سوال ۳۲ کنکور ارشد علوم کامپیوتر ۱۴۰۱

Setup

The curve $x=\cosh t,\; y=\sinh t$ satisfies $x^{2}-y^{2}=\cosh^{2} t-\sinh^{2} t=1,$ so it traces the right branch of the unit hyperbola $x^{2}-y^{2}=1$. As $t$ runs from $0$ to $2$ the point moves from $(1,0)$ to $(\cosh 2,\sinh 2)$.

The described region is the hyperbolic sector: it is bounded by the $x$-axis (from the origin out to $(1,0)$), the hyperbola arc (from $(1,0)$ to $(\cosh 2,\sinh 2)$), and the ray back to the origin.

Sector area by the parametric formula

For a region swept out from the origin by a parametric curve, the area is $A=\frac{1}{2}\int_{t_0}^{t_1}\bigl(x\,y'-y\,x'\bigr)\,dt.$

Here $x'=\sinh t$ and $y'=\cosh t$, so the integrand collapses: $x\,y'-y\,x'=\cosh t\cosh t-\sinh t\sinh t=\cosh^{2} t-\sinh^{2} t=1.$

Therefore $A=\frac{1}{2}\int_{0}^{2} 1\,dt=\frac{1}{2}\cdot 2=1.$

This is exactly the defining geometric fact about hyperbolic functions: the hyperbolic sector for parameter $t$ has area $t/2$, mirroring how a circular sector of angle $\theta$ has area $\theta/2$. With $t=2$ the area is $1$, which is option (2).

Exam trick: Don't integrate the curve as a function of $x$ — the area between a hyperbola arc and a ray is a sector, not an under-the-curve area. The sector formula $\frac12\int(x\,y'-y\,x')\,dt$ collapses to $\frac12\int 1\,dt$ here, instantly giving $t/2$. The decoy options $\frac14\sinh 4$ come from wrongly computing $\int y\,dx$.