Maths

Kindly consider the following figure

We need to find the remainder of (2021)³? ² when divided by 17.
First, find the remainder of 2021 divided by 17.
2021 = 17 * 118 + 15.
So, 2021 ≡ 15 (mod 17).
Also, 15 ≡ -2 (mod 17).
So, (2021)³? ² ≡ (-2)³? ² (mod 17).
(-2)³? ² = 2³? ² = (2? )? ⋅ 2² = 16? ⋅ 4.
Since 16 ≡ -1 (mod 17),
16? ⋅ 4 ≡ (-1)? ⋅ 4 (mod 17).
≡ 1 ⋅ 4 (mod 17)
≡ 4 (mod 17).
The remainder is 4.

A = [0, sin α], [sin α, 0]
A² = A ⋅ A = [0, sin α], [sin α, 0] ⋅ [0, sin α], [sin α, 0]
= [00 + sinαsinα, 0sinα + sinα0], [sinα0 + 0sinα, sinαsinα + 00]
= [sin²α, 0], [0, sin²α] = (sin²α)I
A² - (1/2)I = [sin²α, 0], [0, sin²α] - [1/2, 0], [0, 1/2]
= [sin²α - 1/2, 0], [0, sin²α - 1/2]
det (A² - (1/2)I) = (sin²α - 1/2)² - 0 = 0
sin²α - 1/2 = 0
sin²α = 1/2
sin α = ±1/√2
A possible value for α is π/4.

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

Given r * a = r * b, which means r * a - r * b = 0 ⇒ r * (a - b) = 0.
This implies that vector r is parallel to vector (a - b).
So, r = λ (a - b) for some scalar λ.
a - b = (2i - 3j + 4k) - (7i + j - 6k) = -5i - 4j + 10k.
So, r = λ (-5i - 4j + 10k).
We are also given r ⋅ (i + 2j + k) = -3.
λ (-5i - 4j + 10k) ⋅ (i + 2j + k) = -3
λ (-51 - 42 + 10*1) = -3
λ (-5 - 8 + 10) = -3
λ (-3) = -3 ⇒ λ = 1.
So, r = 1 * (-5i - 4j + 10k) = -5i - 4j + 10k.
We need to find r ⋅ (2i - 3j + k).
(-5i - 4j + 10k) ⋅ (2i - 3j + k) = (-5) (2) + (-4) (-3) + (10) (1)
= -10 + 12 + 10 = 12.

Given equation of tangent is 2x - y + 1 = 0
equation of normal is x + 2y = 12
Solving with x - 2y = 4 we get centre at (6,2) radius = √ (36 + 9) = √45 = 3√5.

We need to evaluate lim (x→0? ) (cos? ¹ (x - [x]²) ⋅ sin? ¹ (x - [x]²) / (x - x²).
As x → 0? , the greatest integer [x] = 0.
So the expression becomes:
lim (x→0? ) (cos? ¹ (x) ⋅ sin? ¹ (x) / (x (1 - x²)
= lim (x→0? ) cos? ¹ (x) * lim (x→0? ) (sin? ¹ (x) / x) * lim (x→0? ) (1 / (1 - x²)
We know lim (x→0) (sin? ¹ (x) / x) = 1.
lim (x→0? ) cos? ¹ (x) = cos? ¹ (0) = π/2.
lim (x→0? ) (1 / (1 - x²) = 1 / (1 - 0) = 1.
So the limit is (π/2) * 1 * 1 = π/2.

Kindly consider the following figure