IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

hsbhatt

2) belongs in Coord Geo not here

3) Consider $f(x) = e^{-x^2} \sin x - 1$

This is a trick question I think, because this equation cannot have real roots. Still the proposition can be proved as if nothing is amiss.

Now suppose x₁and x₂ are two consecutive roots, we have

and similarly

Now consider $g(x) = e^{-x^2} \cos x - 2x$ .

$g(x_1) = \cot x_1 - 2x_1$ and $g(x_2) = \cot x_2 - 2x_2$ again as $\sin x_1 = e^{x^2}$

So, we have g(x_1) = f

Now, if either f'(x₁) = 0 or f'(x₂) = 0, we are done. Let us assume the derivatives are non-zero

So now, a lemma under these conditions:

x₁and x₂ are two consecutive roots, then f'(x₁) f'(x₂)<0 i.e. they are of opposite signs.

As of now I have a crude proof. But please do watch http://www.mathlinks.ro/viewtopic.php?p=1238942#1238942

This means g(x₁) and g(x₂) are of opposite sign and hence g(x) has a root in the interval (x₁, x₂)

Quite an interesting problem. I am not sure what the author intended with this one.

hsbhatt

and of course once its true for consecutive roots the same holds for the interval between any two roots.

1. is pretty straighforward and I will wait for answers

reddevil_2009

Sir question 2 is to be solved using concept of calculus ......................

kaymant

The first family of lines: $(x+3y-7)+\lambda(x+4y-3)=0$ i.e. $(1+\lambda)x + (3+4\lambda)y-(7+3\lambda)=0$

The distance of these lines from origin is

$s_1 = \dfrac{|7+3\lambda|}{\sqrt{(1+\lambda)^2+(3+4\lambda)^2}}$

Now, $s_1$ will be greatest when $s_1^2$ will be greatest. But

$s_1^2 = \dfrac{(7+3\lambda)^2}{(1+\lambda)^2+(3+4\lambda)^2}$

Equating $\dfrac{\mathrm{d}s_1^2}{\mathrm{d}\lambda}$ to zero, we obtain the values of $\lambda$ as $-\dfrac{7}{3}$ and $-\dfrac{61}{80}$ . For the first value $s_1^2=0$ , so the maximum of $s_1^2$ and hence for $s_1$ is obtained for $\lambda = -\dfrac{61}{80}$ . Accordingly, the particular line of the family $(x+3y-7)+\lambda(x+4y-3)=0$ , that is farthest from the origin is

$(x+3y-7)-\dfrac{61}{80}(x+4y-3)=0 \Rightarrow 19x-4y-377=0$

Now consider the second family: $(3x+y-7)+\mu(x+y-3)=0$ i.e.

$(3+\mu)x+(1+\mu)y-(7+3\mu)=0$

Proceeding in a manner similar to the previous case, we find that the line with $\mu=1$ is farthest from the origin, and this line is

The angle between the two lines thus obtained can therefore be found by using the routine formula.

hsbhatt

@Reddevil - I really dont know the reason why your tutors should insist on a calculus method. The intent should be teach you the shortest and most elegant way to get to a solution and here calculus is a horrific choice.

My intent when I said that it is purely coord geo is this: $(x+3y-7) + \lambda (x+4y-3) = 0$ is the family of lines passing thru the intersection of x+3y-7 and x+4y-3 which is (19,-4). Do you need calculus to tell you of these which line will be farthest from the origin? Even if you insist on a mathematical proof, use of the triangle inequality will tell you it is simply the line which is perpendicular to the line joining (0,0) and (19,-4). ditto for the other point.

pramod6990

Q2) looking at the expn we realise that the 2 lines have to be such that the distance of these lines frm the origion has to be the same as that of the corresponding points....or in other words the slopes of these lines have to be perp. to the lines joining the origine to the 2 lines...

so we have P1 =(19,-4) P2=(2,1) M1= -4/19 and m2= 1/2 so the new slopes are 19/4 and -2....so the angle btween lines = tan^-1 ( -27/34)....

give me time for the first one...

Q3) let A and B be the roots of the given eqn.....

so we have for A and B e^(-x2) sinx =1 or sinx -e^(x2) =0 (multiply by e^(x2) throughout)

let f(x) =sinx -e^(x2)

applying rolles theoram...

theres a c belonging to (A, B) where f'(c) =0 or cos x -2xe(x2) =0

multiplying e^(-x2) through out we have that there is atleast one c in (A,B) where

cosxe^(-x2) - 2x =0.....

hsbhatt

wanted to say something about the 2nd solution - nice job. Compare it to my enormously complicated solution

I laughed very hard when I realized this morning that the solution is a whole lot easier but I let it pass. Good you posted this.

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