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New answer posted
7 months agoContributor-Level 10
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.
New answer posted
7 months agoContributor-Level 10
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.
New answer posted
7 months agoContributor-Level 10
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.
New answer posted
7 months agoContributor-Level 10
The expansion is (x + x^ (log? x)?
The (r+1)-th term is T? =? C? * x? * (x^ (log? x)?
The 4th term means r=3.
T? =? C? * x? * (x^ (log? x)³ = 35 * x? * x^ (3 log? x) = 35 * x^ (4 + 3 log? x).
Given T? = 4480.
35 * x^ (4 + 3 log? x) = 4480
x^ (4 + 3 log? x) = 4480 / 35 = 128.
x^ (4 + 3 log? x) = 128.
Take log? on both sides:
log? (x^ (4 + 3 log? x) = log? (128)
(4 + 3 log? x) * (log? x) = 7
Let t = log? x.
(4 + 3t)t = 7
3t² + 4t - 7 = 0
3t² - 3t + 7t - 7 = 0
3t (t-1) + 7 (t-1) = 0
(3t+7) (t-1) = 0
t = 1 or t = -7/3.
log? x = 1 ⇒ x = 2¹ = 2.
log? x = -7/3 ⇒ x = 2^ (-7/3).
Since x ∈ N, x = 2.
New answer posted
7 months agoContributor-Level 10
Circle S? : x² + y² - 10x - 10y + 41 = 0.
Center C? = (5, 5). Radius r? = √ (5² + 5² - 41) = √ (25 + 25 - 41) = √9 = 3.
Circle S? : x² + y² - 16x - 10y + 80 = 0.
Center C? = (8, 5). Radius r? = √ (8² + 5² - 80) = √ (64 + 25 - 80) = √9 = 3.
The solution checks if the center of one circle lies on the other.
Put C? (8, 5) into S? : 8² + 5² - 10 (8) - 10 (5) + 41 = 64 + 25 - 80 - 50 + 41 = 130 - 130 = 0. So C? lies on S?
Put C? (5, 5) into S? : 5² + 5² - 16 (5) - 10 (5) + 80 = 25 + 25 - 80 - 50 + 80 = 130 - 130 = 0. So C? lies on S?
This means both circles pass through the center of each other. So statement (D) is co
New answer posted
7 months agoContributor-Level 10
By property of triangle image of vertex of P is Q about the perpendicular side bisector of triangle Hence according to question X - Y = 0 is a perpendicular side bisector of PQ
Hence solving X - Y = 0 and 2X - y + 2= 0
o (-2, -2)
New answer posted
7 months ago
Contributor-Level 10
P, Q, R represents some students which play all three games. Hence no any option is correct.
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