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

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 $x=\frac{16}{81}and\mathrm{\Delta }x=\frac{1}{81}$
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-x^{\frac{1}{4}}$
$=(\frac{16}{81}+\frac{1}{81}{)}^{\frac{1}{4}}-(\frac{16}{81}{)}^{\frac{1}{4}}$
$(\frac{17}{81}{)}^{\frac{1}{4}}=\mathrm{\Delta }y+\frac{2}{3}$
Now, we know that $\mathrm{\Delta }y$ is approximate equals to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ =\frac{1}{4x^{\frac{3}{4}}}.\frac{1}{81}(\because y=x^{\frac{1}{4}}and\mathrm{\Delta }x=\frac{1}{81}) \\ =\frac{1}{4(\frac{16}{81}{)}^{\frac{3}{4}}}.\frac{1}{81}=\frac{27}{4\times 8}.\frac{1}{81}=\frac{1}{96} \end{gathered}$$
Now,
$(\frac{17}{81}{)}^{\frac{1}{4}}=\mathrm{\Delta }y+\frac{2}{3}=\frac{1}{96}+\frac{2}{3}=\frac{65}{96}=0.677$
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 $x=32and\mathrm{\Delta }x=1$
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{-1}{5}}-x^{\frac{-1}{5}}$
$=(32+1{)}^{\frac{-1}{5}}-(32{)}^{\frac{-1}{5}}$
$(33{)}^{\frac{-1}{4}}=\mathrm{\Delta }y+\frac{1}{2}$
Now, we know that $\mathrm{\Delta }y$ is approximately equals to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ =\frac{-1}{5x^{\frac{6}{5}}}.1(\because y=x^{\frac{-1}{5}}and\mathrm{\Delta }x=1) \\ =\frac{-1}{5(32{)}^{\frac{6}{5}}}.1=\frac{-1}{5\times 64}.1=\frac{-1}{320} \end{gathered}$$
Now,
$(33{)}^{\frac{-1}{5}}=\mathrm{\Delta }y+\frac{1}{2}=\frac{-1}{320}+\frac{1}{2}=\frac{159}{320}=0.497$
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
$f(x)=\frac{\log x}{x}$
$f^{{}^{\prime }}(x)=\frac{1}{x}.\frac{1}{x}+logx\frac{-1}{x^2}=\frac{1}{x^2}(1-\log x)$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{1}{x^2}(1-\log x)=0 \\ \frac{1}{x^2}\ne 0Sologx=1\Rightarrow x=e \end{gathered}$$
Hence, x =e is the critical point
Now,
$$\begin{gathered} f^{{}^{″}}(x)=\frac{-2x}{x^3}(1-\log x)+\frac{1}{x^2}(-\frac{1}{x})=\frac{1}{x^3}(-2x+2xlogx-1) \\ {f}^{{}^{″}(e)}=\frac{-1}{e^3}<0 \end{gathered}$$
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 , $\frac{dx}{dt}=-3cm/s$
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
$\frac{dA}{dx}=\frac{1}{2}b\frac{2b}{\frac{\sqrt{3}b}{2}}.(-3)=-\sqrt{3}b$
Hence, the area decreasing when the two equal sides are equal to the base is $\sqrt{3}b$ $c{m}^2/s$

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

Answer:

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

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

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

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-1=-1(x-2) \\ y+x=3 \end{gathered}$$
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
$Slopeofnormal=\frac{-1}{Slopeoftangent}=\frac{-1}{\tan \theta }$
equation of normal with given points and slope
$$\begin{gathered} y_2-y_1=m(x_2-x_1) \\ y-a\sin \theta +a\theta \cos \theta =\frac{-1}{\tan \theta }(x-a\cos \theta -a\theta \sin \theta ) \\ y\sin \theta -a{\sin }^2\theta +a\theta \cos \theta \sin \theta =-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta \\ y\sin \theta +x\cos \theta =a \end{gathered}$$
Hence, the equation of normal is $y\sin \theta +x\cos \theta =a$
Now perpendicular distance of normal from the origin (0,0) is
$$\begin{gathered} D=\frac{|(0)\sin \theta +(0)\cos \theta -a|}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}=|-a|=a=constant \\ (\because {\sin }^{2}x+{\cos }^{2}x=1) \end{gathered}$$
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
$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$
$f^{{}^{\prime }}(x)=\frac{(4\cos x-2-\cos x+x\sin x)(2+\cos x)-(4\sin x-2x-x\cos x)(-\sin x)}{(2+\cos x{)}^2}$
$=\frac{4\cos x-{\cos }^{2}x}{2+\cos x}$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{4\cos x-{\cos }^{2}x}{2+\cos x}=0 \\ \cos x(4-\cos x)=0 \\ \cos x=0and\cos x=4 \end{gathered}$$
But $\cos x\ne 4$
So,
$$\begin{gathered} \cos x=0 \\ x=\frac{\pi }{2}and\frac{3\pi }{2} \end{gathered}$$
Now three ranges are there $(0,\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2})and(\frac{3\pi }{2},2\pi )$
In interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$ , $f^{{}^{\prime }}(x)>0$

Hence, the given function $f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is increasing in the interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$
in interval $,(\frac{\pi }{2},\frac{3\pi }{2}),f^{{}^{\prime }}(x)<0$ 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
$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$
$f^{{}^{\prime }}(x)=\frac{(4\cos x-2-\cos x+x\sin x)(2+\cos x)-(4\sin x-2x-x\cos x)(-\sin x)}{(2+\cos x{)}^2}$
$=\frac{4\cos x-{\cos }^{2}x}{2+\cos x}$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{4\cos x-{\cos }^{2}x}{2+\cos x}=0 \\ \cos x(4-\cos x)=0 \\ \cos x=0and\cos x=4 \end{gathered}$$
But $\cos x\ne 4$
So,
$$\begin{gathered} \cos x=0 \\ x=\frac{\pi }{2}and\frac{3\pi }{2} \end{gathered}$$
Now three ranges are there $(0,\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2})and(\frac{3\pi }{2},2\pi )$
In interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$ , $f^{{}^{\prime }}(x)>0$

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

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

Increasing

Answer:

Given function is
$f(x)=x^3+\frac{1}{x^3}$
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2+\frac{-3x^2}{x^4} \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2+\frac{-3x^2}{x^4}=0 \\ {x}^4=1 \\ x=\pm 1 \end{gathered}$$
Hence, three intervals are their $(-\mathrm{\infty },-1),(-1,1)and(1,\mathrm{\infty })$
In interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty }),f^{{}^{\prime }})x>0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is increasing in interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty })$
In interval (-1,1) , $f^{{}^{\prime }}(x)<0$
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 $f(x)=x^3+\frac{1}{x^3},x\ne 0$

decreasing

Answer:

Given function is
$f(x)=x^3+\frac{1}{x^3}$
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2+\frac{-3x^2}{x^4} \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2+\frac{-3x^2}{x^4}=0 \\ {x}^4=1 \\ x=\pm 1 \end{gathered}$$
Hence, three intervals are their $(-\mathrm{\infty },-1),(-1,1)and(1,\mathrm{\infty })$
In interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty }),f^{{}^{\prime }})x>0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is increasing in interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty })$
In interval (-1,1) , $f^{{}^{\prime }}(x)<0$
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:

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 = $(-n,\frac{b}{a}.\sqrt{a^2-n^2})$
Coordinates of B = $(-n,-\frac{b}{a}.\sqrt{a^2-n^2})$
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,
$\frac{dA}{dn}=\frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{b{n}^2}{\sqrt{a^2-n^2}}$
Now,
$$\begin{gathered} \frac{dA}{dn}=0 \\ \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{b{n}^2}{\sqrt{a^2-n^2}}=0 \\ -abn+n(a^2-n^2)-b{n}^2=0 \\ \Rightarrow n=-a,\frac{a}{2} \end{gathered}$$
but n cannot be zero
therefore, $n=\frac{a}{2}$
Now, at $n=\frac{a}{2}$
$\frac{d^{2}A}{d{n}^2}<0$
Therefore, $n=\frac{a}{2}$ is the point of maxima
Now,
$b=2\frac{b}{a}.\sqrt{a^2-(\frac{a}{2}{)}^2}=\sqrt{3}b$
$h=(a+n)=a+\frac{a}{2}=\frac{3a}{2}$
Now,
Therefore, Area (A) $=\frac{1}{2}bh=\frac{1}{2}\sqrt{3}b\frac{3a}{2}=\frac{3\sqrt{3}ab}{4}$

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)
$A(b)=4+2h(\frac{4}{b}+b)$
$$\begin{gathered} A^{{}^{\prime }}(b)=2h(\frac{-4}{b^2}+1) \\ {A}^{{}^{\prime }}(b)=0 \\ 2h(\frac{-4}{b^2}+1)=0 \\ {b}^2=4 \\ b=2 \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(b)=2h(\frac{-4\times -2b}{b^3}) \\ {A}^{{}^{″}}(2)=8>0 \end{gathered}$$
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 = $4m^2$
building of tank costs Rs 70 per sq metres for the base
Therefore, for $4m^2$ Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) = $16m^2$
building of tank costs Rs 45 per square metre for sides
Therefore, for $16m^2$ 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) = $\pi r^2+a^2$
$A=\pi r^2+(\frac{k-2\pi r}{4}{)}^2$
$$\begin{gathered} A^{{}^{\prime }}(r)=2\pi r+2(\frac{k-2\pi r}{16})(-2\pi ) \\ {A}^{{}^{\prime }}(r)=0 \\ 2\pi (\frac{8r-k-2\pi r}{8})=0 \\ r=\frac{k}{8-2\pi } \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(r)=2\pi (\frac{8-2\pi }{8})=0 \\ {A}^{{}^{″}}(\frac{k}{8-2\pi })>0 \end{gathered}$$
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

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

$$\begin{gathered} l+2b+\pi \frac{l}{2}=10 \\ l=\frac{2(10-2b)}{2+\pi } \end{gathered}$$
Area of window id given by (A) = $lb+\frac{\pi }{2}{\left(\frac{l}{2}\right)}^2$
$=\frac{2(10-2b)}{2+\pi }b+\frac{\pi }{2}{\left(\frac{10-2b}{2+\pi }\right)}^2$
$A^{{}^{\prime }}(b)=\frac{20-8b}{2+\pi }+\frac{\pi }{2}.2(\frac{10-2b}{2+\pi }).\frac{(-2)}{2+\pi }$
$$\begin{gathered} =\frac{20-8b}{2+\pi }-2\pi (\frac{10-2b}{(2+\pi {)}^2}) \\ {A}^{{}^{\prime }}(b)=0 \\ \frac{20-8b}{2+\pi }=2\pi (\frac{10-2b}{(2+\pi {)}^2}) \\ 40+20\pi -16b-8\pi b=20\pi -4\pi b \\ 40=4b(\pi +4) \\ b=\frac{10}{\pi +4} \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(b)=\frac{-8}{2+\pi }+\frac{4\pi }{(2+\pi {)}^2}=\frac{-16-8\pi +4\pi }{(2+\pi {)}^2}=\frac{-16-4\pi }{(2+\pi {)}^2} \\ {A}^{{}^{″}}(\frac{10}{\pi +4})<0 \end{gathered}$$
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

Let the angle between AC and BC is $\theta$
So, the angle between AD and ED is also $\theta$
Now,
CD = $bcosec\theta$
And
AD = $a\sec \theta$
AC = H = AD + CD
= $a\sec \theta$ + $bcosec\theta$
$$\begin{gathered} \frac{dH}{d\theta }=a\sec \theta \tan \theta -b\cot \theta cosec\theta \\ \frac{dH}{d\theta }=0 \\ a\sec \theta \tan \theta -b\cot \theta cosec\theta =0 \\ a\sec \theta \tan \theta =b\cot \theta cosec\theta \\ a{\sin }^3\theta =b{\cos }^3\theta \\ {\tan }^3\theta =\frac{b}{a} \\ \tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}} \end{gathered}$$
Now,
$\frac{d^{2}H}{d{\theta }^2}>0$
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
$f(x)=(x-2{)}^4(x+1{)}^3$
$$\begin{gathered} f^{{}^{\prime }}(x)=4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4 \\ {f}^{{}^{\prime }}(x)=0 \\ 4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4=0 \\ (x-2{)}^3(x+1{)}^2(4(x+1)+3(x-2)) \\ x=2,x=-1andx=\frac{2}{7} \end{gathered}$$
Now, for value x close to $\frac{2}{7}$ and to the left of $\frac{2}{7}$ , $f^{{}^{\prime }}(x)>0$ ,and for value close to $\frac{2}{7}$ and to the right of $\frac{2}{7}$ $f^{{}^{\prime }}(x)<0$
Thus, point x = $\frac{2}{7}$ is the point of maxima
Now, for value x close to 2 and to the Right of 2 , $f^{{}^{\prime }}(x)>0$ ,and for value close to 2 and to the left of 2 $f^{{}^{\prime }}(x)<0$
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: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:

The volume of a cone (V) = $\frac{1}{3}\pi R^{2}h$
The volume of the sphere with radius r = $\frac{4}{3}\pi r^3$
By Pythagoras theorem in $\mathrm{\Delta }ADC$ we ca say that
$$\begin{gathered} O{D}^2=r^2-R^2 \\ OD=\sqrt{r^2-R^2} \\ h=AD=r+OD=r+\sqrt{r^2-R^2} \end{gathered}$$
V = $$\begin{gathered} \frac{1}{3}\pi R^2(r+\sqrt{r^2+R^2}) \\ =\frac{1}{3}\pi R^{2}r+\frac{1}{3}\pi R^2\sqrt{r^2+R^2} \end{gathered}$$
$$\begin{gathered} \frac{1}{3}\pi R^2(r+\sqrt{r^2-R^2}) \\ {V}^{{}^{\prime }}(R)=\frac{2}{3}\pi Rr+\frac{2}{3}\pi R\sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}} \\ {V}^{{}^{\prime }}(R)=0 \\ \frac{1}{3}\pi R(2r+2\sqrt{r^2-R^2}-\frac{R^2}{\sqrt{r^2-R^2}})=0 \\ \frac{1}{3}\pi R\left(\frac{2r\sqrt{r^2-R^2}+2r^2-2R^2-R^2}{\sqrt{r^2-R^2}}\right)=0 \\ R\ne 0So, \\ 2r\sqrt{r^2-R^2}=3R^2-2r^2 \\ Squarebothsides \\ 4r^4-4r^2{R}^2=9R^4+4r^4-12R^2{r}^2 \\ 9R^4-8R^2{r}^2=0 \\ {R}^2(9R^2-8r^2)=0 \\ R\ne 0So,9R^2=8r^2 \\ R=\frac{2\sqrt{2}r}{3} \end{gathered}$$
Now,
$$\begin{gathered} V^{{}^{″}}(R)=\frac{2}{3}\pi r+\frac{2}{3}\pi \sqrt{r^2-R^2}+\frac{2}{3}\pi R.\frac{-2R}{2\sqrt{r^2-R^2}}-\frac{3\pi R^2}{\sqrt{r^2-R^2}}-\frac{(-1)(-2R)}{(r^2+R^2)\frac{3}{2}} \\ {V}^{{}^{″}}(\frac{2\sqrt{2}r}{3})<0 \end{gathered}$$
Hence, the point $R=\frac{2\sqrt{2}r}{3}$ 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 $xϵ(a,b)$ . Then prove that f is an increasing function on (a, b).