online user counteronline user counter

MATHMAA

Only Search Your Site

16th KVJ Solutions5

Probability Introduction :

Here you find 16th KVJ Solutions5 in detail explanation .

5Q) A semi-circle is drawn outwardly on chord AB of the circle with centre O and unit radius. The perpendicular from O to AB, meets the semi-circle on AB at C. Determine the measure of \(\angle AOB \) and length AB so that OC has maximum length.

Solution :

Let  O be the center, OA be the radius of 1 unit, AB is the chord.

Let length of AB =2\(\sqrt{a}\). Let Perpendicular from O to AB meets at M on AB.

AM=MB=\(\sqrt{a}\).

OM extended to meet semicircle at C.

We know that AM=MB=MC as these are length of radius of semicircle. From right triangle OAM , we get ,

OM= \(\sqrt{1^{2}-(\sqrt{a})^{2}}\)

OM = \(\sqrt{1-a}\)

OC= OM + MC = \(\sqrt{1-a} +\sqrt{a}\)

\( OC^{2} =(\sqrt{1-a} + \sqrt{a})^{2} \)

\( OC^{2}= 1-a + a + 2\sqrt{(1-a)a} \)

\(OC^{2} = 1 + 2 \sqrt{a-a^2}\)

OC is maximum then\(OC^{2}\) also maximum.

\(OC^{2} \) is maximum when a-a^2 is maximum.

We know that \( y = px^{2} +qx + r \) has maximum when p<0 and   it occurs at x = -q/2p .

\(OC^{2} \) is maximum since a^2 coefficient p=-1 < 0 , q=1 .

a= -q/2p = -1/-2 =1/2

OC is maximum when a = 1/2

AB= \(2\sqrt{a}\) = 2\(\sqrt{1/2}\) = \(\sqrt{2}\)

OA =1, AM=\(\frac{1}{\sqrt{2}}\) , From Triangle OAM,

sin (\(\angle AOM \) )= \(\frac{AM}{OA}\)

sin(\(\angle AOM \) =\(\frac{1}{\sqrt{2}}\)

 \(\angle AOM\) = \(45^{\circ}\)

 \( \angle AOB =2 \angle AOM\)

 \( \angle AOB = 90^{\circ}\)


<<BACK<<                          QUESTION PAPER                                               >>NEXT>>




SHARE YOU ENORMOUS EFFORT AND SMART EXAMPLES HERE

!! NEED MORE HELP !!

SBI! Case Studies