PGDBA 2016 Question Paper

The series is 1, 3*1, 7*3*1, 3*7*3*1, 7*3*7*3*1, 3*7*3*7*3*1.........100 terms

The series of the even terms = 3*1, 3*7*3*1, 3*7*3*7*3*1.....

Sum of the first 50 even terms = $$\frac{3(21^{50}-1)}{20}$$

The series of the odd terms = 1, 7*3*1, 7*3*7*3*1.....

Sum of the first 50 even terms = $$\frac{1(21^{50}-1)}{20}$$

Required sum =  $$\frac{3(21^{50}-1)}{20}$$+$$\frac{1(21^{50}-1)}{20}$$

=4*$$\frac{21^{50}-1}{20}$$

=$$\frac{1}{5}(21^{50} - 1)$$

A is the correct answer.

$$4 \sin^2 x - 4 \cos x = 1$$

=> $$4\left(1-\cos\ ^2x\right)-4\cos x-1=0$$

=> $$4\cos\ ^2x+4\cos x+1=4$$

$$\left(2\cos x+1\right)^2=2^2$$

$$2\cos x+1=\pm\ 2$$

$$2\cos x=1\ or\ 2\cos x=-3$$

$$\cos x=\frac{1}{2}\ or\ \cos x=-\frac{3}{2}$$ 

-3/2 is not possible

Hence x can be $$\frac{\pi}{3}\ or\ 2\pi\ -\frac{\pi}{3}$$

Sum = $$2\pi$$

$$\frac{h}{SQ}=\frac{8}{25}$$

$$\frac{h}{RQ}=\frac{17}{25}$$

=> $$SQ^2=\frac{25^2}{8^2}h^2$$

$$RQ^2=\frac{25^2}{17^2}h^2$$

$$SQ^2=RQ^2+375^2$$

Upon solving, h = 136

The system will have a unique solution if x=y=z

Substituting in 1, we get x+x+kx=1 => x(2+k) =1

=> x = 1/2+k = y = z, where k is not equal to -2

Hence it has a unique solution for infinitely many choices of k

Using $$A.M.\ge\ G.M.$$

$$\frac{\left(4^{\sin x}+4^{\cos x}\right)}{2}\ge\ \sqrt{\ 4^{\sin x+\cos x}}$$

or,  $$\left(4^{\sin x}+4^{\cos x}\right)\ge2\ \sqrt{\ 4^{\sin x+\cos x}}$$.

$$\left(4^{\sin x}+4^{\cos x}\right)\ge2\ \sqrt{\ 2^{2\left(\sin x+\cos x\right)}}=2.2^{\sin x+\cos x}=2^{1+\sin x+\cos x}$$

The minimum value of sinx +cosx when x lies in R is $$-\sqrt{\ 2}$$

Thus, the minimum value of the expression is $$2^{1-\sqrt{\ 2}}$$

Numerator is equal to $$2^n$$

Hence $$\sum_{n = 0}^\infty \frac{n_{C_{0}} + n_{C_{1}} + ..... +n_{C_{n}}}{n_{P_{n}}}$$ = $$\sum_{n=0}^{\infty}\frac{2^n}{n!}$$ 

We know $$e^x=\frac{x^n}{n!}$$

Hence e^2

f(x) = $$cosx (x^2*2-x*2x) - x(2sinx *2-2x*tanx) + 1(2sinx *x-tanx*x^2)$$

= $$x^2\tan x-2x\sin x$$

$$\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$$ 

$$=\lim_{x\rightarrow0}\frac{x^2\tan x}{x^2}-\frac{\left(2x\sin x\right)}{x^2}$$

=$$\lim_{x\rightarrow0}\tan x-\frac{2\sin x}{x}$$

= 0-2=-2

B is the correct answer.

det P = $$2\left(2\alpha\ \right)-\alpha\ \left(-\alpha\ ^2\right)+3\left(2\alpha\ -6\right)=\ \alpha\ ^3+10\alpha\ -18$$

cofactor of $$a_{22}=2\alpha\ -9$$

$$\alpha\ ^3+10\alpha\ -18=2\alpha\ -9$$

$$\alpha\ ^3+8\alpha\ -9=0$$

=> $$\alpha\ =1$$

det(P) = $$2\alpha\ -9\ =-7$$

$$\det\left(adj\left(P^{-1}\right)\right)=\left[\det\left(P^{-1}\right)\right]^{3-1}$$   since $$\det\left(adjA\right)=\left(\det A\right)^{n-1}$$

= $$\left[\frac{1}{\det\left(P\right)}\right]^2=\frac{1}{49}$$