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UPSC NDA II 2025 Mathematics Paper

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Question : 8 of 120

Marks: +1, -0

If

\alpha =\frac{-1+\sqrt{-3}}{2}

then what is the value of

(1+\alpha ^{19}-\alpha ^{35})^{100}-(1-3\alpha ^{25}+\alpha ^{38})^{50}?

$-2$
$-1$
0
2

Solution:

Concept:

For

\alpha = \frac{-1+\sqrt{-3}}{2}

, we have

\alpha^3=1

and

1+\alpha+\alpha^2=0

.

Explanation:

Reduce exponents modulo

3

:

\alpha^{19}=\alpha

,

\alpha^{35}=\alpha^2

,

\alpha^{25}=\alpha

,

\alpha^{38}=\alpha^2

.

First bracket:

1+\alpha^{19}-\alpha^{35}=1+\alpha-\alpha^2

.

From

1+\alpha+\alpha^2=0

, we get

1+\alpha=-\alpha^2

, so

1+\alpha-\alpha^2 = -2\alpha^2

.

Thus

( -2\alpha^2)^{100}=2^{100}\alpha^{200}=2^{100}\alpha^2

(since

200\bmod 3=2

).

Second bracket:

1-3\alpha^{25}+\alpha^{38}=1-3\alpha+\alpha^2

.

Using

1+\alpha^2=-\alpha

, we get

1-3\alpha+\alpha^2 = -\alpha-3\alpha = -4\alpha

.

Thus

(-4\alpha)^{50}=4^{50}\alpha^{50}=2^{100}\alpha^{50}=2^{100}\alpha^2

(since

50\bmod 3=2

).

Subtract:

2^{100}\alpha^2-2^{100}\alpha^2=0

.

Answer:

0

(Option C).

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