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Conjugate of a Complex Number

The possible ...

Question

The possible number of extremum point(s) for $f (x) = x^{2} + 2 cos x + 2$ on $(0, \frac{π}{2})$ is:

A

0

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B

1

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C

2

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D

3

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Solution

The correct option is A $0$
Given, $f (x) = x^{2} + 2 cos x + 2$
$\Rightarrow f^{'} (x) = 2 (x - sin x)$
Now, $f^{'} (x) = 0$ iff $sin x = x$
But $sin x \neq x$ for any $(0, \frac{π}{2})$ .
Hence, $f (x)$ has no extremum.

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Conjugate of a Complex Number

Standard XII Mathematics

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