Correct Answer: (c)

Explanation: $a_{n+2} = \; \frac{2}{a_{n+1}} + a_n$ $\Rightarrow a_n a_{n+1} + 1 = a_{n+1} a_{n+2} - 1$ $\Rightarrow a_{n+2} a_{n+1} - a_n \cdot a_{n+1} = 2$ For $n=1 \; a_3 a_2 - a_1 a_2 = 2$ $n=2 \; a_4 a_3 - a_3 a_2 = 2$ $n=3 \; a_5 a_4 - a_4 a_3 = 2$ . . . $n=n \;\; \frac{a_{n+2} a_{n+1} - a_n a_{n+1} = 2}{a_{n+2} a_{n+1} = 2n + a_1 a_2}$ Now, $\frac{a_1 a_2+1}{a_2 a_3} \cdot \; \frac{a_2 a_3+1}{a_3 a_4} \cdot \; \frac{a_3 a_4+1}{a_4 a_5} \cdots \cdot \; \frac{a_{30} a_{31}+1}{a_{31} a_{32}}$ $= \; \frac{3}{4} \cdot \; \frac{5}{6} \cdot \; \frac{7}{8} \cdots \cdot \; \frac{61}{62}$ $= 2^{-60} \binom{61}{31}$

Correct Answer: (d)

Explanation: $x, y > 0$ and $x^3 y^2 = 2^{15}$ Now, $3x + 2y = (x + x + x) + (y + y)$ So, by $A \cdot M \ge$ G.M inequality $\frac{3x+2y}{5} \ge \;\sqrt[5]{x^3 \cdot y^2}$ $\therefore 3x+2y \ge 5 \;\sqrt[5]{2^{15}} \ge 40$ $\therefore$ Least value of $3x + 4y = 40$

Correct Answer: 1251

Explanation: $2040=2^{3} \times 3 \times 5 \times 17$ $n$ should not be multiple of $2,3,5$ and 17 . Sum of all $n=(1+3+5+\dots+99)-(3+9+15+$ $21+\dots+99)-(5+25+35+55$ $+65+85+95)$ $-(17)$ $=2500-\frac{17}{2}(3+99)-365-17$ $=2500-867-365-17$ $=1251$

Correct Answer: (b)

Explanation: We know that $e=1+\;\frac{1}{1!}+\;\frac{1}{2!}+\;\frac{1}{3!}+\cdots$ and $e^{-1}=1-\;\frac{1}{1!}+\;\frac{1}{2!}-\;\frac{1}{3!}+\cdots$ $\therefore e+e^{-1}=2\left[1+\;\frac{1}{2!}+\;\frac{1}{4!}+\cdots\right]$ $\therefore\;\frac{1}{2!}+\;\frac{1}{4!}+\;\frac{1}{6!}+\cdots$ $=\;\frac{e+e^{-1}}{2}-1$ $=\;\frac{e^2+1-2e}{2e}$ $=\;\frac{(e-1)^2}{2e}$

Correct Answer: (c)

Explanation: We already know, Arithmetic mean $\ge$ Geometric mean, Let us take AM and GM of two terms $a^{a^{x}}$ and $a^{1-a^{x}}$ , $\Rightarrow A M=\;\frac{a^{a^{x}}+a^{1-a^{x}}}{2}$ and $G M=\sqrt{a^{a^{x}} \cdot a^{1-a^{x}}}$ $\because\ A M \ge G M \Rightarrow\ \frac{a^{a^{x}}+a^{1-a^{x}}}{2} \ge \sqrt{a^{a^{x}} \cdot a^{1-a^{x}}} \Rightarrow a^{a^{x}}+a^{1-a^{x}} \ge 2 \sqrt{a^{1}}$ $\therefore$ Minimum value of $f(x)=a^{a^{x}}+a^{1-a^{x}}$ is $2 \sqrt{a}$ .