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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