Find the point at which the tangent to the curve y = <span class="math-tex">\(\rm \sqrt{4x-3}-1\)</span> has its slope <span class="math-tex">\(\dfrac{2}{3}\)</span>.
Slope \(= \frac{d y}{d x} = \frac{4}{2 \sqrt{4x - 3}} = \frac{2}{3}\) \[ \sqrt{4x - 3} = 3 \Rightarrow 4x - 3 = 9 \Rightarrow x = 3 \] and \( y = 2 \) Required point is \( (3, 2) \).
Reinforce your conceptual grasp with related questions from NIMCET 2013:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →