Given $x>0$, the value of $f(x)=-3 \cos \sqrt{3+x+x^2}$ lie in the interval .......... .
Given function, $f(x)=-3 \cos \sqrt{3+x+x^2}$
We know that,
$$\begin{aligned} -1 \leq \cos x \leq 1 \\ \Rightarrow \quad -3 \leq 3 \cos x \leq 3 \\ \Rightarrow \quad 3 \geq-3 \cos x \geq-3 \\ \Rightarrow \quad -3 \leq-3 \cos x \leq 3 \end{aligned}$$
So, the value of $f(x)$ lies in $[-3,3]$.
The maximum distance of a point on the graph of the function $y=\sqrt{3} \sin x+\cos x$ from $X$-axis is ......... .
Given that, $y=\sqrt{3} \sin x+\cos x$
$$\begin{aligned} y & =2\left[\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x\right] \\ & =2\left[\sin x \cdot \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}\right] \\ & =2 \sin (x+\pi / 6) \end{aligned}$$
Graph of $y=2\sin x$
Hence, the maximum distance is 2 units.
In each of the questions 68 to 75 , state whether the statements is True or False? Also, give justification.
The equality $\sin A+\sin 2 A+\sin 3 A=3$ holds for some real value of $A$.
$\sin10\Upsilon$ is greater than $\cos10\Upsilon$.