For option (a) $$\cos P = {{{q^2} + {r^2} - {p^2}} \over {2qr}}$$ ..... (i)


$$\because$$ $${{{q^2} + {r^2}} \over 2} \ge \sqrt {{q^2}{r^2}} (AM \ge GM)$$

$$ \Rightarrow {q^2} + {r^2} \ge 2qr$$

From Eq. (i), we get

$$\cos P \ge {{2qr - {p^2}} \over {2qr}}$$

$$ \Rightarrow \cos P \ge 1 - {{{p^2}} \over {2qr}}$$ $$\to$$ option (a) is correct.

For option (b)

$${{(q - r)\cos P + (p - r)\cos Q} \over {p + q}}$$$$ = {{(q\cos P + p\cos Q) - r(\cos P + \cos Q)} \over {p + q}}$$

$$ \Rightarrow 4{\lambda ^2} - 2\lambda = 0$$

$$ \Rightarrow u\,.\,v\, = 0$$ (Rejected)

or $$u\,.\,v = {1 \over 2}$$

$$\therefore$$ $$u\,.\,v = {1 \over 2}$$

$$\therefore$$ $${\left| {3u + 5v} \right|^2} = 9{\left| u \right|^2} + 25{\left| v \right|^2} + 3 \times 5 \times 2 \times u\,.\,v$$

$$ = 9 + 25 + 30 \times {1 \over 2}$$

$$ = 49$$ ($$\because$$ $$\left| u \right| = \left| v \right| = 1$$, given)

$$\therefore$$ $$\left| {3u + 5v} \right| = 7$$

$$ = {{r - r(\cos P + \cos Q)} \over {p + q}} = {{r - r\cos P - r\cos Q} \over {p + q}}$$

$$ = {{r - (q - \cos R) - (p - q\cos R)} \over {p + q}}$$

$$ = {{(r - p - q) + (p + q)\cos R} \over {p + q}}$$

$$ = \cos R + {{r - (p + q)} \over {p + q}} \le \cos R$$ ($$\because$$ $$r < p + q$$)

Hence, option (b) is correct.

For option (c)

$${{q + r} \over p} = {{\sin Q + \sin R} \over {\sin P}} \ge {{2\sqrt {\sin Q.\sin R} } \over {\sin P}}$$

$$\to$$ option (c) is incorrect.

For option (d)

If p < q and p < r, then p is the smallest side and hence one of Q or R can be obtuse.

So, one of cos Q and cos R can be negative.

Therefore, $$\cos Q > {p \over r}$$ and $$\cos R > {p \over q}$$ cannot hold always.

Option (d) is incorrect.

For a $$\Delta $$XYZ, it is given that

$$\tan {X \over 2} + \tan {Z \over 2} = {{2y} \over {x + y + z}}$$

$$ \Rightarrow {\Delta \over {s(s - x)}} + {\Delta \over {s(s - z)}} = {y \over s}$$

$$ \Rightarrow \Delta {{(s - z + s - x)} \over {(s - x)(s - z)}} = y$$

$$ \Rightarrow \Delta = (s - x)(s - z)$$

$$ \Rightarrow s(s - x)(s - y)(s - z) = {(s - x)^2}{(s - z)^2}$$

$$ \Rightarrow {s^2} - sy = {s^2} - (x + z)s + xz$$

$$ \Rightarrow s(x + z - y) = xz$$

$$ \Rightarrow {(x + z)^2} - {y^2} = 2xz$$

$$ \Rightarrow {x^2} + {z^2} = {y^2} \Rightarrow y = {\pi \over 2}$$

$$ \because $$ $$X + Y + Z = \pi \Rightarrow X + Z = {\pi \over 2} = Y$$

$$ \Rightarrow X + Z = Y$$

$$ \because $$ $$\tan {X \over 2} = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} = \sqrt {{{1 - {z \over y}} \over {1 + {z \over y}}}} $$



$$ = \sqrt {{{y - z} \over {y + z}}} = {{\sqrt {{y^2} - {z^2}} } \over {y + z}}$$

$$\tan {X \over 2} = {x \over {y + z}}$$ {$$ \because $$ x² + z² = y²}

Let a non-right angled $$\Delta $$PQR.

Now, by sine rule



$${P \over {\sin P}} = {q \over {\sin Q}} = {r \over {\sin R}} = 2 \times $$ circumradius

$$ \Rightarrow {{\sqrt 3 } \over {\sin P}} = {1 \over {\sin Q}} = {r \over {\sin R}} = 2 \times 1$$

[circumradius = 1 unit]

$$ \Rightarrow $$ $${\sin P}$$ = $${{\sqrt 3 } \over 2}$$ and sinQ = $${1 \over 2}$$

$$ \Rightarrow P = 120^\circ $$ and $$ \Rightarrow Q = 30^\circ $$

($$ \because $$ $$\Delta $$PQR is non-right angled triangle)

So, R = 30$$^\circ $$

$$ \Rightarrow $$ r = 1, so $$\Delta $$PQR is an isosceles triangle. And, since RS and PE are the median of $$\Delta $$PQR, so 'O' is centroid of the $$\Delta $$PQR.

Now,

Option (a),

From Apollonius theorem,

$$2(P{E^2} + Q{E^2}) = P{Q^2} + P{R^2}$$

$$ \Rightarrow 2\left( {P{E^2} + {3 \over 4}} \right) = 1 + 1$$

$$ \Rightarrow P{E^2} = 1 - {3 \over 4} \Rightarrow P{E^2} = {1 \over 4}$$

$$ \Rightarrow PE = {1 \over 2}$$ units

and $$OE = {1 \over 3}PE = {1 \over 6}$$ units

[$$ \because $$ O divides PE is 2 : 1]

Option (b),

Again from Apollonius theorem,

2 (PS² + RS²) = PR² + QR²

$$ \Rightarrow 2\left( {{1 \over 4} + R{S^2}} \right) = 1 + 3$$

$$ \Rightarrow R{S^2} = 2 - {1 \over 4} \Rightarrow R{S^2} = {7 \over 4}$$

$$ \Rightarrow RS = {{\sqrt 7 } \over 2}$$ units

Option (c),

Area of $$\Delta $$SOE = $${1 \over 2}(OE)\,(ST)$$

= $${1 \over 2} \times {1 \over 6}[(PS)sin60^\circ ]$$

= $${1 \over {12}} \times {1 \over 2} \times {{\sqrt 3 } \over 2}$$

= $${{\sqrt 3 } \over {48}}$$ square units

Option (d),

$$ \because $$ Inradius of

$$\Delta PQR = {\Delta \over s} = {{{1 \over 2}pq\sin R} \over {{1 \over 2}(p + q + r)}} = {{{1 \over 2}(\sqrt 2 )(1){1 \over 2}} \over {{1 \over 2}(\sqrt 3 + 1 + 1)}}$$

$$ = {{\sqrt 3 } \over 2}(2 - \sqrt 3 )$$ units

Hence, options (a), (b) and (d) are correct.