Question

The range of the function, $f(x) = {\cot ^{ - 1}}{\log _{0.5}}({x^4} - 2{x^2} + 3)$ is
a)$(0,\pi )$
b)$\left( {0,\dfrac{{3\pi }}{4}} \right]$
c)$\left[ {\dfrac{{3\pi }}{4},\pi } \right)$
d)$\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{4}} \right]$

Answer 1

In this question we have to find the range of the function. In the question, the $f(x)$ is given i.e.
$\Rightarrow f(x) = {\cot ^{ - 1}}{\log _{0.5}}({x^4} - 2{x^2} + 3)$
Here, we can see the $f(x)$ consist of mixed functions i.e. algebraic function, logarithmic function and inverse cotangent function.
To find the range of $f(x)$ first of all, solve the algebraic part and find its range. Now, after that solve the logarithmic part by taken in consider the range of algebraic, we will find the range of logarithmic part and finally solve the inverse cotangent function by taken in consider the range of logarithmic part we can find the range of this function.

Complete step-by-step answer:
We have to find the range of the given function $f(x)$i.e.
$\Rightarrow f(x) = {\cot ^{ - 1}}{\log _{0.5}}({x^4} - 2{x^2} + 3)$
We can see the function $f(x)$ contains three functions. To find the range we will split these functions and solve each function one by one in order.
Start with the algebraic expression as it is inside of the other two functions. So, we are going to solve it first let take it as A i.e.
$\Rightarrow A = {x^4} - 2{x^2} + 3$
Solve this by writing 3 as addition of 2 and 1, it is done so to write the expression in proper square form. By doing so we get,
$\Rightarrow A = {x^4} - 2{x^2} + 1 + 2$
From the formula ${(a - b)^2} = {a^2} + {b^2} - 2ab$, we can write ${x^4} - 2{x^2} + 1$ as a whole square of ${x^2} - 1$. Hence, we get,
$\Rightarrow A = {({x^2} - 1)^2} + 2$
Here we can see the minimum value of ${x^2} - 1$ is 0. As for any value of x the square will always be positive. So, the range of the equation will be greater than or equal to 2. i.e.
$\Rightarrow {({x^2} - 1)^2} + 2 \geqslant 2$
After this we are going to solve the logarithmic part of the function as it is inside the inverse cotangent function. Let take it as B i.e.
$\Rightarrow B = {\log _{0.5}}({x^4} - 2{x^2} + 3)$
Here, we writing 0.5 in the logarithmic as -2 we get,
$\Rightarrow B = {\log _{ - 2}}({x^4} - 2{x^2} + 3)$
By taking negative sign of 2 outside the logarithmic we get,
$\Rightarrow B = - {\log _2}({x^4} - 2{x^2} + 3)$
The range of the algebraic expression in the bracket is greater than 2 so if we put the minimum value of ${x^4} - 2{x^2} + 3$i.e. 2 then ${\log _2}2$is , after that if we put the value of ${x^4} - 2{x^2} + 1$ as 4 then ${\log _2}4$is 4 and so on. So, ${\log _2}({x^4} - 2{x^2} + 3)$ is greater than or equal to 1 but if we take its negative then we get,
$\Rightarrow - {\log _2}({x^4} - 2{x^2} + 3) \leqslant - 1$
$\Rightarrow B \leqslant - 1$
Finally, we are going to solve the inverse cotangent part and it will give the solution for the question i.e. range of $f(x)$.
$\Rightarrow f(x) = {\cot ^{ - 1}}{\log _{0.5}}({x^4} - 2{x^2} + 3)$
As we have taken ${\log _{0.5}}({x^4} - 2{x^2} + 3)$as B. So, above equation can be written as
$\Rightarrow f(x) = {\cot ^{ - 1}}B$
As the range of inverse of cot is $\left[ {0,\pi } \right]$ and if we put the maximum value of B then ${\cot ^{ - 1}}B$i.e. ${\cot ^{ - 1}}( - 1)$is $\dfrac{{3\pi }}{4}$. The minimum value of B is $- \infty$then ${\cot ^{ - 1}}( - \infty )$is 0. Hence, the range of ${\cot ^{ - 1}}B$ will be $\left( {0,\dfrac{{3\pi }}{4}} \right]$. Therefore,
$\Rightarrow {\cot ^{ - 1}}{\log _{0.5}}({x^4} - 2{x^2} + 3) \in \left( {0,\dfrac{{3\pi }}{4}} \right]$
$\Rightarrow f(x) \in \left( {0,\dfrac{{3\pi }}{4}} \right]$
Hence, option b is the correct answer.

Note: To solve this question students can also use another method but this is the easiest method amongst all. Students can get confused by multiple functions but if you do it step by step and consider all the conditions of the function then chances of mistakes may be reduced.
In the algebraic part if we directly place the different values of x then it will be difficult to find range that’s why we convert it into square form.
It is noted that the range depends on the input of the function i.e. ranges if the domain of the function changes. Domain is the possible values of x in which function can be defined and range is the possible outcomes of a function.