Maths

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.

. Let the terms in Arithmetic Progression be a – 2d, a – d, a, a + d, a + 2d.
Sum of terms: (a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 5a.
5a = 25 ⇒ a = 5.
Product of terms: (5 – 2d) (5 – d) (5) (5 + d) (5 + 2d) = 2520.
5 (25 – 4d²) (25 – d²) = 2520.
(25 – 4d²) (25 – d²) = 504.
625 – 25d² – 100d² + 4d? = 504.
4d? – 125d² + 121 = 0.
Factoring the equation: (4d² - 121) (d² - 1) = 0.
So, d² = 1 or d² = 121/4.
d = ±1 or d = ±11/2.
If d = ±1, the terms are 3, 4, 5, 6, 7.
If d = ±11/2, the terms are -6, -1/2, 5, 21/2, 16.
The largest term is 5 + 2d = 5 + 2 (11/2) = 5 + 11 = 16.

Given 2ae = 6 and 2a/e = 12.
From these, ae = 3 and a/e = 6.
Multiplying the two equations: (ae) (a/e) = 3 * 6 => a² = 18.
We know that b² = a² (1 - e²) = a² - a²e² = 18 - (ae)² = 18 - 3² = 18 - 9 = 9.
The length of the latus rectum (L.R.) is 2b²/a.
L.R. = 2 * 9 / √18 = 18 / (3√2) = 6/√2 = 3√2.

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

c = λ (a x b).
a = I + j - k
b = I + 2j + k
a x b = | I j k |
| 1 -1 |
| 1 2 1 |
= I (1 - (-2) - j (1 - (-1) + k (2-1) = 3i - 2j + k.
c = λ (3i - 2j + k).
Given c ⋅ (i + j + 3k) = 8.
λ (3i - 2j + k) ⋅ (i + j + 3k) = 8
λ (3 - 2 + 3) = 8 => 4λ = 8 => λ = 2.
c = 2 (a x b).
We need to find c ⋅ (a x b).
c ⋅ (a x b) = 2 (a x b) ⋅ (a x b) = 2|a x b|².
|a x b|² = 3² + (-2)² + 1² = 9 + 4 + 1 = 14.
So, c ⋅ (a x b) = 2 * 14 = 28.