Question

The slope of tangent to the curve $y=\sqrt{2x^2-3x+2}$ at point(s) where its ordinate is equal to abscissa, can be

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{4}{5}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B), (D)

$\frac{5}{4}$

Given curve : $y=\sqrt{2x^2-3x+2}$
For ordinate to be equal to abscissa
i.e., $y=x$
$\therefore x^2=2x^2-3x+2$
$\Rightarrow x^2-3x+2=0$
$\Rightarrow x=1,2$
Now,
$\frac{dy}{dx}=\frac{4x-3}{2\sqrt{2x^2-3x+2}}$
Slope of tangent at $x=1$
$\frac{dy}{dx}{|}_{x=1}=\frac{1}{2}$
Slope of tangent at $x=2$
$\frac{dy}{dx}{|}_{x=2}=\frac{5}{4}$