Class 12th

Let x be the radius of the sphere ad x be radius of the right circular cone.

Let height of cone = y

Then, in ΔOBA,

(y-r)² + x² = r²

y² + r²- 2ry + x² = r²

x²=2ry - y²

So, the volume V of the cone is

$V=\frac{1}{3}\pi ·x^2·y$ $\left\{=\frac{1}{3}\pi {(radius)}^2*hight\right\}$

$=\frac{1}{3}*\left(2xy-y^2\right)y$

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

So, $\frac{dv}{dy}=\frac{\pi }{3}\left[4xy-3y^2\right]$

And $\frac{d^2}{d{y}^2}=\frac{x}{3}[4x-6y].$

At $\frac{dV}{dy}=0.$

$\Rightarrow \frac{\pi }{3}\left[4xy-3y^2\right]=0$

4x -y- 3y² = 0

$\Rightarrow y=\frac{4x}{3}$ asy> 0.

At y = $\frac{4\pi }{3},\frac{d^{2}v}{d{y}^2}=\frac{\pi }{3}[4x-8x]$ = $-\frac{4\pi r}{3}<0.$

Ø V is maximum when y = $\frac{4\pi }{3}$

We have,

f (x) = cos²x + sin x, x∈ [0, π ].

So, f (x) = 2 cos x ( -sin x) + cos x = cos x (1 - 2 sin x).

At f (x) = 0

cosx (1 - 2 sin x) = 0

cosx = 0  or  1 - 2 sin x = 0

cosx = cos $\frac{\pi }{2}$ or sin x = $\frac{1}{2}$ = sin $\frac{\pi }{6}$ = sin $x-\frac{\pi }{6}$

x= $\frac{\pi }{2}$ , x = $\frac{\pi }{6}$ and x = $\frac{5\pi }{6}$  [0, π ].

So, f $\left(\frac{\pi }{2}\right)$ = cos²$\frac{\pi }{2}$ + sin $\frac{\pi }{2}$ = 1.

Absolute minimum of f (x) = $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and absolute minimum of f (x) = 1.

This is a short answer type question as classified in NCERT Exemplar

Semi permeable membranes are continuous sheets or films (natural or synthetic) that have a network of submicroscopic holes or pores through which small solvent molecules like water may pass but larger molecules of solute cannot. Osmosis is the process of diffusion via this membrane.

This is a short answer type question as classified in NCERT Exemplar

'Depression in the freezing point of water when a nonvolatile solute is dissolved in it' is the phenomenon involved in removing snow-covered roads in hilly places. As a result, when salt is spread over snow-covered roads, snow melts from the surface due to a drop in the freezing point of water, which aids in road cleaning.

Let P be the point on hypotenuse of a triangle. ABC, t angle at B.

Which is at distance a& b from the sides of the triangle.

Let < BAC = < MPC = .

Then, in… ΔANP,

$\Rightarrow ta{n}^3\theta =\frac{a}{b}$

$tan\theta ={\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..}$

At tan Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3,$}\right.}$

$\frac{d^2\stackrel{?}{q}}{d{\theta }^2}=+ve.>0.$ { Øall trigonometric fxⁿ are + ve in I^st quadrant}.

So, z is least for tanØ = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$

As, Sec²Ø = 1 + tan²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$ = $\frac{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}}$

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

And tan²Ø = $\frac{1}{co{t}^2\theta }$

cot²Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

And cosec²Ø = 1 + cot²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ = $\frac{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

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

Let 'x' metre be the radius of the semi-circular opening mounded on the length '2x' side of rectangle. Then, let 'y' be the breadth of the rectangle.

Then, perimeter of the window = 10m

$\Rightarrow \frac{2\pi x}{2}+2x+y+y=10$

x + 2x + 2y + = 10.

$\Rightarrow y=\frac{10-(\pi +2)\pi }{2.}$

Let the area of the window be A.

Then, A = $\frac{1}{2}\left(\pi x^2\right)+2xy.$

$=\frac{\pi x^2}{2}+2x·\left[\frac{10-(x+2)x}{2}\right].$

$=\frac{\pi x^2+20x-2\pi x^2-4x^2}{2}$

= $\frac{1}{2}$ [-πx²- 4x² + 20x].

So, $\frac{d}{dx}=\frac{1}{2}$ [ -2πx - 8x + 20]

And $\frac{d\stackrel{2}{}}{d{x}^2}=\frac{1}{2}$ [ -2π - 8] = -π -4 = -( π+ 4)

At $\frac{d}{dx}=0.$

$\frac{1}{2}$ [ -2πx - 8x + 20] = 0

2x + 8x = 20

x = $\frac{20}{2\pi +8}$ = $\frac{10}{\pi +4}.$

At x = $\frac{10}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}$ = -( π+ 4) < 0

Øx = $\frac{10}{x+4}$ is a point of minima.

And y = $\frac{10-(x+2)*\left(\frac{10}{x+4}\right)}{2}$

$=\frac{10(\pi +4)-(\pi +2)10}{2(\pi +4)}$

$=\frac{10\pi +40-10\pi -20}{2(x+4)}=\frac{10}{\pi +4.}$

Ø Dimensions of the window are

length = 2x = $\frac{20}{\pi +4}m$

breadth = y $\frac{10}{\pi +4}m.$

radius = y = $\frac{10}{\pi +4}m.$

Let r and s be the radius of the circle and length of side of square.

Then, sum of perimeter of circle and square = k

2πr +4s = k

s = $\frac{k-2*r}{4}$

The area A be the total areas of the circle and square.

Then, A = πr² + s²

$=\pi r^2+{\left(\frac{k-2\pi r}{4}\right)}^2=\pi r^2+\frac{x^2+4{\pi }^2{r}^2-4\pi rk}{16}$

$=\frac{16\pi r^2+x^2+4{\pi }^2{r}^2-4\pi k\cdot r}{16}$

$=\frac{1}{16}\left[\left(16\pi +4{\pi }^2\right){\pi }^2-4\pi kr+x^2\right].$

So, $\frac{dA}{dr}=\frac{1}{16}\left[\left(16\pi +4x^2\right)\cdot 2n-4\pi k.\right].$

And $\frac{d^{\mathrm{2A}}}{d{r}^2}=\frac{1}{16}\left[\left(6\pi +4{\pi }^2\right)2\right]=\frac{16\pi +4{\pi }^2}{8}$

At $\frac{dA}{dr}=0$

$\frac{1}{16}\left[\left(16x+4x^2\right)2x-4\pi k\right]=0$

$\Rightarrow \left(16x+4x^2\right)2x-4\pi k=0$

$\Rightarrow 4\pi [4+\pi ]2r=4\pi k.$

$r=\frac{k}{2(4+\pi ).}$

At $r=\frac{x}{2(4+\pi )},\frac{d^{\mathrm{2A}}}{d{x}^2}=\frac{16\pi +4{\pi }^2}{8}>0.$

s = $\frac{2x}{2(4+x).}$

s = 2r.

Hence, proved.