Class 12th

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].

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

p =K_Hx (where p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas in solution) is Henry's law expressed mathematically.

As a result of the aforementioned equation, "the lower the solubility of the gas in the liquid, the greater the value of Henry's law constant K_H at a given pressure."

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

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

The solubility rule "like dissolves like"  is based on the intermolecular forces of that exist in solution as follows:

If the intermolecular interactions in both components are similar, a substance (solute) dissolves in a solvent (ie. solvent and solute particles or molecules). When polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents, this is a regular occurrence. 

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 $y-2=-1\left(x-1\right).$

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