Class 12th

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New answer posted

6 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

Since,  limx0? f (x)x exist f (0)=0

Now,  f' (x)=limh0? f (x+h)-f (x)h=limh0? f (h)+xh2+x2hh ( take y=h)

=limh0? f (h)h+limh.0? (xh)+x2

f' (x)=1+0+x2

f' (3)=10

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

D=1-232111-7a=0a=8

also,  D1=9-23b1124-78=0b=3

hence,  a-b=8-3=5

New question posted

6 months ago

0 Follower 1 View

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

D=14-1315ab6=021a-8b-66=0P:2x-3y+6z=15

so required distance =217=3

New question posted

6 months ago

0 Follower 1 View

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Given 2x2+3x+410=r=020?arxr

replace x by 2x in above identity :-

2102x2+3x+410x20=r=020??ar2rxr210r=020??arxr=r=020??ar2rx(20-r)( from (i))

now, comparing coefficient of x7 from both sides

(take r=7 in L.H.S. and r=13 in R.H.S.)

210a7=a13213a7a13=23=8

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

xdydx-y=x2(xcos?x+sin?x),x>0

dydx-yx=x(xcos?x+sin?x)dydx+Py=Q

So, I.F. =e-1xdx1|x|=1x(x>0)

Thus, yx=1x(x(xcos?x+sin?x))dx

yx=xsin?x+C

?y(π)=πC=1

So, y=x2sin?x+x(y)π/2=π24+π2

Also, dydx=x2cos?x+2xsin?x+1

d2ydx2=-x2sin?x+4xcos?x+2sin?x

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

max {n (A), n (B)}n (AB)n (U)

7676+63-x100

-63-x-39

63x39

New answer posted

6 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

f(x)=13?xdx(1+x)2=13?t.2tdt1+t22 (put x=t )

=-t1+t213+tan-1?t13 [Applying by parts]

=-34-12+π3-π4=12-34+π12

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  [x]2+2 [x+2]-7=0

[x]2+2 [x]+4-7=0 [x]=1, -3x [1,2) [-3, -2)

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