Relations and Functions

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New question posted•a year ago
New question posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

 −1≤x2−3x+2x2+2x+7≤1

⇒0≤2x2−x+9x2+2x+7&−5x−5x2+2x+7≤0

x∈R&−1≤x<∞

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New answer posted•a year ago
S
a year ago

a0 = 0, a1 = 0

an+2 = 3an+1 – 2an + 1

a25 a23 – 2a25a22 – a23a24 + 4a22a24 =?

a2 = 3a1 – 2a0 + 1

a3 = 3a2 – 2a1 + 1

a4 = 3a3 – 2a2 + 1

a5 = 3a4 – 2a3 + 1

an+2  = 3an+1 – 2an + 1

( + ) ⇒ ( a 2 + a 3 + a 4 + . . . . + a n + 1 + a n + 2 )                

⇒ an+2 = 2 (a2 + a3 + …. + an + an+1) –2 (a1 + a2 + ….+ an) + n +

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New answer posted•a year ago
S
a year ago

y = 2x |3x2−5x+2|,0≤x≤1

={−3x2+7x−2,3x2−3x+2,0≤x≤2323<x≤1

3x2 – 5x + 2 = 0

x=+5±25−246

= 5+16=1,23

3x2 – 7x + 3 = 0

x = 7±49−396=7±136

−3x2+7x−2

−7±49−24−6=7+56=2,13

3x2 + 7x – 2 = 1

3x2 – 7x + 1 = 0

x = 7±49−126=7±376

I = ∫06((−2)+1)dx+∫7−37613((−1)+1)dx+∫−137−136(0+1)dx+∫7−13623(1+1)dx+∫231(1+1)dx

=37+13−46

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New answer posted•a year ago
S
a year ago

For R1, take a = 1, b = 0, c = -1

a b ≥ 0 , b c ≥ 0 but AC < 0

So R1 is not an equivalence relation.

For R2, it will not be symmetric.

So R2 is also not an equivalence relation.

 

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

  x 2 − 1 0 x + 9 ≤ 0

  ( x − 1 ) ( x − 9 ) ≤ 0              

A = {1, 2, 3, ……, 9}

for set B,   f ( x ) ≤ ( x − 3 ) 2 + 1

total number of such function = 2 × 1 × 1 × 1 × 2 × 3 × 4 × 5 × 6 = 2 × 6! = 1440

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New answer posted•a year ago
S
a year ago

(x2+1)*1=x*2

⇒ (x2+1)2+1=x2+8⇒x2=2

2sin−1 (x4+x2−2x4+x2+2)=2sin−1 (12)=π3

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