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New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

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.

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

x? = Σf? x? / Σf?
Σf? =? C? +? C? +? C? + . +? C? = 2?
Σf? x? = (0 *? C? ) + (2 *? C? ) + (2² *? C? ) + . + (2? *? C? )
This sum is Σ? ? C? 2? = (Σ? ? C? 2? ) -? C?2? = (1+2)? - 1 = 3? - 1.
x? = (3? - 1)/2?
Given x? = 728 / (something that resolves to 2? ). Assuming it is 728/2?
(3? - 1)/2? = 728/2?
⇒ 3? - 1 = 728
⇒ 3? = 729
⇒ 3? = 3?
⇒ n = 6

New answer posted

7 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

|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.

New answer posted

7 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

|λ-1 3λ+1 2λ|
|λ-1 4λ-2 λ+3| = 0
|2 3λ+1 3 (λ-1)|
R? → R? - R? and R? → R? - R? (from a similar matrix setup, applying operations to simplify)
The provided solution uses a slightly different matrix but let's follow the subsequent steps.
A different matrix from the image is used in the calculation:
|λ-1 3λ+1 2λ|
|0 λ-3 -λ+3|
|3-λ 0 λ-3 |
C? → C? + C?
|3λ-1 3λ+1 2λ |
|3-λ λ-3-λ | = 0
|0 λ-3 |
⇒ (λ-3) [ (3λ-1) (λ-3) - (3λ+1) (3-λ)] = 0
⇒ (λ-3) [ (λ-3) (3λ-1) + (λ-3) (3λ+1)] = 0
⇒ (λ-3)² [3λ-1 + 3λ+1] = 0
⇒ (λ-3)² [6λ] = 0 ⇒ λ = 0, 3
Sum of values of λ = 3

New answer posted

7 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

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

New answer posted

7 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

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]

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