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New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

(p⇒q)∧ (q⇒¬p) ≡ (¬p∨q)∧ (¬q∨¬p) ≡ ¬p∧ (q∨¬q) ≡ ¬p.

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

L: x/1=y/0=z/-1. Let a point on L be (r,0, -r).
Direction ratio of PN is (r-1, -2, -r+1).
PN is perpendicular to L, so (r-1) (1)+ (-2) (0)+ (-r+1) (-1)=0 ⇒ 2r-2=0 ⇒ r=1. N (1,0, -1).
Let Q be (s,0, -s). Direction ratio of PQ is (s-1, -2, -s+1).
PQ is parallel to plane x+y+2z=0, so normal to plane is perp to PQ.
(s-1) (1)+ (-2) (1)+ (-s+1) (2)=0 ⇒ s-1-2-2s+2=0 ⇒ -s-1=0 ⇒ s=-1. Q (-1,0,1).
PN = (0, -2,0). PQ= (-2, -2,2).
cosα = |PN.PQ|/|PN|PQ| = 4/ (2√12) = 1/√3.

New answer posted

2 months ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

New answer posted

2 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

e? - e? - 2e³? - 12e²? + e? + 1 = 0
e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
Let y=e? y? -y? -2y³-12y²+y+1=0.
The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
Either e³? = 4e? + 1 (One Solution)
OR e³? = 1 - 3e? (One Solution)
∴ the equation has total 2 solutions.

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1

New answer posted

2 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

A = {2,4,6,8, . . .50} ⇒ 25 element
A = {7,14,21, . . . .49} ⇒ 7 elements
A ∩ B = {14,28,42} = 3 elements
Required number of elements = 25 + 7 - 3 = 29

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Let A=[a?]?. tr(AA?)=3.
Σa?² = 3.
Possible cases for non-zero elements are (1,1,1) or (-1,-1,-1) or (1,1,-1) etc.
Total combinations = ?C? * 2³ = 84 * 8 = 672.C? * 2³ = 84 * 8 = 672.

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