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Question

Question: If \( {\sin ^{ - 1}}(x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{4} - .....\infty ) + {\cos ^{ - 1}}({x...

If sin1(xx22+x34.....)+cos1(x2x42+x64....)=π2{\sin ^{ - 1}}(x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{4} - .....\infty ) + {\cos ^{ - 1}}({x^2} - \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{4}....\infty ) = \dfrac{\pi }{2} for 0<x<20 < \left| x \right| < \sqrt 2 , then x equals
A.12\dfrac{1}{2}
B.11
C.12\dfrac{{ - 1}}{2}
D.1- 1

Explanation

Solution

Hint : First, identify the geometric progressions in the question and then find their sum using a suitable formula. Using the inverse trigonometric identities find out the relation between the functions of sin and cos. Keep in mind the conditions for x to axis. This way you can find the correct answer.

Complete step-by-step answer :
xx22+x34.....x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{4}..... is a Geometric progression.
Its common ratio, r=x22x=x2r = \dfrac{{\dfrac{{ - {x^2}}}{2}}}{x} = \dfrac{{ - x}}{2}
The sum of an infinite geometric progression is given by the formula, S=a11rS = \dfrac{{{a_1}}}{{1 - r}} , where a1{a_1} is the first term of G.P.
So the sum of given G.P. is
S1=x1(x2)=x1+x2=2x2+x\Rightarrow {S_1} = \dfrac{x}{{1 - ( - \dfrac{x}{2})}} = \dfrac{x}{{1 + \dfrac{x}{2}}} = \dfrac{{2x}}{{2 + x}}
x2x42+x64.....{x^2} - \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{4}..... is also a G.P.
Its common ratio is , r=x42x2=x22r = \dfrac{{ - \dfrac{{{x^4}}}{2}}}{{{x^2}}} = - \dfrac{{{x^2}}}{2}
Sum of this infinite G.P. is
S2=x21(x22)=x21+x22=2x22+x2\Rightarrow {S_2} = \dfrac{{{x^2}}}{{1 - ( - \dfrac{{{x^2}}}{2})}} = \dfrac{{{x^2}}}{{1 + \dfrac{{{x^2}}}{2}}} = \dfrac{{2{x^2}}}{{2 + {x^2}}}
Now, the equation given in the question can be rewritten as,
sin1(2x2+x)+cos1(2x22+x2)=π2{\sin ^{ - 1}}(\dfrac{{2x}}{{2 + x}}) + {\cos ^{ - 1}}(\dfrac{{2{x^2}}}{{2 + {x^2}}}) = \dfrac{\pi }{2}
Now, we know that sin1x+cos1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} when 1x1- 1 \leqslant x \leqslant 1
So using this identity, we get
2x2+x=2x22+x2 2x(2+x2)=2x2(2+x) 4x+2x3=4x2+2x3 4x24x=0 4x(x1)=0   \Rightarrow \dfrac{{2x}}{{2 + x}} = \dfrac{{2{x^2}}}{{2 + {x^2}}} \\\ \Rightarrow 2x(2 + {x^2}) = 2{x^2}(2 + x) \\\ \Rightarrow 4x + 2{x^3} = 4{x^2} + 2{x^3} \\\ \Rightarrow 4{x^2} - 4x = 0 \\\ \Rightarrow 4x(x - 1) = 0 \;
This gives x=0x = 0 or x=1x = 1
But we are given the question that x>0x > 0 .
Therefore, x=0x = 0 is rejected and x=1x = 1 is the right answer.
So, the correct answer is “Option B”.

Note : A geometric progression is a sequence of numbers in which each term is found by the multiplication of its previous term with a non-one number, this non-one number is called the common ratio. For example, the sequence 2, 4, 8, 16,32…. Is a geometric progression or geometric sequence and the common ratio can be obtained by dividing any two consecutive numbers, in the given example the common ratio is 2.