[Shirsendu Roy]

Here we see that she must also take three upsteps and five diagonal steps. Now, a step to the right or an upstep changes the parity of the coordinate sum, and a diagonal step does not change it.

So, between two right steps there must be a upstep and similarly between two upsteps there must be a right step.. We may write HVHVHV

The diagonal steps may be distributed in any fashion before, in between and after HV sequence. The required number is nothing but the number of ways of distributing 5 identical objects into 7 distinct boxes =

{11 \choose 6}.

[Shirsendu Roy]

Let radius of half circles be r where 4r=4 given then r=1

area of half petal=

difference between area of sector-area of triangle in context with quarter sector

=\pir^{2}/4-r^{2}/2=r^{2}(\pi-2)/4

8 such half petals=(8)(r^{2}(\pi-2)/4)=2r^2(\pi-2) equation 1

 

area of inside square with side 2r=(2r)(2r)=4r^2   equation 2

 

area between inside square and half circles in 8 parts=

diffrence as r^2/2-r^2/4(\pi-2)

8 such parts accumulates area=(8)[r^2/2-r^2/4(\pi-2)]=2r^2(4-\pi) equation 3

adding all area from equation 1 2 and 3= required non shaded part=

8r^2=(8)(1)^2=8.

[Shirsendu Roy]

Let y=x-{(a+b)/2}

then the given equation reduces to

16y^{4}+24y^{2}(a-b)^{2}-7(a-b)^{4}=0

which on solving gives

y=+-i\sqrt{7}(a-b)/2 or, y=+-(a-b)/2

which gives two complex roots and two real roots.

 

[Shirsendu Roy]

f'(x)f(x)<0

caseI

f(x)<0, f'(x)>0

f(x) is increasing

|f(x)| taken and found decreasing

case II

f(x)>0, f'(x)<0

f(x) is decreasing

|f(x)| taken and found decreasing

so, |f(x)| must be decreasing.

[Shirsendu Roy]

out of odd number of factors of the number N,

1 is common factor rest of the factors can be taken in pairs

like say x,y,z are rest of the factors ... then N/x , N/y, N/z are also the factors as pair which can be expressed as prime factors each to an even power

 

and suppose that x factor have no pair then x and N/x is unique that is x=N/x

that is N=x^2

that is perfect square.

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