Class 12th

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New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

f ( x ) = a x 2 + b x + c  

g (x) = px + q

  f ( g ( x ) ) = a ( p x + q ) 2 + b ( p ) ( + q ) + c

8 x 2 2 x = a ( p x + q ) 2 + b ( p x + q ) + c

Compare 8 = ap2 …………… (i)

-2 = a (2pq) + bp

0 = aq2 + bq + c

  9 ( f ( x ) ) = p ( a x 2 + b x + c ) + q

=>4x2 + 6x + 1 = apx2 + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x2 + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Fraction of molecules having enough energy to form product  =   e E a / R T

 Fraction of molecules having enough energy to form product =   e 8 0 . 9 * 1 0 3 8 . 3 1 4 * 7 0 0

= e 1 3 . 8 e 1 4

So, x = 14

 

New answer posted

3 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Δ T f = K f * m * i  

0.93 = 1.86 * 1 * i

i =  1 2  

i = 1 + ( 1 n 1 )

n = 2            

            

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

M n O 4 + S 2 O 3 2 O H M n O 2 + S O 4 2

O.S of S in    S O 4 2 = + 6

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

Ratio = n/n = 1

New answer posted

3 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

f (x) is an even function

f ( 1 4 ) = f ( 1 2 ) = f ( 1 2 ) = f ( 1 4 ) = 0  

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

g ( 3 4 ) = g ( 3 4 ) = 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 d d x ( f ( x ) g ' ( x ) ) = 0 will have atleast 4 roots in (-2, 2)

New answer posted

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)           

6x = 5 = 0             x = 5 6  

P (atmost 2H)

P ( O H , 5 T ) + P ( 1 H , 4 T ) + P ( 2 H , 3 T )  

= 1 6 5 ( 1 + 2 5 + 2 5 0 ) = 2 7 6 6 5 = 4 6 6 4  

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given a > b

Area common to x2 + y2 a 2 a n d x 2 a 2 + y 2 b 2 1  

is π a 2 π a b = 3 0 π . . . . . . . . . . . . . . ( i )  

Similarly π a b π b 2 = 1 8 π . . . . . . . . . . . . . . . . . ( i i )  

Equation (i) and equation (ii)  a b = 5 3  

Equation (i) + equation (ii) a 2 b 2 = 4 8  

a2 = 75, b2 = 27

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Seliwanoff's test is used to distinguish carbohydrates while xanthoproteic test is used to distinguish proteins

New answer posted

3 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

a * ( b * c ) = 3 b c = u

b * ( c * a ) = c 2 a = v

c * ( b * a ) = 3 b 2 a = w

u + v = w

so vectors

u , v a n d w

are coplanar, hence their Scalar triple product will be zero.

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