Question

Let $P$ and $T$ be the subsets of $X-Y$ plane defined by
$P=\left\{\right(x,y):x>0,y>0\text{and}{x}^2+y^2=1\}$
$T=\left\{\right(x,y):x>0,y>0\text{and}{x}^8+y^8<1\}$
Then $P\cap T$ is

• A. the void set $\Phi$
• B. $P$
• C. $T$
• D. $P-T^C$

Answer 1

The correct option is B $P$

Let $(h,k)\in P\cap T$.
This means $(h,k)$ satisfies $x^2+y^2=1$ and $x^8+y^8<1$. Also $h>0,k>0$.
So, $h^2+k^2=1...\left(1\right)$
and $h^8+k^8<1...\left(2\right)$
Substituting $k^2$ from (1) to (2)
$h^8+k^8=h^8+(1-h^2{)}^4$
$=2h^8-4h^6+6h^4-4h^2+1$
$=2h^2(h^2-1)(h^4-h^2+2)+1$
$=-2h^2{k}^2(h^4-h^2+2)+1$

Now,
$h^4-h^2+2={(h^2-\frac{1}{2})}^2+\frac{7}{4}>0$
$\Rightarrow -2h^2{k}^2(h^4-h^2+2)+1<1$

Hence, all solution of $x^2+y^2=1$ satisfies $x^8+y^8<1\Rightarrow P\cap T=P$