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New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

x2 = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

now |α8+β8|=2|α8|

=2| (α2)|4

=2|α2|4

=2|12i|4

=2*25

 

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

=|21113214δ|=0δ=3

and Δ1=|711132k43|=0k=6

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Area of ΔABC

12ABBC

=1221

=12

now required area

=04 (222x)dx12

=3223=8212

=1326

New answer posted

4 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

 ac^=α+6+21+4+4

103=8+α3α=2

b*c=i^ (2β8)+j^ (10)+k^ (6+β)

=6i^+10j^+7k^

So = 1

New answer posted

4 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Draw g(t) = t3 – 3t

g'(t) = 3(t2 – 1)

g(1) is maximum in (-2, 2)

So, maximum (t3 – 3t) = {t33t;2<t<12;1<t<2

I=22f(x)dx

21(t33t)dt+122dt

I = 274

again rewrite the f(x)

f(x)={x33x2;x11<x2x2+2x69;2<x<33x<410112x+1;4x<5x=5x>5}

f'(x)={3x23;x<10;1<x<22x+2;2<x<30;3<x<40;4<x<52;x>5}

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

x23=y41=z74=2 (64+282)32+ (1)2+42

Let x23=y41=z74=2813=λ

x=3λ+2y=λ+4z=4λ+7

Now according to the question

(a, b, c) =  (3λ+2, λ+4, 4λ+7)

Now 2a+b+2c=6λ+4λ+4+8λ+14

13λ+22

= 28 + 22 [Use λ=2813 ]

= 6

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

 dydx=yx+y2+16x2x

Put y = Vx

differentiable worst = x

dydx=V+xdVdx

V + xdVdx=V+V2+16

xdVdx=V2+16

apply variable separable method

dVV2+16=dxx+lnC

ln|V+V2+16|=lnCx

yx+y2+16x2x=Cx

Given y(1) = 3 C = 8

Now at x = 2

y2+y2+16.42=8.2

y2 + 64 = (32 – y)2

y = 15

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

=10+4*10? C2104=10+4*10.92104=191000

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Let PT perpendicular to QR

x+12=y+23=z12=λT (2λ1, 3λ2, 2λ+1) therefore

2 (2λ5)+3 (3λ4)+2 (2λ6)=0λ=2

T (3, 4, 5)PT=1+4+4=3QT=269=17

ΔPQR=12*217*3=317

Therefore square of ar (ΔPQR) = 153.

New answer posted

4 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

 f' (x)=4x2? 1x so f (x) is decreasing in  (0, 12)and? ? (12, ? ) ? a=12

Tangent at y2 = 2x is y = mx + 12m it is passing through (4, 3) therefore we get m = 12or? ? 14

So tangent may be y=12x+1? ? or? ? y=14x+2? ? ? but? ? y=12x+1 passes through (-2, 0) so rejected.

Equation of normal x9+y36=1

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