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New answer posted

7 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

We have, 1 - (probability of all shots result in failure) > 1/4
? 1 - (9/10)? > 1/4
? 3/4 > (9/10)? ? n? 3 > (9/10)? ? n? 3

New answer posted

7 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

-5 ≤ [x/2] < 5
I ⇒ [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.

New answer posted

7 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

-5? [x/2] < 5
I? [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.

New answer posted

7 months ago

0 Follower 10 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

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Volume of parallelepiped v = | [a? b? c? ]|
v = | 1 n |
| 2 4 -n| = ±158
| 1 n 3 |
1 (12+n²) - 1 (6+n) + n (2n-4) = ±158
3n²-5n+152=0 or 3n²-5n+164=0
D<0 (no real roots)
n=8, -19/3 ⇒ n=8
then b? ⋅c? = 2+4n-3n=10
a? ⋅c? = 1+n+3n=33

New answer posted

7 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

Equation x²/5 + y²/4 = 1 then P (√5cosθ, 2sinθ)
(PQ)² = 5cos²θ + 4 (sinθ+2)² = cos²θ + 16sinθ + 20
= -sin²θ + 16sinθ + 21
= 85 - (sinθ - 8)²
∴ (PQ)²max = 85 - 49 = 36
? (sinθ - 8)² ∈

New answer posted

7 months ago

0 Follower 3 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

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 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

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