JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Application of Derivatives

First, find the slope of the tangent to $y=5x^2+2x-25$ at $(2,-1)$:

$\frac{dy}{dx}=10x+2$, so at $x=2$: slope $=22$

The tangent line at $(2,-1)$ is: $y+1=22(x-2)$, or $y=22x-45$

For the curve $y=x^3-x^2+x$, we have $\frac{dy}{dx}=3x^2-2x+1$

At point $(a,b)$, the slope must also be 22:
$3a^2-2a+1=22$
$3a^2-2a-21=0$
$(3a+7)(a-3)=0$

So $a=3$ or $a=-\frac{7}{3}$

For $a=3$: $b=3^3-3^2+3=27-9+3=21$

Check: The tangent at $(3,21)$ has equation $y-21=22(x-3)$, which gives $y=22x-45$ ✓

For $a=-\frac{7}{3}$: the tangent would be parallel but not the same line (rejected).

Therefore: $|2a+9b|=|2(3)+9(21)|=|6+189|=195$.