Q. Let$g(x) = \ \left\{ \begin{array}{r} 2e\ \ \ \ \ \ \ \ \ \ \ \ \ if\ x \leq 1 \\ {\ log}(x - 1),\ \ \ if\ \ x > 1 \end{array} \right.\ \$. The equation of the normal to y=$g(x)$at the point (3,log 2), is
A) $y - 2x = 6 + \log 2$
B) $y + 2x = 6 + \log 2$
C) $y + 2x = 6 - \log 2$
D) $y + 2x = - 6 + \log 2$
Explanation
(b)
For the point $(3,\log 2)$, we take $y = g(x) = \log{(x - 1)}$
$\frac{dy}{dx} = \frac{1}{(x - 1)}\ \ \ \ \Rightarrow \ \ \left( \frac{dy}{dx} \right)_{\left( 3,\log 2 \right)} = \frac{1}{2}$
$\therefore$ Equation of normal is
$y - \log{2 = - 2(x - 3)}$
$\Rightarrow \ \ \ y + 2x = 6 + \log 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.