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CTET Paper II (Mathematics and Science) 29 Jan 2023 Solved Paper

Question : 32 of 150

Marks: +1, -0

Value of

(\frac{-1}{3})^{4}+(\frac{-1}{6})^{3}

lies between:

Solution:

Concept:

When a negative number is raised to an even power, the result is positive. When raised to an odd power, the result is negative.

Explanation:

First,

(\frac{-1}{3})^4 = \frac{1}{81}

because

(-1)^4=1

and

3^4=81

Next,

(\frac{-1}{6})^3 = -\frac{1}{216}

because

(-1)^3=-1

and

6^3=216

Add them:

\frac{1}{81} - \frac{1}{216}

Use common denominator 648:

\frac{8}{648} - \frac{3}{648} = \frac{5}{648}

\frac{5}{648} \approx 0.00771

, which is between 0 and 0.5.

Answer:

D. 0 and 0.5

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