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SSC CHSL 2022 PYQ — Question 55

SSC CHSL - 3 June 2022 - Shift 1 • Quantitative Aptitude • Q55

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If a2+b2+1a2+1b2=4,a≠0,b≠0,a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}=4,a\neq0,b\neq0, then the value of a4 + b4 is:

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 2

Given:

a2+b2+1a2+1b2=4,a≠0,b≠0,a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}=4,a\neq0,b\neq0,

Explanation:

Let the value of a and b be 1 and 1 respectively as a2+b2+1a2+1b2a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2} = 1 + 1 + 1 + 1 = 4 

Now, a4 + b4 = 1 + 1 = 2

Hence the required answer is 2.