Class 12th

Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.

A = 1/3 [ 1; 1 ω ω² 1 ω² ω ]
A² = A * A = 1/9 [ . ]
(The calculation in the image shows A² is the identity matrix, let's verify)
A² leads to I (Identity matrix).
So A² = I.
A³ = A² * A = I * A = A.
A? = (A²)² = I² = I.
A³? = (A²)¹? = I¹? = I.

dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
g' (t) = 2/t³ - 12/t? g' (4) < 0, hence maximum.
∴ g (t)max = g (4) = (4 - 2)/4² = 2/16 = 1/8.

The equation of the line is (x-2)/1 = (y-1)/1 = (z-6)/-2.
Let this be equal to k. So, a point on the line is (k+2, k+1, -2k+6).
This point lies on the plane x + y - 2z = 3.
(k+2) + (k+1) - 2 (-2k+6) = 3
2k + 3 + 4k - 12 = 3
6k - 9 = 3
6k = 12 ⇒ k = 2.
The point of intersection is (2+2, 2+1, -2 (2)+6) = (4, 3, 2).

PR (line): r = (3i - j + 2k) + λ (4i - j + 2k) - (I)
QS (line): r = (i + 2j - 4k) + μ (-2i + j - 2k) - (II)
If they intersect at T then:
3 + 4λ = 1 - 2μ
-1 - λ = 2 + μ
2 + 2λ = -4 - 2μ
Solving the first two equations gives λ = 2 & μ = -5. These values satisfy the third equation.
∴ T (11, -3, 6)
Also, OT is coplanar with lines PR and QS.
⇒ TA ⊥ OT
|OT| = √166
|TA| = √5
|OA| = √ (|OT|² + |TA|²) = √171

Given f (g (x) is defined piecewise:
f (g (x) =
x³ + 2 ; x < 0
x? ; 0? x < 1
(3x - 2)² ; x? 1
fog (x) is discontinuous at x = 0.? non differentiable.
fog (x) is not differentiable at x = 0.

Using Lagrange's Mean Value Theorem (LMVT) for x ∈ [−7, -1].
[f (-1) - f (-7)] / [-1 - (-7)] ≤ 2
[f (-1) - (-3)] / 6 ≤ 2
f (-1) + 3 ≤ 12
f (-1) ≤ 9

Using LMVT for x ∈ [−7, 0].
[f (0) - f (-7)] / [0 - (-7)] ≤ 2
[f (0) - (-3)] / 7 ≤ 2
f (0) + 3 ≤ 14
f (0) ≤ 11

Therefore, f (0) + f (-1) ≤ 11 + 9 = 20.

Answer given by NTS is (1) which is wrong.
I = 1/ (a+b) ∫? x [f (x) + f (x+1)]dx . (1)
Using the property x → a + b - x
I = 1/ (a+b) ∫? (a+b-x) [f (a+b-x) + f (a+b+1-x)]dx
Given f (a+b+1-x) = f (x)
I = 1/ (a+b) ∫? (a+b-x) [f (x+1) + f (x)]dx . (2)
Adding (1) and (2):
2I = 1/ (a+b) ∫? (a+b) [f (x) + f (x+1)]dx
2I = ∫? [f (x) + f (x+1)]dx
2I = ∫? f (x)dx + ∫? f (x+1)dx
Let x+1 = t in the second integral, so dx = dt.
When x=a, t=a+1. When x=b, t=b+1.
∫? f (x+1)dx = ∫? ¹ f (t)dt = ∫? ¹ f (x)dx