Match List I with List II

Choose the correct answer from the options given below:

A-II, B-III, C-IV, D-I

A. $y=x^3 - 2x$

$y'=3x^2-2$

Slope at $x = 2$

$y'|_{x=2}=3×2^2-2=10$

B. points $P(0, 2)$ and $Q(5,-6)$

Slope → $\frac{-6-2}{5-0}=-\frac{8}{5}$

C. $y=\sqrt{4x-3}$ for $y'=\frac{2}{3}$

$y'=\frac{1}{2}\frac{4}{\sqrt{4x-3}}⇒\frac{2}{3}=\frac{4}{2\sqrt{4x-3}}$

so $\sqrt{4x-3}=3=y$

$⇒4x-3=9=4x=12⇒x=3$  $⇒(3,3)$

D. $y =\frac{x-2}{x-1}$, so $y'=-\frac{(x-2)+(x-1)}{(x-1)^2}$

$y'=\frac{+1}{(x-1)^2}$  (slope of tangent)

slope of normal = $-(x-1)^2$ at $x = 10$

Slope of normal = $-(10-1)^2=-81$