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NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

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NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:43 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 6.3 Class 12 Maths chapter 6 is related to the topic normal and tangents. The equations of normals and tangent to a curve and questions related to these are discussed in exercise 6.3 Class 12 Maths. There are twenty-seven questions presented in the Class 12 Maths chapter 6 exercise 6.3. These questions in Class 12th Maths chapter 6 exercise 6.3 are solved by our Mathematics subject matter experts and are reliable and according to the CBSE pattern. The NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 along with all other exercises of the NCERT chapter applications of derivatives gives a good idea of concepts discussed in the NCERT Book.

12th class Maths exercise 6.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Application of Derivatives Class 12 Chapter 6 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:2 . Find 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 curve y = 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 curve y = 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 curve y = 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 (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 curve y = 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 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

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 curve y = 2x ^2 + 3 \sin x \: \: at \: \: x = 0 is
(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

More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

As the questions in the Class 12 Maths chapter 6 exercise 6.3 deal with an application of derivatives, it is better for the students to revise the basic derivatives of trigonometric functions, exponential functions and some other special functions and rules and properties related to the derivatives. Out of the 27 problems in the Class 12th Maths chapter, 6 exercise 6. 3 question 26 is to find the slope of the normal to a given curve and question 27 asks to find the points for which a line is a tangent to the given curve.

Also Read| Application of Derivatives Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

  • Solving exercise 6.3 Class 12 Maths will be beneficial for both the CBSE board exam and JEE Main exam.

  • One question from NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 can be expected for the CBSE Class 12 Maths board paper.

  • The applications of tangents will be used in Class 12 Physics and Chemistry also.

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Key Features Of NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.3 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 6.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 6.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 6.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 6.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 6.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. What is the equation of the tangent to a curve y=f(x) at (x0,y0)?

y-y0=f’(x0)(x-x0) 

2. Give the equation of normal to the tangent y-y0=f’(x0)(x-x0)

(y-y0)f’(x0)+(x-x0)=0

3. Equation of a tangent with slope=0 st (x0, y0) is

The slope =0. Therefore the equation of tangent is y=y0

4. The slope of a tangent is infinity, then what is its equation at (x0,y0)?

The equation of tangent with infinite slope is x=x0

5. Normal is ……………...to the tangent

Normal is perpendicular to the tangent

6. What is the relation between the slope of a tangent and the normal to the tangent?

The slope of normal is the negative of the inverse of the slope of the tangent.

7. Give the number of questions discussed in Exercise 6.3 Class 12 Maths

27 questions are answered in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.3. For more questions refer to NCERT exemplar. Following NCERT syllabus will be useful for CBSE board exams.

8. How many solved examples are given in the topic tangents and normals?

There are 7 solved examples explained before the NCERT book Class 12 Maths chapter 6 exercise 6.3.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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