Class 12th

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.

Let x and y meters be the length and breath of the rectangular base of the tank respectively.

Then, volume V of the tank is

V = length * depth * breath.

V = 2xy = 8m³(given).

$\Rightarrow y=\frac{8}{2x}=\frac{4}{x}.$

Let 't' be the total cost of building the tank.

Then, t = cost of base + cost of sides.

= 70xy + 45[4x+4y]     {there are four sides.

= 70xy + 180x+ 180y.

= $=70\cdot x\cdot \frac{4}{x}+180x+180*\frac{4}{x}.$

$t=280+180x+\frac{720}{x}.$

So, $\frac{dz}{dx}=0+180-\frac{720}{x^2}.$

And $\frac{d^{2}t}{d{x}^2}=\frac{1440}{x^3}$

At $\frac{d^2}{dx}=0\Rightarrow 180-\frac{720}{x^2}=0$

$\Rightarrow x^2=\frac{720}{180}=4$

x = ± 2

x = 2, (x = length and it cannot be negative)

At x = 2, $\frac{d^{2}t}{d{x}^2}=\frac{1440}{2^3}=\frac{1440}{8}=180>0.$

x = 2 is point of maxima.

Hence, minimum cost = $280+180*2+\frac{720}{2}$ = 280 + 360 + 360 = 1000.

The given equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  (1)

Let the major axis be along x-axis so, vertex is at $(\pm a,0)$

Let ΔABC be the isosceles triangle inscribed on the

ellipse with one vertex C at (a, 0).

Then, let A have Co-ordinate (x0, yo) from figure.

So, Co-ordinate of B = (x0, y0)

As A and B lies on the ellipse, from equation (i),

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

It's worth noting that when the temperature drops, the values of Henry's law constant (KH) rise. Because of this, the solubility of oxygen in water increases with decreasing temperature at a given pressure. As a result, the presence of more oxygen at lower temperatures makes aquatic organisms feel more at ease in cold water than in warm water.

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].