Find the point at which the tangent to the curve y = \(\rm \sqrt{4x-

1.	Find the point at which the tangent to the curve y = \(\rm \sqrt{4x-3}-1\) has its slope \(\dfrac{2}{3}\).1. (3, 3)2. (3, 2)3. (2, 3)4. (2, 2)
Answer» Correct Answer - Option 2 : (3, 2) Concept: For a given curve y = f(x), the slope (m) of the tangent at a given point x = a is given by m = f'(a). Chain Rule of Derivatives: \(\rm \dfrac{d}{dx}f[g(x)]=\dfrac{d}{d\ g(x)}f[g(x)]\times \dfrac{d}{dx}g(x)\). \(\rm \dfrac{d}{dx}x^n=nx^{n-1}\). Calculation: Let us first find out f'(x) for the curve y = f(x) = \(\rm \sqrt{4x-3}-1\). f'(x) = \(\rm \dfrac{d}{dx}(\sqrt{4x-3}-1)\) = \(\rm \dfrac{d}{d(4x-3)}(\sqrt{4x-3})\times \dfrac{d}{dx}(4x-3)\) = \(\rm \dfrac{1}{2}\left(\dfrac{1}{\sqrt{4x-3}}\right)\times4\) = \(\rm \dfrac{2}{\sqrt{4x-3}}\). Let's say that the tangent at a point (a, b) has slope m = \(\dfrac23\). ∴ m = f'(a) = \(\rm \dfrac23\) ⇒ \(\rm \dfrac{2}{\sqrt{4a-3}}=\dfrac23\) ⇒ \(\rm \sqrt{4a-3}=3\) ⇒ 4a - 3 = 9 ⇒ a = 3. And b = f(a) = f(3) = \(\rm \sqrt{4(3)-3}-1=\sqrt9 -1 = 3-1 =2\). ∴ The required point is (a, b) = (3, 2).

Find the point at which the tangent to the curve y = \(\rm \sqrt{4x-3}-1\) has its slope \(\dfrac{2}{3}\).1. (3, 3)2. (3, 2)3. (2, 3)4. (2, 2)

Answer» Correct Answer - Option 2 : (3, 2)

Concept:

For a given curve y = f(x), the slope (m) of the tangent at a given point x = a is given by m = f'(a).
Chain Rule of Derivatives: \(\rm \dfrac{d}{dx}f[g(x)]=\dfrac{d}{d\ g(x)}f[g(x)]\times \dfrac{d}{dx}g(x)\).
\(\rm \dfrac{d}{dx}x^n=nx^{n-1}\).

Calculation:

Let us first find out f'(x) for the curve y = f(x) = \(\rm \sqrt{4x-3}-1\).

f'(x) = \(\rm \dfrac{d}{dx}(\sqrt{4x-3}-1)\)

= \(\rm \dfrac{d}{d(4x-3)}(\sqrt{4x-3})\times \dfrac{d}{dx}(4x-3)\)

= \(\rm \dfrac{1}{2}\left(\dfrac{1}{\sqrt{4x-3}}\right)\times4\)

= \(\rm \dfrac{2}{\sqrt{4x-3}}\).

Let's say that the tangent at a point (a, b) has slope m = \(\dfrac23\).

∴ m = f'(a) = \(\rm \dfrac23\)

⇒ \(\rm \dfrac{2}{\sqrt{4a-3}}=\dfrac23\)

⇒ \(\rm \sqrt{4a-3}=3\)

⇒ 4a - 3 = 9

⇒ a = 3.

And b = f(a) = f(3) = \(\rm \sqrt{4(3)-3}-1=\sqrt9 -1 = 3-1 =2\).

∴ The required point is (a, b) = (3, 2).