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New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

? given statement is

(AC)B then its negation is  { (AC)B}

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

Letx3=θθ2 (π4, 3π4)

y=tan1 (secθtanθ)

tan1 (1sinθcosθ)

dydx=3x22d2ydx2=3x

x2d2ydx26y+3π2=0x2y116y+3π2=0

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

|a+b+2 (a*b)|=2, θ (0, π)

squaring on both sides, we get = 2π3

where θ is angle between a^andb^.

2|a^*b^|=3=|a^b^|, S1 is correct.

and projection of a^ona^+b^=|a^. (a^+b^)|a^+b^||=12

So (S2) is correct.

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

Let P (x, y, z) be any point on plane P1 then

(x+4)2+ (y2)2+ (z1)2= (x2)2+ (y+2)2+ (z3)2

cosθ=|62+3|14θ=π3

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

S. D. = 13

b1*b2=|i^j^k^23λ145|=i^ (154λ)+j^ (λ10)+k^ (5)

| (i^2j^2k^). { (154λ)i^+ (λ10)j^+5k^}| (154λ)2+ (λ10)2+25=13

λ=16

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

cos (x+π3)cos (π3x)=14cos22x

x=3π, 2π, π, 0, π, 2π, 3π

 total number of solution = 7.

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

 ?  x is a random variable.

k+2k+4k+6k+8k=1k=121

P ( (1<x<4)|x2)=4k7k=47

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

f (x)=x77x2

f' (x)=7x67f' (x)=0, x=±1

f (1)=<0andf (1)>0

Hence number of real roots of f (x) = 0 are 3.

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

Δ=412|1α1α010α1|=4

α=±8

(α, α), (α, α)and (α2, β) are collinear.

|αα1αα1α2β1|=0β=64

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

 x2a2+y24=1

Δ=12*a (1+cosθ).4sinθ

ΔmaxΔ=63a=4

e=32

 

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