Application of Derivatives

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol: 

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthaty=mx+cwilltouchtheellipse\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1if{c}^2=a^2{m}^2+b^2\\ Hereequationofstraightlineisxcos\alpha +ysin\alpha =pandthatofellipseis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ xcos\alpha +ysin\alpha =p\\ \Rightarrow ysin\alpha =-xcos\alpha +p\\ \Rightarrow y=-x\frac{cos\alpha }{sin\alpha }+\frac{p}{sin\alpha }\Rightarrow y=-xcot\alpha +\frac{p}{sin\alpha }\\ Compaingwithy=mx+c,weget\\ m=-cot\alpha andc=\frac{p}{sin\alpha }\\ So,accordingtothecondition,weget{c}^2=a^2{m}^2+b^2\\ \frac{p^2}{si{n}^2\alpha }=a^2{\left(-cot\alpha \right)}^2+b^2\\ \Rightarrow \frac{p^2}{si{n}^2\alpha }=a^2\frac{co{s}^2\alpha }{si{n}^2\alpha }+b^2\Rightarrow p^2=a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha \\ Hence,a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^{2}Henceproved.\end{array}$

Sol:

$\begin{array}{l}LetusconsiderthatthecompanyincreasestheannualsubscriptionbyRs.x\\ So,xisthenumberofsubscriberswhodiscontinuetheservices.\\ \therefore Totalrevenue,R\left(x\right)=\left(500-x\right)\left(300+x\right)\\ =150000+500x-300x-x^2\\ =-x^2+200x+150000\\ Differentiatingbothsidesw.r.t.x,weget{R}^{\text{'}}\left(x\right)=-2x+200\\ Forlocalmaximaandlocalminima,R^{\text{'}}\left(x\right)=0\\ -2x+200=0\Rightarrow x=100\\ R^{\text{'}\text{'}}\left(x\right)=-2<0Maxima\\ So,R\left(x\right)ismaximumatx=100\\ Hence,inordertogetmaximumprofit,thecompanyshouldincreaseitsannualsubscription\\ byRs.100.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetusconsiderthatthecompanyincreasestheannualsubscriptionbyRs.x\\ So,xisthenumberofsubscriberswhodiscontinuetheservices.\\ \therefore Totalrevenue,R\left(x\right)=\left(500-x\right)\left(300+x\right)\\ =150000+500x-300x-x^2\\ =-x^2+200x+150000\\ Differentiatingbothsidesw.r.t.x,weget{R}^{\text{'}}\left(x\right)=-2x+200\\ Forlocalmaximaandlocalminima,R^{\text{'}}\left(x\right)=0\\ -2x+200=0\Rightarrow x=100\\ R^{\text{'}\text{'}}\left(x\right)=-2<0Maxima\\ So,R\left(x\right)ismaximumatx=100\\ Hence,inordertogetmaximumprofit,thecompanyshouldincreaseitsannualsubscription\\ byRs.100.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

The equation of the given curve is $9y^2=x^3$

Differentiate with respect to x, we have:

$\begin{array}{l}9\left(2y\right)\frac{dy}{dx}=3x^2\\ \Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}\end{array}$

The slope of the normal to the given curve at point $\left(x_1,y_1\right)$ is

$\frac{-1}{{\frac{dy}{dx}]}_{\left(x_1,y_1\right)}}=-\frac{6y_1}{{x_1}^2}$

$\therefore$ The equation of the normal to the curve at $\left(x_1,y_1\right)$ is

$\begin{array}{l}y-y_1=-\frac{6y_1}{{x_1}^2}(x-x_1)\\ \Rightarrow {x_1}^{2}y-{x_1}^2{y}_1=-6x{y}_1+6x_1{y}_1\\ \Rightarrow 6x{y}_1+{x_1}^{2}y=6x_1{y}_1+{x_1}^2{y}_1\\ \Rightarrow \frac{6x{y}_1}{6x_1{y}_1+{x_1}^2{y}_1}+\frac{{x_1}^{2}y}{6x_1{y}_1+{x_1}^2{y}_1}=1\\ \Rightarrow \frac{x}{\frac{x_1(6+x_1)}{6}}+\frac{y}{\frac{y_1(6+x_1)}{x_1}}\end{array}$

It is given that the normal makes intercepts with the axes.

Therefore, we have:

$\begin{array}{l}\therefore \frac{x_1(6+x_1)}{6}=\frac{y_1(6+x_1)}{x_1}\\ \Rightarrow \frac{x_1}{6}=\frac{y_1}{x_1}\\ \Rightarrow {x_1}^2=6y_1..........(i)\end{array}$

Also, the point $\left(x_1,y_1\right)$ lies on the curve, so we have

$9{y_1}^2={x_1}^3..........(ii)$

From (i) and (ii), we have:

$9\left(\frac{{x_1}^2}{6}\right)={x_1}^3\Rightarrow \frac{{x_1}^4}{4}={x_1}^3\Rightarrow x_1=4$

From (ii), we have:

$\begin{array}{l}9{y_1}^2={(4)}^3=64\\ \Rightarrow {y_1}^2=\frac{64}{9}\\ \Rightarrow y_1=\pm \frac{8}{3}\end{array}$

Hence, the required points are $\left(4,\pm \frac{8}{3}\right)$ .

Therefore, option (A) is correct.

The equation of the given curve is $x^2=4y$

Differentiating with respect to x, we have:

$\begin{array}{l}2x=4.\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}=\frac{x}{2}\end{array}$

The slope of the normal to the given curve at point (h,k) is given by,

$\frac{-1}{{\frac{dy}{dx}]}_{(h,k)}}=-\frac{2}{h}$

$\therefore$ Equation of the normal at point (h,k) is given as:

$y-k=\frac{-2}{h}(x-h)$

Now, it is given that the normal passes through the point (1,2).

Therefore, we have:

$2-k=\frac{-2}{h}(1-h)or,k=2+\frac{2}{h}(1-h)..........(i)$

Since (h,k) lies on the curve $x^2=4y$ ,we have $h^2=4k$

$\Rightarrow k=\frac{h^2}{4}$

From equation (i), we have:

$\begin{array}{l}\frac{h^2}{4}=2+\frac{2}{h}(1-h)\\ \Rightarrow \frac{h^3}{4}=2h+2-2h=2\end{array}$

$\begin{array}{l}\Rightarrow h^3=8\\ \Rightarrow h=2\\ \therefore k=\frac{h^2}{4}\Rightarrow k=1\end{array}$

Hence, the equation of the normal is given as:

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

Therefore, option (A) is correct.

The equation of the given curve is $2y+x^2=3$ .

Differentiate with respect to x, we have:

$\begin{array}{l}\frac{2dy}{dx}+2x=0\\ \Rightarrow \frac{dy}{dx}=-x\\ \therefore {\frac{dy}{dx}]}_{(1,1)}=-1\end{array}$

The slope of the normal to the given curve at point (1,1) is

$\frac{-1}{{\frac{dy}{dx}]}_{(1,1)}}=-1$

Hence, the equation of the normal to the given curve at (1,1) is given as:

$\begin{array}{l}\Rightarrow y-1=1(x-1)\\ \Rightarrow y-1=x-1\\ \Rightarrow x-y=0\end{array}$

Therefore, option (B) is correct.

The equation of the tangent to the given curve is $y=mx+1.$

Now, substituting $y=mx+1.$ in $y^2=4x,$ we get:

$\begin{array}{l}\Rightarrow {\left(mx+1\right)}^2=4x\\ \Rightarrow m^2{x}^2+1+2mx-4x=0\\ \Rightarrow m^2{x}^2+x(2m-4)+1=0..........(i)\end{array}$

Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:

Discriminant = 0

$\begin{array}{l}{(2m-4)}^2-4{(m)}^2(1)=0\\ \Rightarrow 4m^2+16-16m-4m^2=0\\ \Rightarrow 16-16m=0\\ \Rightarrow m=1\end{array}$