Question : The 12th term of the series $\frac{1}{x}+\frac{x+1}{x}+\frac{2x+1}{x}+...$ is:
Option 1: $\frac{11x+1}{x}$
Option 2: $\frac{12x+1}{x}$
Option 3: $\frac{x+12}{x}$
Option 4: $\frac{x+11}{x}$
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Correct Answer: $\frac{11x+1}{x}$
Solution : The given series is a sequence where each term is of the form $\frac{nx+1}{x}$, where n is the term number starting from 0. So, the 12th term of the series (where n = 11 because we start from 0) would be $\frac{11x+1}{x}$. Hence, the correct answer is $\frac{11x+1}{x}$.
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Question : The simplified form of $(x+3)^2+(x-1)^2$ is:
Option 1: $(x^2+2x+5)$
Option 2: $2(x^2+2x+5)$
Option 3: $(x^2-2x+5)$
Option 4: $2(x^2-2x+5)$
Question : The simplified form of $(x+3)^{2}+(x-1)^{2}$ is:
Option 1: $(x^{2}+2x+5)$
Option 2: $2(x^{2}+2x+5)$
Option 3: $(x^{2}-2x+5)$
Option 4: $2(x^{2}-2x+5)$
Question : If $x^2-11x+1=0$, what is the value of $x^8-14159x^4+11$?
Option 1: 9
Option 2: 10
Option 3: 12
Option 4: 11
Question : Which of the following expressions is equal to the expression $\frac{x^2-3 x+2}{x^2-4}$?
Option 1: $\frac{x+1}{x-2}$
Option 2: $\frac{x-1}{x+2}$
Option 3: $\frac{x+1}{x+2}$
Option 4: $\frac{x-1}{x-2}$
Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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