The range of the function $ f\left( x \right)={{\log }_{\cos e{{c}^{2}}\dfrac{7\pi }{3}}}\left( {{4}^{x}}-{{2}^{x}}+1 \right) $ is equal to
\[\begin{align}
  & A.\left( -\infty ,1 \right] \\
 & B.\left[ -1,\infty \right) \\
 & C.\left( -\infty ,-1 \right] \\
 & D.\left[ 1,\infty \right) \\
\end{align}\]

Hint: In this question, we need to find the range of the function $ f\left( x \right)={{\log }_{\cos e{{c}^{2}}\dfrac{7\pi }{3}}}\left( {{4}^{x}}-{{2}^{x}}+1 \right) $ . For this, we will first evaluate $ \text{cose}{{\text{c}}^{2}}\dfrac{7\pi }{3} $ by formula $ \text{cosec}\left( 2\pi +\theta \right)=\text{cosec}\theta $ . Then we will convert $ {{4}^{x}}-{{2}^{x}}+1 $ into the form of $ {{a}^{2}}-a+1 $ where $ a={{2}^{x}}\text{ }>\text{ }0 $ . Then we will use $ {{a}^{2}}-a+1 $ to find maximum and minimum value of $ {{a}^{2}}-a+1 $ which will give us minimum and maximum value of f(x) and hence our range. For finding the minimum value of any function g(x) we first find g'(x) = 0 and then find the value of x. Then we find g''(x) and put the value of x in it. If g''(x)>0, then x gives minimum value. If g''(x)<0, then x gives maximum value.

Complete step by step answer:
Here we are given the function as,
 $ f\left( x \right)={{\log }_{\cos e{{c}^{2}}\dfrac{7\pi }{3}}}\left( {{4}^{x}}-{{2}^{x}}+1 \right) $ .
We can write $ \dfrac{7\pi }{3} $ as $ 2\pi +\dfrac{\pi }{3} $ so,
 $ \text{cosec}\dfrac{7\pi }{3}=\text{cosec}\left( 2\pi +\dfrac{\pi }{3} \right) $ .
We know that $ \text{cosec}\left( 2\pi +\theta \right)=\text{cosec}\theta $ therefore, $ \text{cosec}\left( 2\pi +\dfrac{\pi }{3} \right)=\text{cosec}\dfrac{\pi }{3} $ .
From trigonometric ratio table, we know that $ \text{cosec}\dfrac{\pi }{3}=\dfrac{2}{\sqrt{3}} $ so $ \text{cosec}\dfrac{7\pi }{3}=\dfrac{2}{\sqrt{3}} $ .
Squaring both sides we get,
 $ \text{cose}{{\text{c}}^{2}}\dfrac{7\pi }{3}={{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}=\dfrac{4}{3} $ .
Hence our function becomes,
 $ f\left( x \right)={{\log }_{\dfrac{4}{3}}}\left( {{4}^{x}}-{{2}^{x}}+1 \right) $ .
Now as we can see $ {{4}^{x}} $ can be written as $ {{\left( {{2}^{2}} \right)}^{x}} $ which can be further written as $ {{\left( {{2}^{x}} \right)}^{2}} $ so we get,
 $ f\left( x \right)={{\log }_{\dfrac{4}{3}}}\left( {{\left( {{2}^{x}} \right)}^{2}}-{{2}^{x}}+1 \right) $ .
Let us put $ {{2}^{x}}=a $ which is greater than 0 (if x is the negative value of log it is negative which is not possible).
So our function becomes,
 $ f\left( x \right)={{\log }_{\dfrac{4}{3}}}\left( {{a}^{2}}-a+1 \right) $ .
As we can see $ {{a}^{2}}-a+1 $ can have value as high as infinity and $ \log \infty =\infty $ . So maximum value of $ {{\log }_{\dfrac{4}{3}}}\left( {{a}^{2}}-a+1 \right) $ is infinity.
For finding the minimum value of f(x) let us find the minimum value of $ {{a}^{2}}-a+1 $ .
Let $ g\left( a \right)={{a}^{2}}-a+1 $ . Taking derivatives on both sides we get, $ g'\left( a \right)=2a-1 $ .
Putting g'(a) = 0, we get $ 2a-1=0\Rightarrow a=\dfrac{1}{2} $ .
Finding g''(a) =, we get 2, $ g''\left( a \right)=2 $ .
For $ a=\dfrac{1}{2},g''\left( \dfrac{1}{2} \right)=2\text{ }>\text{ }0 $ .
So, $ a=\dfrac{1}{2} $ will give a minimum value of $ {{a}^{2}}-a+1 $ .
Putting $ a=\dfrac{1}{2} $ we get,
 $ {{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{1}{2}+1\Rightarrow \dfrac{1}{4}-\dfrac{1}{2}+1\Rightarrow \dfrac{1}{4}+\dfrac{1}{2}\Rightarrow \dfrac{1+2}{4}=\dfrac{3}{4} $ .
So we get minimal value of f(x) as $ \underset{\min }{\mathop{f\left( x \right)}}\,={{\log }_{\dfrac{4}{3}}}\left( \dfrac{3}{4} \right) $ .
Write $ \dfrac{3}{4} $ as $ {{\left( \dfrac{4}{3} \right)}^{-1}} $ we get, $ \underset{\min }{\mathop{f\left( x \right)}}\,={{\log }_{\dfrac{4}{3}}}{{\left( \dfrac{4}{3} \right)}^{-1}} $ .
We know that $ {{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x $ we get, $ \underset{\min }{\mathop{f\left( x \right)}}\,=-{{\log }_{\dfrac{4}{3}}}\left( \dfrac{4}{3} \right) $ .
Now we know that $ {{\log }_{b}}b=1 $ so we get, $ \underset{\min }{\mathop{f\left( x \right)}}\,=-1 $ .
Hence minimum value of f(x) = -1.
Therefore the range of f(x) is $ \left[ -1,\infty \right) $ .
Hence option B is the correct answer.

Note:
 Take care of signs while solving this question. Note that, for $ \log \left( x \right)=y $ x can never be less than 0. Students should keep in mind all the properties of the logarithm for finding the solution. Keep in mind all the values from the trigonometric ratio table.