Maths

Kindly consider the following Image

lim (x?0) (tan (? /4+x)¹/? = e^ (lim (x?0) (tan (? /4+x)-1)/x)
= e^ (lim (x?0) (2tanx/ (1-tanx)/x) = e².

S? =0 ⇒ (a? +a? ) (11/2)=0 ⇒ a? =-a?
2a? +10d=0 ⇒ a? =-5d.
Sum = a? +a? +.+a? = (a? +a? ) (12/2) = 6 (2a? +22d)
= 6 (2a? +22 (-a? /5) = 6 (2a? -22a? /5) = 6 (-12a? /5)=-72a? /5. k=-72/5.

Normal to plane is n= (-4i+5j+7k).
Plane: -4 (x-3)+5 (y-1)+7 (z-1)=0 ⇒ -4x+5y+7z=0.
Passes through (α, -3,5) ⇒ -4α-15+35=0 ⇒ α=5.

f (x)=a (x-3) (x-α)
f (2)=a (2-α)
f (-1)=a (-4) (-1-α)=4a (1+α)
f (-1)+f (2)=0 ⇒ a (2-α+4+4α)=0 ⇒ a≠0 ⇒ 5α=-2 ⇒ α=-0.4
α ∈ (-1,0)

λ=- (sin? θ+cos? θ) = - (sin²θ+cos²θ)²-2sin²θcos²θ)
λ = - (1-½sin²2θ) = ½sin²2θ-1
sin²2θ ∈
λ ∈ [-1, -1/2]

A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3

1. Let t? denotes r+1th term of (? x? +? x? )¹?
   t? = ¹? C? (? x? )¹? (? x? )? = ¹? C? ¹? x? ¹?
   If t? is independent of x
   90-15r=0? r=6
   This differs from the solution.
   Let's follow the solution's powers.
   (10-r)/9 - r/6 = 0? r=4
   maximum value of t? is 10K (given)
   ? ¹? C? is maximum
   By AM? GM (for positive numbers)
   (? ³/2+? ³/2+? ²/2+? ²/2)/4? (? /16)¹/?
   ? ? ? 16
   So, 10K = ¹? C?16
   ? K=336

∫? ² |x-1|-x|dx
Let f (x)=|x-1|-x|
= {|1-2x|, x≤1; 1, x≥1}
A = 1/2+1=3/2
∫? ¹/² (1-2x)dx+∫? /? ¹ (2x-1)+∫? ²1dx
= [x-x²]? ¹/²+ [x]? ² = 3/21dx
= [x-x²]? ¹/²+ [x]? ² = 3/2

Circle x²+y²-2x-4y+4=0
⇒ (x-1)²+ (y-2)²=1
Centre: (1,2) radius=1
line 3x+4y-k=0 intersects the circle at two distinct points.
⇒ distance of centre from the line < radius
⇒ |3*1+4*2-k|/√ (3²+4²) < 1
⇒ |11-k|<5
⇒ 6⇒ k∈ {7,8,9, .,15} since k∈I
Number of K is 9.