) = -2\sin2 x
Now, as
0 < x < \frac{\pi}{2}\\ 0 < 2x < \pi
f^{'}(x) < 0   for 2x in (0,\pi)
Hence, f(x) = \cos 2x is decreasing function in  (0,\frac{\pi}{2})

(C)
   f(x) = \cos 3x \\ f^{'}(x) = -3\sin3 x
Now, as
 0 < x < \frac{\pi}{2}\\ 0 < 3x < \frac{3\pi}{2}
f^{'}(x) < 0  for x \ \epsilon \ \left ( 0,\frac{\pi}{3} \right )   and    f^{'}(x) > 0 \ x \ \epsilon \ \left ( \frac{\pi}{3} , \frac{\pi}{2}\right )
Hence, it is clear that  f(x) = \cos 3x is neither increasing nor decreasing in  (0,\frac{\pi}{2})

(D)
f(x) = \tan x\\ f^{'}(x) = \sec^{2}x
f^{'}(x) > 0   for x in (0,\frac{\pi}{2})
Hence, f(x) = \tan x is strictly increasing function in the interval  (0,\frac{\pi}{2})

So, only (A) and (B) are decreasing functions in  (0,\frac{\pi}{2})    

Question:13 On which of the following intervals is the function f given by f ( x) = x ^{100} + \sin x - 1 decreasing ?
(A) (0,1)                      (B) \frac{\pi}{2},\pi                    (C)  0,\frac{\pi}{2}                      (D) None of these

Answer:

(A) Given function is,
f ( x) = x ^{100} + \sin x - 1
f^{'}(x) = 100x^{99} + \cos x
Now, in interval (0,1)
f^{'}(x) > 0 
Hence, f ( x) = x ^{100} + \sin x - 1  is increasing function in interval  (0,1)

(B)  Now, in interval \left ( \frac{\pi}{2},\pi \right )
100x^{99} > 0 \ but \ \cos x < 0
100x^{99} > \cos x \\ 100x^{99} - \cos x > 0      ,       f^{'}(x) > 0
Hence, f ( x) = x ^{100} + \sin x - 1  is increasing function in interval  \left ( \frac{\pi}{2},\pi \right )

(C)  Now, in interval \left ( 0,\frac{\pi}{2} \right )
100x^{99} > 0 \ and \ \cos x > 0
100x^{99} > \cos x \\ 100x^{99} - \cos x > 0   ,  f^{'}(x) > 0
Hence, f ( x) = x ^{100} + \sin x - 1  is increasing function in interval  \left ( 0,\frac{\pi}{2} \right )

So,  f ( x) = x ^{100} + \sin x - 1 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) = x^2 + ax + 1  is increasing on
[1, 2]?

Answer:

Given function is,
f (x) = x^2 + 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) = x^2 + 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) = 1 - \frac{1}{x^2}
Now,  
f^{'}(x) = 0\\ 1 - \frac{1}{x^2} = 0\\ x^{2} = 1\\ x = \pm1

So, intervals are from  (-\infty,-1), (-1,1) \ and \ (1,\infty)
In interval (-\infty,-1), (1,\infty)  ,  \frac{1}{x^2} < 1 \Rightarrow 1 - \frac{1}{x^2} > 0
f^{'}(x) > 0
Hence, f ( x) = x + 1/x is increasing   in  interval  (-\infty,-1)\cup (1,\infty)
In interval (-1,1) ,  \frac{1}{x^2} > 1 \Rightarrow 1 - \frac{1}{x^2} < 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) = \log \sin x  is increasing on 

\left ( 0 , \pi /2 \right )\: \: and \: \: decreasing \: \: on \: \: \left ( \pi/2 , \pi \right )
Answer:

Given function is,
f (x) = \log \sin x
f^{'}(x) = \frac{1}{\sin x}\cos x = \cot x
Now, we know that  cot x is+ve in the interval  \left ( 0 , \pi /2 \right )   and -ve  in the interval \left ( \pi/2 , \pi \right )
f^{'}(x) > 0 \ in \ \left ( 0,\frac{\pi}{2} \right ) \ and \ f^{'}(x) < 0 \ in \ \left ( \frac{\pi}{2} , \pi \right )
Hence, f (x) = \log \sin x  is increasing in the interval \left ( 0 , \pi /2 \right ) and   decreasing in interval  \left ( \pi/2 , \pi \right )

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

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) = \frac{1}{\cos x}(-\sin x) = -\tan x
We know that  in interval \left ( 0,\frac{\pi}{2} \right )  ,  \tan x > 0 \Rightarrow -\tan x< 0  
f^{'}(x) < 0 
Hence, f(x) =  log|cos x| is decreasing in interval   \left ( 0,\frac{\pi}{2} \right )

We know that  in interval  \left ( \frac{3\pi}{2},2\pi \right )  , \tan x < 0 \Rightarrow -\tan x> 0
f^{'}(x) > 0
Hence, f(x) =  log|cos x| is increasing in interval  \left ( \frac{3\pi}{2},2\pi \right )

Question:18 Prove that the function given by f (x) = x^3 - 3x^2 + 3x - 100 is increasing in R.

Answer:

Given function is,
f (x) = x^3 - 3x^2 + 3x - 100
f^{'}(x) = 3x^2 - 6x + 3
             = 3(x^2 - 2x + 1) = 3(x-1)^2
f^{'}(x) = 3(x-1)^2
We can clearly see that for any value of x in R f^{'}(x) > 0
Hence, f (x) = x^3 - 3x^2 + 3x - 100  is an increasing function in R

Question:19 The interval in which y = x ^2 e ^{-x}is increasing is

(A) ( - \infty , \infty ) (B) ( - 2 , 0 )(C) ( - 2 , \infty ) (D) ( 0, 2 )

Answer:

Given function is,
f(x) \Rightarrow y = x ^2 e ^{-x}
f^{'}(x) \Rightarrow \frac{dy}{dx} = 2x e ^{-x} + -e^{-x}(x^{2})
                            xe ^{-x}(2 -x)
f^{'}(x) = xe ^{-x}(2 -x)
Now, it is clear that f^{'}(x) > 0  only in the interval (0,2)
So,  f(x) \Rightarrow y = x ^2 e ^{-x} is  an increasing function for the interval  (0,2)
Hence, (D) is the answer

CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.3

Question:1. Find the slope of the tangent to the curve y = 3 x ^4 - 4x \: \: at \: \: x \: \: = 4

Answer:

Given curve is,
y = 3 x ^4 - 4x
Now, the slope of the tangent at point  x =4  is given by
\left ( \frac{dy}{dx} \right )_{x=4} = 12x^3 - 4
                      = 12(4)^3-4
                      = 12(64)-4 = 768 - 4 =764

Question:2Find the slope of the tangent to the curve   \frac{x-1}{x-2} , x \neq 2 \: \: at\: \: x = 10

Answer:

Given curve is,

y = \frac{x-1}{x-2}
The slope of the tangent at x = 10 is given by
\left ( \frac{dy}{dx} \right )_{x=10}= \frac{(1)(x-2)-(1)(x-1)}{(x-2)^2} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}
 at x = 10
                       = \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}
hence, slope of tangent at x = 10 is \frac{-1}{64}

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

Answer:

Given curve is,
y = x ^3 - x +1
The slope of the tangent at x = 2 is given by
\left ( \frac{dy}{dx} \right )_{x=2} = 3x^2 - 1 = 3(2)^2 - 1= 3\times 4 - 1 = 12 - 1 = 11
Hence, the slope of the tangent at point x = 2 is 11
 

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

Answer:

Given curve is,
y = x ^3 - 3x +2
The slope of the tangent at x = 3 is given by
\left ( \frac{dy}{dx} \right )_{x=3} = 3x^2 - 3 = 3(3)^2 - 3= 3\times 9 - 3 = 27 - 3 = 24
Hence, the slope of tangent at point x = 3 is 24

Question:5 Find the slope of the normal to the curve x = a \cos ^3 \theta , y = a\sin ^3 \theta \: \: at \: \: \theta = \pi /4

Answer:

The slope of the tangent at a point on a given curve is given by 
 \left ( \frac{dy}{dx} \right )
Now,
\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\cos^2 \theta(-\sin \theta) = 3a(\frac{1}{\sqrt2})^2(-\frac{1}{\sqrt2}) = -\frac{3\sqrt2 a}{4}
Similarly,
\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\sin^2 \theta(\cos \theta) = 3a(\frac{1}{\sqrt2})^2(\frac{1}{\sqrt2}) = \frac{3\sqrt2 a}{4}
\left ( \frac{dy}{dx} \right ) = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{\frac{3\sqrt2 a}{4}}{-\frac{3\sqrt2 a}{4}} = -1
Hence, the slope of the tangent at  \theta = \frac{\pi}{4}      is -1
Now,
Slope of normal =  -\frac{1}{slope \ of \ tangent} = -\frac{1}{-1} = 1
Hence, the slope of normal at \theta = \frac{\pi}{4}       is 1

Question:6 Find the slope of the normal to the curve x = 1- a \sin \theta , y = b \cos ^ 2 \theta \: \: at \: \: \theta = \pi /2

Answer:

The slope of the tangent at a point on given curves is given by 
 \left ( \frac{dy}{dx} \right )
Now,
\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{2}} = -a(\cos \theta)
Similarly,
\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{2}} = 2b\cos \theta(-\sin \theta)
\left ( \frac{dy}{dx} \right )_{x=\frac{\pi}{2}} = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{-2b\cos \theta \sin \theta}{-a\cos \theta} = \frac{2b\sin \theta}{a} = \frac{2b\times1}{a} = \frac{2b}{a}
Hence, the slope of the tangent at  \theta = \frac{\pi}{2} is \frac{2b}{a}
Now,
Slope of normal =  -\frac{1}{slope \ of \ tangent} = -\frac{1}{\frac{2b}{a}} = -\frac{a}{2b}
Hence, the slope of normal at \theta = \frac{\pi}{2}   is  -\frac{a}{2b}

Question:7 Find points at which the tangent to the curvey = x^3 - 3 x^2 - 9x +7 is parallel to the x-axis.

Answer:

We are given  :                   

                                                      y = x^3 - 3 x^2 - 9x +7

Differentiating the equation with respect to x,  we get :

                                                     \frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ -\ 9\ +\ 0

or                                                           =\ 3\left ( x^2\ -\ 2x\ -\ 3 \right )

or                                                \frac{dy}{dx}\ =\ 3\left ( x+1 \right )\left ( x-3 \right )

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

 So, 

                                                        \frac{dy}{dx}\ =\ 0

or                                                    0\ =\ 3\left ( x+1 \right )\left ( x-3 \right )

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 = ( x-2)^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 = \frac{y_2-y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} =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 = ( x-2)^2
\therefore \frac{dy}{dx} = 2(x-2) = 2
(x-2) = 1\\ x = 1+2\\ x=3
Now, when  x=3    y=(3- 2)^2 = (1)^2 = 1
Hence, the coordinates are \left ( 3,1)                 

Question:9 Find the point on the curve y = x^3 - 11x + 5 at which the tangent is y = x -11

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  \frac{dy}{dx}
Given the equation of curve is
y = x^3 - 11x + 5
\frac{dy}{dx} = 3x^2 -11
3x^2 -11 = 1\\ 3x^2 = 12 \\ x^2 = 4 \\ x = \pm2
When x = 2 , y = 2^3 - 11(2) +5 = 8 - 22+5=-9
and 
When x = -2 , y = (-2)^3 - 11(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 = x -11

Question:10 Find the equation of all lines having slope –1 that are tangents to the curve y = \frac{1}{x-1} , x \neq 1

Answer:

We know that the slope of the tangent of at the point of the given curve is given by  \frac{dy}{dx}

Given the equation of curve is
y = \frac{1}{x-1}
\frac{dy}{dx} = \frac{-1}{(1-x)^2}
It is given thta slope is -1
So,
\frac{-1}{(1-x)^2} = -1 \Rightarrow (1-x)^2 = 1 = 1 - x = \pm 1 \\ \\ x = 0 \ and \ x = 2
Now, when x = 0 , y = \frac{1}{x-1} = \frac{1}{0-1} = -1
and 
when x = 2 , y = \frac{1}{x-1} = \frac{1}{(2-1)} = 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 = \frac{1}{x-3} , x \neq 3

Answer:

We know that the slope of the tangent of at the point of the given curve is given by  \frac{dy}{dx}

Given the equation of curve is
y = \frac{1}{x-3}
\frac{dy}{dx} = \frac{-1}{(x-3)^2}
It is given that slope is 2
So,
\frac{-1}{(x-3)^2} = 2 \Rightarrow (x-3)^2 = \frac{-1}{2} = x-3 = \pm \frac{\sqrt-1}{\sqrt2} \\ \\
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2  to the curve  y = \frac{1}{x-3}

Question:12 Find the equations of all lines having slope 0 which are tangent to the curve
y = \frac{1}{x^2 - 2 x +3 }

Answer:

We know that the slope of the tangent  at a point on the given curve is given by  \frac{dy}{dx}

Given the equation of the curve as
y = \frac{1}{x^2 - 2x + 3}
\frac{dy}{dx} = \frac{-(2x-2)}{(x^2-2x+3)^2}
It is given thta slope is 0
So,
\frac{-(2x-2)}{(x^2 - 2x +3)^2} = 0 \Rightarrow 2x-2 = 0 = x = 1
Now, when x = 1 , y = \frac{1}{x^2-2x+3} = \frac{1}{1^2-2(1)+3} = \frac{1}{1-2+3} =\frac{1}{2}

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

Question:13(i) Find points on the curve \frac{x^2 }{9} + \frac{y^2 }{16} = 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  \frac{dy}{dx}
Given the equation of the  curve is 
\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)
18y\frac{dy}{dx} = -32x
\frac{dy}{dx} = \frac{(-32x)}{18y} = 0 \Rightarrow x = 0
From this, we can say that x = 0
Now. when x = 0  ,     \frac{0^2 }{9} + \frac{y^2 }{16} = 1\Rightarrow \frac{y^2}{16} = 1 \Rightarrow y = \pm 4
Hence, the coordinates are (0,4) and (0,-4)

Question:13(ii) Find points on the curve  \frac{x^2}{9} + \frac{y^2}{16} = 1 at which the tangents are parallel to y-axis 

Answer:

Parallel to y-axis means the slope of the tangent is  \infty , 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  \frac{dy}{dx}
Given the equation of the curve is 
\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)
18y\frac{dy}{dx} = 144(1-32x)
\frac{dy}{dx} = \frac{-32x}{18y} = \infty
Slope of normal = -\frac{dx}{dy} = \frac{18y}{32x} = 0
From this we can say that y = 0
Now. when y = 0,  \frac{x^2 }{9} + \frac{0^2 }{16} \Rightarrow 1 = x = \pm 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 = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at\: \: (0, 5)

Answer:

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
y = x^4 - 6x^3 + 13x^2 - 10x + 5
\frac{dy}{dx}= 4x^3 - 18x^2 + 26x- 10
at point (0,5)
\frac{dy}{dx}= 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10
Hence slope of tangent is -10
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-10} = \frac{1}{10}
Now, equation of tangent at point (0,5) with slope = -10 is
y = mx + c\\ 5 = 0 + c\\ c = 5
equation of tangent is
y = -10x + 5\\ y + 10x = 5
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
\\y = mx + c \\5 = 0 + c \\c = 5
equation of normal is
\\y = \frac{1}{10}x+5 \\ 10y - x = 50

Question:14(ii) Find the equations of the tangent and normal to the given curves at the indicated
points:
 y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at \: \: (1, 3)

Answer:

We know that Slope of tangent at a point on given curve is given  by  \frac{dy}{dx}
Given equation of curve
y = x^4 - 6x^3 + 13x^2 - 10x + 5
\frac{dy}{dx}= 4x^3 - 18x^2 + 26x - 10
at point (1,3)
\frac{dy}{dx}= 4(1)^3 - 18(1)^2 + 26(1) - 10 = 2
Hence slope of tangent is 2
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{2}
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 = \frac{-1}{2}\times 1+ c
c = \frac{7}{2}
equation of normal is
y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7

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

 y = x^3\: \: at \: \: (1, 1)

Answer:

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
y = x^3
\frac{dy}{dx}= 3x^2
at point (1,1)
\frac{dy}{dx}= 3(1)^2 = 3
Hence slope of tangent is 3
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{3}
Now, equation of tangent at point (1,1) with slope = 3 is
y = mx + c\\ 1 = 1 \times 3 + c\\ c = 1 - 3 = -2
equation of tangent is
y - 3x + 2 = 0
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
1 = \frac{-1}{3}\times 1+ c
c = \frac{4}{3}
equation of normal is
y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4

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

  y = x^2\: \: at\: \: (0, 0)

Answer:

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
y = x^2
\frac{dy}{dx}= 2x
at point (0,0)
\frac{dy}{dx}= 2(0)^2 = 0
Hence slope of tangent is 0
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{0} = -\infty
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = -\infty is

\\y = x \times -\infty + 0\\ x = \frac{y}{-\infty}\\ x=0

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

 x = \cos t , y = \sin t \: \: at \: \: t = \pi /4

Answer:

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
x = \cos t , y = \sin t
Now,
\frac{dx}{dt} = -\sin t            and           \frac{dy}{dt} = \cos t
Now, 
\left ( \frac{dy}{dx} \right )_{t=\frac{\pi}{4}} = \frac{ \frac{dy}{dt}}{ \frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t = =- \cot \frac{\pi}{4} = -1
Hence slope of the tangent is -1
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-1} = 1
Now, the equation of the tangent at the point t = \frac{\pi}{4} with slope = -1 is
x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}          and       

    y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}
equation of the tangent at

 t = \frac{\pi}{4} i.e. \left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right ) is


y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = -1(x- \frac{1}{\sqrt2})\\ \sqrt2y + \sqrt2x = 2\\ y + x = \sqrt2
Similarly, the equation of normal at t = \frac{\pi}{4}  with slope = 1 is
x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}          and         

  y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}
equation of the tangent at

 t = \frac{\pi}{4} i.e. \left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right ) is
\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y

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

Answer:

Parellel to line 2x - y + 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 \frac{dy}{dx}
Given equation of curve is  
y = x^2 - 2x +7
\frac{dy}{dx} = 2x - 2 = 2\\ \\ x = 2
Now, when x = 2 , y = (2)^2 - 2(2) +7 =4 - 4 + 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 = x^2 -2x +7 which is perpendicular to the line 5y - 15x = 13.

Answer:

Perpendicular to line 5y - 15x = 13.\Rightarrow y = 3x + \frac{13}{5}   means slope \ of \ tangent = \frac{-1}{slope \ 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 = \frac{-1}{slope \ of \ line} = \frac{-1}{3}
Now, we know that the  slope of the tangent at a given point to given curve is given by \frac{dy}{dx}
Given the equation of curve is  
y = x^2 - 2x +7
\frac{dy}{dx} = 2x - 2 = \frac{-1}{3}\\ \\ x = \frac{5}{6}
Now, when x = \frac{5}{6} , y = (\frac{5}{6})^2 - 2(\frac{5}{6}) +7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}
Hence, the coordinates are (\frac{5}{6} ,\frac{217}{36})
Now, the equation of tangent passing through (2,7) and with slope m = \frac{-1}{3}  is
y = mx+ c\\ \frac{217}{36}= \frac{-1}{3}\times \frac{5}{6} + c\\ c = \frac{227}{36}
So,
y = \frac{-1}{3}x+\frac{227}{36}\\ 36y + 12x = 227
Hence, equation of tangent is 36y + 12x = 227

Question:16 Show that the tangents to the curve y = 7x^3 + 11 at the points where x = 2 and x = – 2 are parallel.

Answer:

Slope of tangent = \frac{dy}{dx} = 21x^2 
When x = 2
\frac{dy}{dx} = 21x^2 = 21(2)^{2} = 21 \times4 = 84
When  x = -2
\frac{dy}{dx} = 21x^2 = 21(-2)^{2} = 21 \times4 = 84
Slope is equal when x= 2 and x = - 2 
Hence, we can say that both the tangents to curve y = 7x^3 + 11 is parallel

Question:17 Find the points on the curve y = x ^3 at which the slope of the tangent is equal to the y-coordinate of the point.

Answer:

Given equation of curve is  y = x ^3
Slope of tangent = \frac{dy}{dx} = 3x^2
it is given that  the slope of the tangent is equal to the y-coordinate of the point
3x^2 = y
We have  y = x ^3
3x^2 = x^3\\ 3x^2 - x^3=0\\ x^2(3-x)=0\\ x= 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = 3
So, when x = 0 , y = 0
and when x = 3 , y = x^3 = 3^3 = 27

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

Question:18 For the curve y = 4x ^ 3 - 2x ^5, 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 = 4x ^ 3 - 2x ^5
Slope of tangent =

\frac{dy}{dx} = 12x^2 - 10x^4
Now, equation of tangent is 
Y-y= m(X-x)
at (0,0)  Y =  0 and X = 0
-y= (12x^3-10x^4)(-x)
y= 12x^3-10x^5
and we have y = 4x ^ 3 - 2x ^5
4x^3-2x^5= 12x^3-10x^5
8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1
Now, when x = 0,

  y = 4(0) ^ 3 - 2(0) ^5 = 0
when x = 1 , 

y = 4(1) ^ 3 - 2(1) ^5 = 4-2=2
when x= -1 ,

 y = 4(-1) ^ 3 - 2(-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 x^2 + y^2 - 2x - 3 = 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
x^2 + y^2 - 2x - 3 = 0
Slope of  tangent =
 -2y\frac{dy}{dx} = 2x -2\\ \frac{dy}{dx} = \frac{1-x}{y} = 0\\ x= 1
When x = 1 ,

 -y^2 = x^2 -2x-3= (1)^2-2(1)-3 = 1-5=-4
                      y = \pm 2
Hence, the coordinates are (1,2) and (1,-2)

Question:20 Find the equation of the normal at the point ( am^2 , am^3 ) for the curve ay ^2 = x ^3.

Answer:

Given equation of curve is 
ay ^2 = x ^3\Rightarrow y^2 = \frac{x^3}{a}
Slope of tangent

 2y\frac{dy}{dx} = \frac{3x^2 }{a} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ya}
at point ( am^2 , am^3 )
\frac{dy}{dx} = \frac{3(am^2)^2}{2(am^3)a} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}
Now, we know that 
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-2}{3m}
equation of normal at point ( am^2 , am^3 ) and with slope \frac{-2}{3m}
y-y_1=m(x-x_1)\\ y-am^3 = \frac{-2}{3m}(x-am^2)\\ 3ym - 3am^4 = -2(x-am^2)\\ 3ym +2x= 3am^4+2am^2
Hence, the equation of normal is 3ym +2x= 3am^4+2am^2

Question:21 Find the equation of the normals to the curvey = x^3 + 2x + 6 which are parallel
to the line x + 14y + 4 = 0.

Answer:

Equation of given curve is
y = x^3 + 2x + 6
Parellel to line x + 14y + 4 = 0 \Rightarrow y = \frac{-x}{14} -\frac{4}{14} 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 = \frac{-1}{14}
Slope of tangent = \frac{dy}{dx} = 3x^2+2
We know that
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}
\frac{-1}{3x^2+2} = \frac{-1}{14}
3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 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 = -8-4+6 =-6
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope \frac{-1}{14}
y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254
Similarly,  the equation of at point (-2,-6) with slope \frac{-1}{14}

y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0
Hence, the equation of the normals to the curvey = x^3 + 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 y ^2 = 4 ax at the point (at ^2, 2at).

Answer:

Equation of the given curve is
y ^2 = 4 ax

Slope of tangent = 2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}
at point (at ^2, 2at).
\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}
Now, the equation of tangent with point (at ^2, 2at). and slope \frac{1}{t} is
y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0

We know that
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t
Now, the equation of at point  (at ^2, 2at).  with slope -t
y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0
Hence, the equations of the tangent and normal to the parabola

y ^2 = 4 ax at the point (at ^2, 2at). are
x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively

Question:23 Prove that the curves x = y^2 and xy = k cut at right angles* if \: \: 8k ^ 2 = 1.

Answer:

Let suppose, Curve x = y^2 and xy = k cut at the right  angle
then the slope of their tangent also cut  at the right angle
means,
\left ( \frac{dy}{dx} \right )_a \times \left ( \frac{dy}{dx} \right )_b = -1                                                                   -(i)
2y\left ( \frac{dy}{dx} \right )_a = 1 \Rightarrow \left ( \frac{dy}{dx} \right )_a = \frac{1}{2y}
\left ( \frac{dy}{dx} \right )_b = \frac{-k}{x^2}
Now these values in equation (i)
\frac{1}{2y} \times \frac{-k}{x^2} = -1\\ -k = -2yx^2\\ k =2(xy)(x)\\ k = 2k(k^{\frac{2}{3}}) \ \ \ \ \left ( x = y^2 \Rightarrow y^2y = k \Rightarrow y = k^{\frac{1}{3}} \ and \ x = k^{\frac{2}{3}} \right ) \\ 2(k^{\frac{2}{3}}) = 1\\ \left ( 2(k^{\frac{2}{3}}) \right )^3 = 1^3\\ 8k^2 = 1
Hence proved

Question:24 Find the equations of the tangent and normal to the hyperbola
\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 at the point (x_0 , y_0 )

Answer:

Given equation is 
\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 \Rightarrow y^2a^2 = x^2b^2 -a^2b^2
Now ,we know that
 slope of tangent = 2ya^2\frac{dy}{dx} = 2xb^2 \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}
at point (x_0 , y_0 )
\frac{dy}{dx} = \frac{x_0b^2}{y_0a^2}
equation of tangent at point (x_0 , y_0 ) with slope \frac{xb^2}{ya^2}
y-y_1=m(x-x_1)\\ y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)\\ yy_0a^2-y_0^2a^2 = xx_0b^2-x_0^2b^2\\ xx_0b^2 - yy_0a^2 = x_0^2b^2-y_0^2a^2
Now, divide  both sides by a^2b^2
\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \left ( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right )
                        =1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2 } = 1\right )
 \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1
Hence, the equation of tangent is 

\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1                                                  
We know that
Slope \ of \ normal= \frac{-1}{slope \ of \ tangent } = -\frac{y_0a^2}{x_0b^2}
equation of normal  at the point (x_0 , y_0 ) with slope  -\frac{y_0a^2}{x_0b^2}
y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0

Question:25 Find the equation of the tangent to the curve y = \sqrt{3x-2} which is parallel to the line 4x - 2y + 5 = 0 .

Answer:

Parellel to line 4x - 2y + 5 = 0 \Rightarrow y = 2x + \frac{5}{2} 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 \frac{dy}{dx}
Given the equation of curve is  
y = \sqrt{3x-2}
\frac{dy}{dx} = \frac{1}{2}.\frac{3}{\sqrt{3x-2}}=\frac{3}{2\sqrt{3x-2}}
\frac{3}{2\sqrt{3x-2}} = 2\\ 3^2 = (4\sqrt{3x-2})^2\\ 9 = 16(3x-2)\\ 3x-2=\frac{9}{16}\\ 3x = \frac{9}{16} +2\\ 3x= \frac{41}{16}\\ x = \frac{41}{48}
Now, when

x = \frac{41}{48} ,  y = \sqrt{3x-2} \Rightarrow y = \sqrt{3\times\frac{41}{48}-2 } = \sqrt{\frac{41}{16}-2}=\sqrt\frac{9}{16 } = \pm \frac{3}{4}   

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

\left ( \frac{41}{48},\frac{3}{4} \right )
Now, equation of tangent paasing through

 \left ( \frac{41}{48},\frac{3}{4} \right ) and with slope m = 2 is
y - y_1=m(x-x_1)\\ y-\frac{3}{4}=2(x-\frac{41}{48})\\ 48y-36=2(48x-41)\\ 48x-24y=41-18\\ 48x-24y=23
Hence, equation of tangent paasing through \left ( \frac{41}{48},\frac{3}{4} \right ) and with slope m = 2 is   48x - 24y = 23

Question:26 The slope of the normal to the curvey = 2x ^2 + 3 \sin x \: \: at \: \: x = 0is
(A) 3      (B) 1/3     (C) –3  (D) -1/3

Answer:

Equation of the given curve is 
y = 2x ^2 + 3 \sin x
Slope of tangent = \frac{dy}{dx} = 4x +3 \cos x
at x = 0
\frac{dy}{dx} = 4(0) +3 \cos 0= 0 + 3
\frac{dy}{dx}= 3
Now, we know that
Slope \ of \ normal = \frac{-1}{\ Slope \ of \ tangent} = \frac{-1}{3}
Hence, (D) is the correct option 

Question:27 The line y = x+1 is a tangent to the curve y^2 = 4 x 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
y^2 = 4 x 
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 = 2y\frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}
\frac{dy}{dx} = \frac{2}{y} = 1\\ y = 2
Now, when y = 2, x = \frac{y^2}{4} = \frac{2^2}{4} = \frac{4}{4} = 1
Hence, the coordinates are (1,2)

Hence, (A) is the correct answer

NCERT solutions for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.4

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

Answer:

Lets suppose y = \sqrt x and let x = 25 and \Delta x = 0.3
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{25+0.3} - \sqrt 25
\Delta y = \sqrt{25.3} - 5
\sqrt{25.3} = \Delta y +5
Now, we can say that \Delta y  is approximate equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.3) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.3)\\ dy = \frac{1}{2\sqrt 25}.(0.3)\\ dy = \frac{1}{10}.(0.3)\\ dy = 0.03
Now,
\sqrt{25.3} = \Delta y +5\\ \sqrt {25.3} = 0.03 + 5\\ \sqrt{25.3} = 5.03
Hence, \sqrt{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. 

\sqrt { 49.5 }

Answer:

Lets suppose y = \sqrt x and let x = 49 and \Delta x = 0.5
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{49+0.5} - \sqrt 49
\Delta y = \sqrt{49.5} - 7
\sqrt{49.5} = \Delta y +7
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.5)\\ dy = \frac{1}{2\sqrt 49}.(0.5)\\ dy = \frac{1}{14}.(0.5)\\ dy = 0.035
Now,
\sqrt{49.5} = \Delta y +7\\ \sqrt {49.5} = 0.035 + 7\\ \sqrt{49.5} = 7.035
Hence, \sqrt{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.

  \sqrt {0.6}

Answer:

Lets suppose y = \sqrt x and let x = 1 and \Delta x = -0.4
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{1+(-0.4)} - \sqrt 1
\Delta y = \sqrt{0.6} - 1
\sqrt{0.6} = \Delta y +1
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(-0.4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = -0.4)\\ dy = \frac{1}{2\sqrt 1}.(-0.4)\\ dy = \frac{1}{2}.(-0.4)\\ dy = -0.2
Now,
\sqrt{0.6} = \Delta y +1\\ \sqrt {0.6} = (-0.2) + 1\\ \sqrt{0.6} = 0.8
Hence, \sqrt{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)^{\frac{1}{3}} and let x = 0.008 and \Delta x = 0.001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({0.008+ 0.001})^{\frac{1}{3}} - (0.008)^{\frac{1}{3}}
\Delta y = ({0.009})^{\frac{1}{3}} - 0.2
({0.009})^{\frac{1}{3}} = \Delta y + 0.2
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = 0.001)\\ dy = \frac{1}{3(0.008)^{\frac{2}{3}}}.(0.001)\\ dy = \frac{1}{0.12}.(0.001)\\ dy = 0.008
Now,
(0.009)^{\frac{1}{3}} = \Delta y +0.2\\ (0.009)^{\frac{1}{3}} = (0.008) + 0.2\\ (0.009)^{\frac{1}{3}} = 0.208
Hence, (0.009)^{\frac{1}{3}} 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)^{\frac{1}{10}} and let x = 1 and \Delta x = -0.001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{10}} - (x)^{\frac{1}{10}}
\Delta y = ({1 - 0.001})^{\frac{1}{10}} - (1)^{\frac{1}{10}}
\Delta y = ({0.999})^{\frac{1}{10}} - 1
({0.999})^{\frac{1}{10}} = \Delta y + 1
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{10 (x)^{\frac{9}{10}}}.(-0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{10}} \ and \ \Delta x = -0.001)\\ dy = \frac{1}{10(1)^{\frac{9}{10}}}.(-0.001)\\ dy = \frac{1}{10}.(-0.001)\\ dy = -0.0001
Now,
(0.999)^{\frac{1}{10}} = \Delta y +1\\ (0.999)^{\frac{1}{10}} = (-0.0001) + 1\\ (0.999)^{\frac{1}{10}} = 0.9999 = 0.999 \ upto \ three\ decimal \ place
Hence, (0.999)^{\frac{1}{10}} 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)^{\frac{1}{4}} and let x = 16 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({16 - 1})^{\frac{1}{4}} - (16)^{\frac{1}{4}}
\Delta y = ({15})^{\frac{1}{4}} - 2
({15})^{\frac{1}{4}} = \Delta y + 2
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(16)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 8}.(-1)\\dy = \frac{1}{32}.(-1) \\dy = -0.031
Now,
(15)^{\frac{1}{4}} = \Delta y +2\\ (15)^{\frac{1}{4}} = (-0.031) + 2\\ (15)^{\frac{1}{4}} = 1.969
Hence, (15)^{\frac{1}{4}} 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)^{\frac{1}{3}} and let x = 27 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 1})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26})^{\frac{1}{3}} - 3
({26})^{\frac{1}{3}} = \Delta y + 3
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -1)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-1)\\ dy = \frac{1}{3\times 9}.(-1)\\dy = \frac{1}{27}.(-1) \\dy = -0.037
Now,
(27)^{\frac{1}{3}} = \Delta y +3\\ (27)^{\frac{1}{3}} = (-0.037) + 3\\ (27)^{\frac{1}{3}} = 2.963
Hence, (27)^{\frac{1}{3}} 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)^{\frac{1}{4}} and let x = 256 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({256 - 1})^{\frac{1}{4}} - (256)^{\frac{1}{4}}
\Delta y = ({255})^{\frac{1}{4}} - 4
({255})^{\frac{1}{4}} = \Delta y + 4
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(256)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 64}.(-1)\\dy = \frac{1}{256}.(-1) \\dy = -0.003
Now,
(255)^{\frac{1}{4}} = \Delta y +4\\ (255)^{\frac{1}{4}} = (-0.003) + 4\\ (255)^{\frac{1}{4}} = 3.997
Hence, (255)^{\frac{1}{4}} 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)^{\frac{1}{4}} and let x = 81 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({81 + 1})^{\frac{1}{4}} - (81)^{\frac{1}{4}}
\Delta y = ({82})^{\frac{1}{4}} - 3
({82})^{\frac{1}{4}} = \Delta y + 3
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 1)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(1)\\ dy = \frac{1}{4\times 27}.(1)\\dy = \frac{1}{108}.(1) \\dy = .009
Now,
(82)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.009) + 3\\ (82)^{\frac{1}{4}} = 3.009
Hence, (82)^{\frac{1}{4}} 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)^{\frac{1}{2}} and let x = 400 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({400 + 1})^{\frac{1}{2}} - (400)^{\frac{1}{2}}
\Delta y = ({401})^{\frac{1}{2}} - 20
({401})^{\frac{1}{2}} = \Delta y + 20
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 1)\\ dy = \frac{1}{2(400)^{\frac{1}{2}}}.(1)\\ dy = \frac{1}{2\times 20}.(1)\\dy = \frac{1}{40}.(1) \\dy = 0.025
Now,
(401)^{\frac{1}{2}} = \Delta y +20\\ (401)^{\frac{1}{2}} = (0.025) + 20\\ (401)^{\frac{1}{2}} = 20.025
Hence, (401)^{\frac{1}{2}} 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)^{\frac{1}{2}} and let x = 0.0036 and \Delta x = 0.0001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({0.0036 + 0.0001})^{\frac{1}{2}} - (0.0036)^{\frac{1}{2}}
\Delta y = ({0.0037})^{\frac{1}{2}} - 0.06
({0.0037})^{\frac{1}{2}} = \Delta y + 0.06
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(0.0001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 0.0001)\\ dy = \frac{1}{2(0.0036)^{\frac{1}{2}}}.(0.0001)\\ dy = \frac{1}{2\times 0..06}.(0.0001)\\dy = \frac{1}{0.12}.(0.0001) \\dy = 0.0008
Now,
(0.0037)^{\frac{1}{2}} = \Delta y +0.06\\ (0.0037)^{\frac{1}{2}} = (0.0008) + 0.06\\ (0.0037)^{\frac{1}{2}} = 0.0608
Hence, (0.0037)^{\frac{1}{2}} 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)^{\frac{1}{3}} and let x = 27 and \Delta x = -0.43
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 0.43})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26.57})^{\frac{1}{3}} - 3
({26.57})^{\frac{1}{3}} = \Delta y + 3
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-0.43) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -0.43)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-0.43)\\ dy = \frac{1}{3\times 9}.(-0.43)\\dy = \frac{1}{27}.(-0.43) \\dy = -0.0159 = -0.016 (approx.)
Now,
(26.57)^{\frac{1}{3}} = \Delta y +3\\ (26.57)^{\frac{1}{3}} = (-0.016) + 3\\ (26.57)^{\frac{1}{3}} = 2.984
Hence, (0.0037)^{\frac{1}{2}} 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)^{\frac{1}{4}} and let x = 81 and 0.5
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({81 + 0.5})^{\frac{1}{4}} - (81)^{\frac{1}{4}}
\Delta y = ({81.5})^{\frac{1}{4}} - 3
({81.5})^{\frac{1}{4}} = \Delta y + 3
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 0.5)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(0.5)\\ dy = \frac{1}{4\times 27}.(0.5)\\dy = \frac{1}{108}.(0.5) \\dy = .004
Now,
(81.5)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.004) + 3\\ (82)^{\frac{1}{4}} = 3.004
Hence, (81.5)^{\frac{1}{4}} 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)^{\frac{3}{2}} and let x = 4 and \Delta x = -0.032
Then,
\Delta y = ({x+\Delta x})^{\frac{3}{2}} - (x)^{\frac{3}{2}}
\Delta y = ({4 - 0.032})^{\frac{3}{2}} - (4)^{\frac{3}{2}}
\Delta y = ({3.968})^{\frac{3}{2}} - 8
({3.968})^{\frac{3}{2}} = \Delta y + 8
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{3 (x)^{\frac{1}{2}}}{2}.(-0.032) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{3}{2}} \ and \ \Delta x = -0.032)\\ dy = \frac{3 (4)^{\frac{1}{2}}}{2}.(-0.032)\\ dy = \frac{3\times 2}{2}.(-0.032)\\\\dy = -0.096
Now,
(3.968)^{\frac{3}{2}} = \Delta y +8\\ (3.968)^{\frac{3}{2}} = (-0.096) + 8\\ (3.968)^{\frac{3}{2}} = 7.904
Hence, (3.968)^{\frac{3}{2}} 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)^{\frac{1}{5}} and let x = 32 and \Delta x = 0.15
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{5}} - (x)^{\frac{1}{5}}
\Delta y = ({32 + 0.15})^{\frac{1}{5}} - (32)^{\frac{1}{5}}
\Delta y = ({32.15})^{\frac{1}{5}} - 2
({32.15})^{\frac{1}{5}} = \Delta y + 2
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{5 (x)^{\frac{4}{5}}}.(0.15) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{5}} \ and \ \Delta x = 0.15)\\ dy = \frac{1 }{5 (32)^{\frac{4}{5}}}.(0.15)\\ dy = \frac{1}{5\times16}.(0.15)\\\\dy = \frac{0.15}{80}\\ dy = 0.001
Now,
(32.15)^{\frac{1}{5}} = \Delta y +2\\ (32.15)^{\frac{1}{5}} = (0.001) + 2\\ (32.15)^{\frac{1}{5}} = 2.001
Hence, (32.15)^{\frac{1}{5}} is approximately equal to 2.001

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

Answer:

Let x = 2 and \Delta x = 0.01
f(x+\Delta x) = 4(x+\Delta x)^2 +5(x+\Delta x)+2
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (8x+5).(0.01) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 4x^2+5x+2 \ and \ \Delta x = 0.01)\\ dy = 0.08x+0.05
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.08x + 0.05 + 4x^2 + 5x +2\\ f(x+\Delta x) = 0.08(2)+0.05+4(2)^2+5(2)+2\\ f(x+\Delta x) = 0.16 + 0.05 + 16 + 10 + 2\\ f(x+\Delta x) = 28.21
Hence, the approximate value of f (2.01), where f (x) = 4x^2 + 5x + 2. is 28.21

Question:3 Find the approximate value of f (5.001), where f (x) = x^3 - 7x^2 + 15.

Answer:

Let x = 5 and \Delta x = 0.001
f(x+\Delta x) =(x+\Delta x)^3 - 7(x+\Delta x)^2 +15
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (3x^2 - 14x).(0.001) \ \ \ \ \ \ \ \ \ (\because y = f(x) = x^3-7x^2+15 \ and \ \Delta x = 0.001)\\ dy =0.003x^2 -0.014x
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.003x^2 - 0.014x + x^3 - 7x^2 +15\\ f(x+\Delta x) =0.003(5)^2-0.014(5)+(5)^3-7(5)^2+15\\ f(x+\Delta x) = 0.075-0.07+125-175+15\\ f(x+\Delta x) = -34.995
Hence, the approximate value of f (5.001), where f (x) = x^3 - 7x^2 + 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 = x^3 \ m^3 
we know that \Delta y is approximately equal to dy
So,
dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.01x)\\ dy = 0.03x^3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is 0.03x^3 \ m^3

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%(\Delta x) = -0.01x m
The surface area of cube = 6a^2 \ m^2
We know that, (\Delta y) is approximately equal to dy

dy = \frac{dy}{dx}.\Delta x\\ dy = 12a(-0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = 6a^2 \ and \ \Delta x = -0.01x)\\ dy = 12x(-0.01x)\\ dy=-0.12x^2 \ m^2
Hence,  the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is -0.12x^2 \ m^2

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 (\Delta r) = 0.02 m
Volume of sphere = \frac{4}{3}\pi r^3
Error in volume (\Delta V) 
dV = \frac{dV}{dr}.\Delta r\\ dV = 4\pi r^2 .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because V = \frac{4}{3}\pi r^3, r =7 \ and \ \Delta r = 0.02 )\\ dV = 4\pi (7)^2 (0.02)\\ dV= 4\pi (49) (0.02)\\ dV = 3.92\pi
Hence,  the approximate error in its volume is 3.92\pi \ m^3

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 (\Delta r) = 0.03 m
The surface area of sphere = 4\pi r^2
Error in surface area (\Delta A) 
dA = \frac{dA}{dr}.\Delta r\\ dA = 8\pi r .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because A = 4\pi r^2, r =9 \ and \ \Delta r = 0.03 )\\ dA = 8\pi (9) (0.03)\\ dA= 2.16\pi
Hence,  the approximate error in its surface area is 2.16\pi \ m^2

Question:8 If f(x) = 3x ^2 + 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 \Delta x = 0.02
f(x+\Delta x) = 3(x+\Delta x)^2 +15(x+\Delta x)+5
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (6x+15).(0.02) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 3x^2+15x+5 \ and \ \Delta x = 0.02)\\ dy = 0.12x+0.3
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.12x + 0.3 + 3x^2 + 15x +5\\ f(x+\Delta x) = 0.12(3)+0.3+3(3)^2+15(3)+5\\ f(x+\Delta x) = 0.36+ 0.3 + 27 + 45 + 5\\ f(x+\Delta x) = 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 x^3 \ m^3     (B) 0.6 x^3 \ m^3       (C) 0.09 x^3 \ m^3    (D) 0.9 x^3 \ m^3

Answer:

Side of cube increased by 3% = 0.03x m
The volume of cube = x^3 \ m^3 
we know that \Delta y is approximately equal to dy
So,
dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.03x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.03x)\\ dy = 0.09x^3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is 0.09x^3 \ m^3
Hence, (C) is the correct answer

Solutions of NCERT for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.5

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

Answer:

Given function is,
f (x) = (2x - 1)^2 + 3
(2x - 1)^2 \geq 0\\ (2x-1)^2+3\geq 3
Hence, minimum value occurs when 
(2x-1)=0\\ x = \frac{1}{2}
Hence, the minimum value of function f (x) = (2x - 1)^2 + 3 occurs at  x = \frac{1}{2}
and the minimum value is 
f(\frac{1}{2}) = (2.\frac{1}{2}-1)^2+3\\
            = (1-1)^2+3 \Rightarrow 0+3 = 3
and it is clear that there is no maximum value of f (x) = (2x - 1)^2 + 3 

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

 f (x) = 9x^ 2 + 12x + 2

Answer:

Given function is,
f (x) = 9x^ 2 + 12x + 2
add and subtract 2 in given equation
f (x) = 9x^ 2 + 12x + 2 + 2- 2\\ f(x)= 9x^2 +12x+4-2\\ f(x)= (3x+2)^2 - 2
Now,
(3x+2)^2 \geq 0\\ (3x+2)^2-2\geq -2    for every x \ \epsilon \ R
Hence, minimum value occurs when 
(3x+2)=0\\ x = \frac{-2}{3}
Hence, the minimum value of function f (x) = 9x^2+12x+2 occurs at  x = \frac{-2}{3}
and the minimum value is 
f(\frac{-2}{3}) = 9(\frac{-2}{3})^2+12(\frac{-2}{3})+2=4-8+2 =-2 \\
            
and it is clear that there is no maximum value of  f (x) = 9x^2+12x+2

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

f (x) = - (x -1) ^2 + 10

Answer:

Given function is,
f (x) = - (x -1) ^2 + 10
-(x-1)^2 \leq 0\\ -(x-1)^2+10\leq 10    for every x \ \epsilon \ R
Hence, maximum value occurs when 
(x-1)=0\\ x = 1
Hence, maximum value of function f (x) = - (x -1) ^2 + 10  occurs at  x = 1
and the maximum value is 
f(1) = -(1-1)^2+10=10 \\
            
and it is clear that there is no minimum value of  f (x) = 9x^2+12x+2

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

Answer:

Given function is,
g(x) = x^3 + 1
value of x^3 varies from -\infty < x^3 < \infty
Hence, function  g(x) = x^3 + 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| \geq 0\\ |x+2| - 1 \geq -1       x \ \epsilon \ 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 \ \epsilon \ 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| \leq 0\\ -|x+1| + 3 \leq 3       x \ \epsilon \ 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 t