Answer
Verified
415.2k+ views
Hint: We know that the range of \[{{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}}\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\], so the same would be for \[2{{\cos }^{-1}}x\]. From this, find the range of \[{{\cos }^{-1}}x\] and then find the value of x for which it is true.
Complete step-by-step answer:
We are given an equation \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]. We have to solve it and find all the values of x that satisfy this.
Let us consider the equation i.e. : \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]
We know that the range of \[{{\sin }^{-1}}t\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
So, we also get the range \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] as \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]
Or, \[\dfrac{-\pi }{2}\le {{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\le \dfrac{\pi }{2}\]
Since, we know that \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] then, \[2{{\cos }^{-1}}x\] will also have the same range as \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]. So, we get,
\[\dfrac{-\pi }{2}\le 2{{\cos }^{-1}}x\le \dfrac{\pi }{2}\]
By dividing 2 in the above equation, we get,
\[\dfrac{\dfrac{-\pi }{2}}{2}\le \dfrac{2{{\cos }^{-1}}x}{2}\le \dfrac{\dfrac{\pi }{2}}{2}\]
Or, \[\dfrac{-\pi }{4}\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\]
Now, we will draw the graph of \[{{\cos }^{-1}}x\] to analyze the above inequality.
Since, we can see that the range of \[{{\cos }^{-1}}x\] is \[\left[ 0,\pi \right]\], so we will get the range of \[{{\cos }^{-1}}x\] in this question as, \[0\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\] because \[{{\cos }^{-1}}x\] cannot take negative values.
For \[{{\cos }^{-1}}x=0\], we get x = cos 0 = 1.
And for \[{{\cos }^{-1}}x=\dfrac{\pi }{4}\], we get \[x=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].
So, we get, \[\dfrac{1}{\sqrt{2}}\le x\le 1\].
Hence, option (c) is the right answer.
Note: In this question, we can also consider the graph of \[{{\cos }^{-1}}x\] to solve the inequality
\[0\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\]
From the above graph, we can see that, when \[{{\cos }^{-1}}x\in \left[ 0,\dfrac{\pi }{4} \right]\], then \[x\in \left[ \dfrac{1}{\sqrt{2}},1 \right]\]. In questions involving the inverse trigonometric functions, take the special case of range and domain of the function.
Complete step-by-step answer:
We are given an equation \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]. We have to solve it and find all the values of x that satisfy this.
Let us consider the equation i.e. : \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]
We know that the range of \[{{\sin }^{-1}}t\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
So, we also get the range \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] as \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]
Or, \[\dfrac{-\pi }{2}\le {{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\le \dfrac{\pi }{2}\]
Since, we know that \[2{{\cos }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] then, \[2{{\cos }^{-1}}x\] will also have the same range as \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]. So, we get,
\[\dfrac{-\pi }{2}\le 2{{\cos }^{-1}}x\le \dfrac{\pi }{2}\]
By dividing 2 in the above equation, we get,
\[\dfrac{\dfrac{-\pi }{2}}{2}\le \dfrac{2{{\cos }^{-1}}x}{2}\le \dfrac{\dfrac{\pi }{2}}{2}\]
Or, \[\dfrac{-\pi }{4}\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\]
Now, we will draw the graph of \[{{\cos }^{-1}}x\] to analyze the above inequality.
Since, we can see that the range of \[{{\cos }^{-1}}x\] is \[\left[ 0,\pi \right]\], so we will get the range of \[{{\cos }^{-1}}x\] in this question as, \[0\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\] because \[{{\cos }^{-1}}x\] cannot take negative values.
For \[{{\cos }^{-1}}x=0\], we get x = cos 0 = 1.
And for \[{{\cos }^{-1}}x=\dfrac{\pi }{4}\], we get \[x=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].
So, we get, \[\dfrac{1}{\sqrt{2}}\le x\le 1\].
Hence, option (c) is the right answer.
Note: In this question, we can also consider the graph of \[{{\cos }^{-1}}x\] to solve the inequality
\[0\le {{\cos }^{-1}}x\le \dfrac{\pi }{4}\]
From the above graph, we can see that, when \[{{\cos }^{-1}}x\in \left[ 0,\dfrac{\pi }{4} \right]\], then \[x\in \left[ \dfrac{1}{\sqrt{2}},1 \right]\]. In questions involving the inverse trigonometric functions, take the special case of range and domain of the function.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE