Zadanie 12c, Paper 2, maj 2022, TZ1, Matematyka AA IB HL

(i) With v=y/x>3, separate $\int dv/[(v-3)(v+1)]=\int dx/x$. Partial fractions give $1/[(v-3)(v+1)]=\frac14[1/(v-3)-1/(v+1)]$. Integrating yields $\frac14\ln((v-3)/(v+1))=\ln x+C$. The initial v=4 at x=1 gives C=(ln(1/5))/4. Hence (v−3)/(v+1)=x⁴/5. Substitute v=y/x and rearrange 5y−15x=x⁴y+x⁵, so y(5−x⁴)=15x+x⁵, proving the formula. Its branch is 0<x<5^(1/4), where y>3x. (ii) y(1.4)=22.77127072≈22.8. (iii) The graph becomes very steep near the vertical asymptote x=5^(1/4)≈1.49535. Slopes change rapidly on the last steps, so a step 0.08 gives substantial error (the Euler estimate is about 11.5).

(i) M1:separate variables. A1:correct factored denominator. M1:partial fractions. A1:coefficients 1/4 and −1/4. A1:integrate to logarithms. M1:apply the initial condition. M1:combine logarithms/exponentiate. M1:substitute back v=y/x. M1:rearrange for y. A1:correct equation y(5−x⁴)=15x+x⁵ leading to the stated formula. (ii) A1:22.8. (iii) R1:rapid change of gradient over the interval OR the vertical asymptote at x=5^(1/4). Either source alternative suffices; do not require both or an additional sentence about error.