Class 11th

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New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

S-2, L-2, A, B, Y, U
Required = ²C? ⋅? C? ⋅ 4!/2! = 2 ⋅ 10 ⋅ 24/2 = 240

New answer posted

2 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

2x-y+3=0
4x-2y+α=0 ⇒ 2x-y+α/2=0
6x-3y+β=0 ⇒ 2x-y+β/3=0
d? = |α/2-3|/√ (2²+1²) = 1/√5 ⇒ |α-6|=2 ⇒ α-6=2, -2 ⇒ α=8,4
d? = |β/3-3|/√ (2²+1²) = 2/√5 ⇒ |β-9|=6 ⇒ β-9=6, -6 ⇒ β=15,3
Sum of all value of α and β = 30.

New answer posted

2 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

T? = ²²C? (x? )²²? x? ²?
T? = ²²C? x? ²²? ²?
22m - mr - 2r = 1
22m - 1 = r (m+2)
r = (22m-1)/ (m+2)
r = (22m+44-45)/ (m+2)
r = 22 - 45/ (m+2)
So possible value of m = 1,3,7,13,43
but ²? C? = 1540
only possible condition is m=13

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

 x? = (2+4+10+12+14+x+y)/7 = 8
⇒ 42+x+y = 56 ⇒ x+y = 14
σ² = (Σx? ²/n) - (x? )²
16 = (4+16+100+144+196+x²+y²)/7 - (8)²
⇒ 16+64 = (460+x²+y²)/7
⇒ 560 = 460+x²+y² ⇒ x²+y² = 100
⇒ xy=48
(x-y)² = (x+y)² - 4xy = 4
|x-y| = 2

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

2 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

2 months ago

0 Follower 5 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

2 months ago

0 Follower 2 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

2 months ago

0 Follower 6 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

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

∴ 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

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