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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

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

( 17/81) ^{1/4 }

Answer:

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

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

Answer:

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

Question:2. Show that the function given by f ( x ) = \frac{\log x}{x} has maximum at x = e.

Answer:

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

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

Answer:

It is given that the base of the triangle is b
and let the side of the triangle be x cm , \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

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

Answer:

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

Question:5 . Show 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.

Answer:

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

Question:6(i) Find the intervals in which the function f given by f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x } is

increasing

Answer:

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

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

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

decreasing

Answer:

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 )

Question:7(i) Find the intervals in which the function f given by f (x) = x ^3 + \frac{1}{x^3}, x \neq 0

Increasing

Answer:

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)

Question:7(ii) Find the intervals in which the function f given by f ( x) = x ^3 + \frac{1}{x^3} , x \neq 0

decreasing

Answer:

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
1654666897895 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)

Question:8 Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1 with its vertex at one end of the major axis.

Answer:

1628072034896 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}

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

Answer:

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8 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

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

Answer:

It is given that
the sum of the perimeter of a circle and square is k = 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

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

Answer:

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

= l+2b + \pi \frac{l}{2}
1628072096595
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

Question:12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is ( a ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}

Answer:

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

1628072130108 Let the angle between AC and BC is \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

Question:13 Find the points at which the function f given by f(x) = (x-2)^4(x+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexion

Answer:

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

Question:14 Find the absolute maximum and minimum values of the function f given by
f (x) = \cos ^2 x + \sin x , x \epsilon [ 0 , \pi ]

Answer: Given function is
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

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

Answer:

1628072169934 The volume of a cone (V) = \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}

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

Answer:

Let's do this question by taking an example
suppose
f(x)= 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)

Question:17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \frac{2 R }{\sqrt 3 } . Also, find the maximum volume.

Answer:

1628072213939 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}

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

\frac{4}{27}\pi h ^3 \tan ^2 \alpha

Answer:

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

Volume of cylinder = \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

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

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

Answer:

It is given that
\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
Hence, (A) is correct answer

Question:20 The slope of the tangent to the curve x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 at the point
(2,– 1) is

A ) 22/7

B ) 6/7

C ) 7/6

D ) -6 /7

Answer:

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

Question:21 The line y is equal to is a tangent to the curve if the value of m is
(A) 1

(B) 2

(C) 3

(D)1/2

Answer:

Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by \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
Hence, (A) is correct answer

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

Answer:

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)

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

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

Answer:

Given the equation of the curve
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
Hence, correct answer is (A)

Question:24 The points on the curve 9 y^2 = x ^3 , where the normal to the curve makes equal intercepts with the axes are

A ) \left ( 4 , \pm \frac{8}{3} \right )\\\\ .\: \: \: \: \: B ) \left ( 4 , \frac{-8}{3} \right ) \\\\ . \: \: \: \: \: C) \left ( 4 , \pm \frac{3}{8} \right ) \\\\ . \: \: \: \: D ) \left ( \pm 4 , \frac{3}{8} \right )

Answer:

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.

Also Read| Application of Derivatives Class 12 Notes

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.
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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.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

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.

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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Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

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
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
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
Veterinary Doctor
5 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
Speech Therapist
4 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
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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