Question

Identify the correct option(s):

• A. Number of solutions of $|\sin |x||=x+|x|$ in $[-2\pi ,2\pi ]$ is $3$
• B. Number of solutions of $\tan 4x=\cos x$ in $(0,\pi )$ is $5$
• C. Number of solutions of $2^{\cos x}=|\sin x|$ in $[0,2\pi ]$ is $4$
• D. Number of solutions of $2^{|x^2-12|}=\sqrt{e^{|x|\ln 4}}$ is $2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Number of solutions of $|\sin |x||=x+|x|$ in $[-2\pi ,2\pi ]$ is $3$
B Number of solutions of $\tan 4x=\cos x$ in $(0,\pi )$ is $5$
C Number of solutions of $2^{\cos x}=|\sin x|$ in $[0,2\pi ]$ is $4$

$|\sin |x||=x+|x|$
When $x>0\Rightarrow \text{R.H.S}=2x$
And $x\le 0\Rightarrow \text{R.H.S}=0$


Number of solution is $3$ i.e. $x=-2\pi ,-\pi ,0$

$\tan 4x=\cos x$


Number of solutions is $5.$

$2^{\cos x}=|\sin x|$
Let $f\left(x\right)=2^{\cos x}$
$f^{\prime }\left(x\right)=2^{\cos x}\ln 2(-\sin x)\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow x=0,\pi ,2\pi f^{\prime \prime }\left(x\right)=2^{\cos x}\ln 2[\ln 2\times {\sin }^{2}x-\cos x]f^{\prime \prime }\left(0\right)<0,f^{\prime \prime }\left(\pi \right)>0,f^{\prime \prime }\left(2\pi \right)<0$


Number of solutions is $4.$

$2^{|x^2-12|}=\sqrt{e^{|x|\ln 4}}$
$\Rightarrow 2^{|x^2-12|}=\sqrt{4^{|x|}}$
$\Rightarrow 2^{|x^2-12|}=2^{|x|}$
$\Rightarrow |x^2-12|=|x|$
$\Rightarrow x^2-12=x\text{or}{x}^2-12=-x$
$\Rightarrow x^2-x-12=0\text{or}{x}^2+x-12=0$
$\Rightarrow x=-3,4\text{or}x=3,-4$
$\Rightarrow x=\pm 3,\pm 4$
Number of solutions is $4.$