SATHEE: Functions 2 Question 11

Functions 2 Question 11

12.

Let $f (x) = | x - 1 |$ . Then,

(1983, 1M)

(a) $f (x^{2}) = {f (x)}^{2}$

(b) $f (x + y) = f (x) + f (y)$

(d) None of the above

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Answer:

Correct Answer: 12. (d)

Solution:

Given, $f (x) = | x - 1 |$

$∴ f (x^{2}) = | x^{2} - 1 |$

and ${f (x)}^{2} = (x - 1)^{2}$

$\Rightarrow f (x^{2}) \neq (f (x))^{2}$ , hence (a) is false.

Also, $f (x + y) = | x + y - 1 |$

and $f (x) = | x - 1 |$

$f (y) = | y - 1 |$

$\Rightarrow f (x + y) \neq f (x) + f (y)$ , hence (b) is false.

$f (| x |) = | | x | - 1 |$

and $| f (x) | = | | x - 1 | | = | x - 1 |$

$∴ f (| x |) \neq | f (x) |$ , hence (c) is false.

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Functions 2 Question 11

12.

Answer:

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