Question

solve the questions​

Answer 1

(i)

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}=(x-2)^{x+3}}$

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}=(x-2)^{x+1+2}}$

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}=(x-2)^{x+1}\cdot(x-2)^2}$

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}\cdot(x-2)^2-(x-2)^{x+1}=0}$

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}[(x-2)^2-1]=0}$

This implies,

$\displaystyle\longrightarrow\sf{(x-2)^{x+1}=0}$

$\displaystyle\Longrightarrow\sf{x-2=0}$

$\displaystyle\longrightarrow\sf {x=2}$

Or,

$\displaystyle\longrightarrow\sf{(x-2)^2-1=0}$

$\displaystyle\longrightarrow\sf{(x-2)^2=1}$

$\displaystyle\longrightarrow\sf{x-2=1\quad OR\quad x-2=-1}$

$\displaystyle\longrightarrow\sf{x=1+2\quad OR\quad x=-1+2}$

$\displaystyle\longrightarrow\sf {x=3\quad OR\quad x=1}$

Finally,

$\displaystyle\longrightarrow\sf{x\in X=\{1,\ 2,\ 3\}}$

And hence,

$\displaystyle\longrightarrow\sf {\underline {\underline {n(X)=3}}}$

(ii)

$\displaystyle\longrightarrow\sf{\sin^8x+\cos^8x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{(\sin^4x)^2+(\cos^4x)^2=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{(\sin^4x+\cos^4x)^2-2\sin^4x\cos^4x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{((\sin^2x)^2+(\cos^2x)^2)^2-2\sin^4x\cos^4x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x)^2-2\sin^4x\cos^4x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{(1-2\sin^2x\cos^2x)^2-2\sin^4x\cos^4x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{1-4\sin^2x\cos^2x+4\sin^4x\cos^4x-2\sin^4x\cos^4x=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{2\sin^4x\cos^4x-4\sin^2x\cos^2x+1=\dfrac {97}{128}}$

Let $\displaystyle\sf {\sin^2x\cos^2x=k.}$

$\displaystyle\longrightarrow\sf{2k^2-4k+1=\dfrac {97}{128}}$

$\displaystyle\longrightarrow\sf{256k^2-512k+31=0}$

$\displaystyle\longrightarrow\sf{k=\dfrac {512\pm\sqrt{(-512)^2-4\times256\times31}}{512}}$

$\displaystyle\longrightarrow\sf{k=\dfrac {512\pm480}{512}}$

$\displaystyle\longrightarrow\sf{\sin^2x\cos^2x=\dfrac {31}{16}\quad OR\quad\sin^2x\cos^2x=\dfrac {1}{16}}$

But since $\displaystyle\sf {\sin x,\ \cos x\in [-1,\ 1]}$ for every x,

$\displaystyle\longrightarrow\sf{\sin x\cos x\in[-1,\ 1]}$

$\displaystyle\longrightarrow\sf{\sin^2x\cos^2x\in [0,\ 1]}$

But this is not the actual range!

However it is false that $\displaystyle\sf {\sin^2x\cos^2x=\dfrac {31}{16}.}$

$\displaystyle\longrightarrow\sf{\sin^2x\cos^2x=\dfrac {1}{16}}$

Since $\displaystyle\sf {x\in\left [0,\ \dfrac {\pi}{2}\right],}$

$\displaystyle\longrightarrow\sf{\sin x\cos x=\dfrac {1}{4}}$

$\displaystyle\longrightarrow\sf{\sin (2x)=\dfrac {1}{2}}$

$\displaystyle\longrightarrow\sf{2x=\dfrac {\pi}{6}}$

$\displaystyle\longrightarrow\sf{x=\dfrac {\pi}{12}}$

Hence there exists only one solution.