Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 32 Marks
Question
Show that $f ( x )=\cos ^2 x$ is a decreasing function on $\left(0, \frac{\pi}{2}\right)$.
✓
Answer
Given: $f(x)=\cos ^2 x$
Theorem$:-$ Let $f$ be a differentiable real function defined on an open interval $(a,b).$
i. If $f^\ {\prime}(x);0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
ii. If $f^\ {\prime}(x)<0$ for all,$x \in(a, b)$ then $f(x)$ is decreasing on $(a, b)$
For the value of $x$ obtained in $(ii) f(x)$ is increasing and for remaining points in its domain it is decreasing.
Here we have,
$ f ( x )=\cos ^2 x$
$\Rightarrow f(x)=\frac{d}{d x}\left(\cos ^2 x\right)$
$= f^\ {\prime}( x )=3 \cos x (-\sin x )$
$= f^\ {\prime}( x )=-2 \sin ( x ) \cos ( x )$
$= f^\ {\prime}( x )=-\sin 2 x ;$ as $\sin 2 A=2 \sin A \cos A $
Now, as given
$x \in\left(0, \frac{\pi}{2}\right)$
$=2 x \in(0, \pi)$
$=\operatorname{Sin}(2 x);0$
$=-\operatorname{Sin}(2 x)<0$
$\Rightarrow f^{\prime}(x)<0$
hence, it is the condition for $f(x)$ to be decreasing Thus, $f(x)$ is decreasing on interval $\left(0, \frac{\pi}{2}\right)$
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