PGDBA 2018 Question Paper

With either 0 or 1 as an entry, a sum of 4 is possible only with four 1s and one Zero.

So, each column and row must have four 1s and one 0.

Let us consider the first row, we can put the O in any of the 5 places. 

So, number of ways of placing 1 zero in the first row=5.

In the second row, we cannot put the zero in the same column as that put by the first, as it will result in two )s in the same column. So, the second 0 can be but in any of the remaining 4 places. 

Similarly, for the third zero, we will have 3 ways to do so. And so on.

So, total number of ways to put 5 zeroes in the 5$$\times\ 5$$ matrix is 5*4*3*2*1= 120.

Let the first term of the infinite G.P. be a and the common ratio be r.

So, Sum of infinite GP= $$\ \frac{\ a}{1-r}=14$$ .............. (1)

The cubes of the terms of original GP= $$a^3,\ \left(ar\right)^3,\ \left(ar^2\right)^3...$$= $$a^3,\ a^3r^3,\ a^3r^6...$$

So, in the new infinite GP, first term is $$a^3$$ and the common ratio is $$r^3$$

Therefore, sum to infinite terms= $$\frac{\ a^3\ }{1-r^3}$$ ...............(2)

Cubing the eqn 1 and dividing by eqn 2, we get,

$$\frac{\ \frac{\ a^3}{\left(1-r\right)^3}\ }{\ \frac{\ a^3}{1-r^3}}=\ \ \frac{\ 14\times\ 14\times\ 14}{392}$$

=>$$\ \frac{1-r^3\ }{\left(1-r\right)^3}\ =\ \ \frac{\ 14\times\ 14\times\ 14}{392}$$

=> $$\ \frac{\left(1-r\right)\left(1+r^2+r\right)}{\left(1-r\right)^3}\ =\ \ \ 7$$

=>$$\ \frac{\left(1+r^2+r\right)}{\left(1-r\right)^2}\ =\ \ \ 7$$

=> $$\ \left(1+r^2+r\right)\ =\ \ 7\left(1+r^2-2r\right)$$

=>$$\ 1+r^2+r\ =\ \ 7+7r^2-14r$$

=>$$\ -6-6r^2+15r\ =\ 0$$

=>$$2r^2-5r+2\ =\ 0$$

=>$$2r^2-4r-r+2\ =\ 0$$

=>$$2r\left(r-2\right)-1\left(r-2\right)=0$$

=>(2r-1)(r-2)=0

.'.r= 1/2 or 2.

In v\case of infinite GP, -1<r<1.

So, r= 1/2

Putting this value in eqn 1, we get,

$$\ \frac{\ a}{1-\ \frac{\ 1}{2}}=14$$

=>2a=14

.'. a=7- Option C.

For 3x+4y-24=0 can be re-written as $$3x+4y=24$$

=>$$\frac{3x+4y\ \ }{24}=1$$

=>$$\ \frac{\ x}{8}+\ \frac{\ y}{6}=1$$.

So, this becomes the equation of the straight line of the form $$\ \frac{\ x}{a}+\ \frac{\ y}{b}=1$$, where a and b are x and y intercepts respectively. The X and Y intercepts for this equation are 8 and 6 respectively.

Similarly 3x-4y+24=0 can be re-written as $$\ \frac{\ x}{-8}+\ \frac{\ y}{6}=1$$. The X and Y intercepts for this equation are 8 and -6 respectively.

So, we get the following triangle and the incircle can be drawn as:

So, the triangle has sides of length 10,10 and 16.

Area of the triangle= $$\ \frac{\ 1}{2}\times\ Base\times\ Height$$= $$\ \frac{\ 1}{2}\times\ 16\times\ 6$$= 48

We know that r*s= Area of the triangle, where r is the inradius and s is the semi-perimeter. s= $$\ \frac{\ 1}{2}\left(16+10+10\right)$$= 18

So, 18r= 48

=> Or, r= $$\ \frac{\ 48}{18}=\ \ \frac{\ 8}{3}$$

We will use the formula $$\tan^{-1}\alpha\ -\tan^{-1}\beta\ =\ \tan^{-1}\left(\frac{\alpha\ -\beta\ \ \ }{1+\ \alpha\ \beta\ }\right)$$

The given equation can be re-written as $$\tan^{-1}\left(\ \frac{\ 2-1}{1+1.2}\right)+\tan^{-1}\left(\ \frac{\ 3-2}{1+3.2}\right)+...\ \tan^{-1}\left(\ \frac{\ \left(n+1\right)-n}{1+n\left(n+1\right)}\right)$$

= $$\left(\tan^{-1}2-\tan^{-1}1\right)+\left(\tan^{-1}3-\tan^{-1}2\right)+...\ \left(\tan^{-1}\left(n+1\right)-\tan^{-1}n\right)$$

=$$\tan^{-1}\left(n+1\right)-\tan^{-1}1$$

We will break the function into different parts and then plot the graph.

f(x)= 0; x$$\in\left(-\infty\ ,-1\right)\ $$

$$f\left(x\right)=1+x;\ x\in\left[-1\ ,0\right)\ $$

$$f\left(x\right)=1-x;\ x\in\left[0,1\right]$$

$$f\left(x\right)=1-x;\ x\in\left(1,\infty\ \right)$$

f(x)= 

f(x+2) is same as f(x), but shifted 2 units to the left.

So, f(x+2) can be drawn as:

We can observe that the parts where f(x+2) takes non zero value only where f(x)=0. So, The grapgh of g(x0= f(x)+ f(x+2) will be superimposed upon each other, and there won't be any change in values.

The graph of g(x) will be:

Now, g(x) will be non-differentiable at sharp corners, because at those points Left Hand limit will not equal to Right hand limit.

The graph of g(x) has 5 such sharp edges at (-3,0), (-2,1), (-1,0), (0,1) and (1,0). Therefore g(x) is non-differentiable at 5 points. 

Given $$A = \left(\begin{array}{c}1 & 0\\ -1 & 1\end{array}\right)$$, then $$A^{50}$$

$$\ A^2$$= $$\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}.\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}$$

=$$\begin{bmatrix}(1)(1)+0(-1) & 1(0)+0(1)\\ (1)(-1)+(1)(-1) &(-1)(0)+(1)(1) \end{bmatrix}$$

=$$\begin{bmatrix}1 & 0\\ -2 &1 \end{bmatrix}$$.

Now, $$\ A^3$$= $$\begin{bmatrix}1 & 0\\ -2 &1 \end{bmatrix}.\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}$$

=$$\begin{bmatrix}(1)(1)+0(-1) & 1(0)+0(1)\\ (1)(-2)+(1)(-1) &(-2)(0)+(1)(1) \end{bmatrix}$$

=$$\begin{bmatrix}1 & 0\\ -3 &1 \end{bmatrix}$$.

So, in general $$\ A^n$$= $$\begin{bmatrix}1 & 0\\ -n &1 \end{bmatrix}$$.

Therefore, $$\ A^{50}$$= $$\begin{bmatrix}1 & 0\\ -50 &1 \end{bmatrix}$$, which is given in Option C

It is given that the slope of the curve at any point P is equal to x.

It is also known that, to find the slope at any point P(x,y), we can find $$\ \frac{\ dy}{dx}$$ of the curve at that point P.

For $$\ \frac{\ dy}{dx}$$ of a curve to be equal to x, the original curve must be of the form $$\ ax^2\ +c$$, where a and c are any constants.

Now, y= $$\ ax^2\ +c$$ represents a family of parabolas.

Hence, option B

Given that $$\left|\ \frac{\ df\left(x\right)}{dx}\right|\le\ 1$$

Hence $$-1\le\ \ \frac{\ df\left(x\right)}{dx}\le\ 1$$. Taking definite intergral from [-2 ,2]

-1*(5 - 1) $$\le\ $$ f(2) - f(-2) $$\le\ $$ (5-1)

$$-4\le\ 4\le\ 4$$. It is only possible when $$\ \frac{\ df\left(x\right)}{dx}$$ is always equal to 1. Therefore f(x) is a straight line with slope 1

f(x) = x+3, f(0) = 3

Consider the above Diagram

For option A $$X\cap Z$$ = c+e and $$Y\cap Z$$ = e. Hence option A is False

For Option B $$Y'\cap Z'$$ = a+u  and $$X'\cap Z'$$ = u. Option B is True

For Option C $$X\cap\left(Y\cup Z\right)$$ = c+d+e and $$Y\cup\left(X\cap Z\right)$$ = c+d+e. Option C is True

For option D $$X'\cap Z$$ = b and $$Y'\cap Z$$ = b+c. Option D is False

Given that $$x_0=1\ and\ x_1=2\ $$

From above formula we get $$x_2=3,\ x_3=2,\ x_4=1,\ x_5\ =\ 1\ and\ x_6=2$$

We see that the pattern repeats itself in the period of five with $$x_{5k}=1,\ x_{5k+1}=2,\ x_{5k+2}=3,\ x_{5k+3}\ =\ 2\ and\ x_{5k+4}=1$$

2018 is of form 5k+3 hence $$x_{2018}=2$$