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SSC CHSL 2021 PYQ — Question 69

SSC CHSL - 19 April 2021 - Shift 1 • Quantitative Aptitude • Q69

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If x+1x=7,x+{{1} \over x}=\sqrt7,, then what is the value of (x2+1)÷[x4+(1x2)]?(x^2+1)\div[x^4+({{1} \over x^2})]?

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Correct answer: A

Explanation

The correct answer is Option 1 i.e. 14{{1} \over 4}

x+1x=7x+{{1} \over x}=\sqrt7

x3 + 1/x3 = (7)3−37=47(\sqrt7)^3-3\sqrt7=4\sqrt7

So, the value of (x2+1)÷[x4+(1x2)](x^2+1)\div[x^4+({{1} \over x^2})]

= 1x(x+1x)1x(x3+1x3)\frac{\frac{1}{x}(x+\frac{1}{x})}{\frac{1}{x}(x^3+\frac{1}{x^3})}

= 747=14\frac{\sqrt7}{4\sqrt7}=\frac{1}{4}