Differentiation (Part-38) | S N De

 

In this article, we will discuss few more important Math problems and their solutions. In part-1, $\,10\,$ mathematical problems have been discussed. You can check here. 

11. Prove that the stationary value of $\,\,x^my^nz^p\,\,$ under the condition $\,\,x+y+z=a\,\,$ is : $\,m^mn^np^p \left( \frac{a}{m+n+p} \right)^{m+n+p}$

Sol. Let $ u\equiv f(x,y,z)=x^my^nz^p \cdots (1) \\ \phi \equiv x+y+z-a \cdots (2) \\ \text{Construct the function,} \,\, F=f+\lambda \phi\\= x^my^nz^p+\lambda (x+y+z-a) \\  \frac{\partial F}{\partial x}=0 \Rightarrow mx^{m-1}y^nz^p+ \lambda=0 \cdots (3)\\ \frac{\partial F}{\partial y}=0 \Rightarrow nx^my^{n-1}z^p +\lambda=0 \cdots(4)\\ \frac{\partial F}{\partial z}=0 \Rightarrow px^{m}y^nz^{p-1}+\lambda=0 \cdots (5)$

From (3),(4) and (5), 

$\\ mx^{m-1}y^nz^p = nx^my^{n-1}z^p= px^{m}y^nz^{p-1}\\ \Rightarrow \frac{m}{x}=\frac{n}{y} \\=\frac{p}{z}=t ,\,\,\text{say}\quad [\,\,\text{Dividing by} x^m y^n z^p]\\ \therefore x=\frac{m}{t}; y= \frac{n}{t}; z==\frac{p}{t} \cdots (6)\\ \text{Substituting in (2), we get } \frac{m+n+p}{t}=a \\  \Rightarrow t=\frac{1}{a} (m+n+p). \\ \text{From (6),}\,\,x=\frac{am}{m+n+p}, y=\frac{an}{m+n+p}, z=\frac{ap}{m+n+p}\\ \text{The stationary values of}\,\, u \\=\left( \frac{am}{m+n+p}\right)^m\left( \frac{an}{m+n+p}\right)^n\left( \frac{ap}{m+n+p}\right)^p \\=m^mn^np^p\left(\frac{a}{m+n+p} \right)^{m+n+p}$

12. Solve by using Ferrari Method: $\,\,x^4-12x^3+41x^2-18x-72=0 $

Sol. $\quad x^4-12x^3=-41x^2+18x+72 \\ \Rightarrow (x^2)^2-2.x^2.6x+36x^2 \\=-41x^2+18x+72+36x^2 \\ \Rightarrow (x^2-6x)^2=-5x^2+18x+72 \\ \Rightarrow (x^2-6x +\lambda)^2= -5x^2+18x+72 \\+\lambda^2+2\lambda(x^2-6x ) \\ \Rightarrow (x^2-6x +\lambda)^2= -5x^2+18x+72\\+\lambda^2+2\lambda x^2-12\lambda x \\ \Rightarrow (x^2-6x +\lambda)^2 \\= (-5+2\lambda)x^2+(18-12\lambda)x \\+(\lambda^2+72) \cdots \cdots (1)$

R.H.S. of equation (1) will be a perfect square when        

$\,\,b^2=4ac \\ \Rightarrow (18-12\lambda)^2=4(-5+2\lambda)(\lambda^2+72) \\ \Rightarrow 324-432 \lambda +144 \lambda^2 \\=(-20+8\lambda)(\lambda^2+72) \\ \Rightarrow 324-432 \lambda +144 \lambda^2 \\ =-20 \lambda^2-1440+8\lambda^3 +576 \lambda \\ \Rightarrow 8 \lambda^3-164\lambda^2+1008 \lambda -1764=0 \\ \therefore\lambda=3,7, \frac{21}{2}$

When $\,\, \lambda=3,\,\text{equation (1) becomes}\\  (x^2-6x+3)^2=1.x^2-18x+81=(x-9)^2 \\ (x^2-6x+3)^2-(x-9)^2=0 \\ \Rightarrow (x^2-6x+3+x-9) \\ \times (x^2-6x+3-x+9)=0 \\ \Rightarrow (x^2-5x-6)(x^2-7x+12)=0 \\ \text{Solving we get, } x=6,-1; \,\,\, 4,3.$

13. Solve by using Ferrari Method: $\,\,x^4+12x-5=0 $ 

Sol. $\,\,x^4+12x-5=0\\ \Rightarrow x^4=-12x+5 \\ \Rightarrow (x^2)^2-2.x^2.1+(1)^2 \\=-12x+5-2x^2+1 \\ \Rightarrow (x^2-1)^2=-2x^2-12x+6 \\ \Rightarrow (x^2-1 +\lambda)^2 =x^2(-2+2 \lambda)+x(-12)\\ +(6-2 \lambda+\lambda^2) \cdots \cdots (1)$

R.H.S. of (1) will be a perfect square if    

 $\,\, (-12)^2-4(-2+2\lambda)(6-2\lambda+\lambda^2)=0 \\ \Rightarrow \lambda^3-3\lambda^2+8 \lambda -24=0 \\ \Rightarrow \lambda^2(\lambda-3)+8(\lambda-3)=0 \\ \Rightarrow (\lambda-3)(\lambda^2+8)=0 \\ \Rightarrow \lambda= 3 $

Now for $ \lambda=3,\,\,$ equation(1) becomes                    

$ (x^2-1+3)^2=4x^2-12x+9=(2x-3)^2 \\ \therefore (x^2+3)^2=(2x-3)^2 \\ \Rightarrow x^2+2= \pm (2x-3) \\ \text{Now,}\,\, x^2+2=2x-3 \\ \Rightarrow x^2-2x+5=0 \\ \Rightarrow x=1 \pm 2i \\ \text{Also,}\,\, x^2+2=-(2x-3)\\ \Rightarrow x^2+2x-1=0 \\ \Rightarrow x=-1 \pm \sqrt{2} $

14. Solve by using Ferrari Method: $\,\,x^4+32x-60=0 $ 

Sol.  $\,\,x^4+32x-60=0 \\ \Rightarrow (x^2)^2=-32x+60 \\ \Rightarrow (x^2)^2-2x^2.1+(1)^2\\=-2x^2+1-32x+60 \\ \Rightarrow (x^2-1)^2=-2x^2-32x+61 \\ \therefore (x^2-1 +\lambda)^2=x^2(-2+2\lambda)+x(-32) \\+(61-2\lambda+\lambda^2) \cdots \cdots (1)$ 

The R.H.S. of (1)  will be a perfect square if          

 $ \\(-32)^2-4(-2+2\lambda)(61-2\lambda+\lambda^2)=0 \\ \Rightarrow \lambda^3-3\lambda^2+63\lambda -189=0 \\ \Rightarrow \lambda^2(\lambda-3)+63(\lambda-3)=0 \\ \Rightarrow (\lambda-3)(\lambda^2+63)=0 \\ \Rightarrow \lambda=3 \\ \text{For,}\,\, \lambda=3,\\(x^2+2)^2\\=x^2(-2+6)+x(-32)+(61-6+9)\\=4x^2-32x+64\\=(2x)^2-2.(2x).8+8^2 \\=(2x-8)^2 \\ \therefore x^2+2= \pm(2x-8)\\ \text{Now,}\,\, x^2+2=2x-8 \\ \Rightarrow x^2-2x+10=0 \Rightarrow x=1 \pm 3i \\ x^2+2=-(2x-8) \\ \Rightarrow x^2+2x-6=0 \Rightarrow x=-1 \pm \sqrt{7} $ 

15.Solve by Ferrari's method: $\,\, 2x^4+6x^3-3x^2+2=0$

Sol. $\,\, 2x^4+6x^3-3x^2+2=0 \\ \Rightarrow 2(x^4+3x^3)=3x^2-2 \\ \Rightarrow 2\{(x^2)^2+2x^2.\frac{3x}{2}+\left(\frac{3x}{2}\right)^2\}\\=3x^2-2+2.\left(\frac{3x}{2}\right)^2 \\ \Rightarrow 2\left(x^2+\frac{3x}{2}\right)^2 =3x^2-2+\frac{9x^2}{2}=\frac{15}{2}x^2-2 \\ \Rightarrow \left(x^2+\frac{3x}{2}\right)^2=\frac{15}{4}.x^2-1 \\ \therefore (2x^2+3x)^2=15x^2-4 \\ \Rightarrow (2x^2+3x+\lambda)=(15+4 \lambda)x^2 \\+(6 \lambda)x+(\lambda^2-4)  \cdots(1)$

Now, right hand side of (1) will be a perfect square if               

 $\\ (6 \lambda)^2-4(\lambda^2-4)(15+4 \lambda)=0 \\ \Rightarrow 9 \lambda^2-(\lambda^2-4)(15+4 \lambda)=0 \\ \Rightarrow 2\lambda^3+3 \lambda^2-8\lambda-30=0 \\ \Rightarrow (2 \lambda-5)(\lambda^2+4 \lambda+6)=0 \\ \Rightarrow 2 \lambda-5=0 \Rightarrow \lambda=\frac{5}{2}\\ \text{Now, putting}\,\,\lambda=\frac{5}{2}  \quad $  in (1), we get 

$(2x^2+3x+\frac{5}{2})^2=(5x+\frac{3}{2})^2 \\ \therefore 2x^2+3x+\frac{5}{2}= \pm (5x +\frac{3}{2}) \\ \text{Now, taking} \,\,2x^2+3x+\frac{5}{2}= 5x +\frac{3}{2} \\ 2x^2+3x-5x+\frac{5}{2}-\frac{3}{2}=0 \\ \Rightarrow 2x^2-2x+1=0 \\ \therefore x= \frac{2 \pm \sqrt{4^2-4.2.1}}{2.2}=\frac{1 \pm i}{2}$

Now, taking $\,\,2x^2+3x+\frac{5}{2}=-(5x +\frac{3}{2}) \\ \Rightarrow 2x^2+8x+4=0\\ \Rightarrow x^2+4x+2=0 \\ \therefore x=\frac{-4 \pm \sqrt{4^2-4.1.2}}{2.1}=-2 \pm \sqrt{2}$

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Geometry

16. Find the condition so that two of the lines given by $\,\, ax^3+3bx^2y+3cxy^2+dy^3=0\,\,$ may be at right angles.

Sol. Since the given equation is homogeneous of third degree, it represents three straight lines passing through the origin. Out of these lines , two are at right angles as such one factor giving the 2 lines will be of the form : $\,\,x^2+kxy-y^2=0 \cdots \cdots (1)$

The other factor must be of the form : $\,\,ax-dy=0 \cdots \cdots (2)  $ 

Let $ \,\, ax^3+3bx^2y+3cxy^2+dy^3\\=(x^2+kxy-y^2)(ax-dy)$

Comparing the coefficents of like terms we get, $\,\,\,3b=ak-d,\,\,3c=-a-kd\,\,$

Eliminating k, we get, $\\ \frac{3b+d}{a}=-\frac{3c+a}{d}\\ \Rightarrow d(3b+d)+a(3c+a)=0$

17. If $\,\,ax^2+2hxy+by^2+2gx+2fy+c=0,\,\,$ represents a pair of straight lines and their point of intersection is denoted by P, show that of the bisectors of angles between the lines $\,\,ax^2+2hxy+by^2=0\,\,$ will pass through P if $\,\, h(g^2-f^2)=gf(a-b).$

Sol. Let $f(x,y)=ax^2+2hxy+by^2+2gx+2fy+c \cdots \cdots (1).$   

The coordinates of the point of intersection of the pair of straight lines given by (1)    

can be obtained by solving for $\,x\,$ and $\,y,\,\,$                      

the equations $\,\,\frac{\partial f}{\partial x}=0 \,\,\text{and}\,\,\frac{\partial f}{\partial y}=0  \\ \Rightarrow ax+hy+g=0,\quad hx+by+f=0\,\,$ 

from which we get, $\,\,\frac{x}{hf-bg}=\frac{y}{gh-af}=\frac{1}{ab-h^2} \\ \therefore x=\frac{hf-bg}{ab-h^2},\,\,y=\frac{gh-ab}{ab-h^2}.$

Hence, the coordinates of P, the point of intersection is :  

 $\left( \frac{hf-bg}{ab-h^2},\frac{gh-ab}{ab-h^2} \right).$

Also, the pair of bisectors of angles between the straight lines:   

$\\ ax^2+2hxy+by^2=0\,\,\text{is}\\ \frac{x^2-y^2}{a-b}=\frac{xy}{h}\\ \Rightarrow (x^2-y^2)=(a-b)xy \cdots \cdots (2)$

If any of the bisectors (2) passes through  P,  then  

$ h \left[ \left(\frac{hf-bg}{ab-h^2}\right)^2 -\left(\frac{gh-af}{ab-h^2}\right)^2\right] \\=(a-b).\frac{hf-bg}{ab-h^2}.\frac{gh-af}{ab-h^2}\\ \Rightarrow h\{(hf-bg)^2-(gh-af)^2\}\\=(a-b)(hf-bg)(gh-af) \\ \Rightarrow h(g^2-f^2)=gf(a-b)$

18. If the lines $\,\,ax^2+2hxy+by^2=0\,\,$ be two sides of a parallelogram and the line $\,\,lx+my=1\,\,$ be one of the diagonals, show that the equation of the other diagonal is : $\,\,y(bl-hm)=x(am-hl)$ 

Solution.   Let the seperate equations of the pair $\,\,ax^2+2hxy+by^2=0\,\,$   

 be $\,\,y=m_1x\,\,$and $\,\,y=m_2x\,\,$ intersecting at the point $\,\,A(0,0).$

So, $m_1+m_2=-\frac{2h}{b},\,\, m_1m_2=\frac{a}{b} \cdots (1).$  

The equation of the diagonal BD is : $\,\,lx+my=1\,\,$           

as $\,\,BD\,\,$  does not pass through the origin. 

The coordinates of B and D are obtained by solving                         

$\,\,y=m_1x \,\,\text{and}\,\,y=m_2x$ with                       

 $\,\,lx+my=1\,\,$ and are $\,\, \left( \frac{1}{l+mm_1}, \frac{m_1}{l+mm_1}\right)\,\,\text{and}\,\,\left( \frac{1}{l+mm_2}, \frac{m_2}{l+mm_2}\right)$

Let $\,\,M(h',k')\,\,$ be the middle point of BD.                     

Therefore, $\,\,h'=\frac{1}{2}.\left(\frac{1}{l+mm_1}+\frac{1}{l+mm_2} \right), \\ k'=\frac{1}{2}.\left(\frac{m_1}{l+mm_1}+\frac{m_2}{l+mm_2} \right)\\ \text{so that}\,\, 2h'=\frac{2l+(m_1+m_2)m}{(l+mm_1)(l+mm_2)},\\ 2k'=\frac{l(m_1+m_2)+2mm_1m_2}{(l+mm_1)(l+mm_2)}\\ \Rightarrow h'=\frac{bl-hm}{b(l+mm_1)(l+mm_2)},\,\, k'=\frac{am-hl}{b(l+mm_1)(l+mm_2)}\\ \therefore \frac{h'}{k'}=\frac{am-hl}{bl-hm};$  

The equation of other diagonal AC which passes through                    

 $\,\,(0,0)\, \,$ and $\,\,(h',k')\,$is :                       

 $\,\, y=\frac{k'}{h'}.x \Rightarrow y=\frac{am-hl}{bl-hm}.x \\ \Rightarrow y(bl-hm)=x(am-hl).$

19.  If $\,\, ax^2+2hxy+by^2+2gx+2fy+c=0\,\,$ represents a pair of straight lines , prove that the area of the triangle formed by their bisectors and the axis of $\,\,x\,\,$ is  $\,\,\frac{\sqrt{(a-b)^2+4h^2}}{2h} \times \frac{ca-g^2}{ab-h^2}\quad [\textbf{CU-1995}]$

Solution. Shifting the origin to a point, say $\,\,(\alpha, \beta)\,\,$  

 where the lines given by the general equation of second degree meet          

without changing the direction of axes, so that the equation of given pair of straight lines takes the form : $\,\, ax'^2+2hx'y'+by'^2=0 \cdots \cdots (1)\\$                      

   The pair of bisectors of angles between the lines given by (1) is :       

$\,\, \frac{x'^2-y'^2}{a-b}=\frac{x'y'}{h} \cdots \cdots (2).\\$

Putting $\,\,y'=-\beta\,$in (2), we get the abscissae of the point          

of intersection of (2) with the old x-axis,                      

$\,\,\frac{x'^2-\beta^2}{a-b}=\frac{-\beta x'}{h}\\ \Rightarrow hx'^2+\beta(a-b)x-h\beta^2=0 \cdots\cdots (3).\\$

Let $\,\,x_1',x_2'\,\,$ be the roots of (3). 

So, $ x_1'+x_2'= -\frac{\beta(a-b)}{h},\,\, x_1'.x_2'= -\beta^2  \\ \therefore (x_1'-x_2')^2=(x_1'+x_2')^2-4x_1'.x_2' \\ = \frac{\beta^2(a-b)^2}{h^2}+4 \beta^2 \\=\frac{\beta^2}{h^2} \left[(a-b)^2+4h^2 \right]$

Now $\,\, x_1' \sim x_2'\,\,$  is the intercept made on the old axes              

 by the bisectors and perpendicular from $\,\,(\alpha,\beta)\,\,$ on the axis is $\,\,\beta. $

So, the required area  $=\frac{1}{2}.(x_1' \sim x_2'). \beta \\=\frac{1}{2}. \frac{\beta^2}{h} \sqrt{(a-b)^2+4h^2}\\=\frac{\sqrt{(a-b)^2+4h^2}}{2h}. \frac{g^2-ca}{h^2-ab},\,\,$

where $\,\, \left(\sqrt{\frac{f^2-bc}{h^2-ab}} , \sqrt{\frac{g^2-ac}{h^2-ab}}\right)\,\, $ is the point of intersection of the lines.

20. If $\,\, ax^2+2hxy+by^2+2gx+2fy+c=0\,\,$ represent two straight lines prove that the square of the distance of their points of intersection from the origin is $\,\, \frac{c(a+b)-f^2-g^2}{ab-h^2}.$

Solution. Let the equations of the straight lines represented by                              

 $\,\,ax^2+2hxy+by^2+2gx+2fy+c=0\,\, \text{be}\\ lx+my+n=0 \cdots \cdots (1) \\ l'x+m'y+n=0 \cdots \cdots (2). \\ \therefore ax^2+2hxy+by^2+2gx+2fy+c \\=(lx+my+n)(l'x+m'y+n)\\$ 

Comparing the coefficients, we have                                 

 $\,\, ll'=a,\,\,mm'=b,\,\,nn'=c, \\ m'n+mn'=2f,\,\,ln'+l'n=2g,\,\,lm'+l'm=2h.\\$       

Solving (1) and (2), by the rule of cross-multiplication,                    

$\,\,\frac{x}{mn'-m'n}=\frac{y}{nl'-n'l}=\frac{1}{lm'-l'm}.\\ \text{Now,}\,\, (mn'-m'n)^2=(mn'+m'n)^2-4mm'.nn'  \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4(f^2-bc). \\(nl'-n'l)^2=(nl'+n'l)^2-4ll'.nn'\\=4(g^2-ac)\\ (lm'-l'm)^2=(lm'+l'm)^2-4ll'.mm'\\=4(h^2-ab) $

Hence,  $\,\,\left(\sqrt{\frac{f^2-bc}{h^2-ab}} , \sqrt{\frac{g^2-ac}{h^2-ab}}\right)\,\, $ is the point of intersection of the lines.

So, square of the distance $\,\,(x,y)\,\,$ from the origin                              

$\\=x^2+y^2 \\=\frac{f^2-bc}{h^2-ab}+\frac{g^2-ac}{h^2-ab}\\=\frac{f^2-bc+g^2-ac}{h^2-ab}\\=\frac{f^2+g^2-c(a+b)}{h^2-ab}\\=\frac{c(a+b)-f^2-g^2}{ab-h^2}$