# 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  $\dpi{100} dS/dt$ 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 $\dpi{100} y=f(x)$,then $\dpi{100} \frac{dy}{dx}$ or $\dpi{100} f^{'}(x)$ represents the rate of change of y with respect to x  and $\dpi{100} \dpi{100} \frac{dy}{dx} ]_{x=x_{o}}$ or $\dpi{100} f^{'}(x_{o})$ represents the rate of change of y with respect to x at $\dpi{100} x=x_{o}$.  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 $\dpi{100} A=\pi r^{2}$. Therefore, the rate of change of the area(A) with respect to its radius(r) is given by -                             $\dpi{100} \frac{dA}{dr}=\frac{d}{dr}(\pi r^{2})=2\pi r$                When   $\dpi{100} r=5cm,\:\:\: \frac{dA}{dr}=10\pi$ Thus, the area of the circle is changing at the rate of $\dpi{100} 10\pi \:\:cm^2/s$

## 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

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  $6 \pi$

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  $8 \pi$

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,
$\frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} cm^2/s$

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  $60\pi \ cm^{2}/s$

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  $900 \ cm^{3}/s$

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   $80\pi \ cm^{2}/s$

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 $1.4\pi \ cm/s$

the perimeter of rectangle

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  $2 \ cm/minute$

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  $2 \ cm^{2}/minute$

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  $\frac{1}{\pi} \ cm/s$

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 $400\pi \ cm^{3}/s$

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 $\frac{8}{3} \ cm/s$

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
$(4,11) \ and \ (-4,\frac{-31}{3})$

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   $2\pi \ cm^{3}/s$

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

Given =  $\frac{dV}{dt} = 12 \ cm^{3}/s$      and    $h = \frac{1}{6}r$
To find =  $\frac{dh}{dt}$   at h = 4 cm
Solution:-

Volume of cone(V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt} = \frac{d(\frac{1}{3}\pi (6h)^{2}h)}{dh}.\frac{dh}{dt} = \frac{1}{3}\pi\times36\times3h^{2}.\frac{dh}{dt} = 36\pi \times(4)^{2}.\frac{dh}{dt}$
$\frac{dV}{dt} = 576\pi.\frac{dh}{dt}$

Find the marginal cost when 17 units are produced.

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

Find the marginal revenue when x = 7

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

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

Marginal revenue =  $\frac{dR}{dx}$
$R ( x) = 3 x^2 + 36 x + 5$
$\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)$
at x = 15
$\frac{dR}{dx} = 6(15+6) = 6\times21 = 126$
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is D

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

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

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

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   $\left ( 0,\frac{\pi}{2} \right )$

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

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 $\left ( \frac{\pi}{2},\pi \right )$

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  $\left ( 0,\pi \right )$

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
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   $x \epsilon \left ( \frac{3}{4},\infty \right )$

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
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   $x \epsilon \left ( -\infty ,\frac{3}{4}\right )$

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)

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)

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  $(-1,\infty)$

$10 - 6x - 2x^2$

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  $( -\frac{3}{2},\infty)$

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

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)

$6- 9x - x ^2$

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 $( - \frac{9}{2},\infty )$

$( x+1) ^3 ( x-3) ^3$

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  $(-\infty,-1) \ and \ (-1,1)$

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  $x > -1$

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  $(0,1)\cup (2,\infty)$

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  $\theta \ \epsilon \ [0,\frac{\pi}{2}]$

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  $( 0 , \infty )$

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)

(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  $(0,\frac{\pi}{2})$

(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

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]

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]

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  $\left ( \pi/2 , \pi \right )$

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  $\left ( \frac{3\pi}{2},2\pi \right )$

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

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

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)

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

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

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 $\frac{-1}{64}$

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

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

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

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  $-\frac{a}{2b}$

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).

(4, 4).

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 $\left ( 3,1)$

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 $y = x -11$

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

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  $y = \frac{1}{x-3}$

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
$y = \frac{1}{2}$

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)

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)

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
$\\y = \frac{1}{10}x+5 \\ 10y - x = 50$

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
$y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7$

$y = x^3\: \: at \: \: (1, 1)$

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
$y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4$

$y = x^2\: \: at\: \: (0, 0)$

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

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

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

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
$\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y$

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

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

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

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)

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)

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)

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 $3ym +2x= 3am^4+2am^2$

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 curve$y = x^3 + 2x + 6$ which are parallel
to the line $x + 14y + 4 = 0.$

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

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
$x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively$

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

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  $-\frac{y_0a^2}{x_0b^2}$
$y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0$

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

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

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

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

$\sqrt { 49.5 }$

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

$\sqrt {0.6}$

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

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

$( 0.999) ^{1/10 }$

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)

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

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

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

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

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

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)

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)

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

$( 3.968) ^{3/2 }$

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

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

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

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 $f (x) = x^3 - 7x^2 + 15\ is \ -34.995$

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 $0.03x^3 \ m^3$

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 $-0.12x^2 \ m^2$

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 $3.92\pi \ m^3$

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 $2.16\pi \ m^2$

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

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

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 $f (x) = (2x - 1)^2 + 3$

$f (x) = 9x^ 2 + 12x + 2$

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  $f (x) = 9x^2+12x+2$

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  $f (x) = 9x^2+12x+2$

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

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 $x \ \epsilon \ R$

$g(x) = - | x + 1| + 3$
$-|x+1| \leq 0\\ -|x+1| + 3 \leq 3$       $x \ \epsilon \ R$
$g(-1) = -|-1+1| + 3 = 3$