PGDBA 2017 Question Paper

Consider, $$x^3-1$$ = (x-1)($$x^2 + x + 1 $$)=0

It will have roots as 1, $$\alpha\ $$ and $$\beta\ $$

Now, since both  $$\alpha\ $$ and $$\beta\ $$ satisfy $$x^3-1=0$$, Hence,$$\alpha^{3}-1=0$$ => $$\alpha^3=1$$

and $$\beta^3-1=0$$  => $$\beta^3=1$$

Hence $$\alpha^{2017}+\beta^{2017}$$ = $$\alpha^{3\times\ 672+1}+\beta^{3\times\ 672+1}$$  =  $$\alpha\ +\beta$$

Now, the sum of roots of the equation $$x^3-1$$=0 is zero. 

Hence,  $$1+\alpha\ +\beta\ =0=>\alpha+\beta\ =-1$$

Assume x=a+1, y=b+1, z=c+1

a+1+b+1+c+1=10,

a+b+c=7

The number of non-negative solutions for a, b and c will give the postive integral solution for x,y,z which is $$^{7+3-1}C_{3-1}$$ = $$\ \frac{\ 9\times\ 8}{2}$$ = 36

We have, $$x^2 - y^2 = 2y + 1$$

=> $$x^2 = y^2 + 2y + 1$$

=> $$x^2 - (y+1)^2 $$=0

=> (x-y-1)(x+y+1)=0

Here we have a pair of straight lines: x-y-1=0 and x+y+1=0.

Assume the number of students who failed in exactly 1 subject is x

The number of students who failed in exactly 2 subjects is y

The number of students who failed in exactly 3 subjects is z

Now, x+2y+3z=50+45+40=135

x+y+z=100-1=99

On subtracting the two equations, we get,

y+2z=36

It is given that y=32

Hence, 2z=36-32=4  =>z=2

$$\ \frac{\ dy}{dx}\ =\ 2x-6$$

Slope of the tangent at R(4, 10) = 2

A normal will be perpendicular to tangent, so the slope of the normal = $$\ \frac{\ -1}{2}$$

Equation of the tangent at R(4, 10) = 

y-10 = 2(x-4)

y=2x+2

The above line intersects Y-axis at P (0,2)

Equation of the normal at R(4, 10) =

y-10 = $$\ \frac{\ -1}{2}$$ (x-4)

2y=-x+24

The above line intersects Y-axis at Q (0,12)

Let the equation of the  circle be $$x^2+y^2+2gx+2fy+c=0$$ with centre (-g, -f)

Now we have to find the equation of the circle passing through P (0, 2), Q(0, 12), R(4, 10)

On substituting these values in the equation of the circle, we get g = 0, f = -7, c = 24

Radii of the circle = $$\sqrt{\ \left(g^2+f^2-c\right)}$$

= 5

To make a regular hexagon, three equilateral triangles of side$$\ \frac{\ a}{3}$$ will be cut from the corners to make a regular hexagon of side a.

Hence, this hexagonal can be further divided into 6 equilateral triangles of side $$\ \frac{\ a}{3}$$.

Hence the area of the hexagon = $$\ 6\cdot\ \frac{\ \sqrt{\ 3}}{4}$$*($$\ \ \frac{\ a}{3}$$)$$^2$$ = $$\frac{\sqrt3 a^2}{6}$$

A is the answer.

For the function to be differentiable at x=0, f'(x) =$$\lim\ h\longrightarrow\ 0$$  $$\ \frac{\ f\left(x+h\right)-f\left(x\right)}{h}$$

=> f'(0)=$$\lim\ h\longrightarrow\ 0$$ $$\ \frac{\ f\left(h\right)-f\left(0\right)}{h}$$

Now using the value of f(x), we get f'(0) =$$\lim\ h\longrightarrow\ 0$$   $$\ \frac{\ a_0+a_1\mid h\mid+a_2\mid h\mid^2+a_3\mid h\mid^3-a_0}{h}$$

Now, f'(0)= $$\lim\ h\longrightarrow\ 0$$  $$\ \frac{a_1\mid h\mid+a_2\mid h\mid^2+a_3\mid h\mid^3}{h}$$

For h<0, f'(0)=$$-\ a_1+a_2h-a_3h^2$$

For h>0, f'(0)=$$\ a_1+a_2h+a_3h^2$$

So for $$h\longrightarrow\ 0$$, the two values will only be equal if $$a_1$$ = 0.

Hence, C is the answer.

Taking -1 common from both 2nd and 3rd row of Q, we get

Q = (-1)(-1)$$\begin{bmatrix}-x & a & -p \\-y & b & -q\\-z & c & -r \end{bmatrix}$$

= $$\begin{bmatrix}-x & a & -p \\-y & b & -q\\-z & c & -r \end{bmatrix}$$

Now taking -1 common from 1st and 3rd column of Q, we get

Q= (-1)(-1)($$\begin{bmatrix}x & a & p \\y & b & q\\z & c & r \end{bmatrix}$$

= ($$\begin{bmatrix}x & a & p \\y & b & q\\z & c & r \end{bmatrix}$$

After interchanging the 1st and second column, Q becomes the the transpose of P.

Since the row have been exchange once and the determinant of the transpose matrix is the same, hence det(P)=(-1)$$^1$$*det(Q)

=> $$det(P) = -det(Q)$$

 D is the answer.

The number of ways to select non-empty set out of S = 100C1+100C2+100C3+..........100C100 = $$2^{100}$$-1

Similarly, the number of ways to select non-empty set from (say) P= {1,3,5,7,9,...................99} = $$2^{50}$$-1

Hence, the required number of set = $$2^{100}$$-1 - ($$2^{50}$$-1) = $$2^{50}\left(2^{50}-1\right)$$

In triangle AFJ, $$x_{1}$$+180-a+180-b=180

=> $$x_{1}$$+180=a+b

Similarly, $$x_{2}$$+180=a+e

$$x_{3}$$+180=d+e

$$x_{4}$$+180=c+d

$$x_{5}$$+180=b+c

Adding all the equations,

$$x_{1}$$+$$x_{2}$$+$$x_{3}$$+$$x_{4}$$+$$x_{5}$$+180*5= 2(a+b+c+d+e)

=> Sum of interior angles = 2*(a+b+c+d+e)-900  = 2*540-900=180   (Sum of interior angles of a pentagon = 540)