Q106 / 200 · Manipal (MET) 2013 · Mathematics
If $y=\sqrt{(a-x)(x-b)}-(a-b) \tan ^{-1} \sqrt{\frac{a-x}{x-b}}$, then $\frac{d y}{d x}$ is equal to
- A$1$
- Correct answerB$\sqrt{\frac{a-x}{x-b}}$
- C$\sqrt{(a-x)(x-b)}$
- D$\frac{1}{\sqrt{(a-x)(b-x)}}$
Solution
Let $x=a \cos ^2 \theta+b \sin ^2 \theta$
$\therefore \quad a-x=a-a \cos ^2 \theta-b \sin ^2 \theta=(a-b) \sin ^2 \theta$
and $x-b=a \cos ^2 \theta-b \sin ^2 \theta-b=(a-b) \cos ^2 \theta$
$\therefore \quad y=(a-b) \sin \theta \cdot \cos \theta-(a-b) \tan ^{-1} \tan \theta$
$\therefore \quad y=(a-b) \sin \theta \cdot \cos \theta-(a-b) \tan ^{-1} \tan \theta$
$=\frac{a-b}{2} \sin 2 \theta-(a-b) \theta$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{(a-b) \cos 2 \theta-(a-b)}{(b-a) \sin 2 \theta}$
$=\frac{1-\cos 2 \theta}{\sin \theta}=\tan \theta=\sqrt{\frac{a-x}{x-b}}$
Hence, the correct answer is option: 'B'.