Maths NCERT Exemplar Solutions Class 12th Chapter Six: Overview, Questions, Preparation

1. If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{\pi }{3}$ .

2. Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^5 – 5x⁴ + 5x³ – 1$ Also, find the corresponding local maximum and local minimum values.

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=x^5-5x^4+5x^3-1\\ f^{\text{'}}\left(x\right)=5x^4-20x^3+15x^2\\ Forlocalmaximaandlocalminima,f^{\text{'}}\left(x\right)=0\\ \therefore 5x^4-20x^3+15x^2=0\Rightarrow 5x^2\left(x^2-4x+3\right)=0\\ \Rightarrow 5x^2\left(x^2-3x-x+3\right)=0\Rightarrow x^2\left(x-3\right)\left(x-1\right)=0\\ \therefore x=0,x=1andx=3\\ Now{f}^{\text{'}\text{'}}\left(x\right)=20x^3-60x^2+30x\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=0}=20{\left(0\right)}^3-60{\left(0\right)}^2+30\left(0\right)=0whichisneitherMaximanorMinima.\\ \therefore f\left(x\right)hasthepointofinflextionatx=0\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=1}=20{\left(1\right)}^3-60{\left(1\right)}^2+30\left(1\right)\\ =20-60+30=-10<0Maxima\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=3}=20{\left(3\right)}^3-60{\left(3\right)}^2+30\left(3\right)\\ =540-540+90=90>0Minima\\ Themaximumvalueofthefunctionatx=1\\ f\left(x\right)={\left(1\right)}^5-5{\left(1\right)}^4+5{\left(1\right)}^3-1\\ =1-5+5-1=0\\ Theminimumvalueofthefunctionatx=3\\ f\left(x\right)={\left(3\right)}^5-5{\left(3\right)}^4+5{\left(3\right)}^3-1\\ =243-405+135-1=378-406=-28\\ Hence,thefunctionhasitsmaximaatx=1andthemaximumvalue=0andithasminimum\\ valueatx=3anditsminimumvalue=-28\\ x=0isthepointofinflextion.\end{array}$

3.A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription, and it is believed that for every increase of Re 1/-, one subscriber will discontinue the service. Find what increase will bring maximum profit.

Sol:

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

4. If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1$ , then prove that a² cos²α + b² sin²α =p².

Sol:

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