CUET Preparation Today
CUET
-- Mathematics - Section A
Applications of Derivatives
The slope of tangent to the curve $x=t^2+3 t-8, y=2 t^2-2 t-5$ at the point (2, -1) is:
$\frac{22}{7}$
$\frac{-6}{7}$
$\frac{7}{6}$
$\frac{6}{7}$
The correct answer is Option (4) → $\frac{6}{7}$
