Question

Assertion :

Consider two functions $f\left(x\right)=1+e^{{\cot }^{2}x}$ and $g\left(x\right)=\sqrt{2\left|\sin x\right|-1}+\frac{1-\cos 2x}{1+{\sin }^{4}x}$

Statement I: The solution of the equation $f\left(x\right)=g\left(x\right)$ is given by $x=(2n+1)\frac{\pi }{2},\forall n\in I.$

Reason:

Statement II: If $f\left(x\right)\le k$ and $g\left(x\right)\le k$(where $k\in R),$ then solutions of the equation $f\left(x\right)=g\left(x\right)$ is the solution corresponding to the equation $f\left(x\right)=k$

• A. Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.
• B. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
• C. Statement I is true, Statement II is false.
• D. Statement I is false, Statement II is true.

Answer 1

The correct option is C Statement I is true, Statement II is false.

$e^x\ge 1,whenx\ge 0$

$e^x<1whenx<0$

$\Rightarrow e^{co{t}^{2}x}\ge 1$ $[since,co{t}^{2}x\ge 0]$

$\therefore 1+e^{co{t}^{2}x}\ge 2$

$|sinx|\le 1\Rightarrow 2|sinx|\le 2\Rightarrow 2|sinx|-1\le 1$

$\therefore \sqrt{2|sinx|-1}\le 1$

Now,

$LHS=1+e^{{\cot }^{2}x}\ge 2$

As $\sqrt{2\left|\sin x\right|-1}\le 1$

and $\frac{1-\cos 2x}{1+{\sin }^{4}x}=\frac{2{\sin }^{2}x}{1+{\sin }^{4}x}=\frac{2}{\frac{1}{{\sin }^{2}x}+{\sin }^{2}x}\le 1$

$\therefore RHS=\sqrt{2\left|\sin x\right|-1}+\frac{1-\cos 2x}{1+{\sin }^{4}x}\le 2$

Equation will satisfy, if $LHS=RHS=2$ which is possible when ${\cot }^{2}x=0$ and $\left|\sin x\right|=1$

$\Rightarrow x=(2n+1)\frac{\pi }{2},n\in I$

$\therefore$ Statement I is correct

Statement $II$ is always correct because solution of the equation $f\left(x\right)=g\left(x\right)$ will be solutions corresponding to

$f\left(x\right)=g\left(x\right)=k$ is the domain of $f\left(x\right)$ and $g\left(x\right)$ both

Hence,$\left(c\right)$ is the correct answer