Question

The range of $\sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)...............\infty } } }$ is

$$A. {\text{ }}\left[ {0,1} \right] \\\ B. {\text{ }}\left[ {0,\dfrac{1}{2}} \right] \\\ C. {\text{ }}\left[ {0,2} \right] \\\ D. {\text{ }}\left[ {\dfrac{1}{2},1} \right] \\\$$

Answer 1

Hint: in order to solve the problem first convert the given complicated terms in some simpler form and then try to relate it with some kind of known series and then proceed with the formulas of the series.

Complete step-by-step answer:
The given function is $f\left( x \right) = \sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)...............\infty } } }$
By simplifying the terms and bringing it in terms of $\left( {1 - \cos x} \right)$ , we get
$\Rightarrow f\left( x \right) = {\left( {1 - \cos x} \right)^{\dfrac{1}{2}}}{\left( {1 - \cos x} \right)^{\dfrac{1}{4}}}{\left( {1 - \cos x} \right)^{\dfrac{1}{8}}}...........\infty$
As we know algebraically $\left[ {{x^a}{x^b}{x^c} = {x^{a + b + c}}} \right]$
So we get the above equation as
$\Rightarrow f\left( x \right) = {\left( {1 - \cos x} \right)^{\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ..........\infty }}$
We can see that the series in the power of term $\left( {1 - \cos x} \right)$ on RHS is a G.P.
So let’s find out the sum of G.P. separately.
As we know that for any general G.P. with “a” as its first term and “r” be its common ratio.
Sum of infinite term of G.P. is given by $S = \dfrac{a}{{1 - r}}$
For the above question we have a G.P. that is $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},.........\infty$
So here $a = \dfrac{1}{2}{\text{ and }}r = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}} = \dfrac{1}{2}$
So the sum of this infinite G.P. is given by:
$S = \dfrac{a}{{1 - r}} = \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} = 1$
So the function changes to
$\Rightarrow f\left( x \right) = {\left( {1 - \cos x} \right)^{\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ..........\infty }} = {\left( {1 - \cos x} \right)^1} \\\ \Rightarrow f\left( x \right) = \left( {1 - \cos x} \right) \\\$
As we know the range of cosine function, which is given as
$- 1 \leqslant \cos x \leqslant 1$
Multiplying the inequality by -1
$\Rightarrow - \left( { - 1} \right) \geqslant - \left( {\cos x} \right) \geqslant - \left( 1 \right) \\\ \Rightarrow - 1 \leqslant - \cos x \leqslant 1 \\\$
Now, adding 1 to the inequality we get
$\Rightarrow 1 + \left( { - 1} \right) \leqslant 1 + \left( { - \cos x} \right) \leqslant 1 + \left( 1 \right) \\\ \Rightarrow 0 \leqslant 1 - \cos x \leqslant 2 \\\$
So we have

$$\Rightarrow 0 \leqslant f\left( x \right) \leqslant 2 \\\ \Rightarrow 0 \leqslant \sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)\sqrt {\left( {1 - \cos x} \right)...............\infty } } } \leqslant 2 \\\$$

Hence, the range of the given term is $\left[ {0,2} \right]$
So, option C is the right option.

Note: In order to solve such a complex term, the use of series and its formula is very important. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Also students must notice and keep in mind that when we multiply the inequality by a negative term, the sign of the inequality reverses.