Class 12th

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New answer posted

11 months ago

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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

11 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

11 months ago

0 Follower 4 Views

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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

11 months ago

0 Follower 5 Views

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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

11 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

11 months ago

0 Follower 16 Views

P
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

11 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

S = Ltnr=1nn2 (n2+r2) (n+r)

π8+14ln2

New answer posted

11 months ago

0 Follower 9 Views

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Payal Gupta

Contributor-Level 10

l=π2π2dx (1+ex) (sin6x+cos6x) ……. (i)

=20dt4+t2=2 (tan1 (t2))0=π

New answer posted

11 months ago

0 Follower 57 Views

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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=Lth0f (0h)f (0)h=0

RHD=Lth0f (0+h)f (0)h=2

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

m=2, n=3

New answer posted

11 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

x + y + az = 2………… (i)

3x + y + z = 4 ………… (ii)

x + 2z = 1 ……………. (iii)

x = 1, y = 1, z = 0

(for unique solution a 3  

( α , 1 ) , ( 1 , α ) a n d ( 1 , 1 ) are collinear.

α 1 1 α = 1 α 1 1 1 α 2 = 0 α = ± 1                

Sum of absolute value = 2

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