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SSC Selection Post 2022 PYQ — Question 52

SSC Selection Post - Higher Secondary Level 2022 - 15 March 2022 - Shift 2 • Quantitative Aptitude • Q52

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If 2x+13x+1=0,x2x+{{1} \over 3x}+1=0,x ≠0{{≠0}} ,then what is the value of 27x3+1833\sqrt[3]{{27x^3+{{1} \over 8^3}}}

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Correct answer: A

Explanation

The correct answer is Option 1 i.e. 32{{3} \over 2}

6x2+1+3x=0⇒6x2+1=−3x6x^2 + 1 + 3x = 0 \\ \Rightarrow 6x^2 + 1 = -3x \\

27x3+18x33=(3x)3+(12x)33a3+b3=(a+b)3−3ab(a+b)\sqrt[3]{27x^3 + \frac{1}{8x^3}} = \sqrt[3]{(3x)^3 + \left( \frac{1}{2x} \right)^3} \\ a^3 + b^3 = (a + b)^3 - 3ab(a + b) \\

⇒(3x)3+(12x)33=(3x+12x)3−3⋅3x⋅12x⋅(3x+12x)3⇒(6x2+12x)3−92⋅(6x2+12x)3\Rightarrow \sqrt[3]{(3x)^3 + \left( \frac{1}{2x} \right)^3} = \sqrt[3]{\left( 3x + \frac{1}{2x} \right)^3 - 3 \cdot 3x \cdot \frac{1}{2x} \cdot \left( 3x + \frac{1}{2x} \right)} \\ \Rightarrow \sqrt[3]{\left( \frac{6x^2 + 1}{2x} \right)^3 - \frac{9}{2} \cdot \left( \frac{6x^2 + 1}{2x} \right)} \\

6x2+1=−3x:⇒(−3x2x)3−92⋅(−3x2x)3⇒(−32)3−92⋅(−32)3 6x^2 + 1 = -3x: \\ \Rightarrow \sqrt[3]{\left( \frac{-3x}{2x} \right)^3 - \frac{9}{2} \cdot \left( \frac{-3x}{2x} \right)} \\ \Rightarrow \sqrt[3]{\left( \frac{-3}{2} \right)^3 - \frac{9}{2} \cdot \left( \frac{-3}{2} \right)} \\

⇒−278+2743=2783⇒32\Rightarrow \sqrt[3]{\frac{-27}{8} + \frac{27}{4}} = \sqrt[3]{\frac{27}{8}} \\ \Rightarrow \frac{3}{2}