Maths

We have, f(x) = $x^3+\frac{1}{x^3},x\ne 0.$

$\Rightarrow f^{\prime }(x)=3x^2-\frac{3*1}{x^4}=3\left(x^2-\frac{1}{x^4}\right)=3\left(\frac{x^6-1}{x^4}\right).$

$=\frac{3}{x^4}\left[{\left(x^2\right)}^3-1^3\right]$

$=\frac{3}{x^4}\left(x^2-1\right)\left(x^4+x^2+1\right)$ { $\left(a^3-b^3\right)=$ $(a-b)(a^2+b^2+ab)$

$\Rightarrow f^{\prime }(x)=\frac{3(x-1)}{x^4}(x+1)\left(x^4+x^2+1\right).$ $\{?x^2-b^2=(a-b)$ $(a+b)$

At $f^{\prime }(x)=0.$

$\Rightarrow \frac{3(x-1)(x+1)\left(x^4+x^2+1\right)}{x^4}=0$

$\Rightarrow x=1x=-1. 3\left(x^4+x^2+1\right)\ne 0.$

So we have three disjoint internal i.e.,

$(-\infty ,-1],[-1,1](1,\infty ).$

When, $x\in (-\infty ,-1].$

$f^{\prime }(x)=\frac{(+)ve(-ve)(-ve)(+ve)}{(+ve)}=(+)ve \text{\hspace{0.17em}and}0 \text{\hspace{0.17em}at} x=-1.$

So, f(x) is increasing.

When $x\in [-1,1]$

$f^{\prime }(x)=(+ve)(-v)(+ve)=(-ve) 0atx=\mathrm{1and}-1$

So, f(x) is decreasing.

When $x\in [1,\infty )$

f(x) = $(+ve)(+ve) (+ve)=( +ve)\text{\hspace{0.17em}on} 0at x=1$

So, f(x) is increasing.

f(x) is increasing for x$\in$(∞,1) and [1, ∞] and decreasing for x$\in$[1, 1].

We have, f(x)= $\frac{4sinx-2x-xcosx.}{2+cosx}$

$=\frac{4sinx-x(2+cosx)}{2+cosx}$

$f(x)=\frac{4sinx}{2+cosx}-x.$

So, $f^{\prime }(x)=\frac{(2+cosx)\frac{d}{dx}(4sinx)-4sinx\frac{d}{dx}(2+cosx)}{{(2+cosx)}^2}-\frac{dx}{dx}.$

$=\frac{(2+cosx)(4cosx)-(4sinx)(-sinx)}{{(2+cosx)}^2}-1.$

$=\frac{8cosx+4co{s}^{2}x+4si{n}^{2}x}{{(2+cosx)}^2}-1$

$=\frac{8cosx+4\left(co{s}^{2}x+si{n}^{2}x\right)}{{(2+cosx)}^2}-1.$

$=\frac{8cosx+4}{{(2+cosx)}^2}-1.$

$=\frac{8\cdot cosx+4-{(2+cosx)}^2}{{(2+cosx)}^2}$

$=\frac{8cosx+4-4-4cosx-co{s}^{2}x}{{(2+cosx)}^2}$

$=\frac{4cosx-co{s}^{2}x}{{(2+cosx)}^2}=\frac{cosx(4-cosx)}{{(2+cosx)}^2}.$

Now, ${(2+cosx)}^2>0.$

And, $4-cosx>0$ as cos x lies in [1, 1].

So, (i) for increasing, f(x) ≥ 0.

cosx ≥ 0.

x lies in Ist and IVth quadrant.

i.e., f(x) is increasing for $0\le x\le \frac{x}{2}$ and $\frac{3x}{2}\le x\le 2x.$

(ii) for decreasing, f(x) ≤ 0.

cosx ≤ 0.

x lies in IInd and IIIrd quadrant.

i.e., f(x) is decreasing for $\frac{x}{2}\le x\le \frac{3x}{2}$ .

We have $x=acos\theta +a\theta sin\theta$

$\begin{array}{l}\therefore \frac{dx}{d\theta }=-asin\theta +asin\theta +a\theta cos\theta =a\theta cos\theta \\ y=asin\theta -a\theta cos\theta \\ \therefore \frac{dy}{d\theta }=acos\theta -acos\theta +a\theta sin\theta =a\theta sin\theta \\ \therefore \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\frac{a\theta sin\theta }{a\theta cos\theta }=tan\theta \end{array}$

$\therefore$ Slope of the normal at any point $\theta$ is $-\frac{1}{tan\theta }$

The equation of the normal at a given point $\left(x,y\right)$ is given by,

$\begin{array}{l}y-asin\theta +a\theta cos\theta =\frac{-1}{tan\theta }\left(x-acos\theta -a\theta sin\theta \right)\\ \Rightarrow ysin\theta -asi{n}^2\theta +a\theta sin\theta cos\theta =-xcos\theta +aco{s}^2\theta +a\theta sin\theta cos\theta \\ \Rightarrow xcos\theta +ysin\theta -a\left(si{n}^2\theta +co{s}^2\theta \right)=0\\ \Rightarrow xcos\theta +ysin\theta -a=0\end{array}$

Now, the perpendicular distance of the normal from the origin is

Equation of the curve is $y^2=4x..........(i)$

$\begin{array}{l}2y\frac{dy}{dx}=4\\ \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}\\ \therefore {\frac{dy}{dx}]}_{\left(1,2\right)}=\frac{2}{2}=1\end{array}$

Now, the slope of the normal at point $\left(1,2\right)$ is $\frac{-1}{{\frac{dy}{dx}]}_{\left(1,2\right)}}=\frac{-1}{1}=1$

$\therefore$ Equation of the normal at $\left(1,2\right)$ is $y-2=-1\left(x-1\right).$

$\begin{array}{l}\Rightarrow y-2=-x+1\\ \Rightarrow x+y-3=0\end{array}$

Let 'b' and 'x' be the fixed base and equal side of isosceales triangle.

Then,  $\frac{dx}{dt}=-3$ cm/s (Ø decreasing).

We have, f(x) $\frac{logx}{x},x>0.$

f(x) = $\begin{array}{l}\underset{\_}{x\frac{d}{dx}logx-logx\frac{d}{dx}x}\\ x^2.\end{array}$

$=\frac{x*\frac{1}{x}-logx}{x^2}=\frac{1-logx}{x^2}$

f(x) = $\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{d}{dx}{x}^2}{x^4}$

= $\frac{x^2\left(-\frac{1}{x}\right)-(1-logx)(2x)}{x^4}$

= $\frac{-1-2+2logx}{x^3}$ = $\frac{2logx-3}{x^3}$

At extreme points, f(x) = 0.

$\Rightarrow \frac{1-logx}{x^2}=0.$

$\Rightarrow logx=1=loge.$

$\Rightarrow x=e.$

At x = e, f"(e) = $\frac{2loge-3}{e^3}=\frac{2-3}{e^3}=-\frac{1}{e^3}<0$

x = e is a point of maximum.

(a) Consider $y=x^{\frac{1}{4}}.Let,x=\frac{16}{81}\&?x=\frac{1}{81}.$

$\begin{array}{l}Then,?y={\left(x+?x\right)}^{\frac{1}{4}}-x^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-{\left(\frac{16}{81}\right)}^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-\frac{2}{3}\\ \therefore {\left(\frac{17}{81}\right)}^{\frac{1}{4}}=\frac{2}{3}+?y\end{array}$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)?x=\frac{1}{4{\left(x\right)}^{\frac{3}{4}}}\left(?x\right)\left(as,y=x^{\frac{1}{4}}\right)\\ =\frac{1}{4{\left(\frac{16}{81}\right)}^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4*8}*\frac{1}{81}=\frac{1}{32*3}=\frac{1}{96}=0.010\end{array}$

Hence, the approximate value of ${\left(\frac{17}{81}\right)}^{\frac{1}{4}}$ is $\frac{2}{3}+0.010=0.667+0.010$

=0.677

(b) ${(33)}^{\frac{-1}{5}}$

(b) Consider $y=x^{\frac{1}{5}}.Let,x=32\&?x=1$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)\left(?x\right)=\frac{-1}{5{\left(x\right)}^{\frac{6}{5}}}\left(?x\right)\left(as,y=x^{\frac{1}{5}}\right)\\ =-\frac{-1}{5{\left(x\right)}^6}\left(1\right)=-\frac{1}{320}=-0.003\end{array}$

Hence, the approximate value of ${\mathrm{33}}^{\frac{1}{5}}$
is $\frac{1}{2}+\left(-0.003=0.497\right)$

We have,

$f^{\prime }(x)=[\mathrm{X(x\; -\; 1)}+$
${\mathrm{1]}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},0\le x\le 1$

$f(x)=\frac{1}{3}{\left[x^2-x+1\right]}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{d}{dx}\left(x^2-x+1\right)$

$=\frac{2x-1}{3{\left(x^2-x+1\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

At f(x) = 0.

2x – 1 = 0

$x=\frac{1}{2}\in [0,1]$

$f\left(\frac{1}{2}\right)={\left[\frac{1}{2}\left(\frac{1}{2}-1\right)+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\left[\frac{1}{4}-\frac{1}{2}+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

$={\left[\frac{1-2+4}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

= ${\left[\frac{3}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}?0.90$

$f(0)=[\mathrm{0(0\; -\; 1)}$
$\mathrm{+}{1]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

$f(1)=[1(1\; -\; 1)+\mathrm{1]}$
${}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

Option (B) is

Hence maximum value of f(x) = at x = 1 and x = 0.

Option (c) is correct.

We have,

$f(x)=\frac{1-x+x^2}{1+x+x^2}.$

$\cancel{f}(x)=\frac{\left(1+x+x^2\right)\frac{d}{dx}\left(1-x+x^2\right)-\left(1-x+x^2\right)\frac{d}{dx}(1+x+2)}{{\left(1+x+x^2\right)}^2}$

$\Rightarrow f^{\prime }(x)=\frac{\left(1+x+x^2\right)(-1+2x)-\left(1-x+x^2\right)(1+2x)}{{\left(1+x+x^2\right)}^2}$

$=\frac{-1+2x-x+2x^2-x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{{\left(1+x+x^2\right)}^2}$

$=\frac{-2+2x^2}{{\left(1+x+x^2\right)}^2}=\frac{-2\left(1-x^2\right)}{{\left(1+x+x^2\right)}^2}=-\frac{2(1+x)(1-x)}{{\left(1+x+x^2\right)}^2}$

At f(x) = 0.

$\Rightarrow \frac{-2(1+x)(1-x)}{{\left(1+x+x^2\right)}^2}=0$

x = 1 and x = -1.

At $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$ $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$

At x = -1, $f(-1)=\frac{1+1+1}{1-1+1}=3$

The maximum value of f(x) $=\frac{1}{3}x=1$

Hence, option (D) is correct.

The equation of the given curve is

x² = 2y.

Let p (x, y) be a point on the curve.

The distance of p (x, y) from (0, 5) is say S is given by

$s^2=x^2+{(y-5)}^2$

Let z = s² = x² + y² + 25 – 10y = 2y + y² -10y + 25

z = y² – 8y + 25

So,  $\frac{dz}{dy}=2y-8$

$\frac{d^{2}z}{d{y}^2}=2$