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

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. One should visit the NCERT Solutions for Class 12 Maths chapter 6 miscellaneous exercise.

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. The following exercises are also present in the chapter.

• Application of Derivatives Exercise 6.1

• Application of Derivatives Exercise 6.2

• Application of Derivatives Exercise 6.3

• Application of Derivatives Exercise 6.4

• Application of Derivatives Exercise 6.5

## Application of Derivatives Miscellaneous Exercise

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

$( 17/81) ^{1/4 }$

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:
$( 33) ^{-1/5 }$

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.

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 ?

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).

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.

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

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

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

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

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: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.

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?

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.

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.

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

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 $( a ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}$

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

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

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
$f (x) = \cos ^2 x + \sin x , x \epsilon [ 0 , \pi ]$

$f (x) = \cos ^2 x + \sin x$
$f^{'} (x) = 2\cos x(-\sin x) + \cos x \\ f^{'}(x) = 0\\ -2\cos x\sin x + \cos x=0\\ \cos x(1-2\sin x) = 0\\ either\\ \cos x = 0 \ \ \ \ \ \ and \ \ \ \ \ \ \ \sin x = \frac{1}{2} \\ x = \frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{6} \ \ \ \ \ as \ x \ \epsilon [0,\pi]$
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

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).

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.

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

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

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

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

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

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

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)

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

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

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

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

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

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.

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