Maths

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < <

By using condition of tangency,
c² = a² (m² + 1)
⇒ c² = 5 [ (2)² + 1]
⇒ c² = 25
⇒ c = ±5

y = |x − 1|, y = 3 – |x|

(A graph is shown with vertices A (1, 0), B (2, 1), C (0, 3), D (-1, 2). The lines are y = x - 1, y = 3 - x, y = 3 + x, and y = -x + 1)

AB = √2, BC = 2√2
⇒ Area = 4 sq. units

f' (0) = 0   

h' (x) = f' (x) – 2f (x).f' (x) + 3 (f (x)².f' (x)
= f' (x)/3 (f (x) - 1/3)² + 2/9)
∴ h' (x) ≥ 0 ⇔ f' (x) ≥ 0

g (x) = (f (nf (x) – n)?
g' (x) = n (f (nf (x) – n)? ¹ . f' (nf (x) – n) . n . f' (x)
∴ g' (0) = 0
⇒ 4 = n (f (nf (0) – n)? ¹ . f' (nf (0) – n) . nf' (0)
⇒ 4 = n (f (0)? ¹ . f' (0) . nf' (0)
⇒ 4 = n . 1 . (-1) . n (-1)
n² = 4
⇒ n = 2

There are 7 letters permutation.
The total number of 4 letters
Out of 2A's, 2K's and 3 different letters O, L, T.
= Coefficient of x? in 4! (1 + x + x²/2!)² (1 + x)³
= Coefficient of x? in 4! (1 + x + x²/2)² (1 + x)³
= Coefficient of x? in 4! [ (1 + x)? + (1 + x)? x² + (1 + x)³ + x? /4]
= 4! [? C? +? C? + ³C? /4] = 4! (5 + 6 + 1/4)
= 24 [11 + 1/4]
= 264 + 6 = 270

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0

f (x)= {sinx, 0? x? }
f' (x)= {cosx, 0