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CTET Paper II (Mathematics and Science) 31 Dec 2021 Solved Paper

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Question : 34 of 150

Marks: +1, -0

What is the value of

[(\frac{1}{3})^{-2}-(\frac{1}{4})^{-1}]^{-2}?

25
$\frac{1}{25}$
144
$\frac{1}{144}$

Solution:

Concept:

A negative exponent means reciprocal:

a^{-n} = \frac{1}{a^n}

. Also,

(\frac{1}{a})^{-n} = a^n

.

Explanation:

First simplify each term inside the brackets.

(\frac{1}{3})^{-2} = 3^2 = 9

.

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

.

Now subtract:

9 - 4 = 5

.

The expression becomes

5^{-2}

.

Apply the negative exponent:

5^{-2} = \frac{1}{5^2} = \frac{1}{25}

.

Answer:

\frac{1}{25}

, which is option B.

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