NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

If $f(x) = \tan^{-1} \left[ \frac{\sin x}{1 + \cos x} \right]$ , then what is the first derivative of $f(x)$ ?

Choose the correct answer:

Explanation

Solution

Step 1: Simplify the expression inside the bracket

Using the double angle identities:

$\sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$
$1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)$

Substitute these into the function:

f(x) = \tan^{-1} \left[ \frac{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} \right]

Step 2: Further simplification

Cancel out the common terms ( $2$ and one $\cos\left(\frac{x}{2}\right)$ ):

f(x) = \tan^{-1} \left[ \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \right]

f(x) = \tan^{-1} \left[ \tan\left(\frac{x}{2}\right) \right]

Using the property $\tan^{-1}(\tan \theta) = \theta$ :

f(x) = \frac{x}{2}

Step 3: Differentiate with respect to $x$

Now, find the first derivative $f'(x)$ :

f'(x) = \frac{d}{dx} \left( \frac{x}{2} \right)

f'(x) = \frac{1}{2}

Correct Option: 1. $1/2$