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Maths-

General

Easy

Question

Which of the following functions is one-to -one?

$f (x) = \sin x, x \in [- π, π]$
$f (x) = \sin x, x \in [- \frac{3 π}{2}, - \frac{π}{4}]$
$f (x) = \cos x, x \in [- \frac{π}{2}, \frac{π}{2}]$
$f (x) = \cos x, x \in [π, \frac{3 π}{2}]$

The correct answer is: $f left parenthesis x right parenthesis equals cos invisible function application x comma x element of open square brackets pi comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$

In the given options (a), (b), (c), (e) the curves are decreasing and increasing in the given intervals, so it is not one-to-one function. But in option (d), the curve is only increasing in the given intervals, so it is one-to-one function.

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