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New answer posted

3 months ago

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Payal Gupta

Contributor-Level 10

  (x+1)2λ+5= (y+1)2λ+54=1

length of latus rectum = 2b2a=2 (λ+54)5+λ=4

λ+5=8λ=59

Major axis = 2λ+5=16

New answer posted

3 months ago

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Payal Gupta

Contributor-Level 10

Let A =  [abcdef9hi]

Now ATA

trace will be a2+b2+c2+d2+e2+f2+92+h2=6

total ways = 9C6.26.1.1.1 = 5376

New answer posted

3 months ago

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Payal Gupta

Contributor-Level 10

|f (x)|8002n2n1800

2n2n8010

xSf (x)= (2x2x1)

=2 (192+182+........12+02+12+.....+202)

= 10620

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

y5 – 9xy + 2x = 0

differentiate 5y4 – 9x dydx 9y + 2 = 0

dydx=9y25y49x

For horizontal tangent dydx=0y=29 which does not satisfy the equation so no horizontal

For vertical tangent 5y49x=0

m = 0, N = 2

New answer posted

3 months ago

0 Follower 13 Views

P
Payal Gupta

Contributor-Level 10

sin(2x2).109e(tanx2)dy+4xydx=42.x.(sinx2cosπ4cosx2sinπ4)dx

ln(tanx2)dy+4xsin(2x2)ydx=4x(sinx2cosx2)sin(2x2)dx

Integrate

y.ln(tanx2)=2.ln(sinx2+cosx21sinx2+cosx2+1)+C

x=π6,y=1 calculate C.

New answer posted

3 months ago

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Payal Gupta

Contributor-Level 10

eH=1+6449=1137

eH.eE=12

11349. (64a2)64=14a264=322113

l=2a2b=2 (64+322113).18

113l=1552

New answer posted

3 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is x (4a+2λ)+y (15λ)+z (5λ)=7a+3λ

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line x41=y+12=z1

4a+11λ+7=0 ……. (ii)

From (i) & (ii) a = 1,  λ =1

Equation of plane πx+2y+3z2=0

7P+32P+412P+92=0P=2

New answer posted

3 months ago

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Payal Gupta

Contributor-Level 10

x¯=i=110xi10=15i=110xi210 (x¯)2=15

Σxi=150Σxi2=2400

Actual mean x¯=Σxi+152510=14010=14

Actual variance = Σxi2+15225210 (14)2

=240040010196

σ2=4σ=2

New answer posted

3 months ago

0 Follower 12 Views

P
Payal Gupta

Contributor-Level 10

use sin1x=cos11x2

tan11x2=cot111x2

sin1x1x2=cos112x21x2

Sum of roots b = 1 + 2  (k21)k22

Product of roots 5 = 2  (k21k22)

b = 4, k213

New answer posted

3 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

? x=sin (2tan1α)=2α1+α2 ……. (i)

and

y=sin (12tan143)=sin (sin115)=15

Now,

y 2 = 1 x

α = 2 , 1 2 a S 1 6 α 3 = 1 6 * 2 3 + 1 6 * 1 2 3 = 1 3 0

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