Maths

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.

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)

P, Q, R represents some students which play all three games. Hence no any option is correct.

Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0

Kindly consider the following figure

Kindly consider then following figure

According to the question,  
  (7 * 4) + (n * 6) = 52
6n = 24 ⇒ n = 4

The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.