JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q72

MathematicsSingle correct+4 / -1Question 72 of 90

The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$ -axis passes through the point :

\left(\dfrac12,\dfrac12\right)

\left(\dfrac12,-\dfrac13\right)

\left(\dfrac12,\dfrac13\right)

\left(-\dfrac12,-\dfrac12\right)

Answer and worked solution

Official answer: A

At $x=0$ : $6y=6$ , so $y=1$ . Differentiating, $y'(x-2)(x-3)+y(2x-5)=1$ , which gives $6y'-5=1$ and $y'=1$ .

The normal is $y=1-x$ , which passes through $\left(\frac12,\frac12\right)$ .

JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 72