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UPTET Paper 2 Math and Science 2019 Solved Paper

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Question : 105 of 120

Marks: +1, -0

If

x^{4}+\frac{1}{x^{4}}=14

, then value of

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

3\sqrt{6}

\frac{18}{\sqrt{6}}

9\sqrt{\frac{2}{3}}

3\sqrt{2}

A, C, D
B, C, D
A, B, D
A, B, C

Solution:

Concept:

Use algebraic identities:

(a^2+b^2)

from

(a^4+b^4)

and then

(a+b)^2

to get

x+\frac{1}{x}

, finally apply identity for

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

.

Explanation:

Given

x^{4}+\frac{1}{x^{4}}=14

. Add

2

to both sides:

x^{4}+2+\frac{1}{x^{4}}=16

. This is

(x^{2}+\frac{1}{x^{2}})^{2}=16

, so

x^{2}+\frac{1}{x^{2}}=4

.

Add

2

again:

x^{2}+2+\frac{1}{x^{2}}=6

, which gives

(x+\frac{1}{x})^{2}=6

, so

x+\frac{1}{x}=\sqrt{6}

(positive).

Now use identity:

x^{3}+\frac{1}{x^{3}}=(x+\frac{1}{x})^{3}-3(x+\frac{1}{x})=(\sqrt{6})^{3}-3\sqrt{6}=6\sqrt{6}-3\sqrt{6}=3\sqrt{6}

.

Check options:

3\sqrt{6}

matches A.

B=\frac{18}{\sqrt{6}}=3\sqrt{6}

,

C=9\sqrt{\frac{2}{3}}=3\sqrt{6}

, D is

3\sqrt{2}

. So A, B, C are correct.

Answer:

D. A, B, C

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