Class 12th

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New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

CaO – Basic Oxide

SO3 & SiO2 – Acidic oxides

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

Isotopes are the atom of same element with different atomic mass.

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11 months ago

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P
Payal Gupta

Contributor-Level 10

N=M2+M4+.....+M98

=(α2I)+(α2I)2+....+(α2I)49

=I(α2+α4α6+....α98)

N = I(α2α4+α6.......+α98)

=Iα2(1(α2)49)1(α2)

N = Iα2(1+α98)1+α2

Now (Im2)N=2I

(I+α2I)(Iα2(1+α98)1+α2=2I

? α100 + α2 = 2

? α = ±1

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

f (x) is an even function

f (14)=f (12)=f (12)=f (14)=0

So, f (x) has at least four roots in (-2, 2)

g (34)=g (34)=0

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) g" (x)=f' (x)g' (x)=0

It is same as number of roots of ddx (f (x)g' (x))=0 will have atleast 4 roots in (2, 2)

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11 months ago

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P
Payal Gupta

Contributor-Level 10

Given a > b

Area common to x2 + y2 a2andx2a2+y2b21

is πa2πab=30π.............. (i)

Similarly πabπb2=18π................. (ii)

Equation (i) and equation (ii) ab=53

Equation (i) + equation (ii) a2b2=48

a2 = 75, b2 = 27

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

 dydx+2xx1y=1 (x1)2

IF =e2xx1dx

e2x (x1)2

ye2x (x1)2= {e2x (x1)2 (x1)2dx+C

y = e2x2 (x1)2+C (x1)2

y (2) = 1+e42e4, C=12

y (3) = eα+1βeα=e6+18e6

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

Data contradiction.

a* (b*c)= (ac)b (ab)c

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11 months ago

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P
Payal Gupta

Contributor-Level 10

Mid point of BC is 12 (5i^+ (α2)j^+9k^)

AB¯=i^+ (α4)j^+k^

AC¯=i^+ (2α)j^+k^

For = 1,  AB¯ and AC¯ will be collinear. So for non collinearity

= 2

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

 x11=y22=z12=2 (1+4+216)1+22+22

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

|xyz+136612|=0

2x – z = 1

option (B) satisfies.

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11 months ago

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P
Payal Gupta

Contributor-Level 10

Let AB x2y+1=0

AC 2xy+1=0

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

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