Class 11th

This happens due to the fact that they possess strong C-C and H-H bonds which are non-polar sigma bonds and cannot be broken easily. Also, they lack polar functional groups and are saturated hydrocarbons. These properties combined make them less reactive overall.

F = -dU/dr = - [-12A/r¹³ + 6B/r? ]
F=0 ⇒ r= (2A/B)¹/?
U (at r= (2B/A)¹/? ) = -A²/4B

Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

L.C. = 0.5 mm / 50 = 10? ² mm = 10? m = 10µm

Σ (1/a) (1-rb/a)? ¹ = (1/a)Σ (1+rb/a+r²b²/a²+.)
≈ (1/a)Σ (1+rb/a) = n/a + (b/a²)n (n+1)/2
Compare coeffs: α=1/a, β=b/2a². γ=b²/3a³. This differs from solution.

sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.

S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.

Hyperbola: 16 (x+1)² - 9 (y-2)² = 144. (x+1)²/9 - (y-2)²/16 = 1. Center (-1,2).
Foci (-1±ae, 2). a²=9, b²=16. e²=1+16/9=25/9, e=5/3. ae=5. Foci (4,2), (-6,2).
Centroid (h, k) of P, S, S': P (3secθ-1, 4tanθ+2).
h= (3secθ-1+4-6)/3 = secθ-1. k= (4tanθ+2+2+2)/3 = (4/3)tanθ+2.
(h+1)² - (3 (k-2)/4)² = 1. 16 (x+1)²-9 (y-2)²=16.
16x²+32x+16-9y²+36y-36=16. 16x²-9y²+32x+36y-36=0.

Vertex (2,0), Focus (4,0). Parabola y²=4a (x-h) = 4 (2) (x-2) = 8 (x-2).
Tangents from O (0,0): T²=SS? (y (0)-4 (x+0)+16)²= (0-0+16) (y²-8x+16). No, this is for point on tangent.
Equation of tangent y=mx+a/m = m (x-2)+2/m. Passes through (0,0) so -2m+2/m=0 ⇒ m=±1.
Tangents y=x, y=-x. Points of contact S (4,4), R (4, -4).
Area of ΔSOR = ½ * base * height = ½ * 8 * 4 = 16.