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a year ago

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Payal Gupta

Contributor-Level 10

cos (x+π3)cos (π3−x)=14cos22x

x=−3π, −2π, −π, 0, π, 2π, 3π

∴ total number of solution = 7.

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

 ?  x is a random variable.

∴k+2k+4k+6k+8k=1∴k=121

∴P ( (1<x<4)|x≤2)=4k7k=47

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

f (x)=x7−7x−2

f' (x)=7x6−7⇒f' (x)=0, x=±1

∴f (1)=<0   and  f (−1)>0

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

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

Δ=4⇒12|1α1α010α1|=4

⇒α=±8

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

∴|α−α1−αα1α2β1|=0⇒β=−64

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

 x2a2+y24=1

∴Δ=12*a (1+cosθ).4sinθ

∴ΔmaxΔ=63⇒a=4

∴e=32

 

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

Equation of tangent at P (x, y) is Y = dydx (X−x)

A/q,      2x−ydxdy=0⇒2dyy=dxx

⇒2lny=lnx+lncy2=xc

It passes through (3, 3), c = 3

∴y2=3x        ∴ Length of latus rectum = 3

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

S = Ltn→∞∑r=1nn2 (n2+r2) (n+r)

π8+14ln2

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

l=∫−π2π2dx (1+ex) (sin6x+cos6x) ……. (i)

=2∫0∞dt4+t2=2 (tan−1 (t2))0∞=π

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

f (x)= {sin (x+2)x+2, x∈ (−2, −1)         0, x∈ (−1, 0]       2x, x∈ (0, 1)         1, otherwise

LHD=Lth→0f (0−h)−f (0)−h=0

RHD=Lth→0f (0+h)−f (0)h=2

Hence f (x) is not differentiable at x = 1, 0, 1

∴m=2, n=3

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 x1y>0   and   x3y2=215

AM≥GM

3x+2y5≥ (x3y2)15

⇒3x+2y≥40

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