Ncert Solutions Maths class 12th

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),

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.