PGDBA 2017 Question Paper

$$\frac{Numerator}{n}$$ = $$\Sigma\ n\left(n+1\right)\left(n+2\right)\ =\ \frac{\left(\Sigma\ n^3+3\Sigma n^2+2\Sigma\ n\ \right)}{n}$$

$$\frac{n^2\left(n+1\right)^2}{4n}+\frac{3n\left(n+1\right)\left(2n+1\right)}{6n}+\frac{2n\left(n+1\right)}{2n}$$

Denominator 

$$\Sigma\ n\left(n+1\right)\ =\ \Sigma\ n^2+n\ =\ \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}$$

Numerator/Denominator = $$\frac{3}{4}\left[1+\frac{3}{n}\right]$$

When n -> $$\infty\ ,\ \frac{1}{n}->\ 0$$

= 3/4

The letters except the vowels are P, R, B, B, L, T, Y.

The number of words that can be formed = 7!/2 =2520

A is the correct answer.

$$y =2x^2$$ and $$y = 6$$

6 = $$2x^2$$

x = -$$\sqrt{3}$$ and $$\sqrt{3}$$

$$\int_{-\sqrt{3}}^{\sqrt{3}} 2 *x^2$$

Since x^2 is an even function

$$\int_{-\sqrt{3}}^{\sqrt{3}} 2 *x^2$$ 

= 4$$\int_{0}^{\sqrt{3}} x^2$$

= 4*$$\frac{x^3}{3}$$ where x varies from 0 to $$\sqrt{3}$$

= $$4\sqrt3$$.

Now this is the area between the parabola and the x-axis. The area between the line y=6 and x-axis is $$6\times\ 2\sqrt{\ 3}$$

Hence, the area between the curves is $$8\sqrt{\ 3}$$.

We know that $$lim_{x \rightarrow 0} \frac{\sin x}{x}$$ is 1

=$$\lim_{x\rightarrow\ 0} \frac{sin x^2}{x^2}*\frac{x}{sin x}$$

= 1

We know that $$(1+x)^n$$ = $$^nC_0*1^n+^nC_1*1^{n-1}*x......................................^nC_n*x^n$$

On Integrating, we get

$$\frac{(1+x)^{n+1}}{n+1}$$ = $$\frac{^nC_0*1^n}{1}+\frac{^nC_1*1^{n-1}*x}{2}......................................\frac{^nC_n*x^n}{n+1}$$

When x = 1

$$\frac{2^{31}}{31}$$ =  $$\frac{30_{C_0}}{1}$$ +$$\frac{30_{C_1}}{2} + \frac{30_{C_2}}{3} +\frac{30_{C_3}}{4} + \frac{30_{C_5}}{6} +...+ \frac{30_{C_{29}}}{30} + \frac{30_{C_{30}}}{31}$$   -- Eq 1

When x = -1

0= $$ \frac{30_{C_0}}{1}$$ -$$\frac{30_{C_1}}{2} + \frac{30_{C_2}}{3} -\frac{30_{C_3}}{4} + \frac{30_{C_5}}{6} +...+ \frac{30_{C_{29}}}{30} - \frac{30_{C_{30}}}{31}$$  -- Eq 2

Adding Eq 1 and 2, we get

$$\frac{2^{31}}{31}$$ = 2($$\frac{30_{C_1}}{2} + \frac{30_{C_3}}{4} + \frac{30_{C_5}}{6} +...+ \frac{30_{C_{29}}}{30}$$)

($$\frac{30_{C_1}}{2} + \frac{30_{C_3}}{4} + \frac{30_{C_5}}{6} +...+ \frac{30_{C_{29}}}{30}$$) = $$\frac{2^{30}}{31}$$

Join BD

Area of ABCD = area of ABD + area of BCD

In ABD $$AB^2\ +\ AD^2\ =BD^2$$

Therefore   $$24^{2\ }+\ 18^{2\ }=\ BD^2$$ , on solving BD= 30

also $$30^{2\ }+\ 40^2\ =50^2$$ which implies BCD is also a right-angled triangle

Area of ABD = $$\frac{1}{2}\times\ 24\times\ 18$$ = 216

Area of BCD = $$\frac{1}{2}\times\ 30\times\ 40$$= 600

Area of ABCD = 816

The expression will be $$\cot\left(\frac{\pi}{40}\right)\cot\left(\frac{2\pi}{40}\right)....\cot\left(\frac{19\pi}{40}\right)$$

Now $$\cot A=\tan\left(\frac{\pi}{2}-A\right)$$

Thus $$\cot\frac{\pi}{40}=\tan\left(\frac{\pi}{2}-\frac{\pi}{40}\right)=\tan\frac{19\pi}{40}$$

Also, (cotA)(tanA)=1.

Hence, all the terms other than $$\cot\frac{10\pi}{40}$$ pair up and become 1.

Now $$\cot\frac{10\pi}{40}=\cot\frac{\pi}{4}=1$$

Hence, the value of the expression will be 1.

$$\mid x - 1\mid$$ 

=$$\begin{cases}x-1 & x \geq 1\\1-x & x < 1 \end{cases}$$

x < 1 f(x)= $$\mid x + 1\mid$$ 

x  $$\geq$$ 1  f(x) = $$\mid 3-x\mid$$ 

x < -1 f(x) = -x-1

x  $$\geq$$ -1 f(x) = x+1

x < 3 f(x) = x-3

x  $$\geq$$ 3 f(x) =3-x

Calculate the values of 

f'(-2) = -1, f'(0) = 1,f'(2) = 1, f'(4) = -1

$$f'(-2) + f'(0) + f'(2) + f'(4)$$ = 0

Hence B is the correct answer.

$$det(P)$$ = a(2*-2)-(-3*1) + (-b)[(-1*-2)-(2*1)]+0[(-1*-3)-(2*2)]

=a(-1)-b(0)

=-a

But given $$det(P)$$ = -2

Hence a = 2

Minor of $$M_{22}$$ = $$det\begin{bmatrix}a & 0 \\2 & -2 \end{bmatrix}$$

det of $$\begin{bmatrix}2 & 0 \\2 & -2 \end{bmatrix}$$

= -4 

Hence A is the correct answer.