Question

If \(y = \sqrt{(a-x)(x-b)} - (a-b)\tan^{-1}\sqrt{\frac{a-x}{x-b}}\), then \(\frac{dy}{dx}\) is equal to

• A. \(\sqrt{\frac{a-x}{x-b}}\)
• B. \(\sqrt{\frac{x-b}{a-x}}\)
• C. \(\frac{1}{\sqrt{(a-x)(x-b)}}\)
• D. None of these

Hint:

After substitution, always convert back using \(\tan\theta = \sqrt{\frac{x-b}{a-x}}\).

Answer 1

The Correct Option is B

Step 1: Use substitution}
Let \[ x = a\cos^2\theta + b\sin^2\theta \] Then, \[ a-x = (a-b)\sin^2\theta,\quad x-b = (a-b)\cos^2\theta \] \[ \sqrt{\frac{a-x}{x-b}} = \tan\theta \Rightarrow \theta = \tan^{-1}\sqrt{\frac{a-x}{x-b}} \] Step 2: Simplify \(y\)}
\[ y = \sqrt{(a-x)(x-b)} - (a-b)\theta \] \[ = (a-b)\sin\theta\cos\theta - (a-b)\theta \] \[ = \frac{a-b}{2}\sin 2\theta - (a-b)\theta \] Step 3: Differentiate parametrically}
\[ \frac{dy}{d\theta} = (a-b)\cos 2\theta - (a-b) \] \[ \frac{dx}{d\theta} = (b-a)\sin 2\theta \] \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{(a-b)(\cos 2\theta - 1)}{(b-a)\sin 2\theta} = \frac{1 - \cos 2\theta}{\sin 2\theta} \] Step 4: Use identity}
\[ \frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta \] \[ \Rightarrow \frac{dy}{dx} = \tan\theta \] \[ = \sqrt{\frac{x-b}{a-x}} \] Step 5: Final Answer
\[ \sqrt{\frac{x-b}{a-x}} \]