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RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 20 of 120

Marks: +1, -0

I f

X+\frac{1}{X}=3

X^5+\frac{1}{X^5}

is

123
126
113
129

Solution:

Concept:

Use standard algebraic identities to find higher powers from

x+\frac{1}{x}

.

Explanation:

Given:

x+\frac{1}{x}=3

.

Square both sides:

\left(x+\frac{1}{x}\right)^2=3^2

x^2+\frac{1}{x^2}+2=9

So,

x^2+\frac{1}{x^2}=7

.

Cube both sides:

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

x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=27

Put

x+\frac{1}{x}=3

:

x^3+\frac{1}{x^3}=27-9=18

Now multiply the two results:

\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=7\times18

x^5+\frac{1}{x^5}+x+\frac{1}{x}=126

Use

x+\frac{1}{x}=3

:

x^5+\frac{1}{x^5}=126-3=123

Answer:

123

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