NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

Find the point at which the tangent to the curve $y = \sqrt{4x - 3} - 1$ has its slope $\frac{2}{3}$

Choose the correct answer:

Explanation

Step 1: Find the derivative of the curve

The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx}$ .

y = (4x - 3)^{1/2} - 1

Differentiating with respect to $x$ using the chain rule:

\frac{dy}{dx} = \frac{1}{2}(4x - 3)^{-1/2} \cdot \frac{d}{dx}(4x - 3) - 0

\frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} \cdot 4

\frac{dy}{dx} = \frac{2}{\sqrt{4x - 3}}

Step 2: Equate the derivative to the given slope

The slope is given as $\frac{2}{3}$ .

\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}

\frac{1}{\sqrt{4x - 3}} = \frac{1}{3}

\sqrt{4x - 3} = 3

Step 3: Solve for $x$

Squaring both sides:

4x - 3 = 9

4x = 12

x = 3

Step 4: Find the corresponding $y$ value

Substitute $x = 3$ back into the original equation of the curve:

y = \sqrt{4(3) - 3} - 1

y = \sqrt{12 - 3} - 1

y = \sqrt{9} - 1

y = 3 - 1

y = 2

Final Answer:

The point is (3, 2).

Choose the correct answer:

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