Question

The range of the function $f(x) = {\cot ^{ - 1}}\theta \left( {{{\log }_{0.5}}({x^4} - 2{x^2} + 3)} \right)$

A. $(0,\pi )$

B. $(0,\dfrac{{3\pi }}{4}]$

C. $[\dfrac{{3\pi }}{4},\pi )$

D. $\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{4}} \right]$

Answer 1

For finding the range of a function we’ll first assume the function as a new variable let say y, now we’ll transform the equation in such a way that the equation will become y in terms of x and check for which value of y, x is defined and that set will be the range of the function.

Complete step by step Answer:

Given data: $f(x) = {\cot ^{ - 1}}\left( {{{\log }_{0.5}}({x^4} - 2{x^2} + 3)} \right)$

Let us assume that $y = f(x)$

Substituting the value of f(x)

$\Rightarrow y = {\cot ^{ - 1}}\left( {{{\log }_{0.5}}({x^4} - 2{x^2} + 3)} \right)$

Taking both the sides as the function of the cot

$\Rightarrow \cot y = \cot \left( {{{\cot }^{ - 1}}\left( {{{\log }_{0.5}}({x^4} - 2{x^2} + 3)} \right)} \right)$

Using $\cot \left( {{{\cot }^{ - 1}}A} \right) = A$, we get,

$\Rightarrow \cot y = {\log _{0.5}}({x^4} - 2{x^2} + 3)$

We know that if $c = {\log _a}b$ then, ${a^c} = b$

$\Rightarrow {0.5^{\cot y}} = {x^4} - 2{x^2} + 3$
$\Rightarrow {\left( {\dfrac{1}{2}} \right)^{\cot y}} = {x^4} - 2{x^2} + 3$

On expanding the constant term, we get,

$\Rightarrow {\left( {\dfrac{1}{2}} \right)^{\cot y}} = {x^4} - 2{x^2} + 1 + 2$

Using ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$, we get,

$\Rightarrow {\left( 2 \right)^{ - \cot y}} = {\left( {{x^2} + 1} \right)^2} + 2............(i)$

Since the square of any number is always greater than or equal to zero

Therefore we can say that, ${\left( {{x^2} + 1} \right)^2} \geqslant 0$
Adding 2 on both sides

$\Rightarrow {\left( {{x^2} + 1} \right)^2} + 2 \geqslant 2$

i.e. $\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in [2,\infty )$

from equation(i)

$\Rightarrow {\left( 2 \right)^{ - \cot y}} \in [2,\infty )$

On comparing we can say that $- \cot y \in [1,\infty )$

$\Rightarrow \cot y \in ( - \infty , - 1]$

$\Rightarrow y \in (0,\dfrac{{3\pi }}{4}]$
Therefore the range of the function $f(x) \in (0,\dfrac{{3\pi }}{4}]$
Option(B) is correct.

Note: An alternative method for this solution can be
Since the square of any number is always greater than or equal to zero
Therefore we can say that, ${\left( {{x^2} + 1} \right)^2} \geqslant 0$
Adding 2 on both sides
$\Rightarrow {\left( {{x^2} + 1} \right)^2} + 2 \geqslant 2$
Therefore, $\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in [2,\infty )$
Taking logarithm function with base 0.5 on both sides and since log, with baseless than one is a decreasing function
$\Rightarrow {\log _{0.5}}\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in [{\log _{0.5}}2,{\log _{0.5}}\infty )$
$\Rightarrow {\log _{0.5}}\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in ({\log _{{2^{ - 1}}}}\infty ,{\log _{{2^{ - 1}}}}2]$

Using ${\log _{{a^b}}}c = \dfrac{1}{b}{\log _a}c$

$\Rightarrow {\log _{0.5}}\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in ( - {\log _2}\infty , - {\log _2}2]$
$\Rightarrow {\log _{0.5}}\left[ {{{\left( {{x^2} + 1} \right)}^2} + 2} \right] \in ( - \infty , - 1]$
Now, taking both sides as the function of ${\cot ^{ - 1}}\theta$ and since it is also a decreasing function
$\Rightarrow {\cot ^{ - 1}}\left[ {{{\log }_{0.5}}\left( {{{\left( {{x^2} + 1} \right)}^2} + 2} \right)} \right] \in [{\cot ^{ - 1}} - 1,{\cot ^{ - 1}} - \infty )$
Substituting ${\cot ^{ - 1}}\left[ {{{\log }_{0.5}}\left( {{{\left( {{x^2} + 1} \right)}^2} + 2} \right)} \right] = f(x)$
$\Rightarrow f(x) \in (0,\dfrac{{3\pi }}{4}]$
Option(B) is correct