PGDBA 2018

In RHS of the equation,

The min value is attained is 2 when x = 0 ( from A.M-GM inequality)

For the LHS side the max value is 2 when sin term has value of 1. But it can not be obtained when x=0

Hence, there are no real solution for the given equation

For $$x\le\ 1$$ the graph of the function is an upward parabola that attains its min at x = -1 and value is 0, at x = 1 its value is 16

Similarly for $$x>1$$ we see that its upward parabola which has value of 16 at x= 1 and decreases till x=5 at which its value is 0. After which the values increases again.

It can be seen local maxima is attained at x=1 only

Let us write all the number the numbers from 1 to 2018. Here we find that there are 44 numbers that are perfect squares which are 1,4,9... 1936($$44^2$$). Thus after removing the perfect square it has total of 2018-44 = 1974 terms.

Taking next 44 terms from 2019, it will be (2019,2020,2021..... 2062) in which there is 1 perfect square (2025 which is $$45^2$$) hence it is removed and now there is total of 2017 terms. Next term has to be 2063 which is also $$2018^{th}$$ term

For $$x<0$$ , f(x) = $$2e^x$$ and it is a continuous function with approaching the value 2 as $$x\longrightarrow\ 0$$ 

For $$x\ge\ 0$$ , f(x) = $$2e^{-x}$$ and it is a continuous function with approaching the value 2 as $$x\longrightarrow\ 0$$ 

Therefore the function is continuous.

However, $$x\longrightarrow\ 0^-$$ f'(x) = 2 but $$x\longrightarrow\ 0^+$$ f'(x) = -2

Thus it is not differentiable at exactly 1 point

For n, we consider B as a group that has all 6 women. Total arrangement of B and 5 men is 6!. Further within B women can rearrange themselves in 6! ways.

Thus n = 6! $$\ \times\ $$6!

For m, consider a group $$B_1$$ which has 5 women and $$B_2$$ has remaining 1 woman. Arrangement of $$B_1$$,$$B_2$$ and 5 men can be done in 7! ways. From this we need to subtract the permutation in which $$B_1$$ and $$B_2$$ are together. Which is $$2!\times\ 6!$$ . Total ways $$=\ 7!\ -\ 2!\times 6!$$ .Which equals $$5\times\ 6!$$. Futhermore in $$B_1$$ the selection and arrangement of 5 women can be done in $$_5^6C\times\ \ 5!$$ = 6! ways.

m = $$5\times\ 6!\times\ 6!$$

$$\ \frac{\ m}{n}=5$$

Let the locus of the center be represented by (h,k)

Since it passes through origin. Radius r = $$\sqrt{\ h^2+k^2}$$

Since it cuts off chord of length = 2a on y=c. We draw 1 perpendicular bisector of the chord till the center and make a line which is intersecting point of line y=c and the circle.

We obtain a right triangle that has "r" as its hypotenuse and a and |c-k| as its sides.

Applying Pythogoras theorm, 

$$r^2\ =\ a^2\ +\ \left(c-k\ \right)^2$$

it implies

$$h^2+k^2\ =\ a^2\ +\ \left(c-k\ \right)^2$$

$$h^2+k^2\ =\ a^2\ +\ c^2+k^2\ -2ck$$

$$h^2+2ck\ =\ a^2\ +\ c^2$$

replacing (h,k) by (x,y)

we get $$x^2+2cy\ =\ a^2\ +\ c^2$$

$$l_1 = \lim_{x \rightarrow 0^-}f(x)$$

When $$x\longrightarrow\ 0^-$$ then $$\left[x\right]$$ = -1

f(x) can be written as $$[x]+1-\sin\ \left(\ \frac{\pi\ \ }{2}\left[x\right]\right)$$.

Putting [x] = -1 we get $$l_1$$ = $$-1+1-\sin\ \left(\ \frac{\ \pi\ }{2}\times\ \left(-1\right)\ \right)$$ = $$-\sin\ \left(\ \frac{\ \pi\ }{2}\times\ \left(-1\right)\ \right)$$ = 1

for $$l_2$$, $$x\longrightarrow\ 0^+$$ thus [x] = 0

putting it in f(x). $$l_2$$ = 0+1-0 = 1

Hence $$l_1 =l_2=1$$ 

Let the above equation be written as  Ax=b

where A = $$\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8 \\a & 9 & b & 10 \\6 & 8 & 10 & 13 \end{bmatrix}$$

x = $$\begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4 \end{bmatrix}$$

b =  $$\begin{bmatrix}0 \\0 \\0 \\0 \end{bmatrix}$$

It is evident that $$x_1=x_2=x_3=x_4=\ 0\ $$ is a solution for the question. 

It is stated that the equation has at least 2 distinct solution which implies infinite solution exists. and thus det(A) = 0.

Calculating det(A) we get det(A) = 4(a+b-18)

Equating it with 0, we get a+b-18=0 which is a straight line equation

Since we see that both y and x has even terms, it implies the curve is symmetric along x-axis as well as y-axis.

Required area = 4*(area in quadrant I)

Area in quadrant I is given by 

A =$$_0\int^1\sqrt{\ x^2-x^4}dx\ $$

A = $$_0\int^1x\sqrt{1-x^2}dx$$

take $$1-x^2\ =\ t^2$$, then $$-2x\ dx\ =\ 2\ t\ dt$$. when x =0 then t = 1 and  when x = 1 then t = 0

Updated integral can be written as 

$$\int_0^1t^{2\ }dt$$ = $$\left[\frac{t^3\ }{3}\right]_0^1\ =\ \ \frac{\ 1}{3}$$

Required area = 4*A = $$\ \frac{\ 4}{3}$$

We see that 2,8,18... can be represented as $$2n^2$$ where n$$\in\ N$$

Required sum = $$\Sigma_1^{\infty}\left(\cot^{-1}\left(2n^2\ \right)\right)\ =\ \Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ 1}{2n^2}\ \right)\right)$$

=$$\Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ 2}{4n^2}\ \right)\right)$$

= $$\Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ \left(2n+1\right)-\left(2n-1\right)}{1+\left(4n^2-1\right)}\ \right)\right)$$

= $$\Sigma\ _1^{\infty\ }\ \tan^{-1}\left(2n+1\right)\ -\ \tan^{-1}\left(2n-1\right)$$

= $$\tan^{-1}\infty\ -\ \tan^{-1}1$$

=$$\ \frac{\ \pi\ }{2}-\ \frac{\ \pi\ }{4}\ =\ \ \frac{\ \pi\ }{4}$$