Ncert Solutions Maths class 12th

(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,  $\frac{dz}{dy}=2y-8$

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

At $\frac{dz}{dy}=0\Rightarrow 2y-8=0\Rightarrow y=\frac{8}{2}=4$

At y = 4,  $\frac{d^{2}z}{d{y}^2}=2>0$

y = 4, is point of minimum distance.

So, x² = 2y->x2 = 2 * 4-> x² = 8 $x=\pm 2$

Hence, the point of the nearest distance are $\left(\mathrm{2\surd 2},4\right)$ and

$\left(-\mathrm{2\surd 2},4\right).$

Option (A) is correct.

Kindly go through the solution

Let r, h, l and Ø be the radius, height, slant height and semi-vertical angle respectively of the cone. i.e., r, h, l>0.

Then, Volume V of the cone is

$=\frac{1}{3}\pi r^{2}h.$ 

$=\frac{1}{3}\pi \left(l^2-h^2\right)h$

$=\frac{1}{3}\pi \left(l^{2}h-h^3\right).$

So,  $\frac{\mathrm{dV}}{dh}=\frac{1}{3}\pi \left(l^2-3h^2\right)$

$\frac{d^{\mathrm{2V}}}{d{h}^2}=-2\pi h$

Let r and h be the radius and height of the cone.

The volume V of the cone is.

$=\frac{1}{3}\pi r^{2}h$

$r^{2}h=\frac{\mathrm{3V}}{\pi }=\mathrm{k\; (SAY)}\Rightarrow r^2=\frac{k}{h}.$

And curve surface area S is

Let r and h be the radius and height of the one in scribed in the sphere of radius R.

Then, is ΔOBC, rt angle at B (h-r)² + r² = R²

$\Rightarrow$ h² + R²- 2hR + h² = R²

$\Rightarrow$ r² = 2hR -h²

Then the volume v of the cone is, $V=\frac{1}{3}*r^{2}h$ $=\frac{1}{3}\pi \left(2hR-h^2\right)h$

$=\frac{1}{3}\pi \left(2{\mathrm{Rh}}^2-h^3\right).$

$\Rightarrow \frac{\mathrm{dV}}{dh}=\frac{\pi }{3}\left(4\mathrm{Rh}-3h^2\right).$

$\frac{d^{\mathrm{2V}}}{d{h}^2}=\frac{\pi }{3}(\mathrm{4R}-6h).$

At $\frac{\mathrm{dV}}{dh}=0$

$-\frac{\pi }{3}\left(\mathrm{4R}h-3h^2\right)=0.$

4Rh – 3h² = 0.

h(4R – 3h) = 0.

h = 0 and $h=\frac{4R}{3}$

As h> 0, $h=\frac{4}{3}.$

At $h=\frac{\mathrm{4R}}{3},\frac{d^{\mathrm{2v}}}{d{h}^2}=\frac{\pi }{3}\left[4R-6*\frac{\mathrm{4rR}}{3}\right]$

$=\frac{1}{3}[\mathrm{4R}-8R]=-\frac{4\mathrm{R\pi }}{3}<0$

$h=\frac{\mathrm{4R}}{3}$ is a point of maxima.

and $r^2=2\mathrm{hR}-h^2=2*\frac{\mathrm{4R}}{3}R-{\left(\frac{\mathrm{4R}}{3}\right)}^2$

$=\frac{3*\mathrm{8R}{}^2}{9}-\frac{1\mathrm{6R}{}^2}{9}$

$r^2=\frac{\mathrm{8R}{}^2}{9}$

Hence, Volume of Cone, $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\pi *\frac{\mathrm{8R}{}^2}{9}*\frac{\mathrm{4R}}{3}$

$=\frac{8}{27}*\frac{4}{3}\mathrm{\pi R}{}^3=\frac{8}{27}*$ Volume of sphere.

Let x and y in 'm' be the length of side of the square the radius of the circle respectily

Then, length of wire = perimeter of square + circumference of circle

$\Rightarrow$ 28 = 4x + 2πy

$\Rightarrow$ 2x + πy = 14

$\Rightarrow y=\frac{14-2x}{x}$

The combine area A of the square and the circle is

A = x² + πy²

$=x^2+\pi {\left[\frac{14-2x}{\pi }\right]}^2$

${=x^2+\frac{[2(7-x)}{x}]}^2$

$=x^2+\frac{4}{\pi }{(7-x)}^2.$

So, $\frac{\mathrm{dA}}{dx}=2x+\frac{4\cdot 2}{\pi }(7-x)+(-1)=2x-\frac{8}{\pi }(7-x)$

$\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{8}{\pi }$

At, $\frac{\mathrm{dA}}{dx}=0$

$\Rightarrow 2x-\frac{8(7-x)}{\pi }=0$

$\Rightarrow 2x=\frac{8}{\pi }(7-x)$

$\Rightarrow 2\pi x=56-8x$

$\Rightarrow (2x+8)x=56$

$\Rightarrow x=\frac{56}{2\pi +8}=\frac{28}{x+4}.$

At, $x=\frac{28}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{\gamma }{\pi }>0$

$\therefore x=\frac{28}{\pi +4}$ isa point of minima

Hence, length of square = $4x=4\left(\frac{28}{1+4}\right)=\frac{112}{\pi +4}m$

and length of circle = 2πy

$=2\pi \left(\frac{14-2x}{\pi }\right)$

$=2\pi \left[\frac{14}{\pi }-\frac{2}{1}\left[\frac{28}{\pi +4}\right]\right]$

$=28-4\left[\frac{28}{\pi +4}\right]=28-\frac{112}{1+4}$

$=\frac{28x+112-112}{x+4}$

$=\frac{28\pi }{\pi +4}$

The volume v of a cylinder of height h and radius r is

V = πr²h = 100

$\Rightarrow h=\frac{100}{\pi r^2}.$

Let, s be the surface area then

S = 2r²hr(r + h) = $2x\left(r^2+\frac{100r}{\pi r^2}\right)$

$=2\pi \left(n^2+\frac{100}{\pi n}\right).$

$\therefore \frac{dS}{dr}=2\pi \left[2x-\frac{100}{1r^2}\right]$

$\Rightarrow \frac{d^{2}S}{d{r}^2}=2\pi \left[2+\frac{200}{\pi r^3}\right]$

At, $\frac{ds}{dr}=2\pi \left[2r-\frac{100}{\pi r^2}\right]=0$

$\Rightarrow 2r=\frac{100}{\pi r^2}$

$\Rightarrow M{}^3=\frac{50}{\pi }\Rightarrow M={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},>0.$

At, $x={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},\frac{d^{\mathrm{2S}}}{d{r}^2}=2\pi \left[2+\frac{200}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}*3\right]$

$=2\pi [2+4]=12\pi >0$

$\therefore r={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ isa point of minimum

And $h=\frac{100}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=2{\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$