Question : The value of $\frac{4x^{3}-x}{(2x+1)(6x-3)}$, when $x=9999$ is:
Option 1: 1111
Option 2: 2222
Option 3: 3333
Option 4: 6666
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Correct Answer: 3333
Solution : Given expression = $\frac{4x^3 - x}{(2x + 1)(6x - 3)}$ and $x=9999$ Now, $\frac{x\left(4x^2 - 1\right)}{\left(2x + 1\right)\times3\left(2x - 1\right)}$ Using identity: $a^{2}-b^{2}=(a-b)(a+b)$, = $\frac{x(2x + 1)(2x - 1)}{3(2x + 1)(2x - 1)}$ = $\frac{x}{3}$ = $\frac{9999}{3}$ = 3333 Hence, the correct answer is 3333.
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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: $\frac{5}{3}$
Question : If $2x+\frac{1}{4x}=1$, then the value of $x^{2}+\frac{1}{64x^{2}}$ is:
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{4}$
Option 4: $2$
Question : What is the value of $\frac{4x^2+9y^2+12xy}{144}$?
Option 1: $(\frac{x}{3} + \frac{y}{4})^2$
Option 2: $(\frac{x}{3} + y)^2$
Option 3: $(\frac{x}{4} + \frac{y}{6})^2$
Option 4: $(\frac{x}{6} + \frac{y}{4})^2$
Question : If $2x-3(4-2x)<4x-5<4x+\frac{2x}{3}$, then $x$ can take which of the following values?
Option 1: 2
Option 2: 8
Option 3: 0
Option 4: –8
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