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C? → C? - C?
f (θ) = | -sin²θ -1 |
| -cos²θ -1 |
| 12 -2 -2|
= 4 (cos²θ - sin²θ) = 4 (cos2θ), θ ∈ [π/4, π/2]
f (θ)max = M = 0
f (θ)min = m = -4

f (x) is differentiable then will also continuous then f (π) = -1, f (π? ) = -k?
k? = 1
Now f' (x) = { 2k? (x-π) if x≤π
{ -k? sinx if x>π
then f' (π? ) = f' (π? ) = 0
f' (x) = { 2k? if x≤π
{ -k? cosx if x>π
then 2k? =k?
k? = 1/2

For ellipse x²/16 + y²/9 = 1, a=4, b=3, e = √ (1 - 9/16) = √7/4
A and B are foci then PA + PB = 2a = 2 (4) = 8

The word is 'LETTER'.
Consonants are L, T, R.
Vowels are E, E.
Total number of words (with or without meaning) from the letters of the word 'LETTER' is:
6! / (2! 2!) = 720 / 4 = 180.
Total number of words (with or without meaning) from the letters of the word 'LETTER' if vowels are together:
Treat (EE) as a single unit. We now arrange {L, T, R, (EE)}. This is 5 units.
Number of arrangements = 5! / 2! (for the two T's) = 120 / 2 = 60.
∴ The number of words where vowels are not together = Total words - Words with vowels together
Required = 180 - 60 = 120.

P (x) = 0
x² - x - 2 = 0
(x-2) (x+1) = 0
x = 2, -1 ∴ α = 2
Now lim (x→2? ) (√ (1-cos (x²-x-2) / (x-2)
⇒ lim (x→2? ) (√ (2sin² (x²-x-2)/2) / (x-2)
⇒ lim (x→2? ) (√2 sin (x²-x-2)/2) / (x²-x-2)/2) ⋅ (x²-x-2)/2) ⋅ (1/ (x-2)
⇒ for x→2? , (x²-x-2)/2 → 0?
⇒ lim (x→2? ) √2 ⋅ 1 ⋅ (x-2) (x+1)/ (2 (x-2) = 3/√2

|x + y|² = |x|²
(x+y)· (x+y) = x·x
|x|² + 2x·y + |y|² = |x|²
|y|² + 2x·y = 0 (1)
and (2x + λy)·y = 0
2x·y + λ|y|² = 0 (2)
From (1), 2x·y = -|y|².
Substitute into (2):
-|y|² + λ|y|² = 0
(λ-1)|y|² = 0
Assuming y is a non-zero vector, |y|² ≠ 0, therefore λ=1.

R (-1+2r, 3-2r, -r)
dr's of PR are (2 - 2r, -1+2r, -3+r)
Then 2 (2-2r) + 2 (-1+2r) + 1 (3-r) = 0
9-9r = 0 ⇒ r = 1
R (1,1, -1)
then a+1=2, b+2=2, c-3=-2
a=1, b=0, c=1
∴ a+b+c = 2

Total seat intake for BSc at IHM Chandigarh is 289. To secure a seat in the BSc degree, applicants must successfully crack the NCHMCT JEE exam to be considered. To apply, students must visit the official website of IHM Chandigarh and apply online. Moreover, to secure a seat in the BSc programme, applicants must complete certain eligibility requirements.

Given f (1) = a = 3, and assuming the function form is f (x) = a?
So f (x) = 3?
∑? f (i) = 363
⇒ 3 + 3² + . + 3? = 363
This is a geometric progression. The sum is S? = a (r? -1)/ (r-1).
3 (3? -1)/ (3-1) = 363
3 (3? -1)/2 = 363
3? - 1 = 242
3? = 243
3? = 3? ⇒ n = 5

S = tan? ¹ (1/3) + tan? ¹ (1/7) + tan? ¹ (1/13) + . upto 10 term
S = tan? ¹ (2-1)/ (1+1⋅2) + tan? ¹ (3-2)/ (1+2⋅3) + tan? ¹ (4-3)/ (1+3⋅4) + . + tan? ¹ (11-10)/ (1+11⋅10)
S = (tan? ¹2 - tan? ¹1) + (tan? ¹3 - tan? ¹2) + . + (tan? ¹11 - tan? ¹10)
S = tan? ¹11 - tan? ¹1
S = tan? ¹ (11) - π/4
tan (S) = 5/6

P (A∪B∪C) = P (A) + P (B) + P (C) – P (A∩B) – P (B∩C) – P (C∩A) + P (A∩B∩C)
Given relations lead to: α = 1.4 – P (A∩B) – β ⇒ α + β = 1.4 - P (A∩B)
Again, from P (A∪B) = P (A) + P (B) – P (A∩B), and given values, it is found that P (A∩B) = 0.2.
From (1) and (2), α = 1.2 – β.
Now given 0.85 ≤ α ≤ 0.95
⇒ 0.85 ≤ 1.2 – β ≤ 0.95
⇒ -0.35 ≤ -β ≤ -0.25
⇒ 0.25 ≤ β ≤ 0.35, so β ∈ [0.25, 0.35]

I = ∫ from -π/2 to π/2 (1 / (1+e^ (sin x) dx
I = ∫ from -π/2 to π/2 (e^ (sin x) / (1+e^ (sin x) dx
2I = ∫ from -π/2 to π/2 1dx ⇒ I = 1/2 ∫ from -π/2 to π/2 dx
I = 1/2 [x] from -π/2 to π/2 ⇒ I = π/2

IHM Chandigarh offers a three-year BSc course. The curriculum is spread across six semesters. The first semester of the BSc course spans over six months. The names of the subjects offered in the first semester are as follows:

∴ x² = |x|² = t let
9t² - 18t + 5 = 0
(3t - 1) (3t - 5) = 0
|x| = 1/3, 5/3
Product of roots = (1/3) (-1/3) (5/3) (-5/3) = 25/81

P: n³ - 1 is even, q: n is odd.
The contrapositive of p → q is ~q → ~p.
~q: n is not odd, i.e., n is even.
~p: n³ - 1 is not even, i.e., n³ - 1 is odd.
⇒ "If n is not odd then n³ - 1 is not even"
⇒ "For an integer n, if n is even, then n³ – 1 is odd."