JEE Advanced 2019 Paper 2 Mathematics: Paragraph 1

We have $f'(x)=3(x-1)(x-3)$, so $x=1$ is a local maximum and $x=3$ a local minimum. (P) $f(1)=4$, matching (4). (Q) $f(x)-2=x^3-6x^2+9x-2=(x-2)(x^2-4x+1)$, and $x^2-4x+1$ has discriminant $12>0$ with roots $2\pm\sqrt3$, both different from $2$, so there are $3$ real solutions, matching (3). (R) Since $f(x)=x(x-3)^2\ge0$ on $[0,3]$, the area is $\int_0^3 f\,dx=\left[\frac{x^4}4-2x^3+\frac{9x^2}2\right]_0^3=\frac{81}4-54+\frac{81}2=\frac{27}4$, matching (5). (S) $f''(x)=6x-12$ vanishes at $x=2$, where $f(2)=8-24+18=2$, matching (2). The correct matching is P→4, Q→3, R→5, S→2.

(P) $f(x)-4=x^3-6x^2+9x-4=(x-1)^2(x-4)$, so the equation has the two distinct real solutions $x=1$ and $x=4$ (the root $x=1$ is double, because $f$ has its local maximum value $4$ there): matches (2). (Q) $f''(x)=6x-12$, so $f''(3)=6$: matches (6). (R) $f'(x)=3x^2-12x+9=3(x-2)^2-3$ has least value $-3$ at $x=2$: matches (1). (S) $f'(x)=0$ at $x=1$ and $x=3$, whose sum is $4$: matches (4). Counting solutions of $f(x)=4$ with multiplicity would give $3$, which is the trap in the option beginning P→3. The correct matching is P→2, Q→6, R→1, S→4.