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

Hint: 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
\[\begin{align}
  & \Rightarrow 2x-\dfrac{\pi }{6}=\left( \dfrac{\pi }{2} \right) \\
 & \Rightarrow 2x=\dfrac{\pi }{2}+\dfrac{\pi }{6} \\
 & \Rightarrow 2x=\dfrac{4\pi }{6} \\
\end{align}\]
\[\Rightarrow x=\dfrac{\pi }{3}\] …………………….(10)
At \[x=\dfrac{\pi }{3}\] , the function \[f\left( x \right)\] is maximum, and the maximum value of the function \[f\left( x \right)\]
\[\begin{align}
  & \Rightarrow f\left( x \right)=2\sin \left( 2.\dfrac{\pi }{3}-\dfrac{\pi }{6} \right)+4 \\
 & \Rightarrow f\left( x \right)=2\sin \left( \dfrac{2\pi }{3}-\dfrac{\pi }{6} \right)+4 \\
\end{align}\]
\[\Rightarrow f\left( x \right)=2\sin \left( \dfrac{\pi }{2} \right)+4\] ………………..(11)
We know that \[\sin \left( \dfrac{\pi }{2} \right)=1\] ……………………(12)
From equation (11) and equation (12), we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=2\left( 1 \right)+4 \\
 & \Rightarrow f\left( x \right)=2+4 \\
\end{align}\]
\[\Rightarrow f\left( x \right)=6\]
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
\[\begin{align}
  & \Rightarrow 2x-\dfrac{\pi }{6}=\left( \dfrac{-\pi }{2} \right) \\
 & \Rightarrow 2x=-\dfrac{\pi }{2}+\dfrac{\pi }{6} \\
 & \Rightarrow 2x=\dfrac{-2\pi }{6} \\
\end{align}\]
\[\Rightarrow x=\dfrac{-\pi }{6}\] …………………….(16)
At \[x=\dfrac{-\pi }{6}\] , the function \[f\left( x \right)\] is minimum, and the minimum value of the function \[f\left( x \right)\]
\[\begin{align}
  & \Rightarrow f\left( x \right)=2\sin \left( 2.\dfrac{-\pi }{6}-\dfrac{\pi }{6} \right)+4 \\
 & \Rightarrow f\left( x \right)=2\sin \left( \dfrac{-\pi }{3}-\dfrac{\pi }{6} \right)+4 \\
\end{align}\]
\[\Rightarrow f\left( x \right)=2\sin \left( \dfrac{-\pi }{2} \right)+4\] ………………..(17)
We know that \[\sin \left( \dfrac{-\pi }{2} \right)=-1\] ……………………(18)
From equation (17) and equation (18), we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=2\left( -1 \right)+4 \\
 & \Rightarrow f\left( x \right)=-2+4 \\
\end{align}\]
\[\Rightarrow f\left( x \right)=2\]
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.