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Q.

If $a + \frac{1}{a} = 4$ ; find
(1) $a^{2} + \frac{1}{a^{2}}$

(2) $a^{4} + \frac{1}{a^{4}}$

(3) $a^{3} + \frac{1}{a^{3}}$

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a

12, 190 and 48

b

14, 194 and 52

c

16, 198 and 56

d

18, 202 and 60

answer is B.

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Detailed Solution

Given,

a + \frac{1}{a} = 4

(1) In the first part we have to find

a^{2} + \frac{1}{a^{2}}

Here, we have,

a + \frac{1}{a} = 4

On squaring both sides we get,

{(a + \frac{1}{a})}^{2} = (4)^{2}

a^{2} + \frac{1}{a^{2}} + 2 \cdot a \cdot \frac{1}{a} = 16

(since,

(a + b)^{2} = a^{2} + b^{2} + 2 ab

)

a^{2} + \frac{1}{a^{2}} = 16.2

a^{2} + \frac{1}{a^{2}} = 14

(2) In the second part we have to find

a^{4} + \frac{1}{a^{4}}

.
Square both sides,

{(a^{2} + \frac{1}{a^{2}})}^{2} = (14)^{2}

a^{4} + \frac{1}{a^{4}} + 2 . a^{2} \cdot \frac{1}{a^{2}} = 196

a^{4} + \frac{1}{a^{4}} = 196 - 2

a^{4} + \frac{1}{a^{4}} = 194

(3) In the third part we have to find

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

Cube both sides,

{(a + \frac{1}{a})}^{3} = (4)^{3}

a^{3} + \frac{1}{a^{3}} + 3 \cdot a \cdot \frac{1}{a} (a + \frac{1}{a}) = 64

(since,

(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)

)

a^{3} + \frac{1}{a^{3}} + 3 (a + \frac{1}{a}) = 64

a^{3} + \frac{1}{a^{3}} + 3.4 = 64

a^{3} + \frac{1}{a^{3}} = 64 - 12

a^{3} + \frac{1}{a^{3}} = 52

Thus, the required values of the given expression are 14, 194 and 52.
Correct option is 2.

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