Question

Which of the following is/are correct?

• A. The number of real solution of $2^{x^2}\sec x=1$ is $1$.
• B. The number of real ordered point $(x,y)$ satisfying the system of equations $6^x{\left(\frac{2}{3}\right)}^y-3\cdot 2^{x+y}-8\cdot 3^{x-y}+24=0$ and $xy=2$ is $2$
• C. The number of real solution of ${\log }_2(4\cdot 3^x-6)-{\log }_2(9^x-6)=1$ is $1$
• D. $\tan \left(\frac{\pi }{8}\right)+\cot \left(\frac{\pi }{8}\right)=2\sqrt{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The number of real solution of $2^{x^2}\sec x=1$ is $1$.
C The number of real solution of ${\log }_2(4\cdot 3^x-6)-{\log }_2(9^x-6)=1$ is $1$
D $\tan \left(\frac{\pi }{8}\right)+\cot \left(\frac{\pi }{8}\right)=2\sqrt{2}$

$2^{x^2}\sec x=1\Rightarrow \sec x=2^{-x^2}$
We know that,
$\sec x\in (-\infty ,-1]\cup [1,\infty )2^{-x^2}\in (0,1]$
Therefore, they will be equal only when
$\sec x=1,2^{-x^2}=1\Rightarrow x=0$
Hence, only one possible solution.

Given : $6^x{\left(\frac{2}{3}\right)}^y-3\cdot 2^{x+y}-8\cdot 3^{x-y}+24=0$ and $xy=2$
Now,
$2^{x+y}\cdot 3^{x-y}-3\cdot 2^{x+y}-8\cdot 3^{x-y}+24=0\Rightarrow (2^{x+y}-8)(3^{x-y}-3)=0\Rightarrow 2^{x+y}=8\text{or}{3}^{x-y}=3\Rightarrow x+y=3\text{or}x-y=1$

Therefore, using $xy=2$ and $x+y=3$, we get
$\Rightarrow x^2-3x+2=0\Rightarrow x=1,2\Rightarrow (x,y)=(1,2),(2,1)$
Using $xy=2$ and $x-y=1$, we get
$\Rightarrow x^2-x-2=0\Rightarrow x=-1,2\Rightarrow (x,y)=(-1,-2),(2,1)\therefore (x,y)\in \left\{\right(1,2),(2,1),(-1,-2\left)\right\}$
Hence, the number of possible solutions is $3$.

${\log }_2(4\cdot 3^x-6)-{\log }_2(9^x-6)=1\Rightarrow {\log }_2\frac{(4\cdot 3^x-6)}{(9^x-6)}=1\Rightarrow 4\cdot 3^x-6=2(9^x-6)$
Assuming $3^x=t$, we get
$\Rightarrow 4t-6=2t^2-12\Rightarrow t^2-2t-3=0\Rightarrow (t-3)(t+1)=0\Rightarrow t=3(t>0)\therefore x=1$

$\tan \left(\frac{\pi }{8}\right)+\cot \left(\frac{\pi }{8}\right)=(\sqrt{2}-1)+(\sqrt{2}+1)=2\sqrt{2}$