GATE 2023 EE Q.42: solution and answer

Official answer: A, B, DSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Region and symmetry. The curve bounds the region $R_p:|x_1{|}^p+|x_2{|}^p\le 1$. It is unchanged by $x_1\to -x_1$ and $x_2\to -x_2$, so its area is four times the first-quadrant area: $$A(p)=4{\int }_0^1{\left(1-x_1^p\right)}^{1/p}d{x}_1$$ Also, every point of $R_p$ has $|x_1{|}^p\le 1$ and $|x_2{|}^p\le 1$, so $|x_1|\le 1$ and $|x_2|\le 1$. Every $R_p$ therefore sits inside the square $[-1,1]\times [-1,1]$, and $A(p)\le 4$.

2. (A) $p=2$: $x_1^2+x_2^2=1$ is the unit circle, so $A=\pi (1{)}^2=\pi$. True.

3. (D) $p=1$: $|x_1|+|x_2|=1$ is a square turned through $45^{\circ }$, with vertices $(\pm 1,0)$ and $(0,\pm 1)$. Both diagonals have length 2, so $A=\frac{d_1{d}_2}{2}=\frac{2\cdot 2}{2}=2$. (Check: the first-quadrant part is the triangle $x_1+x_2\le 1$ with area $\frac{1}{2}$, and $4\times \frac{1}{2}=2$.) True.

4. (B) $p\to \infty$: if $|x_1|\le 2^{-1/p}$ and $|x_2|\le 2^{-1/p}$, then $|x_1{|}^p+|x_2{|}^p\le \frac{1}{2}+\frac{1}{2}=1$. So the square of side $2\cdot 2^{-1/p}$ lies inside $R_p$, and by Step 1, $R_p$ lies inside the square of side 2: $$4\cdot 4^{-1/p}\le A(p)\le 4$$ As $p\to \infty$, $4^{-1/p}\to 1$, so $A(p)\to 4$. Geometrically, ${\left(|x_1{|}^p+|x_2{|}^p\right)}^{1/p}\to \max (|x_1|,|x_2|)$, so the curve approaches the square $\max (|x_1|,|x_2|)=1$ with sides $x_1=\pm 1$ and $x_2=\pm 1$. True.

5. (C) $p\to 0$: take any point with $0<|x_1|\le 1$ and $0<|x_2|\le 1$. As $p\to 0$, $|x_1{|}^p\to 1$ and $|x_2{|}^p\to 1$, so $|x_1{|}^p+|x_2{|}^p\to 2>1$ and the point eventually falls outside the curve. Only the two axis segments ($x_2=0,|x_1|\le 1$ and $x_1=0,|x_2|\le 1$) stay inside for every $p$. The region shrinks onto a '+' shape, which has zero area, so the area tends to $0$, not 1. (Step 7 makes this exact.) False.

6. Closed form, as a check. In the integral of Step 1, substitute $u=x_1^p$, so $x_1=u^{1/p}$ and $d{x}_1=\frac{1}{p}{u}^{1/p-1}du$: $${\int }_0^1{\left(1-x_1^p\right)}^{1/p}d{x}_1=\frac{1}{p}{\int }_0^1{u}^{\frac{1}{p}-1}(1-u{)}^{\frac{1}{p}}du=\frac{1}{p}\frac{\Gamma \left(\frac{1}{p}\right)\Gamma \left(1+\frac{1}{p}\right)}{\Gamma \left(1+\frac{2}{p}\right)}$$ Since $\frac{1}{p}\Gamma \left(\frac{1}{p}\right)=\Gamma \left(1+\frac{1}{p}\right)$, $$A(p)=\frac{4\Gamma {\left(1+\frac{1}{p}\right)}^2}{\Gamma \left(1+\frac{2}{p}\right)}$$ This gives $A(1)=\frac{4\cdot 1^2}{2}=2$ and $A(2)=\frac{4{\left(\frac{\sqrt{\pi }}{2}\right)}^2}{1}=\pi$. As $p\to \infty$, $A\to \frac{4\Gamma (1{)}^2}{\Gamma (1)}=4$, which agrees with (A), (D) and (B).

7. The limit $p\to 0$ from the closed form. Let $n=1/p$, so $n\to \infty$. Since $\Gamma (2+2n)=(1+2n)\Gamma (1+2n)$, $$A=4(1+2n)\frac{\Gamma (1+n{)}^2}{\Gamma (2+2n)}=4(1+2n){\int }_0^1{\left[t(1-t)\right]}^{n}dt\le 4(1+2n){\left(\frac{1}{4}\right)}^n$$ The last step uses $t(1-t)\le \frac{1}{4}$ on $[0,1]$. The bound $\frac{4(1+2n)}{4^n}\to 0$, so $A\to 0$. For example, $A\left(\frac{1}{2}\right)=\frac{4(2!{)}^2}{4!}=\frac{2}{3}$ and $A\left(\frac{1}{3}\right)=\frac{4(3!{)}^2}{6!}=\frac{1}{5}$. The area goes to 0, never to 1, so (C) is false.

8. The correct options are (A), (B), (D).