On the interval 0- -2 the function logsinx is

Explanation for the correct option.Finding On the interval (0, π/2) the function logsinxis Given: logsinxf (x) = log (sin x)Differentiating w.r.t.x we get;f'(x) ...

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Standard XII

Mathematics

Continuity of a Function

On the interv...

Question

On the interval $(0, \frac{π}{2})$ the function $\log \sin x$ is

Increasing

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Decreasing

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Neither increasing nor decreasing

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None of the above

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Solution

The correct option is A
Increasing

Explanation for the correct option.
Finding On the interval $(0, \frac{π}{2})$ the function $\log \sin x$ is
Given: $\log \sin x$
$f (x) = \log (\sin x)$
Differentiating w.r.t. $x$ we get;
$f' (x) = \frac{1}{\sin x} \times (\cos x) = c o t x c o t x = positive \forall \in (0, \frac{π}{2}) (increasing)$
Hence, the correct answer is option (A)

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On the interval 0,π2 the function logsinx is

The correct option is A IncreasingExplanation for the correct option.Finding On the interval 0,π2 the function logsinxis Given: logsinxf(x)=log(sinx)Differentiating w.r.t.x we get;f'(x)=1sinx×(cosx)=cotxcotx=positive∀∈0,π2(increasing)Hence, the correct answer is option (A)

On the interval $(0, \frac{π}{2})$ the function $\log \sin x$ is