1) Consider two functions f(x)=1+ e^{\cot ^{2}x} and g(x)= \sqrt{2\left|\sin x \right|-1}+ (\frac{1+\cos 2x}{1+ \sin ^{4}x})
Statement-1: The solutions of the equation f (x) = g (x) is given by x=(2n+1)\frac{\Pi }{2} for all n\in I
Statement-2: If f (x) ≥ ï€ k and g (x) ≤ ï€ k (where k \in R) then solutions of the equation f (x) = g (x)
is the solution corresponding to the equation f (x) = k.
Which option is correct?
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B)Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
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1 Answers
1For statement-1: L.H.S 1+e^{\cot ^{2}x} ≥ 2
As \sqrt{2\left|\sin x \right|-1} ≤ 1
and \frac{1-\cos 2x}{1+\sin ^{4}x}= \frac{2\sin ^{2}x}{1+\sin ^{4}x} = \frac{2}{\frac{1}{\sin ^{2}x}+\sin ^{2}x} ≤ 1
So R.H.S. \sqrt{2\left|\sin x \right|-1}+\frac{1-\cos 2x}{1+\sin ^{4}x} ≤ 2
Equation will satisfy if L.H.S. = R.H.S. = 2
It is possible when \cot ^{2}x=0 and \left|\sin x \right|=1
\Rightarrow x=\left(2n+1 \right)\frac{\Pi }{2} ,n\in I
So statement 1 is correct.
For statement 2: solution of the equation f (x) = g (x) will be solutions corresponding to f (x) = g (x) = k in the domain of f (x) and g (x) both. So, statement 2 is not always correct...