NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 12 - Application of Derivatives

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# NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 12 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 04, 2023 01:17 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Chapter 6 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 6 class 12 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. Students can get the NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise on this page. The Class 12 Maths chapter 6 miscellaneous exercise solutions are designed to make the students understand the applications of concepts studies in the NCERT book. Class 12 Maths chapter 6 miscellaneous solutions covers all the topics discussed in the previous exercises. Miscellaneous exercise chapter 6 Class 12 presents questions to practice the complete chapter.

After completing all the topics of NCERT syllabus Class 12 Maths chapter 6 and solved and unsolved questions of all other exercises the miscellaneous exercise can be attempted. The miscellaneous exercise combines questions from the complete chapter and the level of questions compared to exercises will be a bit higher. Miscellaneous exercise class 12 chapter 5 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 enumerated in NCERT Book together using the link provided below.

## Application of Derivatives Miscellaneous Exercise

Let $y = x^\frac{1}{4}$ and $x = \frac{16}{81} \ and \ \Delta x = \frac{1}{81}$
$\Delta y = (x+\Delta x)^\frac{1}{4}-x^\frac{1}{4}$
$= (\frac{16}{81}+\frac{1}{81})^\frac{1}{4}-(\frac{16}{81})^\frac{1}{4}$
$(\frac{17}{81})^\frac{1}{4} = \Delta y + \frac{2}{3}$
Now, we know that $\Delta y$ is approximate equals to dy
So,
$dy = \frac{dy}{dx}.\Delta x \\ = \frac{1}{4x^\frac{3}{4}}.\frac{1}{81} \ \ \ \ \ \ \ (\because y = x^\frac{1}{4} \ and \ \Delta x = \frac{1}{81})\\ = \frac{1}{4(\frac{16}{81})^\frac{3}{4}}.\frac{1}{81} = \frac{27}{4\times 8}.\frac{1}{81} = \frac{1}{96}$
Now,
$(\frac{17}{81})^\frac{1}{4} = \Delta y + \frac{2}{3} = \frac{1}{96}+\frac{2}{3} = \frac{65}{96} = 0.677$
Hence, $(\frac{17}{81})^\frac{1}{4}$ is approximately equal to 0.677

Let $y = x^\frac{-1}{5}$ and $x = 32 \ and \ \Delta x = 1$
$\Delta y = (x+\Delta x)^\frac{-1}{5}-x^\frac{-1}{5}$
$= (32+1)^\frac{-1}{5}-(32)^\frac{-1}{5}$
$(33)^\frac{-1}{4} = \Delta y + \frac{1}{2}$
Now, we know that $\Delta y$ is approximately equals to dy
So,
$dy = \frac{dy}{dx}.\Delta x \\ = \frac{-1}{5x^\frac{6}{5}}.1 \ \ \ \ \ \ \ (\because y = x^\frac{-1}{5} \ and \ \Delta x = 1)\\ = \frac{-1}{5(32)^\frac{6}{5}}.1 = \frac{-1}{5\times 64}.1= \frac{-1}{320}$
Now,
$(33)^\frac{-1}{5} = \Delta y + \frac{1}{2} = \frac{-1}{320}+\frac{1}{2} = \frac{159}{320} = 0.497$
Hence, $(33)^\frac{-1}{5}$ is approximately equals to 0.497

Given function is
$f ( x ) = \frac{\log x}{x}$
$f^{'}(x) = \frac{1}{x}.\frac{1}{x} + log x\frac{-1}{x^2} = \frac{1}{x^2}(1-\log x)$
$f^{'}(x) =0 \\ \frac{1}{x^2}(1-\log x) = 0\\ \frac{1}{x^2} \neq 0 \ So \ log x = 1\Rightarrow x = e$
Hence, x =e is the critical point
Now,
$f^{''}(x) = \frac{-2x}{x^3}(1-\log x)+\frac{1}{x^2}(-\frac{1}{x}) = \frac{1}{x^3}(-2x+2xlog x-1)\\ f^{''(e)} = \frac{-1}{e^3} < 0$
Hence, x = e is the point of maxima

It is given that the base of the triangle is b
and let the side of the triangle be x cm , $\frac{dx}{dt} = -3 cm/s$
We know that the area of the triangle(A) = $\frac{1}{2}bh$
now, $h = \sqrt{x^2-(\frac{b}{2})^2}$
$A= \frac{1}{2}b \sqrt{x^2-(\frac{b}{2})^2}$
$\frac{dA}{dt}=\frac{dA}{dx}.\frac{dx}{dt}= \frac{1}{2}b\frac{2x}{2\sqrt{x^2-(\frac{b}{2})^2}}.(-3)$
Now at x = b
$\frac{dA}{dx} = \frac{1}{2}b\frac{2b}{\frac{\sqrt3b}{2}}.(-3)=-\sqrt3b$
Hence, the area decreasing when the two equal sides are equal to the base is $\sqrt3b$ $cm^2/s$

Given the equation of the curve
$x^2 = 4 y$
We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
$4\frac{dy}{dx} = 2x \\\ \frac{dy}{dx} = \frac{x}{2}$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x}{2}} = \frac{-2}{x}$
At point (a,b)
$Slope = \frac{-2}{a}$
Now, the equation of normal with point (a,b) and $Slope = \frac{-2}{a}$

$y-y_1=m(x-x_1)\\ y-b=\frac{-2}{a}(x-a)$
It is given that it also passes through the point (1,2)
Therefore,
$2-b=\frac{-2}{a}(1-a)\\ 2a -ba = 2a -2\\ ba = 2\\b =\frac{2}{a}$ -(i)
It also satisfies equation $x^2 = 4 y\Rightarrow b = \frac{a^2}{4}$ -(ii)
By comparing equation (i) and (ii)
$\frac{2}{a} = \frac{a^2}{4}\\ a^3 = 8\\ a = 2$
$b = \frac{2}{a} = \frac{2}{2} = 1$
$Slope = \frac{-2}{a} = \frac{-2}{2} = -1$

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

$y-y_1=m(x-x_1)\\ y-1=-1(x-2)\\ y+x=3$
Hence, equation of normal is x + y - 3 = 0

is at a constant distance from the origin.

We know that the slope of tangent at any point is given by $\frac{dy}{dx}$
Given equations are
$x = a \cos \theta + a \theta \sin \theta , y = a \sin \theta - a\theta \cos\theta$
$\frac{dx}{d\theta} = -a\sin \theta + a\sin \theta -a\theta\cos \theta = -a\theta\cos \theta$
$\frac{dy}{d\theta} =a\cos \theta -a\cos \theta +a\theta (-\sin \theta) = -a\theta\sin \theta$
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\theta\sin \theta}{-a\theta \cos \theta} = \tan \theta$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{\tan \theta}$
equation of normal with given points and slope
$y_2-y_1=m(x_2-x_1)\\ y - a\sin \theta + a\theta\cos\theta = \frac{-1}{\tan \theta}(x-a\cos\theta-a\theta\sin\theta)\\ y\sin\theta - a\sin^2 \theta + a\theta\cos\theta\sin\theta = -x\cos\theta+a\cos^2\theta+a\theta\sin\theta\cos\theta\\ y\sin\theta + x\cos\theta = a$
Hence, the equation of normal is $y\sin\theta + x\cos\theta = a$
Now perpendicular distance of normal from the origin (0,0) is
$D = \frac{|(0)\sin\theta+(0)\cos\theta-a|}{\sqrt{\sin^2\theta+\cos^2\theta}} = |-a| = a = \ constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\ (\because \sin^2x+\cos^2x=1)$
Hence, by this, we can say that

the normal at any point $\theta$ to the curve $x = a \cos \theta + a \theta \sin \theta , y = a \sin \theta - a\theta \cos\theta$

is at a constant distance from the origin

increasing

Given function is
$f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$
$f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}$
$=\frac{4\cos x - \cos^2x}{2+\cos x}$
$f^{'}(x)=0\\ \frac{4\cos x - \cos^2x}{2+\cos x} =0\\ \cos x(4-\cos x) = 0\\ \cos x = 0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \cos x =4$
But $\cos x \neq 4$
So,
$\cos x = 0 \\ x = \frac{\pi}{2} \ and \ \frac{3\pi}{2}$
Now three ranges are there $\left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
In interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$ , $f^{'}(x) > 0$

Hence, the given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is increasing in the interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0$ so function is decreasing in this inter

$f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is

decreasing

Given function is
$f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$
$f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}$
$=\frac{4\cos x - \cos^2x}{2+\cos x}$
$f^{'}(x)=0\\ \frac{4\cos x - \cos^2x}{2+\cos x} =0\\ \cos x(4-\cos x) = 0\\ \cos x = 0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \cos x =4$
But $\cos x \neq 4$
So,
$\cos x = 0 \\ x = \frac{\pi}{2} \ and \ \frac{3\pi}{2}$
Now three ranges are there $\left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
In interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$ , $f^{'}(x) > 0$

Hence, given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is increasing in interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0$
Hence, given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is decreasing in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right )$

Increasing

Given function is
$f (x) = x ^3 + \frac{1}{x^3}$
$f^{'} (x) = 3x ^2 + \frac{-3x^2}{x^4}\\ f^{'}(x) = 0\\ 3x ^2 + \frac{-3x^2}{x^4} = 0\\ x^4 = 1\\ x = \pm1$
Hence, three intervals are their $(-\infty,-1),(-1,1) \ and (1,\infty)$
In interval $(-\infty,-1) \ and \ (1,\infty) , f^{'})x > 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is increasing in interval $(-\infty,-1) \ and \ (1,\infty)$
In interval (-1,1) , $f^{'}(x)< 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is decreasing in interval (-1,1)

decreasing

Given function is
$f (x) = x ^3 + \frac{1}{x^3}$
$f^{'} (x) = 3x ^2 + \frac{-3x^2}{x^4}\\ f^{'}(x) = 0\\ 3x ^2 + \frac{-3x^2}{x^4} = 0\\ x^4 = 1\\ x = \pm1$
Hence, three intervals are their $(-\infty,-1),(-1,1) \ and (1,\infty)$
In interval $(-\infty,-1) \ and \ (1,\infty) , f^{'})x > 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is increasing in interval $(-\infty,-1) \ and \ (1,\infty)$
In interval (-1,1) , $f^{'}(x)< 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is decreasing in interval (-1,1)

Given the equation of the ellipse
$\frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1$
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
$m = \pm \frac{b}{a}.\sqrt{a^2-n^2}$
Therefore, Now
Coordinates of A = $\left ( -n,\frac{b}{a}.\sqrt{a^2-n^2} \right )$
Coordinates of B = $\left ( -n,-\frac{b}{a}.\sqrt{a^2-n^2} \right )$
Now,
Length AB(base) = $2\frac{b}{a}.\sqrt{a^2-n^2}$
And height of triangle ABC = (a+n)
Now,
Area of triangle = $\frac{1}{2}bh$
$A = \frac{1}{2}.\frac{2b}{a}.\sqrt{a^2-n^2}.(a+n)= ab\sqrt{a^2-n^2}+bn\sqrt{a^2-n^2}$
Now,
$\frac{dA}{dn} = \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}}$
Now,
$\frac{dA}{dn} =0\\ \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}} =0\\ -abn + n(a^2-n^2)-bn^2 = 0\\ \Rightarrow n = -a,\frac{a}{2}$
but n cannot be zero
therefore, $n = \frac{a}{2}$
Now, at $n = \frac{a}{2}$
$\frac{d^2A}{dn^2}< 0$
Therefore, $n = \frac{a}{2}$ is the point of maxima
Now,
$b = 2\frac{b}{a}.\sqrt{a^2- (\frac{a}{2})^2}= \sqrt3b$
$h = (a+ n ) = a+ \frac{ a}{2} = \frac{ 3a}{2}$
Now,
Therefore, Area (A) $= \frac{1}{2}bh = \frac{1}{2}\sqrt3 b \frac{3a}{2} = \frac{3\sqrt3ab}{4}$

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8 $m^3$
h = 2m (given)
lb = 4 = $l = \frac{4}{b}$
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
$A(b) = 4 + 2h(\frac{4}{b}+b)$
$A^{'}(b) = 2h(\frac{-4}{b^2}+1)\\ A^{'}(b)=0\\ 2h(\frac{-4}{b^2}+1) = 0\\ b^2= 4\\ b = 2$
Now,
$A^{''}(b) = 2h(\frac{-4\times-2b}{b^3})\\ A^{''}(2) = 8 > 0$
Hence, b = 2 is the point of minima
$l = \frac{4}{b} = \frac{4}{2} = 2$
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 = $4 \ m^2$
building of tank costs Rs 70 per sq metres for the base
Therefore, for $4 \ m^2$ Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) = $16 \ m^2$
building of tank costs Rs 45 per square metre for sides
Therefore, for $16 \ m^2$ Rs = 16 X 45 = 720 Rs
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs

It is given that
the sum of the perimeter of a circle and square is k = $2\pi r + 4a = k\Rightarrow a = \frac{k - 2\pi r}{4}$
Let the sum of the area of a circle and square(A) = $\pi r^2 + a^2$
$A = \pi r^2 + (\frac{k-2\pi r}{4})^2$
$A^{'}(r) = 2\pi r + 2(\frac{k-2\pi r}{16})(- 2\pi)\\ A^{'}(r) = 0\\ 2\pi (\frac{8r-k-2\pi r}{8}) = 0\\ r = \frac{k}{8-2\pi}$
Now,
$A^{''}(r) = 2\pi (\frac{8-2\pi }{8}) = 0\\ A^{''}(\frac{k}{8-2\pi}) > 0$
Hence, $r= \frac{k}{8-2\pi}$ is the point of minima
$a = \frac{k-2\pi r}{4} = \frac{k-2\pi \frac{k}{8-2\pi}}{4}=2 \frac{k}{8-2\pi} = 2r$
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle $(r = \frac{l}{2})$
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle

= $l+2b + \pi \frac{l}{2}$

$l+2b + \pi \frac{l}{2} = 10\\ l = \frac{2(10-2b)}{2+\pi}$
Area of window id given by (A) = $lb + \frac{\pi}{2}\left ( \frac{l}{2} \right )^2$
$= \frac{2(10-2b)}{2+\pi}b + \frac{\pi}{2}\left ( \frac{10-2b}{2+\pi} \right )^2\\$
$A^{'}(b) = \frac{20-8b}{2+\pi}+\frac{\pi}{2}.2(\frac{10-2b}{2+\pi}).\frac{(-2)}{2+\pi}$
$= \frac{20-8b}{2+\pi}-2\pi(\frac{10-2b}{(2+\pi)^2})\\ A^{'}(b) = 0\\ \frac{20-8b}{2+\pi}=2\pi(\frac{10-2b}{(2+\pi)^2})\\ 40 + 20\pi -16b -8\pi b = 20\pi - 4\pi b\\ 40 = 4b(\pi+4)\\b = \frac{10}{\pi+4}$
Now,
$A^{''}(b) = \frac{-8}{2+\pi}+\frac{4\pi}{(2+\pi)^2} = \frac{-16-8\pi+4\pi}{(2+\pi)^2} = \frac{-16-4\pi}{(2+\pi)^2} \\ A^{''}(\frac{10}{\pi+4}) < 0$
Hence, b = 5/2 is the point of maxima
$l = \frac{2(10-2b)}{2+\pi} = \frac{2(10-2.\frac{10}{4+\pi})}{2+\pi} = \frac{20}{4+\pi}$
$r= \frac{l}{2}= \frac{20}{2(4+\pi)}=\frac{10}{4+\pi}$
Hence, these are the dimensions of the window to admit maximum light through the whole opening

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

Let the angle between AC and BC is $\theta$
So, the angle between AD and ED is also $\theta$
Now,
CD = $b \ cosec\theta$
And
AD = $a \sec\theta$
AC = H = AD + CD
= $a \sec\theta$ + $b \ cosec\theta$
$\frac{dH}{d\theta} = a \sec\theta\tan\theta - b\cot\theta cosec \theta\\ \frac{dH}{d\theta} = 0\\ a \sec\theta\tan\theta - b\cot\theta cosec \theta =0\\ a \sec\theta\tan\theta = b\cot\theta cosec \theta\\ a\sin^3\theta = b\cos^3\theta\\ \tan^3\theta = \frac{b}{a} \\ \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Now,
$\frac{d^2H}{d\theta^2} > 0$
When $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Hence, $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$ is the point of minima
$\sec \theta = \frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}}$ and $cosec \theta = \frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$

AC = $\frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}} +$ $\frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$ = $(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$
Hence proved

Given function is
$f(x) = (x-2)^4(x+1)^3$
$f^{'}(x) = 4(x-2)^3(x+1)^3 + 3(x+1)^2(x-2)^4\\ f^{'}(x)= 0\\ 4(x-2)^3(x+1)^3 + 3(x+1)^2(x-2)^4=0\\ (x-2)^3(x+1)^2(4(x+1) + 3(x-2))\\ x = 2 , x = -1 \ and \ x = \frac{2}{7}$
Now, for value x close to $\frac{2}{7}$ and to the left of $\frac{2}{7}$ , $f^{'}(x) > 0$ ,and for value close to $\frac{2}{7}$ and to the right of $\frac{2}{7}$ $f^{'}(x) < 0$
Thus, point x = $\frac{2}{7}$ is the point of maxima
Now, for value x close to 2 and to the Right of 2 , $f^{'}(x) > 0$ ,and for value close to 2 and to the left of 2 $f^{'}(x) < 0$
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point of inflexion

$f (x) = \cos ^2 x + \sin x$
$f^{'} (x) = 2\cos x(-\sin x) + \cos x \\ f^{'}(x) = 0\\ -2\cos x\sin x + \cos x=0\\ \cos x(1-2\sin x) = 0\\ either\\ \cos x = 0 \ \ \ \ \ \ and \ \ \ \ \ \ \ \sin x = \frac{1}{2} \\ x = \frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{6} \ \ \ \ \ as \ x \ \epsilon [0,\pi]$
Now,
$f^{''} (x) = -2(-\sin x)\sin x - 2\cos x\cos x + (-\sin x) \\ f^{''}(x)= 2\sin^2x - 2\cos^2x - \sin x\\ f^{''}(\frac{\pi}{6}) = \frac{-3}{2} < 0$
Hence, the point $x = \frac{\pi}{6}$ is the point of maxima and the maximum value is
$f (\frac{\pi}{6}) = \cos ^2 \frac{\pi}{6} + \sin \frac{\pi}{6} = \frac{3}{4}+\frac{1}{2} = \frac{5}{4}$
And
$f^{''}(\frac{\pi}{2}) = 1 > 0$
Hence, the point $x = \frac{\pi}{2}$ is the point of minima and the minimum value is
$f (\frac{\pi}{2}) = \cos ^2 \frac{\pi}{2} + \sin \frac{\pi}{2} =0 + 1 = 1$

The volume of a cone (V) = $\frac{1}{3}\pi R^2h$
The volume of the sphere with radius r = $\frac{4}{3}\pi r^3$
By Pythagoras theorem in $\Delta ADC$ we ca say that
$OD^2 = r^2 - R^2 \\ OD = \sqrt{r^2 - R^2}\\ h = AD = r + OD = r + \sqrt{r^2 - R^2}$
V = $\frac{1}{3}\pi R^2 (r + \sqrt{r^2+R^2})\\ = \frac{1}{3}\pi R^2r + \frac{1}{3}\pi R^2 \sqrt{r^2+R^2}$
$\frac{1}{3}\pi R^2 (r + \sqrt{r^2-R^2})\\ V^{'}(R)= \frac{2}{3}\pi Rr + \frac{2}{3}\pi R \sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}}\\ V^{'}(R) = 0\\ \frac{1}{3}\pi R \left ( 2r + 2\sqrt{r^2-R^2} - \frac{R^2}{\sqrt{r^2-R^2}} \right ) = 0\\ \frac{1}{3}\pi R \left ( \frac{2r\sqrt{r^2-R^2} + 2r^2-2R^2 -R^2}{\sqrt{r^2-R^2}} \right ) = 0\\ R \neq 0 \ So, \\ 2r\sqrt{r^2-R^2} = 3R^2 - 2r^2\\ Square \ both \ sides\\ 4r^4-4r^2R^2 = 9R^4 + 4r^4 - 12R^2r^2\\ 9R^4-8R^2r^2 = 0\\ R^2(9R^2-8r^2) = 0\\ R \neq 0 \ So, 9R^2 = 8r^2\\ R = \frac{2\sqrt2r}{3}$
Now,
$V^{''}(R)= \frac{2}{3}\pi r + \frac{2}{3}\pi \sqrt{r^2-R^2}+\frac{2}{3}\pi R.\frac{-2R}{2\sqrt{r^2-R^2}}-\frac{3\pi R^2}{\sqrt{r^2-R^2}} - \frac{(-1)(-2R)}{(r^2+R^2)\frac{3}{2}}\\ V^{''}(\frac{2\sqrt2r}{3}) < 0$
Hence, the point $R = \frac{2\sqrt2r}{3}$ is the point of maxima
$h = r + \sqrt{r^2-R^2} = r + \sqrt{r^2-\frac{8r^2}{9}} = r + \frac{r}{3} = \frac{4r}{3}$
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}{3}$

Let's do this question by taking an example
suppose
$f(x)= x^3 > 0 , (a.b)$
Now, also
$f{'}(x)= 3x^2 > 0 , (a,b)$
Hence by this, we can say that f is an increasing function on (a, b)

. Also, find the maximum volume.

The volume of the cylinder (V) = $\pi r^2 h$
By Pythagoras theorem in $\Delta OAB$
$OA = \sqrt{R^2-r^2}$
h = 2OA
$h = 2\sqrt{R^2-r^2}$
$V = 2\pi r^2\sqrt{R^2-r^2}$
$V^{'}(r) = 4\pi r\sqrt{R^2-r^2}+2\pi r^2 . \frac{-2r}{2\sqrt{R^2-r^2}}\\ V^{'}(r) = 0\\ 4\pi r\sqrt{R^2-r^2}- \frac{2\pi r^3}{\sqrt{R^2-r^2}} = 0\\ 4\pi r (R^2-r^2 ) - 2\pi r^3 = 0\\ 6\pi r^3 = 4\pi rR^2\\ r =\frac{\sqrt6R}{3}$
Now,
$V^{''}(r) = 4\pi \sqrt{R^2-r^2}+4\pi r.\frac{-2r}{2\sqrt{R^2-r^2}}- \frac{6\pi r^2}{\sqrt{R^2-r^2}}.\frac{(-1)-2r}{2(R^2-r^2)\frac{3}{2}}\\ V^{''}(\frac{\sqrt6R}{3}) < 0$
Hence, the point $r = \frac{\sqrt6R}{3}$ is the point of maxima
$h = 2\sqrt{R^2-r^2} = = 2\sqrt{R^2 - \frac{2R^2}{3}} =\frac{2R}{\sqrt3}$
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2 R }{\sqrt 3 }$
and maximum volume is
$V = \pi r^2 h = \pi \frac{2R^2}{3}.\frac{2R}{\sqrt3} = \frac{4\pi R^3}{3\sqrt3}$

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

Volume of cylinder = $\pi r^2 h'$
Volume of cone = $\frac{1}{3}\pi R^2 h$
Now, we have
$R = h\tan a$
Now, since $\Delta AOG \and \Delta CEG$ are similar
$\frac{OA}{OG} = \frac{CE}{EG}$
$\frac{h}{R} = \frac{h'}{R-r}$
$h'=\frac{h(R-r)}{R}$
$h'=\frac{h(h\tan a-r)}{h\tan a} = \frac{h\tan a- r}{\tan a}$
Now,
$V = \pi r^ 2 h' = \pi r^2 .\frac{h\tan a-r}{\tan a} = \pi r^2 h - \frac{\pi r^3}{\tan a}$
Now,
$\frac{dV}{dr}= 2\pi rh- \frac{3\pi r^2}{\tan a} \\ \frac{dV}{dr}=0\\ 2\pi rh- \frac{3\pi r^2}{\tan a} = 0\\ 2\pi rh = \frac{3\pi r^2}{\tan a}\\ r = \frac{2h\tan a}{3}$
Now,
$\frac{d^2V}{dr^2}= 2\pi h- \frac{6\pi r}{\tan a}$
at $r = \frac{2h\tan a}{3}$
$\frac{d^2V}{dr^2} = 2\pi h- 4\pi h < 0$
Hence, $r = \frac{2h\tan a}{3}$ is the point of maxima
$h' = \frac{h\tan a-r}{\tan a} = \frac{h\tan a- \frac{2h\tan a}{3}}{\tan a}= \frac{1}{3}h$
Hence proved
Now, Volume (V) at $h' = \frac{1}{3}h$ and $r = \frac{2h\tan a}{3}$ is
$V = \pi r^2 h' = \pi \left ( \frac{2h\tan a}{3} \right )^2.\frac{h}{3}= \frac{4}{27}.\pi h^3\tan^2 a$
hence proved

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

It is given that
$\frac{dV}{dt} = 314 \ m^3/h$
Volume of cylinder (V) = $\pi r^2 h = 100\pi h \ \ \ \ \ \ \ \ \ \ \ (\because r = 10 m)$
$\frac{dV}{dt} = 100\pi \frac{dh}{dt}\\ 314 = 100\pi \frac{dh}{dt}\\ \frac{dh}{dt} = \frac{3.14}{\pi} = 1 \ m/h$

A ) 22/7

B ) 6/7

C ) 7/6

D ) -6 /7

Given curves are
$x = t^2 + 3t - 8 \ and \ y = 2t^2 - 2t - 5$
At point (2,-1)
$t^2 + 3t - 8 = 2\\ t^2+3t-10=0\\ t^2+5t-2t-10=0\\ (t+5)(t-2) = 0\\ t = 2 \ and \ t = 5$
Similarly,
$2t^2-2t-5 = -1\\ 2t^2-2t-4=0\\ 2t^2-4t+2t-4=0\\ (2t+2)(t-2)=0\\ t = -1 \ and \ t = 2$
The common value between two is t = 2
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a given point is given by $\frac{dy}{dx}$
$\frac{dy}{dt} = 4t - 2$
$\frac{dx}{dt} = 2t + 3$
$\left ( \frac{dy}{dx} \right )_{t=2} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t-2}{2t+3} = \frac{8-2}{4+3} = \frac{6}{7}$
Hence, (B) is the correct answer

(B) 2

(C) 3

(D)1/2

Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$y^2 = 4x$
$2y\frac{dy}{dx} = 4\\ \frac{dy}{dx} = \frac{2}{y}$
Put this value of m in the given equation
$y = \frac{2}{y}.\frac{y^2}{4}+1 \ \ \ \ \ \ \ \ \ \ (\because y^2 = 4x \ and \ m =\frac{2}{y})\\ y = \frac{y}{2}+1\\ \frac{y}{2} = 1\\ y = 2$
$m = \frac{2}{y} = \frac{2}{2} = 1$
Hence, value of m is 1

Question:22 The normal at the point (1,1) on the curve $2y + x ^2 = 3$ is
(A) x + y = 0

(B) x – y = 0

(C) x + y +1 = 0

(D) x – y = 1

Given the equation of the curve
$2y + x ^2 = 3$
We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
$2\frac{dy}{dx} = -2x\\ \frac{dy}{dx} = -x$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{-x} = \frac{1}{x}$
At point (1,1)
$Slope = \frac{1}{1} = 1$
Now, the equation of normal with point (1,1) and slope = 1

$y-y_1=m(x-x_1)\\ y-1=1(x-1)\\ x-y = 0$
Hence, the correct answer is (B)

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

Given the equation of the curve
$x^2 = 4 y$
We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
$4\frac{dy}{dx} = 2x \\\ \frac{dy}{dx} = \frac{x}{2}$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x}{2}} = \frac{-2}{x}$
At point (a,b)
$Slope = \frac{-2}{a}$
Now, the equation of normal with point (a,b) and $Slope = \frac{-2}{a}$

?
It is given that it also passes through the point (1,2)
Therefore,
$2-b=\frac{-2}{a}(1-a)\\ 2a -ba = 2a -2\\ ba = 2\\b =\frac{2}{a}$ -(i)
It also satisfies equation $x^2 = 4 y\Rightarrow b = \frac{a^2}{4}$ -(ii)
By comparing equation (i) and (ii)
$\frac{2}{a} = \frac{a^2}{4}\\ a^3 = 8\\ a = 2$
$b = \frac{2}{a} = \frac{2}{2} = 1$
$Slope = \frac{-2}{a} = \frac{-2}{2} = -1$

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

$y-y_1=m(x-x_1)\\ y-1=-1(x-2)\\ y+x=3$

Given the equation of the curve
$9 y^2 = x ^3$
We know that the slope of the tangent at a point on a given curve is given by $\frac{dy}{dx}$
$18y\frac{dy}{dx} = 3x^2\\ \frac{dy}{dx} = \frac{x^2}{6y}$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x^2}{6y}} = \frac{-6y}{x^2}$
At point (a,b)
$Slope = \frac{-6b}{a^2}$
Now, the equation of normal with point (a,b) and $Slope = \frac{-6b}{a^2}$

$y-y_1=m(x-x_1)\\ y-b=\frac{-6b}{a^2}(x-a)\\ ya^2 - ba^2 = -6bx +6ab\\ ya^2+6bx=6ab+a^2b\\ \frac{y}{\frac{6b+ab}{a}}+\frac{x}{\frac{6a+a^2}{6}} = 1$
It is given that normal to the curve makes equal intercepts with the axes
Therefore,
$\frac{6b+ab}{a}=\frac{6a+a^2}{6} \\ 6b(6 + a) =a^2( 6+a)\\ a^2 = 6b$
point(a,b) also satisfy the given equation of the curve
$9 b^2 = a ^3\\ 9(\frac{a^2}{6})^2 = a^3\\ 9.\frac{a^4}{36} = a^3\\ a = 4$
$9b^2 = 4^3\\ 9b^2 =64\\ b = \pm\frac{8}{3}$
Hence, The points on the curve $9 y^2 = x ^3$ , where the normal to the curve makes equal intercepts with the axes are $\left ( 4,\pm\frac{8}{3} \right )$
Hence, the correct answer is (A)

## More About NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise

Practice questions related to all the 5 main topics covered in the Class 12 NCERT Mathematics chapter application of derivatives are covered in Class 12 Maths chapter 6 miscellaneous exercise solutions. All these solutions of miscellaneous exercise are detailed in this page and are solved by Mathematics experts. The NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise are given in detail and step by step manner.

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

• Class 12 Maths chapter 6 miscellaneous solutions can be used to prepare for board exam as well as copetitive exams to the admission for various engineering colleges across India.
• The NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise helps in revising the whole chapter and also some basic derivatives.

## Key Features Of NCERT Solutions For Class 12 Chapter 6 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 6, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 6 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 6 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 6 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 6 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 6 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exemplar Solutions

1. What are the concepts to be covered to solve NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise?

Questions related to the tropics rate of change of quantities, approximation, increasing and decreasing functions, tangents and normals and maxima and minima are covered in Class 12 Maths chapter 6 miscellaneous solutions.

2. What is the number of questions covered in the miscellaneous exercise?

24 questions are present in Class 12 Maths chapter 6 miscellaneous exercise solutions

3. How many miscellaneous solved examples are there in NCERT Class 12 Mathematics book?

10 miscellaneous questions are solved in Class 12 NCERT Mathematics book.

4. How many exercises are covered in the NCERT chapter application of derivatives?

Including miscellaneous, there are 6 exercises. For more questions students can use NCERT exemplar.

5. In total how many solved examples are given in class 12 NCERT Maths chapter 6?

51 solved examples are given in the chapter 6 application of derivatives

6. What is covered in exercise 6.5?

The questions regarding maximum and minimum of a function using derivatives are covered in the class 12 maths exercise 6.5

7. In which exercise the questions related to approximation are covered?

Exercise 6.4 and a few questions of Class 12 Maths chapter 6 miscellaneous solutions covers the concept of approximation.

8. What are the two tests discussed in the topic maxima and minima?

The two tests discussed are the first derivative test and the second derivative test.

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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.

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.

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) Option 2) Option 3) Option 4)

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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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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##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

2 Jobs Available
##### Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

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
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

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
##### 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
##### 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
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

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 cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

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
##### Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

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
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 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
##### 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
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

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
##### 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
##### 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
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available