The function , for , is maximum when:

Explanation: 1. Simplify the Function To make differentiation easier, let's rewrite by expressing as : For easier calculation, we can consider the reciprocal . The function is maximum when is minimum. 2. Find the Derivative Now, differentiate with respect to : 3. Set the Derivative to Zero To find the critical point (where the maximum or minimum occurs), set : Taking the square root on both sides (since is in the first quadrant, values are positive): 4. Solve for x Since , we have: Conclusion The function reaches its extremum when . By checking the second derivative, we can confirm this point provides the minimum for , and consequently, the maximum for . Correct Option: B)

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NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

The function $f (x) = \frac{x}{1 + x t a n x}$ , for $0 \leq x \leq \frac{π}{2}$ , is maximum when:

Choose the correct answer:

Explanation

1. Simplify the Function

To make differentiation easier, let's rewrite $f(x)$ by expressing $\tan x$ as $\frac{\sin x}{\cos x}$ :

f(x) = \frac{x}{1 + x \frac{\sin x}{\cos x}} = \frac{x \cos x}{\cos x + x \sin x}

For easier calculation, we can consider the reciprocal $g(x) = \frac{1}{f(x)}$ . The function $f(x)$ is maximum when $g(x)$ is minimum.

g(x) = \frac{\cos x + x \sin x}{x \cos x} = \frac{\cos x}{x \cos x} + \frac{x \sin x}{x \cos x}

g(x) = \frac{1}{x} + \tan x

2. Find the Derivative

Now, differentiate $g(x)$ with respect to $x$ :

g'(x) = \frac{d}{dx} (x^{-1} + \tan x)

g'(x) = -\frac{1}{x^2} + \sec^2 x

3. Set the Derivative to Zero

To find the critical point (where the maximum or minimum occurs), set $g'(x) = 0$ :

-\frac{1}{x^2} + \sec^2 x = 0

\sec^2 x = \frac{1}{x^2}

Taking the square root on both sides (since $x$ is in the first quadrant, values are positive):

\sec x = \frac{1}{x}

4. Solve for x

Since $\sec x = \frac{1}{\cos x}$ , we have:

\frac{1}{\cos x} = \frac{1}{x}

x = \cos x

Conclusion

The function reaches its extremum when $x = \cos x$ . By checking the second derivative, we can confirm this point provides the minimum for $g(x)$ , and consequently, the maximum for $f(x)$ .

Correct Option: B) $x = \cos x$