Maths

C? → C? - C?
f (θ) = | -sin²θ -1 |
| -cos²θ -1 |
| 12 -2 -2|
= 4 (cos²θ - sin²θ) = 4 (cos2θ), θ ∈ [π/4, π/2]
f (θ)max = M = 0
f (θ)min = m = -4

f (x) is differentiable then will also continuous then f (π) = -1, f (π? ) = -k?
k? = 1
Now f' (x) = { 2k? (x-π) if x≤π
{ -k? sinx if x>π
then f' (π? ) = f' (π? ) = 0
f' (x) = { 2k? if x≤π
{ -k? cosx if x>π
then 2k? =k?
k? = 1/2

For ellipse x²/16 + y²/9 = 1, a=4, b=3, e = √ (1 - 9/16) = √7/4
A and B are foci then PA + PB = 2a = 2 (4) = 8

The word is 'LETTER'.
Consonants are L, T, R.
Vowels are E, E.
Total number of words (with or without meaning) from the letters of the word 'LETTER' is:
6! / (2! 2!) = 720 / 4 = 180.
Total number of words (with or without meaning) from the letters of the word 'LETTER' if vowels are together:
Treat (EE) as a single unit. We now arrange {L, T, R, (EE)}. This is 5 units.
Number of arrangements = 5! / 2! (for the two T's) = 120 / 2 = 60.
∴ The number of words where vowels are not together = Total words - Words with vowels together
Required = 180 - 60 = 120.

P (x) = 0
x² - x - 2 = 0
(x-2) (x+1) = 0
x = 2, -1 ∴ α = 2
Now lim (x→2? ) (√ (1-cos (x²-x-2) / (x-2)
⇒ lim (x→2? ) (√ (2sin² (x²-x-2)/2) / (x-2)
⇒ lim (x→2? ) (√2 sin (x²-x-2)/2) / (x²-x-2)/2) ⋅ (x²-x-2)/2) ⋅ (1/ (x-2)
⇒ for x→2? , (x²-x-2)/2 → 0?
⇒ lim (x→2? ) √2 ⋅ 1 ⋅ (x-2) (x+1)/ (2 (x-2) = 3/√2

|x + y|² = |x|²
(x+y)· (x+y) = x·x
|x|² + 2x·y + |y|² = |x|²
|y|² + 2x·y = 0 (1)
and (2x + λy)·y = 0
2x·y + λ|y|² = 0 (2)
From (1), 2x·y = -|y|².
Substitute into (2):
-|y|² + λ|y|² = 0
(λ-1)|y|² = 0
Assuming y is a non-zero vector, |y|² ≠ 0, therefore λ=1.

R (-1+2r, 3-2r, -r)
dr's of PR are (2 - 2r, -1+2r, -3+r)
Then 2 (2-2r) + 2 (-1+2r) + 1 (3-r) = 0
9-9r = 0 ⇒ r = 1
R (1,1, -1)
then a+1=2, b+2=2, c-3=-2
a=1, b=0, c=1
∴ a+b+c = 2

|λ-1 3λ+1 2λ|
|λ-1 4λ-2 λ+3| = 0
|2 3λ+1 3 (λ-1)|
R? → R? - R? and R? → R? - R? (from a similar matrix setup, applying operations to simplify)
The provided solution uses a slightly different matrix but let's follow the subsequent steps.
A different matrix from the image is used in the calculation:
|λ-1 3λ+1 2λ|
|0 λ-3 -λ+3|
|3-λ 0 λ-3 |
C? → C? + C?
|3λ-1 3λ+1 2λ |
|3-λ λ-3-λ | = 0
|0 λ-3 |
⇒ (λ-3) [ (3λ-1) (λ-3) - (3λ+1) (3-λ)] = 0
⇒ (λ-3) [ (λ-3) (3λ-1) + (λ-3) (3λ+1)] = 0
⇒ (λ-3)² [3λ-1 + 3λ+1] = 0
⇒ (λ-3)² [6λ] = 0 ⇒ λ = 0, 3
Sum of values of λ = 3