AP ICET 16th May 2016

Retain the first term as it is. 

Multiply the numerator and denominator of the second term by x. 
it gets transformed to $$\frac{x}{x+x^2+1}$$

Multiply the numerator and denominator of the third term by x*x
It gets transformed to $$\frac{x^2}{x+x^2+1}$$

As can be seen, the denominators of all three terms are equal. 
The numerators can all be added up directly now. It is seen that numerator and denominator are equal now.
Hence, answer = 1 i.e. option C

Given,  $$2^{2n-1}=\frac{1}{8^{n-3}}$$

$$\Rightarrow$$  $$2^{2n-1}=\frac{1}{2^{3\left(n-3\right)}}$$

$$\Rightarrow$$  $$2^{2n-1}=2^{-3\left(n-3\right)}$$

As bases are equal, equating powers

$$2n-1=-3\left(n-3\right)$$

$$\Rightarrow$$  $$2n-1=-3n+9$$

$$\Rightarrow$$  $$5n=10$$

$$\Rightarrow$$  $$n=2$$

$$\therefore\ $$ $$2^{n-2}=2^{2-2}=2^0=1$$

Given,

$$a+b=7$$ ............(1)

$$b+c=-3$$ .........(2)

$$c+a=-4$$ .........(3)

Adding (1), (2) and (3), we get

$$2a+2b+2c=0$$

$$\Rightarrow$$  $$a+b+c=0$$ ........(4)

From (1) and (4),  $$c=-7$$

From (2) and (4), $$a=3$$

From (3) and (4), $$b=4$$

$$\therefore\ $$ $$a^3+b^3+c^3=3^3+4^3+\left(-7\right)^3=27+64-343=-252$$

Multiplying both numerator and denominator of first term by $$\left(2+\sqrt{\ 3}\right)$$, the term changes to $$\left(2+\sqrt{\ 3}\right)^2$$
Multiplying both numerator and denominator of second term by $$\left(2-\sqrt{\ 3}\right)$$, the term changes to $$\left(2-\sqrt{\ 3}\right)^2$$
Multiplying both numerator and denominator of third term by $$\left(\sqrt{\ 3}-1\right)$$, the term changes to $$\frac{\left(\sqrt{\ 3}-1\right)^2}{2}$$

Opening up the squares and adding up the terms, we get $$16-\sqrt{\ 3}$$, hence, a = 16, b = -1
Hence, a+4b becomes 12