Question

If the real function $f:X\to \left[ 2,6 \right]$ , where $f\left( x \right)=\sqrt{3}\sin 2x-\cos 2x+4$ is one-one onto then possible X among the following is
(A) $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
(B) $\left[ \dfrac{-\pi }{4},\dfrac{\pi }{4} \right]$
(C) $\left[ \dfrac{-\pi }{6},\dfrac{\pi }{3} \right]$
(D) $\left[ -\pi ,\pi \right]$

Answer 1

Transform the given function, $f\left( x \right)=\sqrt{3}\sin 2x-\cos 2x+4$ as $f\left( x \right)=2.\dfrac{1}{2}\left( \sqrt{3}\sin 2x-\cos 2x \right)+4$ . Now, simplify it using the formula $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$ . The maximum and minimum value of the sine function is 1 and -1 respectively. We know that $\sin \left( \dfrac{\pi }{2} \right)=1$ and $\sin \left( \dfrac{-\pi }{2} \right)=-1$ . Using this get the values of x corresponding to which the function $y=f\left( x \right)=2\sin \left( 2x-\dfrac{\pi }{6} \right)+4$ , has the maximum and minimum value. Now, plot the graph of the function and get those intervals of x in which y has different values corresponding to different values of x.

Complete step by step answer:
According to the question, we have the function,
$f\left( x \right)=\sqrt{3}\sin 2x-\cos 2x+4$ ……………………………(1)
It is given that the range of this function is $\left[ 2,6 \right]$ . It means that the minimum value of the function $f\left( x \right)$ is equal to 2 and the maximum value of the function $f\left( x \right)$ is equal to 6.
First of all, we need to simplify the function $f\left( x \right)$ .
Transforming equation (1), we get
$\Rightarrow f\left( x \right)=2.\dfrac{1}{2}\left( \sqrt{3}\sin 2x-\cos 2x \right)+4$
$\Rightarrow f\left( x \right)=2\left( \dfrac{\sqrt{3}}{2}\sin 2x-\dfrac{1}{2}\cos 2x \right)+4$ ………………………………….(2)
We know that, $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$ and $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ …………………………..(3)
Now, from equation (2) and equation (3), we get
\Rightarrow f\left( x \right)=2\left\\{ \sin 2x\cos \left( \dfrac{\pi }{6} \right)-\sin \left( \dfrac{\pi }{6} \right)\cos 2x \right\\}+4 …………………………..(4)
We know the formula, $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$ …………………………..(5)
Now, replacing A by 2x and B by $\left( \dfrac{\pi }{6} \right)$ in equation (5), we get
$sin\left( 2x-\dfrac{\pi }{6} \right)=\sin 2x\cos \left( \dfrac{\pi }{6} \right)-\sin \left( \dfrac{\pi }{6} \right)\cos 2x$ ………………………….(6)
From equation (4) and equation (6), we get
$\Rightarrow f\left( x \right)=2\sin \left( 2x-\dfrac{\pi }{6} \right)+4$ ………………………..(7)
We know that the maximum value of the sine function is 1.
So, the function $f\left( x \right)$ is maximum when the value of $\sin \left( 2x-\dfrac{\pi }{6} \right)$ is equal to 1 and we know that
$\sin \left( \dfrac{\pi }{2} \right)=1$ . Therefore,
$\Rightarrow \sin \left( 2x-\dfrac{\pi }{6} \right)=\sin \left( \dfrac{\pi }{2} \right)$
$\Rightarrow 2x-\dfrac{\pi }{6}={{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right)$ ……………………………………(8)
We know the property, ${{\sin }^{-1}}\left( \sin y \right)=y$ ………………………………..(9)
From equation (8) and equation (9), we get

& \Rightarrow 2x-\dfrac{\pi }{6}=\left( \dfrac{\pi }{2} \right) \\\ & \Rightarrow 2x=\dfrac{\pi }{2}+\dfrac{\pi }{6} \\\ & \Rightarrow 2x=\dfrac{4\pi }{6} \\\ \end{aligned}$$ $$\Rightarrow x=\dfrac{\pi }{3}$$ …………………….(10) At $$x=\dfrac{\pi }{3}$$ , the function $$f\left( x \right)$$ is maximum, and the maximum value of the function $$f\left( x \right)$$ $$\begin{aligned} & \Rightarrow f\left( x \right)=2\sin \left( 2.\dfrac{\pi }{3}-\dfrac{\pi }{6} \right)+4 \\\ & \Rightarrow f\left( x \right)=2\sin \left( \dfrac{2\pi }{3}-\dfrac{\pi }{6} \right)+4 \\\ \end{aligned}$$ $$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{\pi }{2} \right)+4$$ ………………..(11) We know that $$\sin \left( \dfrac{\pi }{2} \right)=1$$ ……………………(12) From equation (11) and equation (12), we get $$\begin{aligned} & \Rightarrow f\left( x \right)=2\left( 1 \right)+4 \\\ & \Rightarrow f\left( x \right)=2+4 \\\ \end{aligned}$$ $$\Rightarrow f\left( x \right)=6$$ The maximum value of the function $$f\left( x \right)$$ is equal to 6 …………………………..(13) We know that the maximum value of sine function is -1. So, the function $$f\left( x \right)$$ is minimum when the value of $$\sin \left( 2x-\dfrac{\pi }{6} \right)$$ is equal to -1 and we know that $$\sin \left( \dfrac{-\pi }{2} \right)=-1$$ . Therefore, $$\Rightarrow \sin \left( 2x-\dfrac{\pi }{6} \right)=\sin \left( \dfrac{-\pi }{2} \right)$$ $$\Rightarrow 2x-\dfrac{\pi }{6}={{\sin }^{-1}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right)$$ ……………………………………(14) We know the property, $${{\sin }^{-1}}\left( \sin y \right)=y$$ ………………………………..(15) From equation (14) and equation (15), we get $$\begin{aligned} & \Rightarrow 2x-\dfrac{\pi }{6}=\left( \dfrac{-\pi }{2} \right) \\\ & \Rightarrow 2x=-\dfrac{\pi }{2}+\dfrac{\pi }{6} \\\ & \Rightarrow 2x=\dfrac{-2\pi }{6} \\\ \end{aligned}$$ $$\Rightarrow x=\dfrac{-\pi }{6}$$ …………………….(16) At $$x=\dfrac{-\pi }{6}$$ , the function $$f\left( x \right)$$ is minimum, and the minimum value of the function $$f\left( x \right)$$ $$\begin{aligned} & \Rightarrow f\left( x \right)=2\sin \left( 2.\dfrac{-\pi }{6}-\dfrac{\pi }{6} \right)+4 \\\ & \Rightarrow f\left( x \right)=2\sin \left( \dfrac{-\pi }{3}-\dfrac{\pi }{6} \right)+4 \\\ \end{aligned}$$ $$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{-\pi }{2} \right)+4$$ ………………..(17) We know that $$\sin \left( \dfrac{-\pi }{2} \right)=-1$$ ……………………(18) From equation (17) and equation (18), we get $$\begin{aligned} & \Rightarrow f\left( x \right)=2\left( -1 \right)+4 \\\ & \Rightarrow f\left( x \right)=-2+4 \\\ \end{aligned}$$ $$\Rightarrow f\left( x \right)=2$$ The minimum value of the function $$f\left( x \right)$$ is equal to 2 …………………………..(19) Now, plotting graph of the function $$y=f\left( x \right)=2\sin \left( 2x-\dfrac{\pi }{6} \right)+4$$ , we get  We know that one-one onto functions doesn’t have the same value of y for different values of x. And we can see in the graph that when $$x\in \left( \dfrac{-\pi }{6},\dfrac{\pi }{3} \right)$$ then, y has different values corresponding to different values of x. The interval $$x\in \left( \dfrac{-\pi }{6},\dfrac{\pi }{3} \right)$$ also includes the range of y. Therefore, the given function is one-one onto when $$x\in \left( \dfrac{-\pi }{6},\dfrac{\pi }{3} \right)$$ . **So, the correct answer is “Option C”.** **Note:** In this question, one might think to take the periodicity of the sine function that is, $$\left( 0,2\pi \right)$$ as the interval for the one-one onto function. This is wrong because in the interval $$\left( 0,2\pi \right)$$ the graph includes the same range on the y-axis between the x-range which is a contradiction for one-one onto function.