IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

p3arth

PARAGRAPH FOR Q.NO 1 TO 3 [4,-1]

CONSIDER TWO POINTS AND . O IS ORIGIN.

A POINT P IS IN THE XY PLANE SATISFYING

d(P,OA) d(P,OB) d(P,AB) = min {3, } , WHERE d(P,OA)= DISTANCE OF POINT P FROM THE LINE OA.LET THE CENTROID OF THE ABO IS G.

Q.1 LEAST VALUE OF PG

(A)1 (B) (C) (D) 0

Q.2 AREA OF REGION IN WHICH P CAN LIE

(A) (B) + (C) (D) NONE

Q.3 IF d(P,OA)=1 THEN THE LEAST VALUE OF d(P,OB) IS

(A) 1 (B) (C) 0 (D)NONE

WHO CAN GET THE 12 MARKS ,IF ANYBODY I WILL GIVE

HIM/HER 20 POINTS.

EXPERTS COME ON...............

p3arth

COME ON EXPERT I WANT TO COMPLTE SOLUTION .

MY ANSWER IS (D), (D),(A) IS IT CORRECT OR NOT?

bladeX

is the distance algebraic or absolute

ab_batra

r u parth khanna

kaymant

It's quite obvious that triangle $OAB$ is equilateral. The equations of the sides:

$OA: y=0$

$OB: y-\sqrt{3}\, x=0$

$AB: y+\sqrt{3}\,x-6=0$

Let $P(h,\,k)$ be any arbitrary point. According to the given condition, we should have

$|k|+\dfrac{|k-\sqrt{3}\,h|}{2}+\dfrac{|k+\sqrt{3}\,h-6|}{2}=\mathrm{min }\{3,\,h^2+k^2\}$

We consider two cases:

Case I: $h^2+k^2\geq 3$ . Here, $\mathrm{min }\{3,\,h^2+k^2\}=3\qquad (1)$

This region is the region of the $xy$ plane lying on or outside the circle $x^2 + y^2 = 3$ . This region is

further divided into different regions ( $I$ , $II$ , $\ldots$ , $VII$ as shown in the figure by the points

which make any of the modulii in the above equation zero. If we consider the equation

$\mathrm{d}(P,\,OA) + \mathrm{d}(P,\,OB) + \mathrm{d}(P,\,AB)=3\qquad$

it's quite clear that we cannot have solutions in the regions $I$ to $VII$ except, possibly, the region $III$ .

That is because, if we consider a region (say) $II$ , one can always find a point in this region whose distance from

$OA$ exceeds 3. The same argument works for all the regions except $III$ . But in this region, $k>0$ , $k-\sqrt{3}\,h<0$ ,

and $k+\sqrt{3}\,h-6<0$ . So, we get

$|k|=k,\quad |k-\sqrt{3}\,h|=\sqrt{3}\,h-k, \quad |k+\sqrt{3}\,h-6|= 6 -k-\sqrt{3}\,h$

So, for any point in the region $III$ ,

$\mathrm{d}(P,\,OA) + \mathrm{d}(P,\,OB) + \mathrm{d}(P,\,AB)$

$=k + \dfrac{\sqrt{3}\,h-k}{2}+\dfrac{6 -k-\sqrt{3}\,h}{2}$

$=3$

which is the RHS of $(2)$

That is all points of the region $III$ satisfy the required condition. And this is the only region when $h^2 + k^2 \geq 3$ .

Case 2: $h^2+k^2<3$ . Then, $\mathrm{min }\{3,\,h^2+k^2\}=h^2+k^2$ . That is now we are talking about

only those points lying within the circle $x^2 + y^2=3$ .

In this case, solving (1) with $\mathrm{min }\{3,\,h^2+k^2\}=h^2+k^2$ , gives no solution.

According the required region in which $P$ can lie is the region shown in green in the figure.

From the figure, it is quite easy to answer the questions 1 and 2: in both cases the answer is (D).

For the last question, note that when $\mathrm{d}(P,\,OA)=1$ , the point $P$ lies on the dashed line --

the part of which lies in the green region. Thus, the minimum value of $\mathrm{d}(P,\,OB)$ will be attained when $P$ lies

on the intersection of the circle and the dashed line. This minimum turns out to be

$\dfrac{\sqrt{6}-1}{2}$ . Thus, again (D) is the answer for question 3.

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CONSIDER TWO POINTS AND . O IS ORIGIN.

d(P,OA) d(P,OB) d(P,AB) = min {3, } , WHERE d(P,OA)= DISTANCE OF POINT P FROM THE LINE OA.LET THE CENTROID OF THE ABO IS G.