Question

Let $\tan \left(2\pi |\sin \theta |\right)=\cot \left(2\pi |\cos \theta |\right)$, where $\theta \in R$ and $f\left(x\right)=(|\sin \theta |+|\cos \theta |{)}^x,x\ge 1$. Then range of $f\left(x\right)$ not include

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given,

$\tan \left(2\pi |\sin \theta |\right)=\cot \left(2\pi |\cos \theta |\right)$

where, $\theta \in R$

and

$f\left(x\right)=(|\sin \theta |+|\cos \theta |{)}^x;x⩾1$

Now,

$\begin{array}{cc}\tan \left(2\pi |\sin \theta |\right)=\tan (\frac{\pi }{2}& -2\pi |\cos \theta |)\\ & \because \cot x=\tan (\frac{\pi }{2}-x)\end{array}$

$\Rightarrow 2\pi |\sin \theta |=\frac{\pi }{2}-2\pi |\cos \theta |+n\pi$

$\Rightarrow 2\pi |\sin \theta |+2\pi |\cos \theta |=\frac{\pi }{2}+n\pi$

$\Rightarrow 2\mathrm{x/}(|\sin \theta |+|\cos \theta |)=\frac{\pi }{2}+n\pi$

$\Rightarrow (|\sin \theta |+|\cos \theta |=\frac{1}{4}+\frac{n}{2}$

Now, putting the value of $n=1$

$(|\sin \theta |+|\cos \theta |\mid =\frac{3}{4}-\left(i\right)$

for $n=2;(|\sin \theta |+|\cos \theta |)=\frac{5}{4}-\left(ii\right)$

Now,

AQ, $(|\sin \theta |+\cos \theta {)}^x$ whae, $x⩾1$

For eqn (i), we get

$|\sin \theta |+|\cos \theta |=\frac{3}{4}⩽1,{\left(\frac{3}{4}\right)}^x<1$

from eq $\left(ii\right),$

$|\sin \theta |+|\cos \theta |=\frac{7}{4}>1,{\left(\frac{7}{4}\right)}^x>1$

$\therefore$. Range of $f\left(x\right)$ does not include 1 .

Option A = 1