Question

The range of $y=\sin \left( 3x \right)+\sin \left( x \right)$ is
(a) $\left[ \dfrac{-8}{3\sqrt{3}},\dfrac{8}{3\sqrt{3}} \right]$
(b) $\left[ \dfrac{-4}{3\sqrt{3}},\dfrac{4}{3\sqrt{3}} \right]$
(c) $\left[ \dfrac{-3}{\sqrt{2}},\dfrac{3}{\sqrt{2}} \right]$
(d) $\left[ \dfrac{-3}{3\sqrt{2}},\dfrac{3}{2\sqrt{2}} \right]$

Answer 1

Hint: We will apply the formula $\sin \left( 3x \right)=3\sin x-4{{\sin }^{3}}x$ to solve the question. Also we will use some formulas of integration which are given as $\dfrac{d}{dx}\left( \sin x \right)=\cos x+c$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}+c$ where c is any constant. By differentiation formulas we will lead to the correct answer.

Complete step-by-step answer:
We will first consider the equation $y=\sin \left( 3x \right)+\sin \left( x \right)$...(i).Now we will use the formula of $\sin \left( 3x \right)=3\sin x-4{{\sin }^{3}}x$ in trigonometric expression (i). Thus we will get $\begin{aligned} & y=3\sin x-4{{\sin }^{3}}x+\sin x \\\ & \Rightarrow y=4\sin x-4{{\sin }^{3}}x \\\ \end{aligned}$
Now, we will take the number 4 out from the equation as a common point between them. Thus we get $y=4\left( \sin x-{{\sin }^{3}}x \right)...(ii)$
Now, we will use differentiation here. This is done by using $\dfrac{d}{dx}$ operation on both the sides of the equation. We will use the formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x+{{c}_{1}}$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}+{{c}_{2}}$ where ${{c}_{1}},{{c}_{2}}$ are any constant. But in this we will not include c. Thus, we get
$\begin{aligned} & \dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left[ 4\left( \sin x-{{\sin }^{3}}x \right) \right] \\\ & \Rightarrow \dfrac{d}{dx}\left( y \right)=4\left( \dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}\left( {{\sin }^{3}}x \right) \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( y \right)=4\left( \dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}{{\left( \sin x \right)}^{3}} \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( y \right)=4\left( \cos x+{{c}_{1}}-\left[ 3\left( {{\sin }^{2}}x \right)\times \dfrac{d}{dx}\left( \sin x \right)+{{c}_{2}} \right] \right) \\\ \end{aligned}$
Now, we will again apply the formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x+{{c}_{3}}$. Therefore, we get
$\begin{aligned} & \dfrac{d}{dx}\left( y \right)=4\left( \cos x+{{c}_{1}}-\left[ 3\left( {{\sin }^{2}}x \right)\times \cos x+{{c}_{2}}+{{c}_{3}} \right] \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( y \right)=4\left( \cos x-3{{\sin }^{2}}x\cos x+0 \right) \\\ \end{aligned}$
Here, ${{c}_{1}}+{{c}_{2}}+{{c}_{3}}=0$ as being these as arbitrary constants. Now by the second derivative with respect to x we get,
$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{d}{dx}\left( y \right) \right)=\dfrac{d}{dx}\left( 4\left( \cos x-3{{\sin }^{2}}x\cos x \right) \right) \\\ & \Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right)=4\left( \dfrac{d}{dx}\left( \cos x \right)-3\dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right) \right) \\\ & \Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right)=4\left( \dfrac{d}{dx}\left( \cos x \right)-3\dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right) \right)...(iii) \\\ \end{aligned}$
By the formula $\left( f.g \right)'=f'g+g'f$ and the substitution $f={{\sin }^{2}}x$ and $g=\cos x$ we get
$\dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)\cos x+\dfrac{d}{dx}\left( \cos x \right){{\sin }^{2}}x$. As we know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x+c$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}+c$ where c is any constant. Thus we get $\begin{aligned} & \dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=\left( \dfrac{d}{dx}\left( {{\sin }^{2}}x \right) \right)\cos x+\left( \dfrac{d}{dx}\left( \cos x \right) \right){{\sin }^{2}}x \\\ & \Rightarrow \dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=\left( 2\sin x\dfrac{d}{dx}\left( \sin x \right) \right)\cos x+\left( -\sin x \right){{\sin }^{2}}x \\\ & \Rightarrow \dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=\left( 2\sin x\dfrac{d}{dx}\left( \sin x \right) \right)\cos x-{{\sin }^{3}}x \\\ \end{aligned}$
Now, with the help of the formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x+c$ we have a new equation as,
$\begin{aligned} & \dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=\left( 2\sin x\left( \cos x \right) \right)\cos x-{{\sin }^{3}}x \\\ & \Rightarrow \dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=2\sin x{{\cos }^{2}}x-{{\sin }^{3}}x \\\ \end{aligned}$
Now we will substitute the value of $\dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right)=2\sin x{{\cos }^{2}}x-{{\sin }^{3}}x$ in equation (iii). Thus, we have
$\begin{aligned} & \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right)=4\left( \dfrac{d}{dx}\left( \cos x \right)-3\dfrac{d}{dx}\left( {{\sin }^{2}}x\cos x \right) \right) \\\ & \Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right)=4\left( \dfrac{d}{dx}\left( \cos x \right)-3\left[ 2\sin x{{\cos }^{2}}x-{{\sin }^{3}}x \right] \right) \\\ \end{aligned}$
By formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x+c$ we get $\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right)=4\left( -\sin x-6\sin x{{\cos }^{2}}x+3{{\sin }^{3}}x \right)$....(iv).
Now we will substitute $\dfrac{dy}{dx}=0$. This is due to the fact that by doing this we can get the maximum and minimum values of the range y. And thus we will be able to find the range of the equation (i). Thus, we get
$\begin{aligned} & \dfrac{d}{dx}\left( y \right)=0 \\\ & \Rightarrow 4\left( \cos x-3{{\sin }^{2}}x\cos x \right)=0 \\\ & \Rightarrow \cos x-3{{\sin }^{2}}x\cos x=0 \\\ & \Rightarrow \cos x\left( 1-3{{\sin }^{2}}x \right)=0 \\\ \end{aligned}$
Therefore, we get $\cos x=0$ and $\left( 1-3{{\sin }^{2}}x \right)=0$. In $\cos x=0$we will use the formula in which $\cos x=0$ results into $x=n\pi \pm y$ where y is the angle given by $\cos \left( \dfrac{\pi }{2} \right)=0$. Therefore, y is $\dfrac{\pi }{2}$.
Now we will consider $\left( 1-3{{\sin }^{2}}x \right)=0$. After simplifying it we get
$\begin{aligned} & \left( 1-3{{\sin }^{2}}x \right)=0 \\\ & \Rightarrow 1=3{{\sin }^{2}}x \\\ & \Rightarrow \dfrac{1}{3}={{\sin }^{2}}x \\\ & \Rightarrow \pm \sqrt{\dfrac{1}{3}}=\sin x \\\ & \Rightarrow \pm \dfrac{1}{\sqrt{3}}=\sin x \\\ \end{aligned}$
Therefore we have $+\dfrac{1}{\sqrt{3}}=\sin x$ and $-\dfrac{1}{\sqrt{3}}=\sin x$.
Now we have three intervals to deal with; these are given by $\left( -\infty \right.,\left. -\dfrac{1}{\sqrt{3}} \right],\left[ -\dfrac{1}{\sqrt{3}}, \right.\left. \dfrac{1}{\sqrt{3}} \right]$ and $\left[ \dfrac{1}{\sqrt{3}}, \right.\left. +\infty \right)$. By putting the value of $\sin x=\dfrac{1}{2}$ from $\left( -\infty \right.,\left. -\dfrac{1}{\sqrt{3}} \right]$ interval we will have $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ which is further $\sin \left( \dfrac{\pi }{6} \right)=\sin x$. Thus, by formula $\sin x=\sin y$ resulting into $x=n\pi +y$ we get $x=n\pi +\dfrac{\pi }{6}$ or by n = 0, $x=\dfrac{\pi }{6}$. After putting this value and $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2},\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2},\sqrt{3}=1.73$ values in equation (iv) we get
$\begin{aligned} & {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -\sin x-6\sin x{{\cos }^{2}}x+3{{\sin }^{3}}x \right) \\\ & \Rightarrow {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -\left( \dfrac{1}{2} \right)-6\left( \dfrac{1}{2} \right){{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+3{{\left( \dfrac{1}{2} \right)}^{3}} \right) \\\ & \Rightarrow {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -\dfrac{1}{2}-6\left( \dfrac{1}{\sqrt{3}} \right)\left( \dfrac{3}{4} \right)+3\times \dfrac{1}{8} \right) \\\ & \Rightarrow {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -\dfrac{1}{2}-\dfrac{3\sqrt{3}}{2}+\dfrac{3}{8} \right) \\\ & \Rightarrow {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -0.5-2.59+0.375 \right) \\\ & \Rightarrow {{\left[ \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( y \right) \right]}_{\left( x=\dfrac{\pi }{6} \right)}}=4\left( -2.715 \right) \\\ \end{aligned}$
Since, this is less than zero so we have it as minimum value in this interval. Similarly we will get maximum in $\left[ -\dfrac{1}{\sqrt{3}}, \right.\left. \dfrac{1}{\sqrt{3}} \right]$ and maximum in $\left[ \dfrac{1}{\sqrt{3}}, \right.\left. +\infty \right)$ interval.
Now we will substitute $+\dfrac{1}{\sqrt{3}}=\sin x$ belonging to $\left[ \dfrac{1}{\sqrt{3}}, \right.\left. +\infty \right)$ in equation (ii) we will get maximum value as,
$\begin{aligned} & y=4\left( \sin x-{{\sin }^{3}}x \right) \\\ & \Rightarrow y=4\left( \dfrac{1}{\sqrt{3}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{3}} \right) \\\ & \Rightarrow y=4\left( \dfrac{1}{\sqrt{3}}-\dfrac{1}{3\sqrt{3}} \right) \\\ & \Rightarrow y=4\left( \dfrac{3}{3\sqrt{3}}-\dfrac{1}{3\sqrt{3}} \right) \\\ & \Rightarrow y=4\left( \dfrac{2}{3\sqrt{3}} \right) \\\ & \Rightarrow y=\dfrac{8}{3\sqrt{3}} \\\ \end{aligned}$
And this will be regarded as the maximum value of range y.
Also, after we put $-\dfrac{1}{\sqrt{3}}=\sin x$ belonging to $\left( -\infty \right.,\left. -\dfrac{1}{\sqrt{3}} \right]$ and in equation (ii) we will get minimum value as,
$\begin{aligned} & y=4\left( \sin x-{{\sin }^{3}}x \right) \\\ & \Rightarrow y=4\left( -\dfrac{1}{\sqrt{3}}-{{\left( -\dfrac{1}{\sqrt{3}} \right)}^{3}} \right) \\\ & \Rightarrow y=4\left( -\dfrac{1}{\sqrt{3}}-\dfrac{1}{3\sqrt{3}} \right) \\\ & \Rightarrow y=4\left( -\dfrac{3}{3\sqrt{3}}-\dfrac{1}{3\sqrt{3}} \right) \\\ & \Rightarrow y=4\left( \dfrac{-2}{3\sqrt{3}} \right) \\\ & \Rightarrow y=-\dfrac{8}{3\sqrt{3}} \\\ \end{aligned}$
And this will be regarded as the minimum value of range y. Therefore, the range of y is given by the interval $\left[ \dfrac{-8}{3\sqrt{3}},\dfrac{8}{3\sqrt{3}} \right]$.
Hence, the correct option is (a).

Note: Using the formula $\sin \left( A \right)+\sin \left( B \right)=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ in the equation will lead to nowhere. Instead we have used the formula of $\sin \left( 3x \right)=3\sin x-4{{\sin }^{3}}x$ to solve the question. Solving for the second derivative here is important here as by substituting the values in this we will be able to figure out the minimum and maximum points. As this question carries so many steps, one needs to focus on this question completely for solving it in a correct way leading to the right answer. Notice where we are taking the closed interval and where we are taking the open interval. Semi- open and semi-closed intervals are also used here.